10.2-1
Gas Solubility Solubility in Aqueous Aqueous Solution. At 303 K the the concentration concentration of CO CO 2 in water is 0.90 x
Variables that can be set
10-4 kg CO2/kg water. Using the Henry’s law constant from Appendix Appendix A.3, what partial pressure of CO2 must be kept in the ags to keep the CO 2 from vaporizing from the
SOLUTION:
aqueous solution? F = C – C – P P + 2 F = 3 – 3 – 2 2 + 2 F=3
Given: T = 303 K
Variables that can be set: -4
x A = 0.9 x 10 kg CO2 /kg H2O
1.
Required: P A of CO2 Solution:
10.3-2
total pressure
2.
temperature
3.
mole fraction composition x A of SO2
Equilibrium Swtage Contact for Gas-Liquid Gas-Liquid System. System. A gas gas mixture mixture at 2.026 x 10 5 Pa total pressure containing air and SO 2 is contacted in a single-stage equilibrium mixer with pure water at 293 K. The partial pressure of SO2 in the original gas is 1.52 x 10 4 Pa.
From A.3-18 A.3-18 for Henry’s law c onstant (Geankoplis (Geankoplis p. 884) H = 0.186 x 10 4 atm/mol frac.
The inlet gas contains 5.70 total kg mol and the inlet water 2.20 total kg mol. The exit
P A= Hx A
gas and liquid leaving are in equilibrium. Calculate the amounts amounts and compositions of the outlet phases. Use equilibrium data from Fig.10.2-1. Fig.10.2-1.
P A = 0.186 x 104 atm x 0. 0.9 x 10 -4
kgCO2 18 kgmolH2O x 0.06848 atm kg H2O 44 kg kg mo mo l CO2
GIVEN: Use equilibrium data in Fig. 10.2-1 PT = 2.026 x 10 5 Pa =
5
P A = 0.0684 0.06848 8 atm x
1.01325 x 10
3
= 6.939 6.939 x 10 Pa
1 atm
T = 293 K
P A of SO2 = 1.52 x 10 4 Pa = .15 atm Inlet gas = 5.70 kg mol
10.3-1
Phase Rule for a Gas-Liquid System. For the systen SO 2-air-water, then total pressure is set at 1 atm abs and the partial pressure of SO 2 in the vapor is set at 0.20 atm. Calculat e the number of degrees of freedom, F. What variables are unspecifi ed that can be arbitrarily set? GIVEN: SO2 – air – air – – H H2O system P AT = 1 atm P A of SO2 = 0.2 atm REQUIRED: Degrees of freedom, F
Inlet H2O = 2.20 total kgmol REQUIRED X A1, y A1, L1 V1
SOLUTION: x Ao = 0 amount of entering acetone = y AN+1v AN+1 = 0.01(30) = 0.30 = 29.7 kgmol/air h acetone leaving in Vi = 0.10(0.30) = 0.30 kgmol/h acetone leaving in Ln = 0.9 (0.30) = 0.27 kgmol/h
V1 = 29.7 + 0.03 = 29.73 kgmolH 2O + acetone/hr
y A1 =
GIVEN: LO = 108 kg mol H 2O/hr
0.030 =0 =0.00 .00101 29.73
Y AN + 1 = 0.01 VN + 1 = 30.0 kg mol/h
Ln = 108 + 0.27 = 108.27 kgmol H 2O + acetone/hr
0.27
X AN =
108.27
T = 300 K PT = 101.3 kPa
0 ..0 00 24 249 3
y A = 2.53 x A REQUIRED: Theoretical stages graphically using analytical Kremser equation.
Using equation:
A1 =
L Lo = mv mv1
SOLUTION:
108 =1.4358 =1.4358 2.53 (29.73)
x Ao = 0 amount of enetering acetone = y AN + 1 VN + 1 = 0.01(30) = 0.30
AN =
LN mvN+1
=
108.27 2.53 (30)
entering air = (1-y AN + 1) VN + 1 = (1 – (1 – 0.01)(30) 0.01)(30) = 29.7 kg mol/air h
=1.4265
acetone leaving in V 1 = 0.10(0.30) = 0.30 kg mol/h acetone leaving in L n = 0.9(0.30) = 0.27 kg mol/h V1 = 29.7 + 0.03 = 29.73 kg mol air + acetone /hr
A= A1AN
1.4358 1.4358(1 (1.426 .4265) 5)
1.4311 1.4311
Ln = 108 + 0.27 = 108.27 kg mol H 2O + acetone/hr 10.4-1
1 1 log yN+1 - mxo 1 A A N=
Interface Concentrations Concentrations and and Overall Overall Mass-Transfer Mass-Transfer Coefficients. Use the same equilibrium data and film coefficients k’ y and k’x as in Example 10.4-1. However, use bulk concentrations of y AG = 0.25 and x AL = 0.05. Calculate the following: following:
log A
N=
0.01 - 2.53(0) 1 1 log 1 1.4311 1.4311 0 . 0 0 1 0 1 2 . 5 3 ( 0 ) log 1.4311
3.7662 stages
a.
Interface concentrations y Ai and x Ai and flux N A.
b.
Overall mass transfer coefficients K’y and K y and flux N A.
c.
Overall mass transfer coefficient K’ x and flux N A.
GIVEN: Equilibrium data y Ao = 0.380 mol fraction x AL = 0.10
10.3-3
Absorption in a Countercurrent Stage Tower. Repeat example 10.3-2 using the same conditions but with the following change. change. Use a pure water flow to the tower of 108 kg mol H2O/h, that is, 20% above the 90 used in Example 10.3-2. Determine the number of stages required graphically. Repeat using the analytical Kremser equation.
T = 298 K P = 1.013 x 10 5 Pa ky = 1.465 x 10 -3 kg mol/s.m 3 mol fraction kx = 1.967 x 10 -3 kg mol A/s.m 2 mol fraction
V1 = 29.7 + 0.03 = 29.73 kgmolH 2O + acetone/hr
y A1 =
GIVEN: LO = 108 kg mol H 2O/hr
0.030 =0 =0.00 .00101 29.73
Y AN + 1 = 0.01 VN + 1 = 30.0 kg mol/h
Ln = 108 + 0.27 = 108.27 kgmol H 2O + acetone/hr
0.27
X AN =
108.27
T = 300 K PT = 101.3 kPa
0 ..0 00 24 249 3
y A = 2.53 x A REQUIRED: Theoretical stages graphically using analytical Kremser equation.
Using equation:
A1 =
L Lo = mv mv1
SOLUTION:
108 =1.4358 =1.4358 2.53 (29.73)
x Ao = 0 amount of enetering acetone = y AN + 1 VN + 1 = 0.01(30) = 0.30
AN =
LN mvN+1
=
108.27 2.53 (30)
entering air = (1-y AN + 1) VN + 1 = (1 – (1 – 0.01)(30) 0.01)(30) = 29.7 kg mol/air h
=1.4265
acetone leaving in V 1 = 0.10(0.30) = 0.30 kg mol/h acetone leaving in L n = 0.9(0.30) = 0.27 kg mol/h V1 = 29.7 + 0.03 = 29.73 kg mol air + acetone /hr
A= A1AN
1.4358 1.4358(1 (1.426 .4265) 5)
1.4311 1.4311
Ln = 108 + 0.27 = 108.27 kg mol H 2O + acetone/hr 10.4-1
1 1 log yN+1 - mxo 1 A A N=
Interface Concentrations Concentrations and and Overall Overall Mass-Transfer Mass-Transfer Coefficients. Use the same equilibrium data and film coefficients k’ y and k’x as in Example 10.4-1. However, use bulk concentrations of y AG = 0.25 and x AL = 0.05. Calculate the following: following:
log A
N=
0.01 - 2.53(0) 1 1 log 1 1.4311 1.4311 0 . 0 0 1 0 1 2 . 5 3 ( 0 ) log 1.4311
3.7662 stages
a.
Interface concentrations y Ai and x Ai and flux N A.
b.
Overall mass transfer coefficients K’y and K y and flux N A.
c.
Overall mass transfer coefficient K’ x and flux N A.
GIVEN: Equilibrium data y Ao = 0.380 mol fraction x AL = 0.10
10.3-3
Absorption in a Countercurrent Stage Tower. Repeat example 10.3-2 using the same conditions but with the following change. change. Use a pure water flow to the tower of 108 kg mol H2O/h, that is, 20% above the 90 used in Example 10.3-2. Determine the number of stages required graphically. Repeat using the analytical Kremser equation.
T = 298 K P = 1.013 x 10 5 Pa ky = 1.465 x 10 -3 kg mol/s.m 3 mol fraction kx = 1.967 x 10 -3 kg mol A/s.m 2 mol fraction
REQUIRED: K’X, KX
(1-y A)iM =
(1-0.194) - (1-0.380)
1-0.194 ln 1-0.380
SOLUTION: Trial 1:
k'x/(1-x A)iM k'y/(1-y A)iM
1.967 x 10 1.465 x 10
-3 -3
/ 1. 1.0 / 1. 1.0
- 1.342
(1-x A)iM =
y = mx + b 0.38 = -0.10 (1.342) + b
= 0.7089
(1-0.1) (1-0.1) - (1-0 (1-0.254 .254)) (1-0.1) (1-0.1) - (1-0.24 (1-0.247) 7) = = 0.8206 1-0.1 1-0.1 ln ln 1-.254 1-0.247
b = 0.5142 x A1 = 0.247 y Ai = 0.183
(1-y A)iM =
k'x/(1-x A)iM k'y/(1-y A)iM
(1-y 1-y Ai) - (1-y 1-yAG) (1-0. 1-0.18 183) 3) - (1(1-0. 0.38 380 0) = = 0.715 1-y Ai 1-0.183 ln ln AG 1 y 1-0.380
(1(1-x AL) - (1(1-x xAi) (1(1-0.1) 0.1) - (1-0 (1-0.2 .247 47)) (1-x A)iM = = = 0.825 1-x AL 1-0.1 ln ln Ai 1 x 1-0.247
-1.967 x 10 -3 / 0 .8 .8206 -3 1.465 x 10 / 0. 0.7089
-1.163
- 1.160
-1.60
x Ai = 0.254 y Ai = 0.194
m'' =
y AG - yAi 0.38 - 0.194 = * x A - x Ai 0.346 - 0.254
2.0217
1 1 1 = + K'x m''k'y k'x
Trial 2:
k'x/(1-x A)iM -1.967 x 10 -3 / 0. 0.825 - 1.163 -3 k'y/(1-y A)iM 1.465 x 10 / 0. 0.715 x Ai = 0.254 y Ai = 0.194 0.38 = - 1.163 (0.10) + b b = 0.4963
1 1 1 = + -3 K'x 2.0217 (1.465 x 10 ) 1.967 x 10 -3 K'x = 0.00118 kgmol/s m2 frac
K'x =
1 K'x/(1-x A)*M
K'x (1- x A) * M
1 2.0217(1.465 x 10 -3 / 0 .7089
1 1.967 x 10 -3 /0.8206
= 656.5320 (1- x A) * M =
(1- x AL) -(1- x A * )
(1- x AL) (1- x A )
ln
% resistance in the gas film
*
= (1- x A) * M =
(1-0.10)- (1-0.346)
(1- 0.10) (1- 0.346)
0.7705
ln
Kx =
239.3425 (100) 36.46% 656.5392
% resistance in the liquid film
=
0.00118 0.00153 0.7705
417.1835 (100) 63.54% 656.54
10.4-2 Use the same equilibrium data and film coefficients k’y and k’x as in Example 10.4 -1.
K'x * N A (xA -xAL) A *M (1-x ) 0.00118 N A = (0.346-0.10) 0.7705 N A = 0.000377 kgmol/sm2
However, use bulk concentrations of y AG = 0.25 and x AL = 0.05. Calculate the following. (a) Interface concentrations y Ai and x Ai and flux N A. (b) Overall mass-transfer coefficients K’y and Ky and flux N A. (c) Overall mass-transfer coefficient K’x and flux N A. GIVEN: y AG = 0.25
1 K'x/(1-x A)*M
1 1 m'' /(1-yA)*M K'x/(1-xA)*M
(1-y A)iM =
(1-y A)iM =
(1-y Ai) - (1-yAG ) (1-y Ai) ln (1-y AG) (1-0.194) - (1-0.38) 0.7089 (1-0.194) ln (1-038)
x AL = 0.05 ky = 1.465 x10 –3 kgmol A / sm 3 mol fraction kx = 1.967 x10 –3 kgmol A / sm2 mol fraction REQUIRED: a.) Interface concentrations y Ai & x Ai & N A SOLUTION: Trial 1:
k'x/(1-x A)iM
(1-x A)iM
(1-x AL) - (1-xAi ) (1-0.1) - (1-0.254) 0.8206 (1-x AL) (1-0.1) ln ln (1-x Ai) (1-0.254)
k'y/(1-y A)iM
1.967 x 10
-3
/ 1.0
1.465 x 10
-3
/ 1.0
- 1.342
y = mx + b 0.25 = -1.342 (0.05) + b b = 0.3171
x Ai = 0.1634
y Ai = 0.1010
Trial 2:
(1-y Ai) - (1-yAG) (1-0.1010) - (1-0.25) (1-y A)iM = = = 0.8223 1-y Ai 1-0.1010 ln ln 1-y AG 1-0.25
(1-x A)iM =
(1-x AL) - (1-xAi ) (1- 0.05) - (1- 0.1634) = = 0.8921 1-x AL 1- 0.05 ln ln 1-x Ai 1- 0.1634
k'x/(1-x A)iM k'y/(1-y A)iM
1.967 x 10 -3 / 0 .8921 0.8223 1.465 x 10 -3 / 1.0
- 1.2376
(yAG -yAi) (1- y A)iM 1.967 x 10 -3 N A = (0.25 - 0.1054) 0.8202 N A
k'y
N A = 2.583 x 10-4 kgmol/sm2 or,
k'x (xAi -xAL) (1- x A)iM 1.967 x 10 -3 N A = (0.1686 - 0.05) 0.8894 N A
y = mx + b 0.25 = -1.2376 (0.05) + b b = 0.3119
x Ai = 0.1686
y Ai = 0.1054
N A = 2.62 x 10-4 kgmol/sm2
Trial 3:
(1-y A)iM =
(1-x A)iM =
-
(1-y Ai) - (1-yAG) (1-0.1054) - (1-0.25) = = 0.8202 1-y Ai 1-0.1054 ln ln 1-y AG 1-0.25 (1-x AL) - (1-xAi) (1- 0.05) - (1- 0.1686) = = 0.889 1-x AL 1- 0.05 ln ln 1-x Ai 1- 0.1686
k'x/(1-x A)iM k'y/(1-y A)iM
-1. 96 7 x 10 1.465 x 10
-3 -3
/ 0 .8894 / 0 .8202
0.8223 - 1.2382
-1.2382 1.2376
x Ai = 0.1686
yAi = 0.1054
b)
m' =
y Ai - yA * 0.1054 - 0.0243 0.6838 = x Ai - x AL 0.1686 - 0.05
(1-y A)*M =
(1-y A * ) - (1-yAG) (1-y A * ) ln (1-y AG)
y A * = 0.0243
(1-y A)*M =
(1-0.0243) - (1-0.25 0.8579 (1- 0.0243) ln (1- 0.25)
10.6-6. - 10.6-7. A gas stream contains 4.0 mol % NH 3 and its ammonia content is reduced to 0.5 mol % in a packed absorption tower at 293 K and 1.013 x 10 5. The inlet pure water flow is 68.0 kg mol/h and the total inlet gas flow is 57.8 kgmol/h. The tower diameter is 0.747 m. The film masstransfer coefficients are k’ya = 0.0739 kgmol.sm 3 mol frac and k’xa = 0.169 kgmol/sm3 mol frac. Using the design methods for dilute gas mixtures, do as follows.
1 1 m' K'y/(1-y A)*M k'y /(1-yA)iM k'x /(1-xA)iM =
Repeat Example 10.6-2, using the overall liquid mass- transfer coefficient K’xa to calculate the tower height.
1 0.6838 1.4 65 x 10 -3 / ( 0.82 02 ) 1 .96 7 X 10-3 / 0 .8 89 4
(a) Calculate the tower height using k’ya (b) Calculate the tower height using K’ya.
K'y = 9.872 x 10-4 Solution:
K'y 9.872 x 10-4 0.00115 = (1 y A)* M 0.8579
Ky =
(a) La = 68 kgmol/h ya = 0.005
xa = 0
9.872 x 10-4 K'y (yAG-yA * ) (0.25 - 0.0243) 0.8579 (1 y A)* M
N A =
N A = 2.5972 x 10 - 4 kgmol/sm2
(b) Vb = 57.8 kgmol/h
c)
m" =
yb = 0.04
y AG - yAi 0.25 - 0.1054 = 1.1814 * Ai 0.291 - 0.1686 x A - x *
(1-x A)*M =
1 K'x/(1-x A)*M 1 K'x/0.8236
(1- x A L ) - (1-xA )
(1-x AL ) ln * (1-x A )
(1- 0.05) - (1-0.291) 0.8236 (1-0.05) ln (1-0.291)
1 m"k'y/(1-yA)*M
1 k'x /(1-xA) *M
1 1 1.1814(1-y A )*M 1.967 x 10-3 / 0.8894
K'x = 7.8775 x 10-4
V' = V (1 - yb ) L' = L (1 - xb )
57.8(1 0.04)
55.488 kgmol/h
68 (1 - 0) = 68 kgmol/h
V' (yb -ya ) = L'(xb - xa ) 55.488 (0.04-0.005) = 68x b
xb
0.029
YN+1
L' 68 0.005 Xn + Yn Xn + V' 55.488' 0.995 YN+1 1.2255 Xn + 0.0050 if Xn 0,
X=
x 1-x
Y 1+Y
Nx =
2
4
dx xi - x
y A1 -y A * x A1 -x AL
K'y =
K'y =
k'y k'x k'x + m' k'y
(0.747)
KyaS
=
V5 =
57.8 1- 0.040
56.7834
56.7864 / 3600 0.0559 / 0.4383
68 68 1- 0 68 + 70.0309 L= 2
4
= 0.4383 m 3
m" =
=0.6438
KxaS = 0.6438
y AG -y Ai
K'x =
L
=
2.2796 m
L5 =
Hox =
x A *-x Ai
68 1- 0.029
70.0309
69.0155
69.0155 / 3600 0.0457 / 0.4383
z = 0.9571 (2.3209) =
=0.9571
2.2213 m
m" k'y k'x k'x + m''k'y
(0.0739)(0.169) 0.169 + 0.8287(0.0739)
K'y Ky = KyaS
V
L1 =
height = z = Nx Hx = 0.2250 (8.6692) = 2.2106 m
m' =
2
z = 0.6438 (3.54009) =
2
=
55.7668
= 0.6438 y = mx + b
L kxaS V' V= 1- y d
1-0.005 55.7668 + 57.8
Hoy =
k'xa/(1-x) k'ya/(1-y)
Hx =
S = Area =
V=
55.488
YN+1 = 0.0050
y=
slope = m = -
V1 =
K'x Kx = (1- X)
0.0450
X
y
X
Y
m
b
yi
0
0.0050
0
0.005
-2.2754
0.005
0.0014
0.01
0.0170
0.0101
0.0173
-2.2707
0.0397
0.0096
0.015
0.0229
0.0152
0.0234
-2.2685
0.0569
0.0141
0.02
0.0287
0.0204
0.0295
-2.2667
0.0740
0.0188
0.029
0.0400
0.0299
0.0417
-2.2610
0.0108
0.0272
xi
1/(y – 1/(y – y yi)
1/(x – 1/(x – x xi)
0.0018
277.7780
555.5556
0.0132
135.1351
312.50
0.0190
113.6364
250
0.0246
101.0101
217.3913
0.0348
78.125
172.4138
10.6-10. Repeat Example 10.6-2 but use transfer units and calculate H L, NL, and tower height.
1/(y – 1/(y – y yi) ave
∆y
∆x
206.4565
0.012
0.01
124.3858
0.0059
0.005
107.3233
0.0058
0.005
89.5678
0.0113
0.009
Given: Acetone-H2O system A = 0.186 m2 T = 293 K P = 101.32 kPa y1 = 0.026 y = 0.005
dy/(y-yi) Ave
dx/(x-xi) Ave
2.4775
4.3403
0.7339
1.4063
0.6225
1.1685
1.0121
1.7541
Ky=k’ya/(1-y) Ky=k’ya/(1-y)
Kx =k’xa/(1-x) k’xa/(1-x)
V = 13.65 kgmol inert air / h L = 45.36 kgmol H 2O / h
0.0743
0.169
0.0752
0.1707
0.0756
0.1716
0.0761
0.1724
0.0756
0.1740
k’ya = 3.78 x 10 -2 kgmol / sm 3 mol frac k’xa = 6.16 x 10 -2 kgmol / s m3 mol frac REQUIRED:
HL, NL, z
SOLUTION: Ny = 4.8549
Nx = 8.6692
ky Ave = 0.0756
kx Ave = 0.1715
X
y
m’
m” m”
Ky
Kx
0.0072
0
0.7778
0.6667
0.0590
0.0450
0.0224
0.0072
0.7500
0.8043
0.0555
0.0455
0.0292
0.011
0.8158
0.8627
0.0559
0.0457
0.0360
0.0148
0.8696
0.8684
0.0562
0.0459
0.0484
0.0224
0.8276
0.9412
0.0569
0.0463
AVE: 0.8082 0.8082
Ave: 0.8287
1/(y – y )
1/(x – x)
200
138.8889
102.0408
80.6452
84.0336
70.4225
71.9424
62.5
56.8182
51.5464
1/(y – y )ave
151.0204
1/(x – x)ave
109.7671
93.0372
75.5339
77.9880
66.4613
64.3803
57.0203
dy/(y – y )ave
dx/(x – x)ave
1.8122
1.0977
0.5481
0.3777
0.4523
0.3323
0.7275
0.5132
Noy= 3.5409
Nox= 2.3209
y1 y xo x + V' 1 - y1 L' 1 - x V' 1 - y o 1 x 0 0.026 x1 45.36 + 13.65 1 - 0.026 45.36 1 - x1 +13.65 1 - 0
L'
0.005 1 - 0.005
x1 = 0.00648
1h 45.36 45.36 kgmolH kgmolH2O / h L 3600 s HL = = = 1.0264 -2 3 K'xa K'xa S 6.6 6.6 x 10 kgmo kgmoll / sm mol mol frac frac.. (0.1 (0.186 86 m2 ) from the graph: xi1 = 0.0136
xi2 = 0.0019
(xi - x)M =
(xi1 - x1) - (xi2 - x 2)
(xi1 - x1) ln (xi2 - x2)
=
(0.0132132- 0. 0.0064 0648) - (0.001 .00199- 0) 0)
(0.0132 - 0.00648) (0.0019)
=0.0038
ln
y2 y1 x2 x1 + V' 1 - y1 L' 1 - x1 V' 1 - y 1 - x2 2
L'
0.2 x1 0.02 0 -4 -2 -4 + 6.53x10 4.20x10 +6.23x10 1 - 0 1 - 0.2 1 - x1 1 - 0.02
4.20 x 102
x1 = 0.00355 0.00355
NL =
dx 0.00648 1.7053 53 1.70 xi - x 0.0038
z = NL HL = 1.70 1.7053 53 (1.0 (1.0264 264))
6.53 x 10 -4 ( 29)kg air / s + 6.53 x 10 -4
y
1 y
Gy =
64.1
0.0929
z = 1.75 m
if y = 0.20
0.2 64.1 1 0.2
6.53 x 10 -4 ( 29)kg air / s + 6.53 x 10 -4
10.7-1 Liquid Film Coefficients and Design of SO 2 Tower. Using the data of example example 10.7-1,
Gy =
calculate the height of the tower using Eq. (10.6-15), which is based on the liquid film
0.0929
= 0.3164kg / sm
mass-transfer mass-transfer cofficient k’ xa. [Note: the interface interface values xi have alre4ady been obtained.
2
Usi a graphical integration of Eq. (10.6-15). Given:
V=
V' 6.53 x 10-4 8.16 x 104 8.16 1-y 1 0.2
L=
L' 4.20 x 10-2 = 0.0421 1- x 1 - 0.00355
A = 0.0929 m 2 T= 293 K P = 1.013 x 10 5 y1= 0.20 y1 = 0.02
Gx =
x1 = 0 V’ = 6.53 x 10 -4 kgmol air/ s L’ = 4.20 x 10 -2 kgmol / s k’ya = 0.0594 Gy 0.7Gx0.25 k’xa = 0.152 Gx
x 64.1 1 x
4.20 x 10-2 18 + 4.20 x 10-2
0.82
0.0929 0.00355 4.20 x 10-2 + 4.20 x 10-2 64.1 1 0.00355 8.2410 0.0929 k'xa = 0.15 0.152 2 Gx 0.82 = 0.15 0.152 2 (8.24 8.2410 10))0.82 0.85 0.8569 69 k'y k'ya = 0.0594G 94Gy0.7 Gx0.25 = 0.0594 (0.2130)0.7 (8.2410 410)0.25 = 0.0341
REQUIRED: SOLUTION:
Z using kxa
HL
L 0.04207 0.5319 = kxa kxa (1-x (1-x))iM 0.85 0.8535 356( 6(0. 0.99 9975 75)( )(0. 0.09 0929 29))
z = NLHL 3.051 (0 (0 .5319) =
1.6228
y
x
V
L
Gy
Gx
k’ya
0.02
0
6.66 x 10-4
0.042
0.2130
8.138
0.03398
0.04
0.000332
6.68 x 10-4
0.0421
0.2226
8.147
0.03504
SOLVED PROBLEMS:
0.07
0.000855
7.02 x 10
-4
0.04203
0.2378
8.162
0.03673
12.8-1. Effective Diffusivity in Leaching Particles. In Example 12.8-1 a time of leaching of the
0.13
0.00201
7.51 x 10-4
0.04208
0.2718
8.196
0.04032
solid particle of 3.11 h is needed to remove 80% of the solute. Do the following
0.20
0.00355
8.16 x 10-4
0.04215
0.3164
8.241
0.04496
calculations.
Leaching
(a)
Using the experimental experimental data, data, calculate the effective diffusivity, diffusivity, D Aeff .
(b)
Predict the time time to leach 90% 90% of the solute solute from the 2.0 2.0 mm particle.
GIVEN: k’xa
m
b
xi
yi
m’
b
0.848
24.4567
0.02
0.00046
0.0088
24.7476
0.02
0.849
23.2680
0.04772
0.00109
0.026
23.6252
0.04784
0.850
21.5403
0.08842
0.00188
0.04824
22.2705
0.08904
0.853
18.4426
0.1671
0.00356
0.1016
19.0742
0.1683
0.857
15.3034
0.2543
0.00572
0.1663
15.9829
0.2567
80 % efficiency t = 3.11 h REQUIRED: (a) D Aeff (b) T if 90% 90% efficient with same same diameter diameter
xi
yi
m”
b
0.000459
0.0087
24.7501
------
0.00104
0.02416
23.6687
0.04762
0.00186
0.0476
22.0814
0.08888
0.00354
0.1004
19.0993
0.1684
0.00570
0.1642
16.0037
0.2568
SOLUTION:
∆x ∆x
dy/(xi – x) – x)
0.000332
0.5897
0.000523
0.6247
D Aeff t
0.001155
0.9687
a
0.00154
0.8679
(a)
2
D Aeff
0.112
0.112 a
t
NL = 3.051
2 2
0.112
2
mm
3.11 hr
D Aeff
2
2
1hr r 1h 3600 s
1.0004 x 10-5 mm2 /s
V2 = 80 kg solvent (b)
x A2 = 0.03 xC2 = 0.97
D Aeff t a
2
x A1 = 0.22
0.18
N = 1.5 kg insoluble solid/kg solution REQUIRED:
t
0.18 a
2
(a) amount and composition of overflow, V 1 (b) amount and composition of the underflow, L 1
D Aeff SOLUTION:
2 2
2
0.18 t
1.0004 x 10
mm -5
B 100 (1 - 0.22) 78 kg insoluble solid
2
2
mm /s
t
1 hr 18000 s 3600 s
t
5 hr
Lo 100 - 78 22 kg of A
B 78 kg solid 3.5455 N o kg solution Lo 22 y Ao
12.9-1.
1.0
Leaching of Oil from Soybeans in a Single Stage. Repeat Example 12.9-1 for single stage leaching of oil from soybeans. The 100 kg of soybeans contains 22 wt % oil and the solvent feed is 80 kg of solvent containing 3 wt % soybean oil. V1, x1 Lo, No, yo, B
V2, x2 L1, N1, y1, B
For Overall Material Balance:
Lo
V2
L1
Lo
V2
M 22 80
GIVEN:
V1
M 102 For Solute Balance:
M
L o y Ao
V2 xA 2
Mx AM
20.1. Roasted copper ore containing the copper as CuSO 4 is to be extracted in a countercurrent stage extractor. Each hour a charge consisting of 10 tons of gangue, 1.2 tons of copper sulfate, and 0.5 ton of water is to be treated. The strong solution produced is to consist
22(1.0) 80(0.03) 102x AM
of 90 percent H 2O and 10 percent CuSO 4 by weight. The recovery of CuSO 4 is to be 98 percent of that in the ore. Pure water is to be used as the fresh solvent. After each
x AM
stage, 1 ton of inert gangue retains 2 tons of water plus the copper sulfate dissolved in
0.2392
To find NM
that water. Equilibrium is attained in each stage. How many stages are required? GIVEN:
B N o Lo N M M
10 tons of inert solids/hr 1.2 tons CuSO 4 and 0.5 ton H 20 is to be treated per hour Strong solution – 90% H2O, 10% CuSO4
78 N M (102)
CuSO4 recovery of 98% of that in ore yb = 0
N M
0.7647
after each stage, 1 ton inert solids retains 2 tons of H 2O + CuSO4 dissolved REQUIRED:
To solve for the value of exit underflow
N1L1 N M M (1.5)L1 (0.7647)(102)
Stages required SOLUTION: yb = 0
L1 51.9996 52 kg
= 1.7 tons solution/hr
For exit Overflow:
L1
V1
52 V1
ya = 0.1 La = 1.2 tons CuSO 4 + 0.5 ton H20
M
102
xa
1.2
0.7059
1.7
CuSO4 recovery = 0.98(1.2) = 1.176 tons
V1 50 kg
CuSO4 retained = 1.2 – 1.176 = 0.024 tons
solution retained
2 tons H 2 0 0.024 tons CuSO 4
20.4. Oil is to be extracted from halibut livers by means of ether in a countercurrent extraction battery. The entrainment of solution by the granulated liver mass was found by
1 ton inert solid
experiment to be as shown in Table 20.5. In the extraction battery, the charge per cell is to be 100 lb, based on completely exhausted livers. The unextracted livers contain
2.024 tons solution
L b
L b
2.024 tons solution
0.043 gal of oil per pound of exhausted material. A 95 percent` recovery of oil is desired. The final extract is to contain 0.65 gal of oil per gallon of extract.
1 ton inert solid
1 ton inert solid
10tons solid
Solution retained by 1
Solution
Solution retained by
Solution
lb exhausted livers, gal
concentration, gal
1 lb exhausted livers,
concentration, gal
20.24 tons solution/h r
Solute Balance:
La x a
V b y b
L b x b
oil/gal solution
gal
oil/gal solution
0.035
0
0.068
0.4
0.042
0.1
0.081
0.5
0.050
0.2
0.099
0.6
0.058
0.3
0.120
0.68
Va y a The ether fed to the system is oil free. (a) How many gallons of ether are needed per charge of livers? (b) How many extractors are needed?
1.7 (0.7059) V b (0) 20.24(0.00 12) 0.1(Va ) GIVEN:
Va
Charge per cell = 100 lbs
1.1757 tons /hr
95% oil recovery y A = 0.65
By Material Balance:
La
yB = 0 assuming x A = 1, L A = 0.043 gal oil/lb exhaust liver(100)
Vb Lb Va
1.7 Vb 20.24 1.1757
L A = 4.3 gal REQUIRED: (a) gal of ether needed per charge of liver
Vb 19.7157 tons H2O
SOLUTION:
from the graph :
No. of stages
(b)
3.70
For XB:
number of extractors needed
Oil retained = (0.043 gal oil/ lb exhausted liver) (0.05) (100) = 0.215 gal oil
LB = 0.0394 x 100 = 3.94
For the solution in the spent solids: (by trial & error)
xB
0.251
3.94
0.0637
Trial 1:
Let XB = 0.1
0.0637 (close to 0.0634)
Solution retained = 0.042 LB = 0.042 x 100 = 4.2 gal
xB
assuming X A = 1, L A = 0.043 gal oil/lb exhaust liver x 100
0.251
4.2
L A = 4.3
0.051
By solute balance: Trial 2:
Lax A +VBYB = LBXB +V AY A XB = 0.051, by linear regression: Solution retained = 0.0386
4.3(1) + VB(0) = 3.94(0.0637) +V a(0.65)
LB = 0.0386 x 100 = 3.86 gal VA = 6.2293 gal
xB
0.251
3.86
0.065
By OMB: L A + VB = LB +V A
Trial 3
4.3 + V B = 3.94 + 6.2293 XB = 0.065, by linear regression Solution retained = 0.0396 LB = 0.0396 x 100 = 3.96
xB
3.96
If X1 = ya = 0.65, solution retained = 0.1121 L1 = 0.1121 x 100
0.251
V B = 5.8693 g al
0.0634 L1 = 11.21 gal By OMB:
Trial 4 xB = 0.0634, by linear regression solution retained = 0.0394
La + V2 = L1 + Va
V2 = 11.21 +6.2293-4.3 Number of stages = 6.7813 V2 = 13.1393 gal Geankoplis 11.2-1 Single-Stage Contact of Vapor-Liquid System. A mixture of 100 mol containing 60 mol% By Solute Balance:
n-pentane and 40 mol% n-heptane is vaporized at 101.32 kPa abs pressure until 40 mol of vapor and 60 mol of liquid in equilibrium with each other is produced. This occurs in a single-stage
Lax + V2y2 = L1x1 + V aya
system and the vapor and liquid are kept in contace with other until the vaporization is complete. The equilibrium data are given in Example 11.3-2. Calculate the composition of the vapor and the
4.3(1)+ 13.1393y 2 = 11.21(0.65) + 6.2293(0.65)
liquid.
y2 = 0.5355
Given:
If xN = 0.4, s olution retained = 0.068 LN = 0.068 x 100
x
y
x
y
1.000
1.000
0.254
0.701
0.867
0.984
0.145
0.521
0.594
0.925
0.159
0.271
0.398
0.836
0
0
LN = 6.8 gal Solution: By OMB:
C5H12 Balance:
0.60(100) YA (40) X A (60)
La + VN+1 = LN +Va
YA 4.3 + V N+1 = 6.8 + 6.2293
Laxa +VN+1 yN+1 = LNxN + Vaya 4.3(1) + 8.8293y N+1 = (6.8) (0.4) + (6.2293) (0.65) yN+1 = 0.2828 From the Graph:
1.5 1.5X A
when X A = 0.6 , Y A = 0.6……….pt 1 when X A = 0.4, Y A = 0.9………...pt 2
VN+1 = 8.7293 By Solute Balance:
1) Create a graph of the equilibrium curve for n-pentane. 2) Plot points at pt.2. 3) Get the value of X A and Y A, on the point of intersection between the equilibrium curve and the segment joining the two points. 4) Get Y B and XB. Answers: From the graph: X A = 0.43 and Y A= 0.855 XB = 1 - 0.43 = 0.57 and Y B = 1 – 0.855 = 0.14
11.3-2 Comparison of Differential and Flash Distillation. A mixture of 100 kg mol which contains 60 mol % n-pentane (A) and 40 mol % n-heptane (B) is vaporized at 101.32 kPa pressure
A4
2.42 2.3 0.0425 0.1003 2
under differential conditions until 40 kg mol are distilled. Used equilibrium data from Example 11.32. (a)
What is average composition of the total vapor distilled and the composition of the liquid
2.42 0.05 0.1250 2
A3
2.58
left.
____________
(b) If this same vaporization is done in an equilibrium or flash distillation and 40 kg mol are
AT = 0.5048
distilled, what is the composition of the vapor distill ed and of the liquid left?
therefore X2 = 0.4075
Given:
Answers:
x
y
1/ ( y-x )
1.000
1.000
0.867
0.984
8.547
0.594
0.925
3.021
11.3-3 Difeerential Distillation of Benzene-Toluene. A mixture containing 70 mol % benzene and
0.398
0.836
2.283
30 mol % toluene is distilled under differential conditions at 101.32 kPa (1 atm). A total of one third
0.254
0.701
2.237
of the moles in the feed is vaporized. Calculate the average composition of the distillate and the
0.145
0.521
2.660
composition of the remaining liquid. Use equilibrium data Table 11.1-1.
0.059
0.271
4.717
0
0
X A = 0.4075 y A = 0.845
Given: F = 0.7 benzene and 0.3 toluene
Solution:
L 1 ln L 2 ln
X2
dx y x X1
L 1 100kmol ln 0.5108 L 2 60kmol
from the graph:
A1
3.04 2.78 0.05 0.1455 2
A2
2.78 2.58 0.05 0.1340 2
Solution: OMB: B = F – D = F – 1/3F = 2/3F
ln
ln
L 1 L 2 L 1 L 2
X2
dx
y x
X1
ln
F 2 3
from the graph:
F
0.4055
A1
A2
6.68 5.92 0.05 0.3150 2 5.92 5.66 0.0175 2
R= 4.52
R
Ln D
0.10
______________ AT = 0.4163 Therefore x2 = 0.6325 Answer: x A = 0.6325 and y A = 0.7975 xB = 1 – 0.6325 = 0.3675 and y B = 1 – 0.7975 = 0.2025
R 1 Xn R 1 X D 1 R 4.52 1 Yn 1 5.52 Xn 5.52 0.9 Yn1
if x = 0 ,Yn+1 = 0.16 ( 0,0.16)..........plotted on the graph Answer:
11.4-1 Distillation Using McCabe-Thiele Method. A rectification column is fed 100 kg mol/h of a mixture of 50 mol % benzene and 50 mol % toluene at 101.32 kPa abs.pressure. The feed is liquid at the boling point. The distillate is to contain 90 mol % benzene and the bottoms 10 mol % benzene. The reflux ratio is 4.52:1. calculate the kg mol/h distillate, kg mol/ h bottoms and the theoretic al number of trays needed using the McCabe –Thiele method. Given: F = 100 kg mol/h
4.9 trays + a reboiler
11.4-2. Rectification of a Heptane-Ethyl Benzene Mixture . A saturated liquid feed of 200 mol/h at the boiling point containing 42 mol% heptane and 58 mol% ethyl benzene is to be fractionated at 101.32 kPa abs to give a distillate containing 97 mol% heptane and a bottoms containing 1.1 mol% heptane. The reflux ratio used is 2.5:1. Calculate the mol/h distillate. Mol/h bottoms, theoretical number of trays, and the feed tray number. Equilibrium data are given below at 101.32 kPa abs pressure for the mole fraction n-heptane x H and y H.
0.5 benzene 0.5 toluene R = Ln/D = 4.52 XD = 0.9 XB = 0.10
Temperature K
ºC
xH
yH
K
ºC
xH
yH
136.2
0
0
383.8
110.6
0.485
0.730
402.6
129.4
0. 08
0.230
376.0
102.8
0.790
0.904
392.6
119.4
0.250
0.154
371.5
98.3
1.000
1.000
F=D+B 100 = D + B XF F = DXD + BXB 50 = 0.9D – 0.1D + 10 40 = 0.8D D = 50 kg mol/h B = 50 kg mol/h
Temperature
409.3
Solution:
100 ( 0.5 ) = 0.9D + 0.1 ( 100 – D )
( see preceeding graph )
Given: F = 200 mol/h 0.42 heptane 0.58 ethyl benzene XD = 0.97 heptane XB = 0.011 heptane
R = Ln/D = 2.5
Rm = 0.88 (b)
4 trays + a reboiler ( see preceeding graph )
Solution: F=D+B
11.4-4. Minimum number of Theoretical Plates and Minimum Reflux Ratio. Determine the
200 = D + B
minimum reflux ratio Rm and the minimum number of theoretical plates at total reflux for the
XF F = DXD + BXB
rectification of a mixture of heptane and ethyl benzene as given in Problem 11.4-2. Do this
200( 0.42 ) = 0.97D + 0.011 ( 200 – D )
graphical methods of Mc Cabe-Thiele.
84 = 0.97D - 0.011 + 2.2 81.8 = 0.959D
Given:
D = 85.2972 mol/h
y’ = 0.68
B = 114.7028 mol/h
x’ = 0.42
R = 2.5
Solution:
Yn1
1 2.5 X n X D 3.3 3.5
Rm Rm
1
0.97
0.68
0.97
0.42
Rm = 0.5273 ( Rm + 1 )
= 0.277
0.4227Rm = 0.5273
Answer: 10.77 theoretical stages or 9.77 trays + a reboiler
Rm = 1.12
feed enters at 6yh tray from the top Minimum number of trays: 7.64 theoretical stages or 6.64 trays + a reboiler
11.4-3. Graphical Solution for Minimum Reflux Ratio and Total Reflux . For the rectification in problem 11.4-1, where an equimolar liquid feed of benzene and toluene is being distilled to give a distillate of composition XD = 0.90 and a bottoms of composition X W = 0.10, calculate the following using graphical methods. (a) Minimum reflux ratio Rm. (b) Minimum number of theoretical plates at total reflux. Solution: From the graph x’ = 0.5 and y’ = 0.7125
Rm Rm 1
x D
y '
x D
x'
0.96 0.5 0.96 0.7125
Foust 7.27. A mixture containing 30 mol% benzene and 70 mole% toluene is to be fractionated at normal atmospheric pressure in a column with a total condenser and a still from which the bottoms are withdrawn. The distil late is to contain 95 mole percent benzene and the bottoms 4 mole percent
(a)
Rm – 0.4687512Rm = 0.46875
benzene. The feed is at its dew point.
(a) What is the minimum reflux ratio ( L O / D )
(c) What is the minimum refux ratio?
(b) What is the minimum number of equilibrium stages in the column required at total reflux?
(d) What percentage of the feed ethanol is recovered in the distillate?
(c) How many equilibrium stages are required at a reflux ratio of 8? (d) How many equilibrium stages would be required at a reflux ratio of 8 if the feed were a liquid at
SOLVED PROBLEMS:
its bubble point? 1. An air stream at Given:
90 C having a humidity of 0.03
kg H2O
is contacted in adiabatic
kg dry air
humidifier. It is cooled & humidified at 90 % RH. Determine the temperature of the humidified air &
F = 0.3 benzene and 0.7 toluene
3
the make up H 2O for every m of inlet air.
xD = 0.95 benzene xW = 0.04 benzene
GIVEN:
q=0
COOLER
Solution:
Rm
(a)
Rm 1
x D
y '
x D
x'
0.95 0.3 0.95 0.155
T1 = 90 OC H= 0.03
Rm = 0.8176Rm + 0.8176
kg H O 2 kg dry air
RH2 = 90 %
Rm = 4.4827 (b) Minimum number of equilibrium stages @ Total Reflux = 6.89 stages
REQUIRED:
T2
(c) R = 8 Yn 1
R X 1 R
n
R
X 1
1
D
8 1 Yn1 (X n ) 0.95 9 9
L in every m 3 of air SOLUTION:
Step 1: From Psychrometric Chart:
@ Xn = 0 , Yn+1 = 0.1056 @ Xn = 0. 95 , Yn+1 = 0.95 Number of equilibrium stages @ reflux = 8 :
Step 2: Compute for humid volume. 8.79 stages
(d) R = 8, feed is l iquid at boiling point Number of equilibrium stages = 8.75 stages
H2O balance:
7.35. An equimolar mixture of ethanol and water is to be fractionally distilled to produce a distillate
L=W
of composition 0.80 mole fraction ethanol and bottoms of 0.05 mole fraction of ethanol. The feed is saturated liquid and there is a total condenser and a total reboiler. (a) At areflux ratio of 2.0, how many equilibrium stages are required? (b) How many stages are required at total reflux?
VH = =
2.83 x
H2 10
3
H1
4.56
x
10
3
H
T K
2.83 x 10
3
4.56
x
10
3
0.03
90
273
T1 = 300C
T2 = 50 0C
3
VH = 1.0769
Basis: 1 m
HEATER
m kg da
3
RH = 80% Vf =200 m3/h W=1 m
3
kg da 3 1.0679 m
REQUIRED:
qH
OLUTION: Step 1: From Psychrometric Chart
W = 0.9285 kg da
L=W
H2
Step 2: Compute for the Humid Volume
H1
= 0.9285 kg da ( 0.51-.0.03)
kg H2O kg da
VH = ( 2.83 x 10 –3 + 4.56 x 10 –3 H)(T+273) = [2.83 x 10 –3 + (4.56 x 10 –3 x 0.022)] (30+273) VH = 0.8879 m 3/ kg dry air
Step 3: Calculate Mass flowrate of air L = 0.0195 kg H2O
3 m 1kg dry air W = 200 3 hr 0.8879m 2. W = 225.25 kg dry air/hr
3. 200 m 3/h of air at 300C and 80% relative humidity is to be heated to 50 0C using a heater. Step 4: Calculate the specific heat
Calculate the heat load of the heater in KW.
cs = (1.005 + 1.88 H)
GIVEN: qH
cs = [1.005 + 1.88 ( 0.022) ]
c
s
= 1.046
kJ kg k
25 0C
15 0C
Step 5: Calculate the heat load of the heater
L
kg kJ q H= 225.25 1.0464 hr kg K REQUIRED: qc
qH 4714.032
4.
kJ hr
An air condition ing unit must keep the air inside the room at 25 0C and 90% relative 0
humidity. The air coming from the air conditioning unit has a T of 15 C. Calculate the hp rating of the air conditioning unit if the room space is 75m 3, the air in the room is changed every 15 min.
SOLUTION: Step 1: Compute for h 1 & h2 , then vH
3
h1 1.005 + 1.88H1 x10 t1 0 + 2.501
6
x10 H1
3
h1 1.005 + 1.88 0.0184 x10 25 0 +
2.501
GIVEN:
AC
6
h1 72,008.2
J kg
x10 0.0189
3
6
h2 1.005 + 1.88H2 x10 t1 0 + 2.501 x10 H2
qH
ROOM
h2 1.005
2.501
25 0C 90%RH
6
qc
vH 2.83x10 3
COOLER
3
x10 15 0 +
x10 0.0097
h2 39,608.79
+ 1.88 0.0097
J kg
4.56x10 3H1 t1
273
vH 2.83x10 3 4.56x10 3 0.0184 (25 + 273)
4.187x10 3
3
vH 0.8683
m kg dry air
kJ (15-25) K kg K
q C 1.132x10
Step 2: Compute for W.
kg dry air 60 min hr 3 1 min 0.8683 m
hr
x
q c 4.2168
3
W
1 hr
7 J
3600 s
hp
75 m
15
5.
A chemical plant in Baguio is going to build a system that will contain ambient air to 200
0F and 115 0F wet bulb. The air in Baguio is foggy and at an average T of 70 0F, the analysis shows that 0.0008 lb H2O per cubic feet of air is entrained. The system shall consist of a heater,
W=345.5027
kg dry air hr
then an adiabatic humidifier and a reheater system. The exhaust of adiabatic humidifier has a humidity of 90% RH. What is the T of air leaving the preheater and humidifier. GIVEN:
Step 3: Solve for L, and finally q c. H1=H2
L W H1 H2
L 86.3757
kg H2O kg dry air 0.184 0.0097 hr kg dry air
L=3.0059
kg H2O hr
qC W h1 h2 - Lc pL t2 t1
q C 345.5027
H2 72,008.2 39,608.79 kg dry air 3.0059 kg hr J hr kg
TW2=TW3
H3=H4 90% RH
PREHEATER
ADIABATIC HUMIDIFIER
70 0F
REHEATER
TD = 200 0F TW = 115 0F
0.0008 lbH2O 3
ft air
REQUIRED: T of air leaving the preheater and humidifier
SOLUTION: Step 1: From Psychrometric Chart
L = W (0.048 – H2)
W
1
Tw = 115 oF H3=H4 0.0008=
0.048 lb H2O/lb da
vH
eqn. 2
0.048 H2 70 460 0.0252 0.0405H2
H2 =
Td = 200 oF
eqn. 1
0.0367 lb H2O lb dry air
Step 3: Use H 2 to read T 2 from Psychrometric Chart.
90 % RH Tw2 =103.8 0F Tw2=Tw3= 103.8 oF
H3=0.048 lbH2O/lb da
6. Step 2: Solve for H 2.
T3= 100 oF
H2 = 0.0367 LbH2O/lb da
T2 =152 oF
In a plant lab having a floor area of 100 m 2 and a ceiling height of 3 m, the T and RH are
kept at 23.9 0F and 80 % respectively. The closed loop air conditioning unit installed for the purpose has an air capacity to change the air in the room, which 80 % is void space, every 10 minutes. The air leaving the condenser f the aircon unit has a T of 18.3
H2O Balance at Humidifier:
of condensate which has to be dashed from the aircon unit in kg/hr WH2 = WH1 + L L = W( H3 – H2)
0
C. Calculate the quantity
GIVEN:
H=0.0151
AC
H=0.0106
18.3 oC
ROOM A=100 m2 H=3m
23.9 oC
Step 2: Compute for v h.
23.9 oC 80%RH
vH 2.83x10 3
80% void 10 min
4.56x10 3 0.0151 (23.9 + 273) 3
vH =
0.8607 m kg dry air
qH 23.9 oC 80 % RH
Step 3: Solve for W and then L, the condensate.
18.3 oC COOLER
1 2 kg dry air W 100m x3m 0.8607 0.8
REQUIRED: L
L
435.71 kg dry air 60 min 10 min 1hr
W
SOLUTION: Step 1: From Psychrometric Chart.
W=2614026 kg dry air/ hr
L = W (H1-H2)
80% RH
L =
2614.26 kg da 0.0151 0.0106 kg H2O hr kg da
L=
H=
11.7642 kg H2O hr
PA PT PA
x
MA MB
KPa H = 3.59
101.3
GEANKOPLIS:
H = 0.0228
18
3.59
KPa
X
29
kg H2O kg dry air
9.3-1. Humidity from Vapor Pressure The air in a room is at 37.8 oC and a total pressure of 101.3 kPa abs containing water
Step 2: Compute Hs & Hp.
vapor with a partial pressure p A = 3.59 kPa. Calculate: From steam table, by interpolation:
a. Humidity b. Saturation humidity and percentage humidity
P As of H2O = 6.59 KPa
c. Percentage relative humidity GIVEN:
Hs =
ROOM T= 37.8 C PT = 101.3 KPa PA = 3.59 Pa
Hs =
MA p As x P T P As MB
6.59 KPa
101.3 6.59 KPa
HS = 0.0432
REQUIRED: Humidity, Saturation humidity & Percentage Humidity,
X
kg H2O kg dry air
Percentage Relative Humidity
SOLUTION: Step 1: Calculate H.
H p
H =
Hs
x 100
18 29
kg H2O kg dry air kg H2O 0.0432 kg dry air
Percentage Humidity & Percentage Relative Humidity
0.0228
H p
=
H p
x 100
SOLUTION: Step 1: Compute for P A.
PA
= 57.78 %
H=
Step 3: Solve for H R.
PT PA
PA
HR
=
P As
0.021 =
101.3 PA
61.69 - 0.609 =
HR
MA MB
PA x 100
3.59 KPa
HR
x
6.59 KPa
29
P A = 18 P A
61.69 P A = 18.609
x 100
P A = 3.32 KPa
= 54.48 %
9.3-2. Percentage and Relative Humidity
18 x
Step 2: Compute for Hs to solve H P & HR.
The air in a room has a humidity H of 0.021 kg H 2O/ kg dry air at 32.2 oC and 101.3 kPa From steam table, by interpolation:
abs. Pressure. Calculate: a.
Percentage humidity H p
b.
Percentage relative humidity H R
P As of H2O = 4.82 KPa
GIVEN:
Hs =
ROOM T= 32.2 C PT = 101.3 KPa H = 0.021 kg H 2O/ kg da REQUIRED:
Hs =
MA p As x P T P As MB
18 4.82 KPa X 101.3 4.82 KPa 29
HS = 0.031
kg H2O kg dry air
H p
Td = 65.6 C Dp = 15.6 C
H =
Hs
x 100 REQUIRED: Actual humidity and % humidity
H p
kg H2O 0.021 kg dry air = x 100 kg H2O 0.031 kg dry air H p
Humid volume & humid heat in SI & English units. SOLUTION: Step 1: From Psychrometric Chart.
= 67.74 %
Dp = 60 F
HR
HR =
PA =
P As
5.3 % RH
x 100
3.32 KPa 4.82 KPa
HR
Td = 150 F
x 100
= 68.88 % Step 2: Compute for humid heat.
CS
= 1.005 + 1.88H
9.3-3. Use of the Humidity Chart The air entering a dryer has a temperature of 65.6 oC (150 oF) and a dew point of 15.6 oC
= 1.005 + 1.88 (0.011)
(60 oF). Using the humidity chart, determine the actual humidity and percentage humidity. Calculate the humid volume of this mixture and also calculate C S using SI and English units.
CS
GIVEN:
DRYER
H = 0.011 kg H2O/ kg da
= 1.026
KJ kg dry air K
( SI )
CS
= 0.24 + 0.45H
GIVEN:
= 0.24 + 0.45 (0.011)
CS
= 0.245
lbm
Btu dry air
DRYER
F
T= 52.7 C H = 0.030 kg H 2O/kg da
( English )
Step 3: Solve for V H.
vH = 2.83 x 10
3
4.56
x 10
3
H
T K
REQUIRED: Percentage humidity, saturation humidity at 57.2 oC, dew point,
vH = 2.83 x 10
3
4.56
x 10
3
0.011
humid heat & humid volume ( 65.6 + 273 ) SOLUTION:
3
vH = 0.975
m kg dry air
Step 1: From Psychrometric Chart.
( SI )
Step 2: Look for P As at 57.2 oC and solve for H S & HP. From steam table, by interpolation: vH =
0.0204
0.0405 H
T
F
vH = ( 0.0204 + 0.0405(0.011) ) ( 150 + 460 )
P As of H2O at 57.2 oC = 17.60 Kpa
Hs =
MA p As x P T P As MB
3
vH = 15.64
lbm
ft ( English ) dry air
An air-water vapor mixture going to a drying process has a dry bulb temperature of 57.2 o
C and a humidity of 0.030 kg H 2O/kg dry air. Using the humidity chart and appropriate equations,
humid volume.
o
C, dew point, humid heat and
18
101.3 17.60 KPa HS
9.3-4. Properties of Air to a Dryer.
determine the percentage humidity, saturation humidity at 57.2
17.60 KPa =
= 0.1305
x
29
kg H2O kg dry air
HP
3
H =
vH = 0.9796
HS
m kg dry air
( SI )
0.030 =
0.1305
HP
x 100
vH =
0.0204
0.0405 H
T
F
23 % vH = ( 0.0204 + 0.0405(0.030) ) ( 134.96 + 460 )
3
Step 3: Compute for humid heat in SI and English.
vH = 15.72
= 1.005 + 1.88H
CS
= 1.005 + 1.88 (0.030)
lbm
ft ( English ) dry air
9.3-5. Adiabatic Saturation Temperature Air at 82.2 oC and having a humidity H =0.0655 kg H 2O/kg dry air is contacted in an
C S = 1.0614
KJ kg dry air K
adiabatic saturator with water. It leaves at 80% saturation. ( SI )
= 0.24 + 0.45H
CS
a.
What are the final values of H and T oC?
b.
For 100% saturation, what would be the values of H and T?
GIVEN:
= 0.24 + 0.45 (0.030)
Btu lbm dry air
C S = 0.2535
T= 82.2 oC
( English )
F
H = 0.0655 kg H2O/kg da
ADIABATIC SATURATOR
Step 4: Solve for V H. REQUIRED: vH =
vH =
2.83 x 10
3
2.83 x 10
3
4.56
4.56
x 10
3
x 10
3
H
0.030
Final values of H and T oC
T K
( 57.2 + 273 )
Values of H and T at 100 % saturation SOLUTION: Step 1: From Psychrometric Chart.
80%
80 % RH Tw = 120 oF
H2 = 0.079 kg H2O/kg da o H1REQUIRED: = 0.0655 Dp = 40.6 C kg H2O/kg da Final values for H and T oC
T2 = 52.78 oC or 127 oF
T1 = 82.2 oC or 180 oF 4 SOLUTION:
Step 2: For 100 % saturation, take readings in the Psychrometric Chart.
Step 1: From Psychrometric Chart.
Tw = 113 oF
90 % RH
H2 = 0.0802 kg H2O/kg da H1 = 0.0655 kg H2O/kg da
T2 = 49 oC or 120 oF
Dp = 105 oF or 40.6 oC
H2 = kg H2 T2 = 116 oF or 46.7 oC
T1 = 82.2 oC or 180 oF
9.3-6 Adiabatic Saturation of Air
9.3-7 Humidity from Wet and Dry Bulb Temperature o
Air enters an adiabatic saturator having a temperature of 76.7 C and a dew-point o
temperature of 40.6 C. It leaves the saturator 90 % saturated. What are the final values of H and T
An air-water vapor mixture has a dry bulb temperature of o
temperature of 32.2 C. What is the humidity of the mixture?
o
C? GIVEN: Td = 65.6 oC & Tw = 32.2 oC
GIVEN: REQUIRED:
Td = 76.7 oC
T1 = 140 oF or 76.7 oC
90% RH
Humidity
65.6 oC and a wet bulb
SOLUTION: Step 1: From Psychrometric Chart.
Tw = 32.2 C or 90 F
Td = 140 F or 60 C H = 0.0175 kg H2O/kg da 9.3-10
Cooling and Dehumidifying Air Air is entering an adiabatic cooling chamber has a temperature of 32.2
o
C and a
percentage humidity of 65%. It is cooled by a cold water spray and saturated with water vapor in the chamber. After leaving, it is heated to 23.9 oC. The final air has a percentage humidity of 40%. (a) What is the initial humidity of the air?
Td = 65.6 oC
(b) What is the final humidity after heating?
or 149.9 F GIVEN:
9.3-8
Humidity and Wet Bulb Temperature
ADIABATIC COOLING CHAMBER
The humidity of an air-water vapor mixture is H = 0.030 kg H 2O/kg dry air. The wet bulb temperature of the mixture of 60 oC. What is the wet bulb temperature? GIVEN:
T1 = 32.2 C RH1 = 65 %
H = 0.030 kg H 2O/kg da & Td = 60 C REQUIRED: Wet bulb temperature, Tw
T3 = 23.9 C RH1 = 40 %
REQUIRED: Initial and final humidity
SOLUTION:
SOLUTION:
Step 1: From Psychrometric Chart.
Step 1: Obtain H 1 from Psychrometric Chart.
Tw = 98 F or 36.67 C
65 % RH H = 0.030 kg H2O/kg da
HEATER
H1 = 0.02 kg H2O/ kg da
The initial free moisture was X 1 = 0.55 and the critical moisture content Xc = 0.22 kg free moisture/ kg dry solid. Calculate the time to dry a batch of this material from X 1 = 0.45 to X 2 = 0.3 using the same drying conditions but a thickness of 50.8 mm, with drying from the top and bottom surface. Given:
T1 = 32.2 C
R = 2.05 kg H 2O/h-m2 Ls/A = 24.4 kg d.s/m 2 X1= 0.45 kg free moisture/kg d.s X1 = 0.55 kg free moisture/kg d.s Xc = 0.22 kg free moisture/kg d.s
Step 2: Obtain H 3 from Psychrometric Chart.
X2 = 0.3 kg free moisture/kg d.s Required: Time (t)
40 % RH Solution:
H3 = 0.0075 kg H2O/kg da
t
Ls A
XC X1 X2 XC ln X2
For Ls/A of 2 nd condition:
T3 = 23.9 C
Since
PROBLEMS
(GEANKOPLIS) 9.6-1) Time for Drying in Constant-Rate Period. A batch of wet solid was dried on a tray dryer using constant drying conditions and a thickness of material on the tray of 25.4 mm. Only the top surface was exposed. The drying rate during the constant rate period was R = 2.05 kg H 2O/ h – m2 ( 0.42 lb H2O/ h – ft2 ). The ratio Ls/A used was 24.4 kg dry solid/ m 2 exposed surface ( 5 lbm dry solid/ft 2 ).
Ls A
VA A
V(m) vol
V(m) bA
So,
From example 9.6-3 A = 0.457 x 0.457 m b = 25.4 mm
Ls Ls A 1 A 2 V (m ) 25.4 A V (m ) 25.4 A
v = 6.1 m/s T db = 65.6ºC
V (m )
1 (50.8) A 2 V (m )
H = 0.010 kg H 2O/kg d.a Required:
25.4 A
a.
Rc if v = 3.05 m/s
b.
Rc with T = 76.6ºC
Therefore, Solution:
kg d.s Ls Ls 24.4 m2 A 1 A 2
a) From ex. 9.6-3
1.037
t
Ls
AR C 24.4
t
X1
X2
kg d.s m2 0.45 0.3 kg H2O kg H2O kg d.s
2.05
h m
2
G
kg m3
v 3.05
m s kg 3600 1.037 3 s hr m
G = 11386.26 kg m 2/h
t = 1.7854 hrs
9.6-2) Prediction of Effect of Process Variables on Drying Rate. Using the conditions in example 9.6-3 for the constant rate drying period, do as follows: a. Predict the effect on Rc if the air velocity is only 3.05 m/s b.
Predict the effect if the gas temperature is raised tom 76.7ºC and it remains the same.
Given:
h 0.0204G
0.8
0.8
0.0204 11386.26
h
W
h 35.8703
0.8
0.0204G
0.0204 22056.1447
2
m K h
h
Rc
W
W
60.8771 m
2
K
T T W
35.8703
0.8
2433 1000
65.6 28.93600
Rc
h W
T T W
60.8771
2433 1000
2
Rc = 1.9479 kg/ m -h
76.7 28.93600
Rc = 4.21 kg/h-m 2 b) @ T = 76.7ºC H = 0.010 kg H 2O/kg d.a
9.6-3) Prediction in Constant-Rate Drying Region. A granular insoluble solid material wet with water is being dried in the constant rate period in a pan 0.61 m x 0.61 m and the depth of material
VH
2.83x10 4.56x10 HT 2.83x10 4.56x10 0.010273 3
-3
is 25.4 mm. The sides and bottom are insulated. Air flows parallel to the top drying surface at a
3
3
velocity of 3.05 m/s and has fry bulb temperature of 60 C and wet bulb temperature of 29.4 C .
76.7
The pan contains 11.34 kg of dry solid having a free moisture content of 0.35 kg H2O per kg dry solid and the material is to be dried in the constant-rate period to 0.22 kg H2O / kg dry solid. a. Predict the drying rate and time i n hours needed.
3
VH = 1.0056 m /kg da
b. Predict the time needed if the depth of mat’l is increased to 44.5 mm
1.0 0.010 1.0056
1.0044
kg 3
m
Given: A = 0.61m x 0.61m b = 25.4 mm v = 3.05 m/s
G v
6.1
Tdb = 60C
m s kg 3600 1.0044 3 s hr m
G = 22056.1447 kg/ h-m
2
Twb = 29.4C X1 = 0.35 kg H2O/ kg d.s X2 = 0.22 kg H 2O/ kg d.s Ls = 11.34 kg d.s
R c = 1.6437 kg / h- m 2
Required: t, Rc
t
Solution:
a.) @ Tdb = 60C, H = 0.0141 kg H 2O/ Kg d.a
t
VH
2.83x10 4.56x10 HT 2.83x10 4.56x10 0.014160
VH
0.9638 m3 / kg d.a
VH
3
3
Ls AR c
3
273
b.) by ratio & Proportion:
t1
t2
1.0141
t2
0.9638 1.0522 kg/m
0.35 0.22
t = 2.4103 hrs
m 1.0 kg d.a 0.0141 kg H 2 O m m 1.0141 VH
2
0.61x0.611.6437
3
1
11.34
X X
b1 b2
t1b2 b1
2.4103hr 44.5mm 25.4mm
3
t 2 = 4.2228 hrs
9.6-4) Drying a Filter Cake in the constant-rate region . A wet filter cake in a pan 1 ft x 1 ft square
G v(3600 ) G 3.051.0522 (3600 ) G 11,553.031kg/h - m 2
and 1 in. thick is dried on the top surface with air at wet bulb temperature of 80 F and a dry bulb of 120F flowing parallel to the surface at a velocity of 2.5 ft/s. The dry density of the cake is 120 lb m/ ft3 and the critical free moisture content is 0.09 lb H 2O / lb dry solid. How long will it take to dry the material from a free moisture content of 0.20 lb H 2O / lb dry material to the critical moisture content?
h
h
h
0.08
0.0204 G
0.0204(11,553.031)0.08 36.29 W/m
2
K
From steam table: @ T wb = 29.4C , hw = 2432.14 x 10 3
Given A = 1 ft2 t = 1in Twb = 80F Tdb = 120 F
RC
RC
h hw
V = 2.5 ft/s
T T 3600 w
= 120 lbm/ft 3
Xc = 0.90 lbm H 2O/ lbm d.s
36.29 3
2432 14x10
60 24.9 3600
Rc
X1 = 0.20 lbm H 2O/ lbm d.s
h w
T T 3600 w
Required: t
tc
Solution:
tc
VH 0.0252 0.0405 TR
VH
14.92
0.0827
under steady state conditions from a free moisture content of X 1 = 0.4 kg H 2O/ kg d.s to X 2 = 0.02 kg H2O/ kg d.s. The dry solid weight is 99.8 kg dry solid and the top surface area for drying is 4.645
lbmda
m2. The drying rate curve can be represented by Fig. 9-.5-1b.
1
Calculate the time for drying, but use a straight line through the origin for the d rying rate
VH
14.92
120 0.20 0.09
9.7-1) Graphical Integration for Drying in Falling-Rate Region. A wet solid is to be dried in a tray
3
1 0.013
12
tc = 13.30 hrs
0.0252 0.0405 0.013580 ft
Rc 1
@ Twb = 80F Tdb = 120F : H = 0.013 kg H2O/ kg da
VH
sX Xc
0.0679
in the falling rate period.
lbm ft
3
Given: X1 = 0.4
G G 2.53600 0.0679 1,080,000
X2 = 0.02
lbm h ft
Ms = 99.8 kg d.s
2
A = 4.645 m2 Required: T using straight line through the origin
h 0.0128 G 0.8
Solution:
h 0.0128 656.1 2.1681
Btu h ft
2
F
Using Fig 9-5-1b; Rc = 1.51 kg H 2O/h-m 2 Xc = 0.1950
@ Twb = 80F ; w = 1048.4 (from steam table)
2.2950
Rc
Rc
1048 .4 0.0827
120
ft
tC
ARc
X
1
3
XC
99.8
4.645 1.51
tC = 2.9169 hrs
lbm h
80 3600
mS
tC
0.4
0.1950
tF = 6.3185 hrs
tF
tF
mS
ARC
X C ln
X2
99.8
SOLUTION:
XC
4.645 1.51
0.1950 ln
0.1950 0.02
Getting the free moisture content: (at the given time and weight) Free moisture content wet sample weight- equilibrium moisture content-bone dry
sample weight
Free moisture c ontent: tT
tC
tF
tT
2.9169
.267
6.3185
0.2470 0.2266 0.1976
t T = 9.2354 hrs 0.1591 9.7-2) Drying Tests with a Foodstuff. In order to test the feasibility of drying a certain foodstuff,
0.1193
drying data were obtained in a tray dryer with air flow over the top exposed surface having an area
0.07596
of 0.186m2. The bone-dry sample weight was 3.765 kg dry solid. At equilibrium after a long period,
0.0518
the wet sample weight was 3.955kg H2O + solid. Hence, 3.955-3.765, or 0.190, kg of equilibrium moisture was present. The following sample weights versus time were obtained in the drying test.
0.01699 ______________________________________________________________________
601088x10-3
Time(h) Weight(kg)
0
0
Time(hr) Weight(kg)
Time(hr) Weight(kg)__
4.944
2.2
4.554
7.0
4.885
3.0
4.404
9.0
4.808
4.2
4.241
12.0
4.699
5.0
4.150
4.019 0.4 3.978 0.8
9.8-2) Drying when radiation, Conduction, and Convection are present. A material is granular and
3.955 1.4
wet with water and is being dried in a layer 25.4 mm deep in a batch-tray dryer pan. The pan has a metal bottom having a thermal conductivity of km=43.3 W/m. K and a thickness of 1.59mm. The thermal conductivity of the solid is Ks= 1.125 W/m.K. The air flows parallel to the top exposed surface and the bottom metal at a velocity of 3.05m/s and a temperature of 60oC and humidity
______________________________________________________________________
H=0.010kgH2O/kg dry solid. Direct radiation heat from steam pipes having a surface temperature of 104.4oC falls on to exposed top surface, whose emissivity is 0.94. Estimate the surface
Calculat e the free moisture content X kg H2O/kg d.s. for each data point and plot X versus time. (Hint: For 0h, 4.944-0.190-3.765=0.989) kg free moisture in 3.765kg dry solid, Hence, X=0.989/3.765)
temperature and the drying rate for the constant-rate period.
GIVEN:
Zs
0.0254 m
Km
43.3W / mK
Zm
0.00159 m
Ks
hc
G v
1.125W / mK
VH
V 3.05m / s
VH
H 0.010kgH 2 O / kgd.s.
0.94
(2.83 x10
-3
4.56 x103 )(T
273)
3
4.56 x10 (0.010 )](60 273)
3
0.9576 m / kgda
m2K
s 2423 .37KJ / kg hc / k yMB ~ Cs Cs [1.005 1.88H]103
1 0.010
[1.005 1.88(0.010)]103 1023.8 J/kgK
1.0547kg / m3 0.9576 3600 s G 3.05 (1.0547 ) h 11,580.606
UK
kg
1
h.m2
W 2
mK
from psychrometric chart
60o C;H 0.01kgH2O / kgd.a. Tw 28.8889 o C
1 hc
0.0204(11580 .606)0.8 36.3593
W
at Ts 33o Cfrom steam table :
Solution:
@ Td
3
hR 8.6076
Ts and Rc
hc
VH
104.4 0 C ~ 377.4K
TR
Required:
hc
1 H
[(2.83x10
T 60 o C
0.0204G0.8
Zm Km
Zs Ks 1
Tsshould be Tw Ts 33 C ~ 306K o
4
TR Ts 100 100 hR (5.676) TR Ts 4
4
377.4 306 100 100 hR 0.94(5.676) 377.4 306
4
1 36.3593
0.00159
43.3
0.0254
1.125
19.9530
W m2K
by psychrometric chart : @ Ts 33o C, Hs 0.0328
kgH2O kgd.a.
for trial 1
U h 1 k T Ts R TR Ts hc / kyHB he hc 8.6076 0.0328 0.012423 .371000 19.9530 (104.4 Ts ) 1 60 Ts 1023 .8 36.3593 36.3593
Hs - Hs
Ts 31.8355 o C 33 o C 31.8355 o C TRIAL 2 : Ts 32.3 305.3K
4
377.4 305.3 100 100 hR 0.94(5.676) 377.4 305.3 hR 8.5832 W / m2K
4
Trial 3 : Ts 32.6o C 305.6K
@ Ts 32.3 C o
s 2425 .036KJ / Kg
4
@ Ts 32.3o C;Hs 0.0312kgH2O / Kgd.a. (0.0312 0.01)2425 .036 8.5832 1.5488(60 Ts ) 104.4 Ts 1023 .8 36.3593 50.2156 1.5488 (60 Ts ) 0.2361(104.4 Ts ) TS 33.68O C 32.3o C 33.68o C
377.4 305.6 100 100 hR 0.94(5.676) 377.4 305.6
4
Rc = 3.1937 kg/h-m 2
9.9-1) Diffusion Drying of Wood. Repeat example 9.9-1 using the physical properties given but the following changes.
hR 8.3936 W / m2K X* = 0.04 kg H2O/kg dry wood x1 = 0.0127 m
@ Ts 32.6o C, s 2424 .322KJ / Kg (0.0318 0.01)2424 .322(1000 ) 1023 .8
1.5488(60 Ts )
8.5936 36.3593
a. Xt1 = 0.22
(104.4 Ts )
Xt = 0.13
b. Xt1 = 0.29 Xt = 0.09 x1 = 12.7 mm
Ts 32.9908
Required:
TRIAL 4 : @Ts 32.7o C 305.7O K
Solution:
Hs 0.032; s 2424 .084
a.)
a.
time using fig. 5.3-13
b.
time
X = Xt - X* 4
377.4 305.7 100 100 hR 0.94(5.676) 377.4 305.7 hR 8.5971
4
= 0.13 – 0.04 = 0.09 X1 = Xt1 - X* = 0.22 – 0.04
(0.032 0.01)2424 .084(1000 ) 1023 .28
1.5488(60 Ts )
8.5971 36.3593
= 0.18
(104.4 Ts )
Ea = X/ X1 = 0.09/0.18 = 0.50
Ts 32.75o C Rc
(hc Uk )(T Ts ) hR (TR Ts )
s
From Fig.5.3-13
(3600 )
(36.3593 19.9538)(6 0 - 32.75) [8.5991(1. 4.4 - 32.75)] 2424 .084x103
DLt/x12 = 0.20
(3600 ) 2
3.1937 a.
t
Calculat e the time needed to dry the wood from a total moisture of 0.22 to 0.13. Use Fig. 5.3-
13. b.
Calculate the time needed to dry planks of wood 12.7 mm thick from X t1 = 0.29 to Xt = 0.09.
Compare with the time needed for 25.4 mm thickness.
x1 (0.20) DL
0.0127 2 0.20 2.97x10
t= 10.86 hrs
Given: From Ex.9.9-1 DL = 2.97 x 10-6
b.) X1 = Xt1 –X* = 0.29-0.04 = 0.25
6
X = Xt –X* –X* = 0.09-0.04 = 0.05
Given:
x1 = 12.7/ 2 (1000) = 6.35 x 10 -3 m Equation 9.9-6
Required: total time 2
t
4 x1
2 D L
8 X1
ln
4 6.35x10
2
Solution:
2 X -3
2
2.97x10 6
ln
0.05
8 0.25
2
t = 7.70 hrs
by psychrometric chart : @T1
9.10-1) Drying a Bed of Solids by Through Circulation. Repeat example 9.10-1 for drying of a packed bed of wet cylinders by through circulation of the drying air. Use the same conditions except that the air velocity is 0.381m/s
V
0.381m / s
X 1 1.0kgH 2 O / kgdrysolid X 0.01 ds 1602 kg / m
3
bed thickness 50.8mm 0.0508m s
641kg / m 3
H1
0.04kgH 2 O / kgd.s.
for solid cylinder : Dc
X1 X t1 X*
0.00635 m
h 0.0254 m
Xc
1 0.01
0.99kgH2O / kgd.s.
X
X tc
X*
0.50 - 0.01 0.49kgH 2O / kgd.a. Xt
X*
0.10 - 0.01
0.09kgH 2O / kgd.a.
Tw
121.1o C; H
47.2; H w
0.04
0.074
if radiation and conduction are neglected, solid temp is at Tw.
(2.83 x10 3 4.56 x10 3 H)(T oK ) VH [2.83 x10 3 4.56 x10 3 (0.04)](121.1 273 ) VH 1.1872 m3 / kgd.a. VH
1 0.04 1.1872
0.8760 kgd.a. H2O / m3
1 0.381m / s(0.8760 kg / m3 ) (3600 s / h)(1/ 1.04 ) 1.0 0.04 G 1155 .3092kgd.a. / hm 2 G
0.040 and Hw 0.074 assume H2 0.05 sin ce H1
1155 .3092 1155 .3092(0.05) kgair H2O GT 1161 .0747 2 GT
h.m
for dry solid with 641kg present and
ds 1602kg / m3 volume
641kg 1602 kg / m3
1 0.4 0.6 a
4(1 )(h 0.5Dc ) Dch
0.04001m3 .
4(1 0.6)[0.0254
a
0.5(0.00635 )]
0.0254(0.00635 )
a 283.4646m2 D
(Dch
G
0.3209 kgd.a. / s.m2
0.5Dc )
0.5(0.0063 5)2 ]
D 0.0135m
for time of drying at constant rate period
t
t
for heat - transfer coefficien t :
air 2.15 x10 5 kg / m.s x
s
641(2.389 x10 6 )(0.0508 )(0.99 0.49)
61.0555 x 283.4646 x 0.0508 0.3209 x1.099 x10 3
hr
7.74x10 kg / m.h
ax
3600s
-2
s w x1( X1 Xc ) h / GC GC s (T1 Tw )(1 e )
0.3209 (1.099 x10 3 )(121.1 47.2)1 e
assuming Tair 93.3 o C
DGT
3600 s / h
1 2 2
[0.00635(0 .0254)
Nre
1155 .3092kgd.a. / h.m2
G
t
h 0.4519 h 0.9230 3600 s 1626 s
0.0135 (1161 .0747 ) 7.74 x10 2
202.5130
using eq' n 9.10 - 14;
@ falling - rate period :
X s w x1Xc ln c X t h X GC GCs ( T1 Tw ) 1 e a
1
s
h 0.214
GT
0.49
D
0.51
0.214(1161 .0747 )0.49 (0.0135 )0.51
T=1.2024 hrs 9.10-3) Through Circulation Drying in the Constant-Rate Period. Spherical wet catalyst pellets
from steam table : Tw 47.2o C, w 2389
having a diameter of 12.7mm are being dried in a through circulation dryer. The pellets are in a bed
KJ kg
2.389 x10 6 J / kg
Cs 1.005 1.88H Cs 1.005 1.88(0.05 ) Cs 1.099KJ / kgd.a.K or 1.099x10 3 J / kgd.a.K
63.5mm thick on a screen. The solids are being dried by air entering with a superficial velocity of 0.914m/s at 82.2 oC and having a humidity H=0.01 kgH 2O/kg dry air. The dry solid density is determined as 1522kg/m 3, and the void fraction in the bed is 0.35. The initial free moisture content is 0.90 kgH 2O/kg solid and the solids are to be dried to a free moisture content of 0.45, which is above the critical-moisture content. Calculate the time for drying in this constant-rate period.
Given: DP=12.7
Basis:1 kg d.a.
Z=63.5
10.01
V=0.914 T=821.2C
V H
H=0.01kgH2O/kgd.a. P=1522kg/m 3
VH=[2.83x10-3+4.56x10-3H]TK
E=0.35
=[2.86x10-3+4.56x10-3(0.1)](82.2+273)
X1=0.90
=1.02m3/kg d.a.
X2-0.45
So,
Required:
1.01
tCRP
1.02
Solution:
=0.99 kg/m 3 tCRP =P s (X1-X2)
NRE=DPV P/u
ah (T-Tw)
0.49 641(2.389 x10 )(0.0508 )0.49 ln 0.09 t 61.055 x 283.4646 x 0.0508 3 0.3209 x1.099 x10 3 0.3209(1.099 x10 )(121.1 47.2) 1 e 6
t 2701 .83735 x
h 3600 s
0.7505h
N Re
12.7 / 1000 0.914 0.99
0.019 x 103
=604.83 > 350
h
G 0.151 t
0.59
0.49 D P
total time t (0.4519 0.7505)h
1.2024hrs Gt=(0.914m/s)0.99 =0.905kg/m 2s setting X2=Xc ; z=X1m @T =82.2 C & H= 0.01 from P.C. Tw=32.2 C @82.2 C u=0.019x10-3
h
0.151
0.905 0.59 12.7 / 10000.49
=0.853W/m 2K
9.10-4) Material and Heat Balances on a Continuous Dryer. Repeat Example 9.10-2 making heat and material balances, but with the following changes. The solid enters at 15.6 oC and leaves at
1
60oC. The gas enters at 87.8 oC and leaves at 32.2 oC. Heat losses from the dryer are estimated as
S
293
Given:
S
1 1522
(1 - 0.35 )
989.3 kg/m
Q=2931W
3
TG1,H1
Gas
a=6(1-E)/DP solid
a
6(1 ) Dp
61 0.35 12.7 / 1000
G,TG2,H2
Ls,Ts1,X1
Ts2,X2
=307.09 Ls=453.6 @Tw=32.2 C ;
t CRP
w =2425.27kJ/kg(by interpolation)
G L S
1
L S A K Y MB
ln
HW HW
HC
Ts1=15.6 C
X1=0.04
Ts2=60C
X2=0.002
TG2=87.8 C
H2=0.01
TG1=32.2 C
CpA=4.187 kJ/kgK
CpB=1.465 kJ/kgK
H1 Required:
t CRP
a.)G
307.090.85382.2 32.23600 989.3 2425.27 0.9 0.45 1000
b.)H1 Solution:
t CR P
=7.75 h
Material Balance: GH2 + LsX1 = GH1 + LsX2 G(0.01) +453.6(0.04) =GH 1 + 453.6(0.002) G(0.01-H1)=-17.24 Heat Balance : @TG2=87.8C & To=0C H’G2= CS(TG2-To) + H2
(1)
=[1.005 + 1.88(0.01)](87.8-0) + 0.01(2501)
9.10-5 ) Drying in a Continuous Tunnel Dryer . A rate of feed of 700lbm dry solid/h containing a free
=114.9 kJ/kgd.a.
moisture content of X 1=0.4133lb H2O/lb dry solid is to be dried to X=0.0374 lbH 2O/lb d.s. in a continuous counterflow tunnel dryer. A flow of 13280lbm/dry air/h enters at 203 oF with an H2=0.0562lbH2O/lb dry air. The stock enters at the wet bulb temperature of 119 oF and remains
For the exit gas’
essentially constant in temperature in the dryer. The saturation humidity at 119 oF from the humidity
H’G1=CS(HG1-To) + H1 =[1.005 + 1.88H 1](32.2-0) +H1(2501)
chart is Hw=0.0786 lb H2O/lbd.a. The surface area available for drying is (A/L s)=0.30ft2/lbmd.s..
=32.36 +2561.54H 1
A small batch experiment was performed using constant drying conditions, air velocity, and temperature of the solid approximately the same as in the continuous dryer. The equilibrium
For the entering solid,
critical moisture content was found to be X c=0.0959lbH2O/lb d.s., and the experimental value of kyMB was found as 30.15lbm air/hr.ft2. In the falling-rate was directly proportional to X.
H’S1=Cps(Ts1-To) + X1Cpa(Ts1-To) =1.456(15.6-) + 0.04(4.187)(15.6-0)
For the continuous dryer, calculate the time in the dryer in the constant-rate zone and in
=25.33kJ/kgd.a.
the falling-rate zone.
H’S2=Cps(Ts2-To) + X2Cpa(TS2-To) =1.456(60-0) + 0.002(4.187)(60-0)
Given:
=87.86 kJ/kgd.s. Ls=700lbmds/h Heat Balance on dryer,
T w=119F
X1=0.4133
GH’G2+LsH’S1=GH’G1+LsH’S2+Q G(114.9)+453.6(25.33)=G(32.36+2561.54H 1)+39853.3+10551.6 G(82.54-2561.54H 1)=38915.21
(2)
Hw=0.0786lbH2O/lbda
X2=0.0374
A/Ls=0.30ft2/lbds
G=13280lbda/h
Xc=0.0959
T G2 =203F
K y MB =30.15lbmair/hft2
H2=0.0562lbH2O/lbda From Eq 1, G=-17.24/(0.01-H1)
Required a.)tCRP
From Eq 2,
b.)tFRP
-17.24/(0.01-H1) (82.54-2561.54H1)=38915.21 -1422.99+444160.95H 1=389.15-38915.21H 1
Solution:
83076.16H 1=1812.14 So,
Material Balance: H1=0.0218kgH2O/kgd.a.
Ls(Xc-X2)=G(Hc-H2) 700(0.0959-0.0374)=13280(Hc-0.0562) Hc=0.0593lbH2O/lbda
Subs. To Eq1, G=-17.24/(0.01-0.0218kgH2O/kgd.a.)
t CRP G=1461.02kgd.a./h
G LS
1
LS A K YMB
ln
HW HW
HC
H1
Required: GH2 + LsX1 = GH1 + LsX2
a.
fresh air
b.
% air leaving
13280(0.0562)+700(0.4133)=13280H 1+700(0.0374)
c.
heat added (heater)
H1=0.076 lbH2O/lbda
d.
heat loss (dryer)
So,
Solution:
t CRP
13280
700
1
1
0.3 30.15
ln
0.0786
0.0786
H2O Bal. On the heater
0.0593
0.076 G1H1 + G6H2 = ( G1 +G6 ) H4 G1 (0.007) + G6 (0.020) = G 1 (0.010) +0.01G 6
t CR P =4.204h
0.01
G6 = 0.003 G 1 ………….(1)
For Falling rate zone,
t FRP
G LS
XC
1
LS A K YMB HW
G / L
H2
ln S
X2
H
XC HW
H2
X2
HC
W
H2O Bal. On the dryer (G1 +G6)H4 + LsX1 = (G1 + G6)H2 +LsX2
t
13280 FRP
700
1 0.0959 0.3 30.15
1
0.086 0.0562 13281 / 700 0.0374
ln
0.0562 0.0374 0.0786 0.0593 0.059 0.0786
for Ls
t FR P =0.47h
Ls = L ( 1 – X ) = 907.2 (1 – 0.27) = 662.256 kg d.s
9.10-6) Air Recirculation in a Continuous Dryer . The wet feed material to a continuous dryer
dry basis:
contains 50 wt% water on a wet basis and is dried to 27wt% by countercurrent air flow. The dried
since G3 = G4 = G1+G6 = 41728.75 d.a/h
product leaves at the rate of 907.2 kg/h. Fresh air to the system is at 25.6ºC and has a humidity of H = 0.007 kg H2O/kg d.a. The moist leaves the dryer at 37.8ºC and H = 0.020 and part of it is
X1
recirculated and mixed air enters the dryer at 65.6ºC and H = 0.01. Calculate the fresh air flow, the percent air leaving the dryer that is recycled, the heat added in the heater, and the heat loss from
X2
the dryer. Given:
at junction A
0.5 1 0.5 0.27
1.0
1 027
0.37
G1H1 + G6H6 = G4H4 From eqn 1:
41728.75 G 0.02 41728.750.01 G 0.007 834.57 0.02G 417.29 G 0.013 417.289 G1 0.007 1
1
1
G1 G6
1
G6
G3
41728 .75 32099 .15
9629.6 kg/h
G 1 = 32,099.15 kg fresh d.a /h
Heat Balance on Heater
% recycled G6H'G6 G1H'G1 G4H'G4 Q
G6 G4
9629.6
41728.75
% recycled = 23.06% recycled air
H'G Cs( TG TD ) H o H'G 4 Cs 4 (TG4 TD ) H4 o
Q = 440.605 KW
Cs 1.005 1.88(0.01) 1.0238 H'G 4 1.0238 ( 65.6 0) 0.012501
Amount of Heat loss from the dryer Heat bal.
92.17 kJ/kg d.a H'G1 Cs1( TG1 T0 ) H1 o
Q = 32.06 KW
1.0182(25. 6 - 0) 0.007(2501 ) 43.57 KJ/kg d.a H'G6 Cs6 ( TG6 T0 ) H6 o
1.0426(37. 8 - 0) 0.02(2501) 89.43 KJ/kg d.a subs to equation - Q G6H'G6 G1H'G1 G4H'G 4
9629.60(89 .43) 32099.15(4 3.57) - 41722.38(9 2.17)
MC CABE
0.2306
24.1) Fluorspar (CaF2) is to be dried from 6 to 0.4 percent moisture (dry basis) in a countercurrent
a.)
Assuming Eq.(24.8) applies, what would be the diameter and length of the dryer if
G4H'G4 LsHS1 G2H'G2 LsHS2 Q G2 G4 41722 .38 kg d.a/h H'G2 Cs2 (TG2 T0 ) H2o Whe re CS2 1.0426 KJ/kg d.a H'G2 1.0426(37.8 0) 0.022501
89.43 KJ/kg d.a H'S2 1.63781 H'S1 subs to the eqn : 41722 .38(92.17) 662.26(H'S2 1.63781) 41722 .38(89.43) 662.26H'S2 Q
Q 115403 .98
KJ hr hr 3600s
adiabatic rotary dryer at a rate of 18,000 lb/hr
Nt= 2.2? Is this reasonable design? o
of bone-dry solids. The heating air enters at 1000 F with a humidity of 0.03 and wet bulb
b.)
temperature of 150 oF. The solids have a specific heat of 0.48 Btu/lb oF; they enter the dryer at 70 oF and leave at 200 oF. The maximum allowable mass velocity of the air is 2,000 lb/ft 2.h.
Nt
ln
Thb Tha
Twb Twb
Twb
inlet Tw
Twa
outlet Tw
Thb
outlet Tb
Tha
outlet Tb
for Tha
Given:
Repeat part (a) with Nt=1.8.
Solution:
also, xa
e2.2
850
Thc
0.06
150
CsB
2.2
4 xe10 (3Tha 150) 850 Tha 244.18o F Ms 18000 xb
0.24 0.45
0.2535
0.24 0.45(0.03)
quantities needed so, the other rate of mass transfer : are : Mv Ms@150 ( x a xobF) 1008 .1Btu ( App7) 3 lb 18000 (0.06 4 x10 ) specific heats in Btu/lb oF are, Mv 1008 lb / h Cps 0.48 Cpv 0.45 (App14) CpL 1 The heat duty is found from subs the eq(24.1) qT
M' s
Cps(Tsb
Tsa )
x acpL ( Tv
Tsa ) ( x a
xb ) xbcpL (Tsb
Tv ) ( x a
x b )cpv (Tv a Tv )
where : Tsa Tv
feedT
vapor ' nT
Tsb
final solids T
Tv a
final vapor T
heat
of vapor L = 35.842 ft
Cps , CpL , Cpv sp. heat of solid. liquid and vapor
0.48(200 - 70) 0.06(1)(15 0 - 70) (0.06 - 4x10)1008. 1 4x10 - 3 1 (200 150) (0.06 4x10 3 )(0.45)(244.18 150) qt Ms
123.7111
Btu lb
so, qT
123.7111(18000 )
2.2268 x106
Btu h
mg(1 Hb )
qt Csb (Thb Tha ) 2.2268 x106 (1000 244.18)0.2535
24.2)A porous solid is dried in a batch dryer under constant drying conditions. Seven hours are
11622.10699
required to reduce the moisture content from 35 to 10 percent. The critical moisture content was found to be 20 percent and the equilibrium moistur e content is 4 percent. All moisture contents are
since, Hb 0.03;mg 11622.10699 / 1 0.03 mg 11283 .59902
on the dry basis. Assuming that the rate of drying during the falling rate period is proportional to
lb
the free moisture content, how long should it take to dry a sample of the same solid from 35 to 5
h
percent under the same conditions? Given
the outlet humidity Ha Hb mv / mg
0.03
tT = 7 hrs
1008 11283.59902
0.1193
lb lb
for D A
G so,D 2.72ft
2nd condition
x1= 35
x1 = 35
x2 = 10
x2 = 5
xc = 20
1/ 2
4A mg (1 Hb )
1st condition
x* = 4
11622 .10699 2000
5.8105 ft 2
Required: tT @ second condition Solution:
The dryer length is given by eq' n (24.23) : L
0.125 DG0.67 T
T TL
Thb Twb (Tha Twa ) ln[( Thb Twb ) /( Tha Twa )]
since, Twb Twa
1000 150 (244.18 180)
Total moisture – equilibrium moisture = Free moisture 1st condition X1: 35 – 4 = 31 X2: 10 – 4 =6 Xc: 20 – 4 = 16
ln[(1000 150) /( 244.18 150)] 785.82 ln 9.0253
2nd condition
357.1863 o F
x2 = 5 – 4 =1
so, L
Basis: 100 total moisture
qT
2.226x106 0.125(2.72)(2000)0.67 (357.1863 )
at: FOUST
31 t
c
16
dx Rc
31 16
Rc
15 Rc
18.10
A rotary countercurrent dryer is fed with wet sand containing 50 percent moisture and is
discharging sand containing 3 percent moisture. The entering air is at 120Cand has an absolute humidity of 0.007g H2O/g air. The wet sand enters at 25C and leaves at 40C.The air leaves at
t
f
X
X
c
ln
Rc
X
c
2
16
ln
Rc
16 Rc
45C.The wet sand input is 10 k g/s. Radiation amounts to 5 cal/g dry air. Calculate the pounds of dry air passing through the dryer and the humidity of the air leaving the dryer. Latent heat of H2O at 40C = 575
t t T c f Rc = 4.3848 kg H2O/ h – m2 t
At 2nd condition: X1 =31 ; X2 = 1
Cp for dry sand
= 0.21 cal/gC
Cp for dry air
= 0.238 cal/gC
Cp for H2O vapor
= o.48cal/gC
Given:
31 t
t
c
f
1
dx Rc
31 1
6.84
hrs
X c
X c
Rc
ln
X 2
Rc
G1=?,TG2=120C,H2=0.007
4.3848
16 4.3848
10.12 hrs
TG1=45C,H1=?
30
ln
16 1
Ls = 10 kg/s
Ts2 = 49C
Ts1 = 25C
X2 = 0.03
X1 = 0.5 Required: G, H1
t
T
6.84 10.12 tT = 16.36 hrs
Solution: GH2 + LsX1 = H1 + LsX2 G(0.007) + 10 x3(0.5) = GH 1+10 x 3(0.003)
0.007G + 4.7 x103 = GH 1
(1)
EVAPORATION H’G2 = Cs(TG2- To) + H2 = [1.005 + 1.88(0.007)](120-0) + 0.007(2501)
1.
H’G2 = 139.6862 kJ/kg d.a.
It is a unit operation which means to concentrate a solution considering a non-volatile solute and volatile solvent.
For the exit gas, H’G1 = Cs(TG1- To) + H1 = [(1.005 + 1.88H1)(45-0) + H 1 (2501)
2.
H’G1 = 45.225 + 2585.6 H 1
a.
condensation
b.
evaporation
c.
precipitation
d.
extraction
It is conducted by vaporization of a portion of the solvent to produce a concentrated solution
For the entering solid, H’s1 = Cps(Ts1-To) + X1 Cpa(Ts1-To) = 0.221(25-0) + 0.5(0.48)(25-0)
3.
a.
condensation
b.
evaporation
c.
precipitation
d.
extraction
Which of the following is not a processing factor in evaporation?
H’s1 = 11.25 cal/g
a.
H’s2 = Cps(Ts2-To) + X2Cpa(Ts2-To)
b.
temperature
= 0.21(400-0) + 0.03(0.48)(40-0)
c.
frothing
d.
specific heat
H’s2 = 8.976 cal/g 4.
solubility
Which of the following is not a proper assumption in evaporator calculations?
Substituting into Eq,
a.
The standard temperature difference is based on the temp of the boiling liquid at
G H’G2 +Ls H’s1 = GH’G1+Ls H’s2 + Q
b.
The standard temperature difference is based on the temp of the liquid at the liquid
c.
The log mean temperature difference is based on the temp of the boiling liquid at
the liquid vapor interface G(139.6862) +10(11.25) = G(45.225 + 2585.6 H 1) + 10(8.976) +5
94.4612G –17.74 = 2585.6GH 1
interface the liquid vapor interface
(2) d.
The log mean temperature difference is based on the temp of the liquid at the liquid interface
Using (1) &(2),
G=159140.7017g/h~159.1407 kg/h H 1=0.03653 g H2O/g air
5.
The heat required to vaporize 1lb of the solvent is taken as the ____ at the exposed surface temperature of the solution a.
latent heat of vaporization
b.
specific heat
c.
heat transfer
d.
specific heat of condensation
6.
7.
8.
9.
What is the usual solvent in an evaporation process?
b.
a.
water
c.
capacity percent recovery
b.
alcohol
d.
efficiency
c.
benzene
d.
ether
13. This can be determined by the measurement of the total heat transferred to the evaporating mixture
For feed of organic salts in water, the heat capacity may be assumed ___ to that of
a.
economy
water alone
b.
capacity
a.
greater
c.
percent recovery
b.
lesser
d.
efficiency
c.
equal
d.
none of the above
14. It is the most important single factor in evaporator design, since the heating surface represents the largest part of the evaporator cost.
The effect of boiling point rise due to _____ is commonly neglected in evaporation
a.
temperature
calculations
b.
temperature difference
a.
heat transfer
c.
heat transfer
b.
temperature change
d.
pressure
c.
hydrostatic head
d.
change in velocity
15. The greatest increase in steam economy is achieved by reusing the vaporized solvent, which is done in the___
It is the difference in temperature of the liquid at the heat source and the liquid at the top
a.
single effect evaporator
evaporating liquid
b.
double effect evaporator
a.
temperature change
c.
multiple effect evaporator
b.
heat transfer
d.
multiple effect backward evaporator
c.
boiling point rise
d.
log mean temperature
10. When two or more evaporators are connected, it is called
16. In the ___, the desirability of producing crystals of a definite uniform size usually limits the choice to evaporators having a positive means of circulation. a.
multiple effect evaporator
a.
single effect evaporator
b.
single effect evaporator
b.
double effect evaporator
c.
forced circulation evaporator
d.
crystallizing evaporator
c.
multiple effect evaporator
d.
multiple effect forward evaporator
11. ___ is equivalent to the number of effects but it will never reach it unless the temp of the
17. ___ which is the growth on body and heating surface walls of a material having a solubility that increases with increase in temperature
feed is higher than the boiling point of the liquid in the evaporator body
a.
salting
a.
economy
b.
scaling
b.
capacity
c.
crystals
c.
percent recovery
d.
fouling
d.
efficiency
12. ___ of the evaporator is the total amount of evaporation it is capable of producing per
18. It is the deposition and growth on body walls, and especially on heating surfaces, of a material undergoing an irreversible chemical reaction in the evaporator
unit time
a.
salting
a.
b.
scaling
economy
c.
crystals
d.
fouling
19. It is the formation of deposits other than salt or scale and maybe due to corrosion
25. this usually results from the presence in the evaporating liquid of colloids or of surfacetension depressants a.
splashing loss
a.
salting
b.
entrainment losses
b.
scaling
c.
heat losses
c.
crystals
d.
foaming losses
d.
foulingThis evaporator is suitable for the widest variety of evaporator applications
a.
multiple effect evaporator
level and the top of the vapor head
b.
Short-tube vertical evaporators
a.
splashing losses
c.
forced circulation evaporator
b.
entrainment losses
d.
crystallizing evaporator
c.
heat losses
d.
foaming losses
20. This is one of the earliest types still in widespread commercial uses, and its principal use is in the c ane industry
26. ___ are usually insignificant if a reasonable height has been provided between the liquid
27. These are frequently encountered in an evaporator if feed is above the boiling point
a.
multiple effect evaporator
a.
splashing losses
b.
Short-tube vertical evaporators
b.
entrainment losses
c.
forced circulation evaporator
c.
heat losses
d.
crystallizing evaporator
d.
foaming losses
21. When the ratio of feed to evaporation is low, it is desirable to provide for
28. These are frequently used to reduce product losses
a.
recirculation of product
a.
knitted wire mesh
b.
steam recycling
b.
mechanical thermo compressors
c.
used of multiple effect evaporator
c.
steam jacket
d.
high powered evaporator
d.
entrainment separators
22. This is normally the cheapest per unit capacity type of evaporator
29. This employ reciprocating, rotary positive-displacement , centrifugal or axial flow
a.
long-tube evaporator
direction
b.
single effect evaporator
a.
knitted wire mesh
c.
short-tube evaporator
b.
mechanical thermo compressors
d.
falling film long tube evaporator
c.
steam jacket
d.
entrainment separators
23. The principal problem in the falling film long tube evaporator is the a.
feed distribution
b.
temperature control
development of an evaporator employing a ___
c.
high pressure
a.
steam economy
d.
heat loss
b.
multiple effect
24. The falling film long tube evaporator is widely used for concentrating ___
30. the requirement of low temperature for fruit-juice concentration has led to the
c.
temperature distribution
d.
secondary fluid
a.
heat sensitive materials
b.
highly soluble materials
c.
high boiling point rise
but once built, the evaporator establishes its own equilibrium.
d.
corrosive materials
a.
31. The approximate ____ in a multiple effect evaporator is under the control of the designer, recirculation of feed
b.
multiple effect
a.
crystallization
c.
temperature distribution
b.
seeding
d.
secondary fluid
c.
solubility
d.
frothing
32. The ___ of a multiple effect evaporator will increase in proportion to the number of effects a.
steam economy
b.
multiple effect
c.
temperature distribution
d.
secondary fluid
33. It may influence the evaporator selection, since the advantages of evaporators having high heat transfer coefficients are more apparent when expensive materials of constructions are i ndicated a.
heat sensitivity
b.
corrosion
c.
fouling factor
d.
scaling
34. ____ usually range from a minimum of about 1.2 m/s a.
horizontal tube velocity
b.
tube velocity
c.
tube rate
d.
none of the above
35. s solutions are heated and concentration of the solute increases, ___ limit of the material may be exceeded and crystals may form.
36. The simplest form of evaporator consists of an open pan in which the liquid is boiled
d.
Mollier chart
a.
open kettle
b.
vertical type natural circulation
evaporation in each effect are ___
c.
falling film type evaporator
a.
d.
agitated film evaporator
b.
lesser
37. A variation of the long-tube evaporator
c.
constant
d.
equal
a.
forced circulation type evaporator
b.
vertical type natural circulation
43. When a multiple effect evaporat or is at steady state operation, the flow rates and rate of greater
44. In commercial practice, the areas in all effects for a multiple effect evaporator are
c.
falling film type evaporator
a.
d.
agitated film evaporator
b.
lesser
c.
constant
d.
equal
38. This is done in a modified falling film evaporator
greater
a.
forced circulation type evaporation
b.
vertical type natural circulation
c.
falling film type evaporation
must be satisfied, and if the final solution is concentrated, the feed rate will be ______.
d.
agitated film evaporation
a.
increased
b.
decreased
39. In this operation, the fresh feed is added to the first effect and flows to the next in the
45. The over-all material balance made over the whole system of a multiple effect evaporator
same direction as the vapor flow
c.
held constant
a.
Feed-forward
d.
neglected
b.
Feed-backward
c.
Parallel-feed
a.
heat transfer
d.
None of the above
b.
latent heat
40. Involves the adding of fresh feed and the withdrawal of concentrated product from each effect
46. ___ of the vapor is recovered and reuse in a multiple effect evaporator
c.
specific heat
d.
sensible heat
a.
Feed-forward
48. The increase in steam economy obtained by using multiple-effect evaporator is at the
b.
Feed-backward
expense of ___.
c.
Parallel-feed
a.
reduced capacity
d.
None of the above
b.
product quality
41. The fresh feed enters at the coldest effect
c.
evaporator’s efficiency
d.
percent yield
a.
Feed-forward
b.
Feed-backward
c.
Parallel-feed
a.
Thermal recompression
d.
None of the above
b.
Liquid compression
42. For strong solutions of dissolved solutes the boiling point rise due to the solutes in the solution cannot be predicted but the empirical law known as ___ is used
49. Evaporation is also sometimes called as___
c.
Water distillation
d.
Vapor compression
a.
Enthalpy concentration chart
50. In a ____ system, the vapor is compressed by acting on it with high-pressure steam in a
b.
Duhring chart
jet ejector.
c.
Equilibrium chart
a.
Thermal recompression
b.
Liquid compression
c.
Water distillation
d.
Vapor compression
47. 48. 49. 50.
b a c a
Unit Operations Economics Answer key Evaporation (objective questions) 1. b 2. b 3. d 4. a 5. a 6. a 7. c 8. c 9. c 10. c 11. a 12. b 13. b 14. c 15. c 16. d 17. a 18. b 19. d 20. c 21. b 22. a 23. a 24. a 25. a 26. d 27. a 28. b 29. d 30. b 31. d 32. c 33. a 34. b 35. b 36. c 37. a 38. c 39. d 40. a 41. c 42. b 43. b 44. c 45. d 46. a
1. ____ is the money returned to the owners of capital for use of their capital. a. Interest b. Cash c. Loan d. credit 2. Any profit obtained through the uses of capital a. Interest b. Cash c. Loan d. credit 3. In economic terminology, the amount of capital on which interest is paid is designated as the principal, and the rate of interest is defined as the amount of interest earned by a unit of principal in a unit of time a. simple interest b. compound interest c. nominal interest d. interest on loan 4. When an interest period of less than 1 year is involved a. simple interest b. ordinary interest c. nominal interest d .compound interest 5. This interest stipulates that interest is due regularly at the end of each interest period. a. simple interest b. ordinary interest c. nominal interest d.compoundinterest 6. In ______interest rate, other time units are employed butthe interest rate is still expressed on an annual basis. a. simple interest b. ordinary interest c. nominal interest d.compoundinterest 7. The concept of ______is that the cost or income due to interest flows regularly, and this is just as reasonable an assumption for most cases as the concept of interest accumulating only at discrete intervals. a. continuous Interest b. Cash flow c. Loan d. credit 8. The present worth of a future amount is the present ___which must be deposited at a given interest yield tom yield the desired amount at some future date. a. principal b. Cash flow c. capital d. credit
9. An ____is a series of equal payments occurring at equal time intervals. a. principal b. interest c. capital d. annuity 10. The amount of an ___is the sum of all the payments plus interest if allowed to accumulate at a definite rate of interest from the time of initial payment to the end of the term. a. principal b. interest c. capital d. annuity 11. Common type of ____which involves payments, which occur at the end of each interest period a. Perpetuity b. Present worth of annuity c. capital d. annuity 12. Defined as the principal which would have to be invested at the present time at compound interest rate to yield a total a. Perpetuity b. Present worth of annuity c. capital d. annuity 13. An annuity in which the first payment is due after a definite number of years. a. Perpetuity b. Present worth of annuity c. deferred interest d. deferred annuity 14. It is an annuity in which the periodic payments continue indifferently. a. Perpetuity b. Present worth of annuity c. deferred interest d.deferredannuity 15. Defined as the original cost of the equipment plus the present value of the renewable perpetuity. a. Perpetuity b. Present worth of annuity c. deferred interest d.capitalizedcost 16. An ____design could be based on conditions giving the least cost per unit of time or the maximum profit per unit of production. a. optimum economic b. plant c. simple economic d.plant economic 17. A determination of an “____condition” serves as a base point for a cost or design analyses, and it can often be quantitized in specific mathematical form. a. optimum b. plant design c. simple d.plant economic 18. The ___has one distinct advantage over the analytical method. a. optimum b. graphical c. simple d.economical
19. The point where total product cost equals total income represents the_____, and the optimum production schedule must be at a production rate higher than that. a. optimum b. break-even point c. annuity d. balanced 20. The total product cost per unit of time may be divided into the two classifications of operating costs and ____costs. a. optimum operating costs b. Organization costs c. annuity costs d. superproduction costs 21. This depends on the rate of production and include expenses for direct labor, raw materials, power, heat, supplies, and similar items which are function of the amount of material produced. a. operating costs b. Organization costs c. annuity costs d. superproduction costs 22. ______are due to expenses for directive personnel, physical equipment, and other services or facilities which must be maintained irrespective of the amount of material produced. These are independent of the rate of production. a. operating costs b. Organization costs c. annuity costs d. superproduction costs 23. Extra expenses due to increasing the rate of production are known as ____. a. operating costs b. Organization costs c. annuity costs d. superproduction costs 24. In a true ___operation, no product is obtained until the unit is shut down for discharging. a. batch b. continuous c. semi-batch d. semi-continuous 25. In _____cyclic operations, product is delivered continuously while the unit is in operation, but the rate of delivery decreases with time. a. batch b. continuous c. semi-batch d. semi-continuous 26. _____cyclic operations are often encountered in the chemical industry, and the design engineer should understand the methods for determining optimum cycle times in this type of operation a. batch b. continuous c. semi-batch d. semi-continuous 27. It is a mathematical technique, for determining optimum conditions for allocation of resources and operating capabilities to attain a definite objective. a. economics b. programming c. linear programming d. dynamic programming
28. The concept of this is based on converting an overall decision situation involving many variables into a series of simpler individual problems with each of these involving a small number of total variables. a. economics b. programming c. linear programming d. dynamic programming 29. When quality constraints or restrictions on certain variables exist in an optimization situation, a powerful analytical technique is the use of : a. Lagrange Multipliers b. Geometric Programming c. steepest Ascent or Descent d. dynamic programming 30. For the optimization situation in which two or more independent variables are involved, response surface can often be prepared to show the relationship among the variables one of the early methods proposed for establishing optimum conditions from response surfaces is known as the method: a. Lagrange Multipliers b. Geometric Programming c. linear programming d. steepest Ascent or Descent 31. A technique for optimization, based on the inequality relating the arithmetic mean to geometric mean, with this method the basic idea is to start by finding the optimum way to distribute the total cost a. Lagrange Multipliers b. Geometric Programming c. linear programming d. steepest Ascent or Descent 32. ____(CPM) and program evaluation and review technique) have received particular attention and have shown the desirability of applying mathematical and graphical analysis a. critical path method b. critical planning method c. careful planning and management d. none of the above 33. The word ____is used as a general term for the measure of the amount of profit that can be obtained from a given situation. a. profitability b. income c. net income d. sales 34. It is the c ommon denominator for all business activities before capital is invested to a project or enterprise a. profitability b. income c. net income d. sales 35. ____on investments is ordinarily expressed on an annual percentage basis. The yearly profit divided the by the total initial investment necessary represents the fractional return: a. rate of return b. interest income c. profitability d. annuity 36. ___is defined as the difference between income and expense. a. profit b. gross income
c. operating expenses d. sales 37. This is a function of the quantity of goods and services produced and the selling price. a. sales b. gross income c. operating expenses d. profit 38. The method of approach for a profitabilit y evaluation is by ____, which takes into account the time value of the money and is based on the amount of the investment that is unreturned at the end of each year during the estimated life of the project. a. marketing analysis b. gross income c. discounted cash flow d. law of demand 39. In the net present worth study, the procedure has involved the determination of an index or interest rate which discounts the annual ____to a zero present value when properly compared to the initial investment. a. marketing analysis b. gross income c. cash flow d. selling price 40. The ____concept is useful for comparing alternatives which exist as possible investment choices within a single overall project. a. marketing analysis b. sales inventory c. discounted cash flow d. capitalized cost profitability 41. ____ is defined as the minimum length of time theoretically necessary to recover the original capital investment in the form of cash flow to the project based on total income minus all cost except depreciation. a. deadline b. due date c. payout period d. maturity date 42. The term ___ refers to a special type of alternative in which facilities are currently in existence and it may be desirable to replace these facilities with different ones. a. reserves b. replacement c. substitute d. none of the above 43. In the formula, P = F(1+I ) -N , where I = interest rate per period and N the number of interest periods is known as a. sinking fund factor b. single payment present worth factor c. uniform series d. capital recovery factor 44. Which of the following relationships between compound interest factors is not correct? a. Single payment compound amount factor and single payment present worth factor are reciprocals b. Capital recovery factor and uniform series present worth are reciprocals c. Capital recovery factor equals the sinking fund factor plus the interest rate d. capital recovery factor and sinking fund factor are reciprocals 45. An “annuity” is defined as a. Earned interest due at the end of each interest period b. Cost of producing a product or rendering a service
c. Amount of interest earned by a unit principal in a unit of time d. A series of equal payments made at equal interval of time 46. Which one of the following method is not a method of depreciating plant equipment for accounting and engineering economic analysis purposes? a. double entry method b. fixed percentage method c. sum-of-years digit method d. sinking fund method 47. ___ is a science which deals with the attainment of the maximum fulfillment of society’s unlimited demands for goods and services. a. Political science b. Economics c. Engineering Economy d. Social Science 48. It is the branch of economics which deals with the application of economics laws and theories involving engineering and technical projects or equipments. a. Political science b. Economics c. Engineering Economy d. Social Science 49. This refers to the products or services that are directly used by people to satisfy their wants. a. Producer goods and services b. Consumer goods and services c. Luxury Products d. Supply 50. Those that are used to produce the goods and services for the consumer are called a. Producer goods and services b. Consumer goods and services c. Machineries and Technology d. Supply 51. This refers to the satisfaction or pleasure derived from the consumer goods and services. a. Product Satisfaction b. Consumer’s contentment c. Luxury Products d. Utility 52. These are products that have an income-elasticity of demand greater than one. a. Basic Commodities b. Consumer goods c. Luxury Products d.Supply 53. If the income elasticity of demand is greater than one, this means that a. as income decreases, more income will be spent on the basic commodities b. as income increases, more expenses will be made on the basic commodity products c. as income increases, more income will be spent on luxury items d. as income decreases, less will be spent on the basic commodity products 54. It is the amount of goods and products that are available for sale by the suppliers a. Producer goods and services b. Consumer goods and services c. Machineries and Technology d. Supply 55. The want or desire or need for a product using money to purchase it. a. Supply b. Consumer goods and services c. Luxury Products d. Demand
56. The law of supply and demand, states that a. “When free competition exists, the price of the product will be that value where supply is equal to the demand” b. “When perfect competition exists, the price of the product will be that value where supply is equal to the demand” c. “When free competition exists, the price of the product will be that value where supply is greater to the demand” d. “When perfect competition exists, the price of the product will be that value where supply is greater to the demand” 57. This is a form of market structure where the number of suppliers is used to determine the type of market. a. Market Structures b. Competition c. Free competition market d. Perfect Competition 58. This is a market situation wherein a given product is supplied by a very large number of vendors and there is no restriction of any additional vendor from entering the market. a. Market Structures b. Competition c. Free competition market d. Perfect Competition 59. A place where the vendors or the sellers and vendees or the buyers come together is called a. Market Structures b. Market c. Free competition market d. Perfect Competition market 60. This is defined as the interest on a loan or principal that is based on the original amount of the loan or principal. a. Ordinary Simple Interest b. Simple Interest c. Exact Simple Interest d. Compound Interest 61. This is based on one banker’s year. a. Ordinary Simple Interest b. Simple Interest c. Exact Simple Interest d. Compound Interest 62. __ is based on the exact number of days in a given year. a. Ordinary Simple Interest b. Simple Interest c. Exact Simple Interest d. Compound Interest 63. One banker’s year is equ ivalent to a. 365 days b. 350 days c. 12 months and 30 days d. 12 months and 15 days 64. This is defined as the interest of loan or principal which is based on the amount of the original loan and the previous accumulated interest a. Ordinary Simple Interest b. Simple Interest c. Exact Simple Interest d. Compound Interest
65. It is the amount of money or payment for the use of a borrowed money or capital a. Annuity b. Interest c. Simple Interest d. Simple Annuity 66. This is defined as the basic annual rate of interest a. effective rate of annuity b. effective rate of interest c. nominal rate or annuity d. nominal rate of interest 67. It is defined as the actual or exact rate of interest earned on a principal during1 year period a. effective rate of annuity b. effective rate of interest c. nominal rate or annuity d. exact simple interest 68. This refers to the difference between the future worth of a negotiable paper and its present worth. a. Depreciation b. Market value c. Discount d. Fair value 69. This refers to the sales of stock or share at reduced price a. Depreciation b. Market value c. Discount d. Fair value 70. This is the deduction from the published price of services or goods. a. Depreciation b. Salvage value c. Discount d. Fair value 71. “Annuity” is simply defined as a. series of equal payments occurring at equal time interval of time b. series of payments occurring at equal time interval c. series of equal payments occurring at certain time interval d. series of divided payments at pay periods 72. An annuity of a fixed time span is also called as a. Ordinary annuity b. Annuity Certain c. Annuity Due d. Deferred Annuity 73. __ is the type of annuity where the payments are made at the beginning of each period starting from the first period a. Ordinary annuity b. Annuity Certain c. Annuity Due d. Deferred Annuity 74. An annuity where the payments are made at the end of each period beginning on the first period a. Ordinary annuity b. Annuity Certain c. Annuity Due d. Deferred Annuity 75. It is when the first payment does not begin until some later date in the cash flow
a. Ordinary annuity b. Perpetuity c. Annuity Due d. Deferred Annuity 76. When an annuity does not have a fixed time span but continues indefinitely, then it is referred to as a. Ordinary annuity b. Perpetuity c. Annuity Due d. Deferred Annuity 77. This refers to the rate of interest that is quoted in the bond. a. bond b. bond rate c. bond value d. interest rate of bond value 78. It is the present worth of the future payments that will be received a. book value b. bond value c. fair value d. salvage value 79. It is the amount a willing buyer will pay to a willing seller for a property where each has equal advantage and neither one of them is under compulsion to buy or sell a. Book value b. Market value c. Use value d. Fair value 80. The amount of the property which the owner believed to be its worth as an operating unit is called a. value b. Market value c. Use value d. Fair value 81. ___ is the amount obtained from the sale of the property. a. book value b. bond value c. market value d. salvage value 82. It is the worth of the property determined by a disinterested person a. Book value b. Market value c. Use value d. Fair value 83. It is the reduction or fall in the value of an asset or physical property during the course of its working life due to the passage of time a. Depreciation b. Depleting value c. Discounted value d. Fair value 84. __ is the money worth of an asset or product a. Price b. Market value c. value d. Fair value 85. This refers to the present worth of all profits that are to be received through ownership of a particular property
a. Price b. Market value c. value d. capitalized cost 86. Refers to the sum of its first cost and cost of perpetual maintenance of a property a. Price b. Market value c. perpetual value d. capitalized cost 87. __ is a long term note or a financial security issued by businesses or corporation and guaranteed on certain assets of the corporation or its subsidiaries. a. Contract b. Receipt c. bond value d. bond 88. This value implies that the property will still be use for the purpose it is intended a. salvage value b. book value c. use value d. depreciation 89. A depreciation computation method which is also known as Diminishing Balance Method a. Sinking Fund Method b. Straight-line Method c. Declining Balance method d. Break-even analysis 90. Also called as Constant- Percentage Method in depreciation method a. Sinking Fund Method b. Straight-line Method c. Declining Balance method d. Break-even analysis 91. This is also known as “Resale value” a. salvage value b. book value c. use value d. depreciation 92. ___ is considered as the simplest type of business organization wherein the firm is controlled and owned by a single person a. sole-proprietorship b. Partnership c. Corporation d. Entrepreneurship 93. __ is a firm owned and controlled by two or more persons who are bind to an agreement a. sole-proprietorship b. Partnership c. Corporation d. Entrepreneurship 94. It is a firm owned by a group of ordinary shareholders and the capital of which is divided up to the number of shares. a. sole-proprietorship b. Partnership c. Corporation d. Entrepreneurship
95. It is defined as a distinct legal entity separate from the individuals who owns it and can engage in any business transaction which a real person could do. a. sole-proprietorship b. Partnership c. Corporation d. Entrepreneurship 96. Also known as joint-stock company or a cooperative a. Incorporation b. Partnership c. Corporation d. Entrepreneurship 97. This refers to the situation where the sales generated are just enough to cover the fixed and variable cost. a. break-even b. Break even chart c. break even point d. none of the above 98. The break even chart shows a. the relationship between income and fixed cost b. the relationship between the volume and fixed cost, variable costs, and i ncome c. the relationship between the assets and fixed cost, variable costs, and income d. the relationship between the liabilities and fixed cost, variable costs, and income 99. The level of production where the total income is equal to the total expenses is known as a. break-even b. Break even chart c. break even point d. none of the above 100. The sum of perpetuity is a/an a. book value b. zero value c. infinite value d. bond value
Board Problems: 1. Design based on conditions giving the least cost per unit time and maximum profit per unit production a. battery limit b. break-even point c. optimum economic design d. plant design 2. A flow diagram, indicating the flow of materials, unit operations involved, equipment necessary and special information on operating temperature and pressure is a a. schematic diagram b. qualitative flow diagram c. process flow chart d. quantitative flow diagram 3. In plant design implementation, soil testing is done to determine a. pH b. load bearing capacity c. porosity d, viscosity
4. This includes all engineering aspects involved in the development of either a new, modified or expanded industrial plant is called a. plant design b. optimum design c. process design d. engineering design 5. A chemical engineering specializing in the economic aspects of design is called a. plant engineer b. cost engineer c. design engineer d. process engineer 6. This refers to the actual design of the equipment and facilities for carrying out the process a. process engineering b. plant design c. process design d. optimum design 7. The final step before construction plans for plant and includes complete specifications for all components of the plant and accurate costs based on quoted prices are obtained a. preliminary design b. quick estimate design c. firm process design d. detailed design estimate 8. A thorough and systematic analysis of all factors that affect the possibility of success of a proposed undertaking usually dealing with the market, technical, financial, socio-economic and management aspects is called a. pr5oject feasibility study b. plant design c. project development and research d. product development 9. Discusses the nature of the product line, the technology necessary for production, its availability, the proper mix of production resources and the optimum production volume a. market feasibility b. socio-economic feasibility c. technical feasibility d. management feasibility 10. Discusses the nature of the unsatisfied demand which the project seeks to meet growth and the manner in which it is to be met a. financial feasibility b. management feasibility c. market feasibility d. technical feasibility 11. A multiple effect evaporator produces 10, 000 kg of salt from a 20% brine solution per day. One kg of steam evaporates 0.7 N kg of water in N effects at a cost of P25 per 1000 kg of steam. The cost of the first effect is P450, 000 and the additional effects at P300, 000 each. The life of the evaporator is 10 years with no salvage value. The annual average cost of repair and maintenance is 10% and taxes and insurance is 5%. The optimum number of effects for minimum annual cost is a. 3 effect b. 5 effects c. 4 effects d. 2 effects 12. A process requires 20, 000 lb/hr of saturated steam at 115 psig. This is purchased from a neighboring plant at P18. 00 per short ton and the total energy content rate (mechanical) in the steam may be valued at P7. 5 x 10 -6 per Btu. Hours of operation per year are 7200. The friction loss in the line is given by the following equation
F = 187.5 Lq
1.8
mc 0.20 in ft-lbs/lbm d 0.20 Di 4.8
Cf = 1.44 Di 1.5 L , in p/yr Where:
L = length of straight pipe, ft. q = steam flow rate, cu. ft. per sec mc = steam viscosity cp. d = steam density, lb per ft 3 Di = inside diameter of pipe, inches The optimum pipe diameter that should be used for transporting the above steam is a. 6 in b. 4 in c. 3 in d. 5 in o 13. A smelting furnace operating at 2, 400 F is to be insulated on the outside to reduce heat losses and same on energy. The furnace wall consists of a ½ inch steel plate and 4 inch thick refractory inner lining. During operation without other insulation, the outer surface of the steel plate exposed to air has a temperature of 300 oF. Ambient air temperature is at 90 oF. Operation is 300 days per year. Thermal conductivities in Btu per hr-ft oF are: steel plate = 26; refractory =0.1; insulation to be installed = 0.025. The combined radiation and convection loss to air irrespective of material exposed is 3 Btu per hr-ft 2-oF, annual fixed charge is 20 % of the initial insulation cost. If heat energy is P5.0 per 10, 000 Btu and installed cost of insulation is P100 per inch ft 2 of area, the optimum thickness of the outer insulation that should be installed is a. 8 in b. 12 in c. 10 in d. 6 in 14. An organic chemical is produced by a batch process. In this process chemical X and Y react to form Z. Since the reaction rate is very high, the total time required per batch has been found to be independent of the amount of materials and each batch requires 2 hr, including time for charging, heating, and dumping. The following equation shows the relation between the pound of Z produced (lbz) and the pound of X (lbx) and Y (lby) supplied: lbz = 1.5(1.1lbxlbz +1.3lbylbz – lbxlby)0.5 Chemical X costs P0.09 per pound, chemical Y costs P0.04 per pound and chemical Z sells for P0.8 per pound. If half of the selling price for chemical Z is due to cost other than raw materials, the maximum profit obtainable per pound of chemical Z is a. P0.3 per lbz b. P0.5 per lbz c. P0.12 per lbz d. P0.25 per lbz 15. One hundred gram moles of R are to be produced hourly from a feed consisting of a saturated solution of A. (Cao = 0.1 gmole per liter). The reaction is A R with rate ГR = (0.2/hr) Ca. Cost of reactant at Cao = 0.1 gram mol per liter is P3.75/g-mole A; cost of backmix reactor, installed complete with auxiliary equipment, instrumentation, overhead, labor depreciation, etc. is P0.075 per hr-liter. The conversion that should be used for optimum operation is a. 45% b. 60% c. 50% d. 40% 16. One hundred lb moles of reactant A at a concentration of 0.01 lbmole per cuft is to be reacted with reactant B to produce R and S. The reaction follows the aqueous phase elementary reaction: A + B ------- R +S where k = 500 cuft/(lbmole-hr). The amount of R required is 95 lbmoles/hr. Data: * in extracting R from the reacted mixture, A and B are destroyed
* B costs P15 per lbmole in crystalline form, and is very soluble in aqueous such that even when present in large amount does not affect A in solution * capital and operating cost for backmix reactor is P0.10 per(cuft-hr). The optimum backmix reactor size is: a. 23, 900 ft 3 b. 24, 200 ft 3 c. 25, 900 ft 3 d. 23, 500 ft 3
solutions
17. April 1992. A unit of welding machine cost P45, 000 with an estimated life of 5 years. Its salvage value id P2, 500. Find its depreciation rate by straight-line method. a. 17.75% b. 19.88% c. 18.89% d.15.56% 18. April 1997. A machine has an initial cost of P50, 000 and a salvage value of P10, 000 after 10 years. Find the book value after 5 years of using straight-line depreciation. a. P12, 500 b.P30, 000 c. P16, 400 d.P22, 300 19. October 1992. The initial cost of a paint sand mill, including its installation, is P800, 000. The BIR approved life of this machine is 10 years for depreciation. The estimated salvage value of the mill is P50, 000 and the cost of dismantling is estimated is estimated to be P15, 000. Using straightline depreciation, what is the annual depreciation charge and what is the book value of the machine at the end of six years. a. P74, 500 ; P340, 250 b. P76, 500 ; P341, 000 c. P76, 500 ; P 342, 500 d. P77, 500 ; P343, 250 20. November 1997. The cost of the equipment is P500, 000.00 and the cost of installation is P30, 000. If the salvage value is 10% of the cost of equipment at the end of five years, determine the book value at the end of the fourth year. Use straight-line method. a. P155, 000 b.P140, 000 c. P146, 000 d. P132,600 21. April 1998. An equipment costs P10, 000 with a salvage value of P500 at the end of 10 years. Calculate the annual depreciation cost by sinking fund method at 4% interest. a. P791.26 b.P950.00 c. P971.12 d. P845.32 22. November 1995. A machine costing P720, 000 is estimated to have a book value of P40, 545.73 when retired at the end of 10 years. Depreciation cost is computed using a constant percentage of the declining book value. What is the annual rate of depreciation in %? a.28 b. 25 c. 16 d.30 23. November 1998. AMD Corporation makes it a policy that for any new equipment purchased; the annual depreciation cost should not exceed 20 % of the first cost at any time with no salvage value. Determine the length of service life is necessary if the depreciation used is the SYD method. a.9 years b. 10 years c. 12 years d.19 years 24. November 1996. At 6%, find the capitalized cost of a bridged whose cost is P250M and the life is 20 years, i f the bridge must be partially rebuilt at a cost of P100M at the end of each 20 years.
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a. P275.3M b.P265.5M c. P295.3M d. P282.1M October 1995 A company must relocate one of its factories in three years. Equipment for the loading dock is being considered for purchase. The original cost is P20, 000; the salvage value of the equipment after three years is P8, 000. The company’s rate return on th e money is 10% . Determine the capital recovery per years. a. P5, 115 b.P4, 946 c. P5, 625 d. P4, 805 October 1998. The annual maintenance cost of a machine shop is P69, 994. if the cost of making a forging is P56 per unit and its selling price is P135 per forged unit, find the number of units to be forged to break-even. a.886 units b. 885 units c. 688 units d.668 units October 1990. Compute for the number of locks that an ice plant must be able to sell per month to break even based on the following data: Cost of electricity P20.00 Tax to be paid per block 2.00 Real Estate Tax 3, 500.00 per month Salaries and wages 25, 000.00 per month Others 12, 000.00 per month Selling price of Ice 55.00 per block a. 1228 b. 1285 c. 1373 d.1312 April 1998 XYX Corporation manufactures bookcases that sell for P65.00 each. It costs XYZ Corporation P35, 000 per year to operate its plant. This sum includes rent, depreciation charges on equipment, and salary payments. If the cost to produce one bookcase is P50.00, how many cases must be sold each year for XYZ to avoid taking a loss? a. 2334 b. 539 c. 750 d.2333
Answer Key Board Problems 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.
c b b a b c c a c c d a c c c a c b
19. 20. 21. 22. 23. 24. 25. 26. 27. 28.
b c a b a c c a a a
Answer Key Objective Problems 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41.
a a a b d c a a d d d b d a d a a a b b a b d a d d c a d b a a a a a d c c d c a
42. 43. 44. 45. 46. 47. b 48. c 49. b 50. a 51. d 52. c 53. c 54. d 55. d 56. a 57. b 58. d 59. b 60. b 61. a 62. c 63. c 64. d 65. b 66. d 67. b 68. c 69. c 70. c 71. a 72. b 73. c 74. a 75. d 76. b 77. b 78. b 79. b 80. c 81. d 82. d 83. a 84. c 85. c 86. d 87. d 88. a 89. c 90. c 91. a 92. a 93. b 94. c 95. c 96. c
a b d d a
97. a 98. b 99. c 100. c
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Problem Solving 1.
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The purchased cost of a shell-and-tube heat exchanger (floating head and carbon-steel tubes) with 100 ft 2 of heating surface was $3,000 in 1980. What will be the purchased cost of a similar heat exchanger with 200 ft 2 of heating surface in 1980 if purchased-costcapacity exponent is 0.60 for surface area ranging from 100-400 ft 2? a. $4 547.15 b. $4 127.25 c. $4 567.10 d. $4523.00 efer to problem no. 1. If the purchased-cost-capacity exponent for this type of exchanger is 0.81 for surface areas ranging from 400-2,000 ft 2, what will be the purchased cost of a heat exchanger with1, 000 ft 2 of heating surface in 1985? a. $20 423.38 b. $18 527.12 c. $14 547.00 d. $24 542.05 If the purchased cost of a shell- and-tube heat exchanger (floating head and carbon-s teel tubes) with 100 ft 2 of heating surface was $3,000 in 1980. Note that the purchase-costcapacity exponent is not constant over range surface area requested. What will be the purchased cost of a heat exchanger with 700 ft 2 of heating surface in 1985? a. $15 398.79 b. $15 298.79 c. $16 498.79 d. $16 598.79 From the preceding question, what will be the purchased cost of the heat exchanger with 800 ft2 of heating surface in 1985? a. $18 046.32 b. $18 146.32 c. $17 046.32 d. $17 146.32 Refer to problem no. 3. What will be the purchased cost of the heat exchanger with 1500 ft 2 of heating surface in 1985? a. $38 363.62 b. $48 363.62 c. $18 363.62 d. $28363.62 The purchase and installation cost of some pieces of equipment are given as a function of weight rather than capacity. An example of this is the installed cost of large tanks. The 1980 cost for an installed aluminum tank weighing 100,000 lb was $390,000. For a size range from 200,000 to 1,000,000 ld, the installed cost-weight exponent for aluminum tanks is 0.93. If an aluminum tank weighing 700,000 lb is required, what is the present capital investment needed? a. $3 160 100 b. $3 060 000 c. $6 160 100 d. $6 060 000
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The purchased equipment cost for a plant which produces pentaerythritol (solid-fuel processing plant) is $300,000. The plant is to be an addition to an existing formaldehyde plant. The major part of the building cost will be for indoor construction, and the contractor’s fee will be 7% of the direct plant cost. All other costs are close to the average values found for typical chemical plants. On the basis of this information, estimate the total direct plant cost. a. $879 000 b. $879 253 c. $890 560 d. $825 020 The purchased equipment cost for a plant which produces pentaerythritol (solid-fuel processing plant) is $300,000. The plant is to be an addition to an existing formaldehyde plant. The major part of the building cost will be for indoor construction, and the contractor’s fee will be 7% of the direct plant cost. All other costs are close to the average values found for typical chemical plants. On the basis of this information, estimate the fixed capital investment. a. $ 1 246 530 b. $ 2 336 230 c. $ 2 546 531 d. $ 1 246 830
Refer to the previous problem and on the basis of this information, what is the total capital investment? a. $1 542 125 b. $1 478 570 c. $1 532 120 d. $1 468 530 10. The total capital investment for a chemical plant is $1,500,000 and the plant process 3M kg of product annually. The selling price of the product is $0.82/kg. Working capital amounts to 15% of the total capital investment. The investment is from company funds, and no interest is charged. Raw-material costs for the product are $0.09/kg, labor $0.08/kg, utilities $0.05/kg, and packaging $0.008/kg. Distribution costs are 5% of the total product cost. Determine the percent change in total cost. a. 36.16% b. 45.32% c. 12.00% d. 35.61% 11. It is desired to have a $ 9000 available from 12 years from now. If $ 5000 is available for investment at the present time, what is discrete annual rate of compound interest on the investment would be necessary to give the desired amount? a. 6.02% b. 5.02% c. 4.02% d. 3.05% 12. An original loan of $2000 was made at 6 percent simple interest per year for 4 years. At the end of this time, no interest had been paid and the loan was extended for 6 more years at a new, effective, compound-interest rate of 8 percent per year. What is the total amount owned at the end of ten years if no intermediate payments are made? a. $4 935.45 b. $3 945.45 c. $4 945.45 d. $3935.45 The original cost for a distillation tower is $24,000 and the useful life of the tower is estimated to be 8 years. The sinking fund method for determining the arte of depreciation is used, and the effective annual interest for the depreciation
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fund is 6 percent. If the scrap value of the distillation tower is $ 4000, determine the asset value at the end of 5 years. a. $ 3,547.928 b. $2 235.623 c. $2 547.928 d. $3 235.623 An annuity is due to being used to accumulate money. Interest is compounded at an effective annual rate of 8 %, and $1000 is deposited at the beginning of each year. What will be the total amount of annuity due be after 5 years? a. $1233.24 b. $1173.32 c. $3573.2 d. $183.23 For the total year payments of $5000 for ten years, what will be the compound amount accumulated at the end of ten years if the payment is at the end of the year? The effective (annual) interest is 20% and payments are uniform. a. $154 793.41 b. $456 793.41 c. $129 793.41 d. $209 793.41 Referring to the previous problem, estimate the compound amount accumulate d at the end of ten years, if the payment is made weekly? a. $143 951.49 b. $243 951.49 c. $443 951.49 d. $643 951.49 A multiple – effect evaporator is to be used for evaporating 400,000 lb of water per day from a salt solution. The total initial cost for the 1 st effect is $18,000 and each additional effect costs $15,000. The life period is estimated to be 10 years, and the scrap value at the end of the life period may be assumed to be zero. The straight-line depreciation method is used. Fixed charges minus depreciation are 15% yearly based on the first cost of the equipment. Steam cost $1.50 per 1000 lb. Annual maintenance charges are 5% of the initial equipment cost. All other cost is independent of the number of effects. The unit will operate 300 days per year. If the lb of water evaporated per pound of steam equals 0.85 x numbers of effects, determine the optimum number of effects for minimum annual cost. a. 2 effects b. 3 effects c. 5 effects d. 4 effects Determine the optimum economic thickness of insulation that should be used under the following conditions: Standard steam is being passed continuously through a steel pipe with an outside diameter of 10.75 in. The temperature of the steam is 400 F, and the steam is valued at $1.80 per 1000 lb. The pipe is to be insulated with material that has a thermal conductivity of 0.03 Btu/h-ft 2-F/ft. The cost of installed insulation per foot of pipe length is $4.5xI t, where I t is the thickness of the insulation in inches. Annual fixed charges including maintenance amount to 20% of the initial installed cost. The total length of the pipe is 1000 ft, and the average temperature of the surrounding may be taken as 70F. Heat transfer resistance due to the steam film, scale and pipe wall are negligible. The air –film coefficient at the outside of the insulation may be assumed constant at 2.0 Btu/h-ft 2-F for all insulation thickness. a. $3257 b. $4057 c. $3057 d. $4257
18. A proposed chemical plant will require a fixed capital investment of $10 million. It is estimated that the working capital will amount to 25% of the total investment and annual depreciation costs are estimated to be 10 percent of the fixed capital investment. If the annual profit will be $3 million, determine the standard percent return on the total investment a. 22.5% b. 11.25% c. 34.5% d. 15.98% 19. If a plant will require a fixed capital investment of $10 million and the working capital will amount to 25% of the total investment and annual depreciation costs are estimated to be 10 percent of the fixed capital investment. If the annual profit will be $3 million, what is the minimum payout period (POP)? a. 2.5 years b. 4 years c. 3.5 years d. 2 years 20. An annual investigation of a proposed investment has been made. The following result has been presented to management. The minimum payout period based on capital recovery using a minimum annual return of 10 percent as a fictitious expense is 10 years; annual depreciation costs amount top 8 percent of the total investment. Using this information, determine the standard rate of return on the investment. a. 21% b. 10.5% c. 56.93% d. 11.5% 21. The information given in the previous problem, applies to conditions before income taxes. If 34% percent of all profits must be paid out for income taxes, determine the standard rate of return after taxes using the figures given in the previous problem. a. 7.93% b. 6.93% c. 4.93% d. 5.93% 22. A capitalized cost for a piece of equipment has been found to be $55,000. This cost is based on the original cost plus the present value of an indefinite number of renewals. An annual interest rate of 12% was used in determining the capitalized cost. The salvage value of the equipment at the end of the service life was estimated to be 10 years. Under these conditions, what would be the original cost of the equipment? a. $57 391.47 b. $27 291.47 c. $47 391.47 d. $37 291.47 For problems 24- 26.On Aug. 1, a concern had 10,000 lb of raw material on hand, which was purchased at a cost of $0.030 per pound. In order to build up the reserve, 8000 lb of additional raw material was purchased on Aug. 15 at a cost of $0.028 per pound. If none of the raw material was used until after the last purchase. 23. Determine the total cost of 12,000 lb of the raw material on an inventory or cost-of-sales account for the month of August by current average method.
a. $0.00297/lb b. $0.0297/lb c. $0.2297/lb d. $0.2970/lb 24. Determine the total cost of 12,000 lb of the raw material on an inventory or cost-of-sales account for the month of August by “FiFo” method. a. $0.281/lb b. $0.028/lb c. $0.038/lb d. $0.128/lb 25. Determine the total cost of 12,000 lb of the raw material on an inventory or cost-of-sales account for the month of August by lifo method. a. $0.031/lb b. $0.041/lb c. $2.031/lb d. $1.041/lb For question nos. 27 – 31. The following are the data gathered from the AMD Food Corporation: Cash Accounts payable: B Company C Company Accounts receivable Inventories Mortgage payable Common stock sold Machinery and equipment (at present value) Furniture and fixtures (at present value) Government bonds Surplus
$20,000 2,000 8,000 6,000 15,000 5,000 50,000 18,000 5,000 3,000 2,000
26. From the data given above, determine the total asset of the AMD Food Corporation. a. $ 47, 050 b. $ 67, 000 c. $ 87, 050 d. $ 57, 000 27. Determine the total current assets of the AMD Food Corporation. a. $ 11,000 b. $ 44,000 c. $ 22,000 d. $ 33,000 28. What is the total amount of the Current liabilities of the AMD Corporation? a. $ 45,097 b. $ 85,097 c. $ 25,000 d. $ 15,000
29. What is the total amount of the Fixed Assets of the AMD Food Corporation? a. $ 23,000 b. $ 33,050 c. $ 32,000 d. $ 23,050 During the month of October, the following information was obtained in the AC antifreeze retailing company: Salaries Delivery expenses Rent Sales Antifreeze available for sale during October (at cost) Antifreeze inventory on Oct. 31 (at cost) Other expenses Earned surplus before income taxes as of Sept.30
$ 3,000 700 400 15,100 20,000 11,000 1,200 800
30. Prepare an income statement for the month of October to determine the net income is for the month of October. a. $ 29,100 b. $ 19,200 c. $ 29,200 d. $ 19,100 31. From the data above, determine also the total gross income of the AC Antifreeze Company. a. $26,900 b. $36,100 c. $26,100 d. $46,900 The following information applies to MADSteel Company on a given date: Long-term debts Debts due within 1 year Accounts payable Machinery and equipment (at cost) Cash in bank Prepaid rent Government bonds Social security taxes payable Reserve for depreciation Reserve for expansion Inventory Accounts receivable
$ 1,600 1,000 2,300 10,000 3,100 300 3,000 240 600 1,200 1,600 1,700
32. Determine the cash asset for the MADSteel Company at the given date.
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a. $6,100 b. $7,100 c. $7,900 d. $6,900 Determine the current asset for MADSteel Company a. $8, 700 b. $9, 500 c. $8, 500 d. $9, 700 Determine the current liabilities of MADSteel Company a. $5, 540 b. $1, 540 c. $3, 540 d. $2, 540 Determine the Quick ratio for MADSteel Company a. 2.29 b. 9.87 c. 2.56 d. 9.56 Determine the current ratio of MADSteel Company a. 2.74 b. 2.56 c. 2.64 d. 1.98 A reactor of special design is the major item of equipment in a small chemical plant. The initial cost of a completely installed reactor is $60,000, and the salvage value at the end of the useful life is estimated to be $10,000. Excluding depreciation costs for the reactor, the total annual expenses for the plant are $100,000. How many years of useful life should be estimated for the reactor if 12 % of the total annual expenses for the plant are due to the cost for the reactor depreciation? The straight-line method for determining depreciation should be used. a. 4 years b. 2 years c. 8years d. 1year The initial installed cost for a new piece of equipment is $10,000, and its scrap value at the end of its useful life is estimated to be $2,000. The useful life is estimated to be 10 years. After the equipment has been in use for 4 years, it is sold for $7,000. The company which originally owned the equipment employs the straight-line method for determining depreciation costs. If the company had used an alternative method for determining depreciation cost, the asset (or book) value for the piece of equipment at the end of 4 years would have been $5240. The total income-tax rate for the company is 34% of all gross earnings. Capital-gains taxes amount to 34% of the gain. How much net saving after taxes would the company have achieved by using the alternative (in this case, reducing-balance) depreciation method instead of the straight-line depreciation method? a. $2161.60 b. $1261.07 c. $1171.70 d. $1161.60 A piece of equipment is originall y costing $40,000 was put into use 12 years ago. At the time the equipment was put into use, the service life was estimated to be 20 years and the salvage and scrap value at the end of the service life were assumed to be zero. On this basis, the straight-line depreciation fund was set up. The equipment can now be sold for $10,000, and a more advanced model can be installed for $55,000. Assuming the
depreciation fund is available for use, how much new capital must be supplied to make the purchase? a. $22,000 b. $21,000 c. $12,000 d. $11,000 40. The original investment for an asset was $10,000, and the asset was assumed to have a service life of 12 years with $2,000 salvage value at the end of the service life. After the asset has been in use for 5 years, the remaining service life and the final salvage value are reestimated at 10 years and $ 1,000, respectively. Under these conditions, what is the depreciation cost during the sixth year of the total life is straight-line depreciation is used? a. $596.65/yr b. $561.66/yr c. $678.65/yr d. $566.66/yr A piece of equipment having a negligible salvage and scrap value is estimated to have a service life of 10 years. The original cost of the equipment was $40,000. Determine the following: 41. Based on the above data, determine the depreciation charge for the fifth year if doubledeclining balance depreciation is used. a. $ 3,567.2 b. $ 3,276.80 c. $ 4, 245.80 d. $ 4,252.80 42. The depreciation charge for the fifth year if sum-of-the-years-digits depreciation is used. a. $ 4,363.64 b. $ 5,572.14 c. $ 8,265.52 d. $ 9,356.45 43. The percent of the original investment paid off in the first half of the service life using the double-declining balance method. a. 32.8% b. 67.4% c. 23.4% d. 54.2% 44. The percent of the original investment paid off in the first half of the service life using the sum-of-the-years-digits method. a. 45.45% b. 54.54% c. 23.23% d. 35.53% 45. The original cost of the property is $30,000, and it is depreciated by a 6 percent sinkingfund method. What is the annual depreciation charge if the book value of the property after 10 years is the same as if it had depreciated at $2,500/year by the straight-line method? a. $ 167.67/yr b. $ 267.67/yr c. $ 289.67/yr d. $ 189.67/yr 46. A concern has a total income of $1 million/year, and all expenses except depreciation amount to $600,000/year. At the start of the first year of the concern’s operation, a composite account of all depreciable items show a value of $850,000, and the overall service life is estimated to be 20 years. The total salvage value at the end of the service life is estimated to be $50,000. Thirty percent of all profits before taxed must be paid out as income taxes. What would be the reduction in income-tax charges for the first year of
operation if the sum-of-the-years-digits methods were used for depreciation accounting instead of the straight-line method? a. $16,356.72 b. $26,285.72 c. $26,356.72 d. $16,285.72 47. The total value of anew plant is $2 million . A certifi cate of necessity has been obtained permitting a write-off of 60 percent of the initial value ay 5 years. The balance of the plant requires a write-off period of 15 years. Using the straight-line method and assuming negligible salvage and scrap value, determine the total depreciation cost during the first year. a. $ 35, 555. 33/ yr b. $ 55, 333. 33/ yr c. $ 35, 333. 33/ yr d. $ 53, 333. 33/ yr 50. A profit-producing property has an initial value of $50,000, a service life of 10 years, and zero salvage and scraps value. By how much would annual profits before taxes be increased if a 5 percent sinking-fund method were used to determine depreciation costs instead of straight-line method? a. $ 3 556.24 b. $ 3 354.35 c. $ 3 484.31 d. $ 3.034.35 51. In order to make it worthwhile to purchase a new piece of equipment, the annual depreciation costs for the equipment cannot exceed $3,000 at any time. The original cost of the equipment is $30,000, and it has a zero salvage and scrap value. Determine the length of service life necessary if the equipment is depreciated by the sum=of-the-yearsdigits method by the straight-line method. a. 19 years b. 23 years c. 17 years d. 18 years 52. Referring to the previous number, determine the length of service life necessary if the equipment is depreciated by the straight-line method. a. 8 years b.10 years c. 12 years d. 9 years A materials-testing machine was purchased for $20,000 and was to be used for 5 years with an expected residual salvage value of $5,000. Graph the annual depreciation charges and year-end book values obtained by using: 53. By using Straight-line depreciation a. $ 6,000 b. $ 5,000 c. $ 9,000 d. $ 7,000 54.By using Sum-of-digits depreciation a. $ 6,000 b. $ 5,000 c. $ 9,000 d. $ 7,000 55. By using Double-declining balance depreciation a. $ 625
b. $ 675 c. $278 d. $955 56. An asset with an original cost of $10,000 and no salvage value has a depreciation charge of $2381 during its second year of service when depreciated by the sum-of-digits method. What is its expected useful life? a. 6 years b. 4 years c. 9 years d. 2 years 57. An electronic balance costs P90, 000 and has an estimated salvage value of P8, 000 at the end of its 10 years lifetime. What would be the book value after 3 years, using straight-line method in solving for the depreciation? a. P 65,200 b. P 76,466 c. P 24, 674 d. P 21,758 A broadcasting corporation purchased equipment for P53, 000 and paid P1, 5000 for freight and delivery charges top the job sites. The equipment has a normal life of 10 year with a trade-in value of P5, 000 against the purchase of anew equipment at the end of the life. 58. Referring to the problem above, determine the annual depreciation by straight-line method. a. P4, 950 b. P5, 950 c. P7, 950 d. P3, 950 59. From the preceding number, determine annual depreciation by sinking fund method. Assuming interest 6 ½% compounded annually. a. P 3,668 b. P 1,575 c. P 4,245 d. P6,258 A firm brought equipment for P56, 000. Other expenses including installation amounted to P4, 000. The equipment is expected to have a life of 16 years with a salvage value of 10% of the original cost. 60. Determine the book value at the end of 12 years by SLM: a. P19, 500 b. P20,555 c. P15,582 d. P18, 562 61. Determine the book value at the end of 12 years by SFM: a. P29, 520 b. P29, 520 c. P29, 520 d. P29, 520 For numbers 62 – 64. A certain type of machine losses 10% of its value each year. The machine cost P2, 000 originally. Make cut a schedule showing the following: 62. By yearly depreciation a.131.32 b.132.12 c.123.23 d. 213.20
63. The total depreciation after 5 years. a. P819.12 b. P562.23 c. P456.26 d. P895.23 64. Estimate the book value at the end year for 5 years. a. P1, 180.98 b. P5, 125.25 c. P1, 256.32 d. P5, 235.00 65. Determine the rate of depreciation, the total depreciation up to the end of 8 th year and book value at the end of 8 years for an asset that costs P15,000 new and has an estimated scrap value of P2,000 at the end of 10 years by DBM. a. P 12,108 b. P12, 008 c. P11,355 d. P10,245 66. Determine the rate of depreciation, the total depreciation up to the end of 8 th year and book value at the end of 8 years for an asset that costs P15, 000 new and has an estimated scrap value of P2, 000 at the end of 10 years by DDBM. a. P 12,583 b. P 10,483 c. P 12,483 d. P 10,583 67. Mr. Dim bought a calciner for P220, 000 and used it for 10 years, the life span of the equipment. What is the book value of the calciner after 5 years of use? Assume a scrap value of P20, 000 for SLM; a. P 120,566 b. P 120,452 c. P 120,000 d. P 130,000 68. Refer to the previous problem. What is the book value of the calciner after 5 years of use? Assume a scrap value of P22, 000 for textbook for DBM. a. P 96, 570.00 b. P 69, 235.00 c. P 59, 235.00 d. P 69, 570.00 69. Refer to the previous problem. What is the book value of the calciner after 5 years of use? Assume a scrap value of P20, 000 for textbook for DDBM. a. P72 091 b. P72 356 c. P71 190 d. P72 090 70. A structure costs P12, 000 new. It is estimated to have a life of 5 years with a salvage value at the end of life of P1, 000. Determine the book value at the end of three years. a. 733 b. 562 c. 252 d. 377 71. Operator A produces 120 spindle/hr on a lathe. His hourly rate is $1.80. Operator B, using an identical lathe, is able to produce 150 identically units/hr. The overhead charge for a lathe is fixed at $2.50/hr. Determine operator B’s hourly rate so that his cost per piece is identical to A’s.
72.
73.
74.
75.
76.
77.
78.
a. $2.88 b. $3.88 c. 4.88 d. 1.88 In a type 1 warehouse, initial cost will be $24, 000.This warehouse has adequate capacity for the near future, but 12 years from now an addition will be required that costs $15, 000. A type 2 warehouse costs $34, 000. This type has the same capacity as the type 1 warehouse with its addition. What will be the present cost of the type 1 warehouse? Which of these should be built, assuming that depreciation is negligible and that the interest rate is 7%? a. $30, 540 b. $30,660 c. $23, 548 d. $15,235 Determine the equal (year-end) payments that will be available for the next four years if we invest $4, 000 at 6%. a. $2533.23 b. $2322.25 c. $1123.25 d. $1154.36 A new snow removal machine costs $50, 000. The new machine will operate at a reputed savings of $400 per day over the present equipment in terms of time and efficiency. If interest is at 5% and the machine’s life is assumed to be 10 years with zero salvage, how many days per year must the machine be used to make the investment economical? a. 15 days b. 16 days c. 14 days d. 12 days Based on the sinking fund method and using the data in the previous problem, what number of days must the machine be used if the amount to be accumulated in 10 years is a. $50, 000 b. $60,000 c. $55,000 d. $45, 000 Based on the sinking fund method and using the data in the previous problem, what number of days must the machine be used if the amount to be accumulated in 10 years is $50, 000? a. 12 days b. 10 days c. 14 days d. 13 days A consulting engineer decides to set up an educational fund for his son that will provide $3000 per year for 6 years starting in 16 years. The best interest rate he can expect to get is 5% compounded quarterly. He wants to accumulate the necessary capital by making quarterly deposits until his son starts college. What will be his needed quarterly deposit? a. $15, 210 b. $16,213 c. $15, 213 d. $16, 210 A low carbon steel machine part, costing $350 installed, lasts 6 years when operating in a corrosive atmosphere. An identically shaped part, but treated for corrosion resistance, would cost $650 installed. How long would the corrosion resistance part have to last to be at least as good investment as the untreated part? Assume money is worth 7%
a. 11 years b. 12 years c. 14 years d. 13 years 79. The total cost of a cast product consists of (1) the raw material cost that is directly proportional to the weight, of the casting, (2) the machining cost that varies inversely as the weight, and (3) overhead cost that remains constant per unit produced regardless of weight. Find the weight giving the minimum cost per casting. a. W = (k1/k2)1/4 lb b. W = (k2/k1)1/2 lb c. W = (k2/k1)1/4 lb d. W = (k1/k2)1/2 lb 80. Based on the previous problem, what is the minimum total cost? a. Ct = [k2(k2 / k1)1/2 = k2 (k1 / k2) 1/4+ Co] b. Ct = [k1(k2 / k1)1/2 = k2 (k1 / k2) ½ + Co] c. Ct = [k1(k2 / k1)1/4 = k2 (k2 / k1) ½ + Co] d. Ct = [k2(k1 / k1)1/4 = k2 (k1 / k2) ½ + Co] 81. Methyl alcohol condensed at 148 F is to be cooled to 100 F for storage at a rate of 10, 000gal/hr by water available at 75 F in a countercurrent heat exchanger. The over-all heat transfer coefficient is constant and estimated at 200 Btu/ft 2-hr-F. Heat exchanger annual costs including operation are estimated at $2 per ft 2 including depreciation. The cooler is to operate 5, 000 hr/year, and the value of heat utilized is estimated at $5x10 -7 per Btu. What is the estimated optimum cost of the heat exchanger if the cost for surface is $9 per ft 2? a. $4, 014 b. $4,123 c. $5,545 d. $5,123 82. What is the most economical number of effects to use in the recovery of black liquor in a paper plant if the following cost data are available? The annual fixed costs increase essentially linearly with each effect (except for condensing, feeding, and other equipment costs for multiple units which may be considered to balance each other). If a fixed amount of evaporation is to be obtained and each units to have 1, 000 ft 2 of heating surface with a service life of five years, the annual fixed costs Cf would be (using cost data of $25, 000 for a single evaporator of 5, 000 ft 2, employing the 0.6 factor, and neglecting the interest). Cf =( 1, 000) 0.6 25, 000 N ……. dollars per year 5, 000 5 where N is the number of effects. Because of the steam economy in multiple-effect operation, the direct costs for steam will decrease and the total of all annual direct costs, C D, has been established for this type of operation as CD = 65, 000 N -0.95 dollars a. 4 effects b. 2 effects c. 6 effects d. 5 effects 83. A capitalized cost for a piece of equipment has been found to be $55, 000. This cost is based on the original cost plus the present value of an indefinite number of renewals. An annual interest rate of 12% was used in determining the capitalized cost. The salvage value of the equipment at the end of the service life was estimated to be 10 years. Under these conditions, what would be the original cost of the equipment? a. $36, 861.47 b. $32, 251.47 c. $27, 291.47
d. $37, 291.47 84. A heat treating furnace is used to preheat small steel parts. The furnace uses fuel oil consisting $0.04 per gallon, with a heating value of 142, 000Btu/gal. The furnace has a firebrick lining, the outside temperature of which is 1210 F, this is to be covered with insulation costing $300 per 1000 board feet. The air temperature is 110 F. Operations is 7200 hr/yr. Conductivity is 0.028 for insulation in Btu/hr-ft 2-F. Calculate the most economical thickness of insulation. Furnace life is 8 years. Assume negligible temperature drop from insulation to air. a. 5.52 in b. 5.24 in c. 5.48 in d. 5.45 in 85. A batch inorganic chemical operations gives product C from two chemicals A and B according to the following empirical relation: C = 2.8 (AB – AC – 1.2 BC + 0.5C 2 )0.5 where A, B and C are pounds of respective components. The reaction rate is sufficiently high to be neglected, and the time to make any batch is essentially the charging and discharging time, including heating up, which totals 1 hr. If A costs $0.10 per lb and B costs $0.05 per lb, what is the ratio of B to A to give the minimum costs of raw materials per lb of product . what is the cost per lb of C? a. $4.08 b. $3.23 c. $4.23 d. $3.08 86. Based from the previous problem, what is the cost per lb of C? a. $4.08 b. $3.23 c. $4.23 d. $3.08 87. Seven million pounds of water per year is to be obtained from 8 percent solids slurry to be filtered on a leaf filter to produce a cake containing 40 percent solids. The area of the filter is 200 ft 2. Tests show a value of 2 x 104 for k in pound units. The cake is not washed. The dumping and cleaning time is 3 hr and costs $39 each cycle. Filtration costs are $14 per hr, and inventory charges maybe neglected. What is the cycle time for minimum costs? a. 0.014 hrs b. 0.015 hrs c. 1 hr d. 0.25 hrs 88. Referring to the previous problem, calculate the cycle time for maximum production. a. 3.04 hrs. b. 2.04 hrs c. 3.52 hrs d. 1.02 hrs For nos. 89- 91. In processing 500 ton/day of ore assaying 50% mineral, 300 tons of concentrate containing 66.7 % are obtained at a cost of sales ( all fixed operating cost are excluded) of $15 per ton concentrate. An investment of $200, 000 of concentrate that will assay 71% mineral. If the plant operates 200 days/year, equipment must pay out in 5 years with interest at 15% and no salvage value and no additional labor or repair costs need to be considered. 89. Based on the stated problem above, calculate the additional cost per ton of concentrat e for capital recovery on the new equipment.
90.
91.
92.
93.
94.
95.
96.
a. $74.6 b. $56.2 c. $15.2 d. $14.0 Determine the selling price in dollars per ton (100% mineral basis) required for which the cost of the n new equipment is justified. a. $ 123.12 b. $ 123.50 c. $ 263.25 d. $ 263.13 What is the % increase in recovery and rejection for the new process based on mineral and gauged? a. 42.567% b. 22.224% c. 52.677% d. 12.235% A ties on a plant railroad sliding are to be replaced. Untreated ties consisting $ 2.50 installed have a life of 7 years. If created ties have a life of 10 years, what is the maximum installed cost that should be paid for treated ties if money is worth 8 percent? a. $3.15 b. $3.10 c. $3.25 d. $3.20 Powdered coal having a heating value of 13, 500 Btu/ lb is to be compared with fuel oil worth $2.00 per bbl (42 gal) having a heating value of 130,000 Btu/gal as a source of fuel in the processing plant. If the efficiency of the conversion of the fuel is 64% for coal and 72 % for oil, with all other costs being equal, what is the maximum allowable selling price for coal per ton? a. $15.13 / ton b. $11.13 / ton c. $21.13 / ton d. $25.13 / ton A steam boiler is purchased on the basis of guaranteed performance. However, initial tests indicate that the opening (income) cost will be P400 more per year than guaranteed. If the expected life is 25 years and money is worth 10 %, what deduction from the purchase price would compensate the buyer for the additional operating cost? a. 3, 635.82 pesos b. 3, 660.82 pesos c. 3,560.82 pesos d. 3, 630.82 pesos If the sum of P12, 000 is deposited in an account earning interest at the rate of 9% compounded quarterly, what will it become at the end of 8 years? a. 14, 857.24 b. 34, 457.24 c. 14, 527.24 d. 24, 457.24 At a certain interest rate compounded quarterly, P1, 000 will amount to P4, 500 in 15 years. What is the amount at the end of 10 years? a. 2, 125.17 pesos b. 2, 725.17 pesos c. 2, 625.17 pesos
d. 2, 845.17 pesos 97. A one bagger concrete mixer can be purchased with a down payment of P8, 000 and equal installments of P600 each paid at the end of every month for the next 12 months. If the money is worth 12% compounded monthly, determine the equivalent cash prize of the mixer. a. 4, 753.05 pesos b. 34, 753.05 pesos c. 24, 753.05 pesos d. 14, 753.05 pesos 98. A certain company makes it the policy that for any new piece of equipment, the annual depreciation cost should not exceed 10% of the original cost at any time with no salvage or scrap value. Determine the length of service life necessary if the depreciation method use is straight line formula. a. 10 years b. 12 years c. 18 years d. 8 years 99. Solve the previous problem with the sinking fund formula at 8% a. 8 years b. 10 years c. 6 years d. 12 years 100. Determine the ordinary simple interest on $10, 000 for 9 months and 10 days if the rate of interest is 12%. a. $433.33 b. $633.33 c. $333.33 d. $933.33 Answers: 1. The purchased cost of a shell-and-tube heat exchanger (floating head and carbon-steel tubes) with 100 ft 2 of heating surface was $3,000 in 1980. What will be the purchased cost of a similar heat exchanger with 200 ft 2 of heating surface in 1980 if purchased-cost-capacity exponent is 0.60 for surface area ranging from 100-400 ft 2? GIVEN: A1 = 100 ft 2 i100-400 = 0.60 P1980 = $3, 000 i 400-1,000 = 0.81 A2 = 200 ft 2 REQUIRED: $ of 200 ft 2 in 1980 SOLUTION:
200 ft2 2 100 ft
0.6
cost200 ft2 = $3, 000
cost 200 ft2 = $4, 547.15 2. Refer to problem no. 1. If the purchased-cost-capacity exponent for this type of exchanger is 0.81 for surface areas ranging from 400-2,000 ft 2, what will be the purchased cost of a heat exchanger with1, 000 ft 2 of heating surface in 1985? GIVEN: A1 = 100 ft 2 i100-400 = 0.60
P1980 = S3,000 i 400-1,000 = 0.81 A2 = 200 ft 2 REQUIRED: $ of 1,000 ft 2 in 1985 SOLUTION: using table 3, page 163 (Cost indexes as annual average) where: 1980 = 560, and 1985 = 790
400 ft2 cost = $3,000 2 100 ft
400 300
C1985 = $9,722.91 For 500 ft 2
500 400
0.6
cost1985
C1985 = $11,649.13 For 600 ft 2
0.81
790 560
600 C1985 = $11,649.13 500
3. If the purchased cost of a shell-an d-tube heat exchanger (floating head and carbon-s teel tubes) with 100 ft 2 of heating surface was $3,000 in 1980. Note that the purchase-cost-capacity exponent is not constant over range surface area requested. What will be the purchased cost of a heat exchanger with 700 ft 2 of heating surface in 1985? GIVEN: A = 100 to 2,000 ft2 Present year = 1980 P1980 = $3,000 REQUIRED: 1985 purchase cost with 700 ft 2 SOLUTION: Using table 3, page 163 (Cost indexes as annual average) Where: 1980 = 560, and 1985 = 790 For 200 ft 2
C1985
0.60
0.81
C1985 = $13,503.00 For 700 ft 2
700 600
cost1985 = $20, 4 23.38
200 = $3, 000 100
0.81
C1985 = $9, 722.91
cost = $6, 892.19
1,000 ft2 = $6,892.19 2 400ft
0.60
C1985 = $8,181.50
0.81
C1985 = $13,503.00 C1985 = $15,298.79
4. From the preceding question, what will be the purchased cost of the heat exchanger with 800 ft of heating surface in 1985? GIVEN: A = 100 to 2,000 ft2 Present year = 1980 P1980 = $3,000 REQUIRED: 1985 purchase cost with 800 ft 2 SOLUTION: For 700 ft 2
700 600
2
0.81
C1985 = $13,503.00
790 560
C1985 = $15,298.79
C1985 = $6, 414.73 For 300 ft 2
300 200
C1985 = $6,414.73 C1985 = $8,181.50 For 400 ft 2
0.60
For 800 ft 2
800 700
0.81
C1985 = $15,298.79 C1985 = $17,046.32
5. Refer to problem no. 3. What will be the purchased cost of the heat exchanger with 1500 ft 2 of heating surface in 1985? GIVEN: A = 100 to 2,000 ft2
Present year = 1980 P1980 = $3,000 REQUIRED: 1985 purchase cost with 1500 ft 2 SOLUTION: For 800 ft 2
of the building cost will be for indoor construction, and the contractor’s fee will be 7% of the direct plant cost. All other costs are close to the average values found for typical chemical plants. On the basis of this information, estimate the total direct plant cost. GIVEN: Equipment Cost = $300,000 Contractor’s Fee = 7% of the direct plant cost
800 700
0.81
REQUIRED: Total direct plant cost SOLUTION: Purchased Equipment Purchased Installation $117,000 Instrumentation $39,000 Piping $93,000 Electrical Building Yard Improvements Service Facilities Land $18,000
C1985 = $15,298.79 C1985 = $17,046.32 For 1,000 ft
2
1,000 800
0.81
C1985 = $17,046.32 C1985 = $20,423.39 For 1,500 ft 2
1,500 C1985 = $20,423.39 1,000
0.81
C1985 = $28,363.62
13% ($300,000)
=
31% ($300,000)
=
10% ($300,000) = 29% ($300,000) = 10% ($300,000) = 55% ($300,000) = 6% ($300,000)
$30,000 $87,000 $30,000 $165,000 =
Total Direct Plant Cost = $879, 000
6. The purchase and installati on cost of some pieces of equipment are given as a function of weight rather than capacity. An example of this is the installed cost of large tanks. The 1980 cost for an installed aluminum tank weighing 100,000 lb was $390,000. For a size range from 200,000 to 1,000,000 ld, the installed cost-weight exponent for aluminum tanks is 0.93. If an aluminum tank weighing 700,000 lb is required, what is the present capital investment needed? GIVEN: W1980 = 100,000 lb i 200, 000-1,000,000 = 0.93 C1980 = $390,000 REQUIRED: C in $ for 700,000 lb SOLUTION:
200,000 C = $390,000 1,000,000
100% ($300,000)= $300,000 39% ($300,000) =
0.60
700,000 200,000
0.93
C = $1,900,000 Present Cost using table 3, page 163 (Cost indexes as annual average) where: 1980 = 560, and 1985 = 904 904 C1990 = $1, 90 0, 000 560
C1990 = $3, 060, 000 7. The purchased equipment cost for a plant which produces pentaerythritol (solid-fuel processing plant) is $300,000. The plant is to be an addition to an existing formaldehyde plant. The major part
8. The purchased equipment cost for a plant which produces pentaerythritol (solid-fuel processing plant) is $300,000. The plant is to be an addition to an existing formaldehyde plant. The major part of the building cost will be for indoor construction, and the contractor’s fee will be 7% of the direct plant cost. All other costs are close to the average values found for typical chemical plants. On the basis of this information, estimate the fixed capital investment. GIVEN: Equipment Cost = $300,000 Contractor’s Fee = 7% of the direct plant cost REQUIRED: Fixed capital investment SOLUTION: Purchased Equipment Purchased Installation $117,000 Instrumentation $39,000 Piping $93,000 Electrical Building Yard Improvements Service Facilities Land $18,000 Engineering and Supervision
100% ($300,000)= $300,000 39% ($300,000) = 13% ($300,000)
=
31% ($300,000)
=
10% ($300,000) = 29% ($300,000) = 10% ($300,000) = 55% ($300,000) = 6% ($300,000)
$30,000 $87,000 $30,000 $165,000 =
32% ($300,000)
$96,000
=
Construction Expense $102,000
34% ($300,000)
= after producing only 500,000 kg of product TPC = 0.35 + 0.20 + 0.084 = 0.6384
$198,000 Constructor’s Fee Contingencies $108,000
7% ($879,000) = 36% ($879,000)
$61,530 =
% change = 100 - 63.84 % change = 36.16%
$169,530
Fixed Capital Investment = $1, 246, 530 9. Refer to the previous problem and on the basis of this information, what is the total capital investment? GIVEN: Equipment Cost = $300,000 Contractor’s Fee = 7% of the direct plant cost Fixed capital investment = $1, 246, 530 REQUIRED: Total capital investment SOLUTION: Fixed Capital Investment $1, 246, 530 Working Capital 74% ($300,000) = $222,000
Total Capital Investment = $1,468, 530 10. The total capital investment for a chemical plant is $1,500,000 and the plant process 3M kg of product annually. The selling price of the product is $0.82/kg. Working capital amounts to 15% of the total capital investment. The investment is from company funds, and no interest is charged. Raw-material costs for the product are $0.09/kg, labor $0.08/kg, utilities $0.05/kg, and packaging $0.008/kg. Distribution costs are 5% of the total product cost. Determine the percent change in total cost. GIVEN: Total Capital Investment $1,500,000 Product 3,000,000 kg annually Selling Price $0.82/kg Working Capital 15% of the total capital Raw Materials Cost $0.09/kg Labor $0.08/kg Utilities $0.05/kg Packaging $0.008/kg Distribution Cost 5% of total capital cost REQUIRED: % change in total cost SOLUTION: TPC = FC + WC +VO TPC = 0.35 + 0.040 + 0.25 TPC = 1or 100%
x by ratio and proportion
25
11. It is desired to have a $ 9000 available from 12 years from now. If $ 5000 is available for investment at the present time, what is discrete annual rate of compound interest on the investment would be necessary to give the desired amount? GIVEN: S = $ 9000 P = $ 5000 n = 12 REQUIRED: Interest, i SOLUTION: S = P (1 + i ) n $9000 = $5000(1 + i) 12 i = 0.05 or 5.02 % 12. An original loan of $2000 was made at 6 percent simple interest per year for 4 years. At the end of this time, no interest had been paid and the loan was extended for 6 more years at a new, effective , compound-interes t rate of 8 percent per year. What is the total amount owned at the end of ten years if no intermediate payments are made? GIVEN: P = $2000 i = 6% n = 4 years REQUIRED: SC SOLUTION: SS = P (1 + in) = 2000 [1 + 0.06(4)] SS = $2480 Extended for 6 years SC = $2480 (1 + 0.08) 6 SC = $ 3,935.45
13. The original cost for a distillation tower is $24,000 and the useful life of the tower is estimated to be 8 years. The sinking fund method for determini ng the arte of depreciation is used, and the effective annual interest for the depreciation fund is 6 percent. If the scrap value of the distill ation tower is $ 4000, determine the asset value at the end of 5 years. GIVEN: n = 8 years V = $ 24,000 VS = $4000 REQUIRED: R
1.5
500,000 1,000,000
=
x = 8.84%
SOLUTION:
R
(V VS )
i (1 i)n 1
For new ieff = [1+ r/m) m-1 Ieff = [1 + 0.2/52) -1 = 0.2209 Solving for S, S = R[1+ ieff) – 1]/ieff S = $5000 [1.02209) 10-1]/0.2209
0.06 R (24000 4000) (1 0.6)5 1 R = $ 3,547.928 14. An annuity is due to being used to accumulate money. Interest is compounded at an effective annual rate of 8 %, and $1000 is deposited at the beginning of each year. What will be the total amount of annuity due be after 5 years? GIVEN: R = $1000 N = 5 yrs. I=8% REQUIRED: S5 SOLUTION: If no interest: S5 = R [ ( 1+ i) n – 1]/i S5= $5000 But with interest: S5 = 1000[(1 + 0.08) 5-1]/0.08 S5 = $ 5866.6009 Per year:
S = $5866.6009/5 = $ 1173.32
15. For the total year payments of $5000 for ten years, what will be the compound amount accumulated at the end of ten years if the payment is at the end of the year? The effective (annual) interest is 20% and payments are uniform. GIVEN: R = $5000 20% annual interest REQUIRED: S at the end of the year SOLUTION: S = R [(1+ i) n – 1]/i S = $5000[(1 +0.2) 10-1]/0.2 S = $ 129793.41
16. Referring to the previous problem, estimate the compound amount accumulated at the end of ten years, if the payment is made weekly? GIVEN: R = $5000 20% annual interest REQUIRED: S at the end of the year SOLUTION: For weekly i ≠ 0.2
S = $ 143951.4873 17. A multiple – effect evaporator is to be used for evaporating 400,000 lb of water per day from a salt solution. The total initial cost for the 1 st effect is $18,000 and each additional effect costs $15,000. The life period is estimated to be 10 years, and the scrap value at the end of the life period may be assumed to be zero. The straight-line depreciation method is used. Fixed charges minus depreciation are 15% yearly based on the first cost of the equipment. Steam cost $1.50 per 1000 lb. Annual maintenance charges are 5% of the initial equipment cost. All other cost is independent of the number of effects. The unit will operate 300 days per year. If the lb of water evaporated per pound of steam equals 0.85 x numbers of effects, determine the optimum number of effects for mi nimum annual cost. GIVEN: Initial cost = $18000 Lb of water = 400000 lb Cost of additional effect = $15000 Life period = 10 Salvage value = 0 Fixed charges = 15% of first cost Steam cost = $1.50/1000 lb Maintenance charges = 5% of first cost Operation = 300 days/year Lb of steam/lb water = 0.85 x no. of effects REQUIRED: Optimum number of effects for minimum annual cost SOLUTION: Let x = additional no. of effects Cost of equipment = 18000 + 15000x FC = 0.15 (18000 +15000x) = 2700 + 2250 x Annual maintenance = 0.05(18000+15000x) = 900 + 750x Depreciation/yr = (18000+15000x)/10 d = 1800 + 1500x lb of steam = 400000lbH 2O/day (300 days/yr) 0.85(x+1) = 1.41176 x 10 8 (x+1) 1.41176x10 Steam cost / yr = 1.50
1000
x
8
1
= 70,588.23 / (x+1)
Total Cost per year = CT = FC + Steam cost + depreciation + maintenance cost
CT = (2700 + 2250x) + (1800+1500x) + (900+750x) + (70588.23/x+1) CT = 5400 + 4500x + 70588/(x+1) 2
dC T dx
4500
A = DoL =
12
70588 (x 1) 2
Uo
2
0 = 4500(x +2x+1) – 70588 0 = 4500x 2 + 9000x – 66088
9000
1
ho
1 x w Do
kDln
Do do 2 10.75 2I t (10.75) Dln 10.75 I t 2 1 Uo 1 I t (10.75 2I t ) 2 0.03(10.75 I t )(12)
Using quadratic equation: x=
10.75 2I t (1000)
Dln
9000 44500 66088 2
2(4500) x = 2.96
Use total effect = 3
18. Determine the optimum economic thickness of insulation that should be used under the following conditions: Standard steam is being passed continuously through a steel pipe with an outside diameter of 10.75 in. The temperature of the steam is 400 F, and the steam Cost is valued at of steam= $1.80 per 1000 lb. The pipe is to be insulated with material that has a thermal conductivity of 0.03 dTAC Btu/h-ft2-F/ft. The cost of installed insulation per foot of pipe length is $4.5xI t, where It is the 900I t 458.8(10.75 2It) dI t thickness of the insulation in inches. Annual fixed charges including maintenance amount to 20% of the initial installed cost. The total length of the pipe is 1000 ft, and the average temperature of the surrounding may be taken as 70 F. Heat transfer resistance due to the steam film, scale and pipe wall are negligible. The air –film coefficient at the outside of the insulation may be assumed
1 1 I t (10.75 2I t ) 0.03(10.75 It )(12) 2 TAC = 900 I t 2.31 ( 10.75 I t )(10.75 2I t )
constant at 2.0 Btu/h-ft 2-F for all insulation thickness. GIVEN: Steam 400F do = 10.75 in Do (with insulation) = 10.75 + 2 I t Cost of steam = $1.80/1000lb Thermal conductivity = 0.03 Btu/h-ft 2-F/ft Cost of insulation/ft = $4.5 I t Annual fixed charges + maintenance = 20% of Initial cost Length = 1000 ft T surrounding = 70 F h = 2 btu/h-ft2 F REQUIRED:
(10.75 I t )(2I t ) 2 0 = 900 I t 2.31 ( 10.75 I t )(10.75 2I t ) (10.75 I t )(2I t ) 2 0.935
It = 1.4 TAC = $3057 19. A proposed chemical plant will require a fixed capital investment of $10 million. It is estimated that the working capital will amount to 25% of the total investment and annual depreciation costs are estimated to be 10 percent of the fixed capital investment. If the annual profit will be $3 million, determine the standard percent return on the total investment GIVEN: FCI = $10 M Working capital = 25% TCI Annual depreciation costs = 10% FCI Annual profit = $3 M
Cost of insulation = (4.5 I t) (1000 ft ) = 4500 It FC = 0.20 (4500 I t) FC = 500 It cost of steam = m = UAo∆T ∆T = 400 – 70 =330 F = 826 btu/lb
0.935
It = 1 TAC = $3228 = 1.5 TAC = $3054
Optimum economic thickness insulation SOLUTION:
REQUIRED: % return on investment (ROR) SOLUTION:
ROR =
TCI
10 =
But TCI = FCI + WC Let x = TCI
(x 0.10TCI) 0.8TCI
$3M
ROR =
$13.33M
%rate of return = annualprof it x100
TCI
20. If a plant will require a fixed capital investment of $10 million and the working capital will amount to 25% of the total investment and annual depreciation costs are estimated to be 10 percent of the fixed capital investment. If the annual profit will be $3 million, what is the minimum payout period (POP)? GIVEN: FCI = $10 M Working capital = 25% TCI Annual depreciation costs = 10% FCI Annual profit = $3 M REQUIRED: minimum payout period SOLUTION:
depreciabl eFCI profit dep' n yr yr
22. The information given in the previous problem, applies to conditions before income taxes. If 34% percent of all profits must be paid out for income taxes, determine the standard rate of return after taxes using the figures given in the previous problem. GIVEN: Payout period = 10 years Minimum annual return = 0.10 of fictitious expense Annual depreciation cost = 8% TCI 34% of all profits must be paid out for income taxes REQUIRED Standard rate of return after taxes SOLUTION: Profit = 0.105TCI (before taxes) Profit = 0.105TCI – 0.34(0.105TCI) (after taxes) Profit = 0.0693TCI ROR = annualprof it x100
TCI
TCI
ROR
$3M 0.1($10M) yr POP = 2.5 years 21. An annual investigation of a proposed investment has been made. The following result has been presented to management. The minimum payout period based on capital recovery using a minimum annual return of 10 percent as a fictitious expense is 10 years; annual depreciation costs amount top 8 percent of the total investment. Using this i nformation, determine the standard rate of return on the investment. GIVEN Payout period = 10 years Minimum annual return = 0.10 of fictitious expense Annual depreciation cost = 8% TCI REQUIRED Standard rate of return SOLUTION
FCI averagepro fit avedep' n yr
= 10.50%
= 0.0693TCI x100
10M
Payout period =
ROR
x100
ROR = 22.5%
=
0.85TCI
x = 0.105 TCI
x = 10 M + 0.25x x = TCI = $13.33M
POP =
= x – 0.10TCI
annualprof it
Ave. profit = annual profit – expenses
yr
= 6.93%
23. A capitalized cost for a piece of equipment has been found to be $55,000. This cost is based on the original cost plus the present value of an indefinite number of renewals. An annual interest rate of 12% was used in determining the capitalized cost. The salvage value of the equipment at the end of the service life was estimated to be 10 years. Under these conditions, what would be the original cost of the equipment? GIVEN: k = $55,000 i = 12% Vs = 0 n = 10 yrs. REQUIRED: Cv SOLUTION k = CR (1 i)
(1 i)
n
n
1
Vs
Vs = 0 10
55 000 = CR (1 0.12)
101
(1 0.12) CR = $37 291.47
k= C v
CR (1 i)
n
Current Assets
1 10
(1 0.12)
$ 20,000
Accounts Receivable
37291.47
Cv = 55 000-
Current Liabilities
Cash
1
Inventori es
Accounts Payable
6,000
MC Company
$ 2,000
15,000
MD Company
8,000
Cv =$37 291.47 Government Bonds For problems 24- 26.On Aug. 1, a concern had 10,000 lb of raw material on hand, which was purchased at a cost of $0.030 per pound. In order to build up the reserve, 8000 lb of a dditional raw material was purchased on Aug. 15 at a cost of $0.028 per pound. If none of the raw material was used until after the last purchase. 24. Determine the total cost of 12,000 lb of the raw material on an inventory or cost-of-sales account for the month of August by current average method. SOLUTION: Ave = $ (0.030 + 0.028 + 0.031) / 3
3,000 Total
Mortgage Payable
$ 44,000
Fixed Assets
5,000 Total
$ 15,000
Total
$ 50,000
Total
$ 2,000
Total Equities
$ 67,000
Stockholder’s Equity
Machinery & Equipment
18,000
Furniture & Fixtures
Common Stocks Sold
5,000 Total
Total Assets
$ 23,000 $ 67,000
50,000
Surplus
2,000
Ave = $ 0.0297/ lb Total Assets: $ 67, 000 25. Determine the total cost of 12,000 lb of the raw material on an inventory or cost-of-sales account for the month of August by “FiFo” method. SOLUTION: 10,000 lb = $ 0.030
28. Determine the total current assets of the AMD Food Corporation. ANSWER: From the balance sheet; Total Current Asset = $ 44,000
other 2000 lb cost = $ 0.028/lb 26. Determine the total cost of 12,000 lb of the raw material on an inventory or cost-of-sales account for the month of August by lifo method. ANSWER: recent price = $ 0.031/lb For question nos. 27 – 31. The following are the data gathered from the AMD Food Corporation: Cash
$20,000
Accounts payable: B Company C Company
2,000 8,000
Accounts receivable Inventories Mortgage payable Common stock sold Machinery and equipment (at present value) Furniture and fixtures (at present value) Government bonds Surplus
6,000 15,000 5,000 50,000 18,000 5,000 3,000 2,000
27. From the data given above, determine the total asset of the AMD Food Corporation. ANSWER: BALANCE SHEET ASSETS
EQUITIES
29. What is the total amount of the Current liabilities of the AMD Corporation? ANSWER: From the balance sheet; Total Current Liabilities = $ 15,000 30. What is the total amount of the Fi xed Assets of the AMD Food Corporation? ANSWER: From the balance sheet; Total Fixed Assets = $ 23,000 During the month of October, the following information was obtained in the AC antifreeze retailing company: Salaries Delivery expenses Rent Sales Antifreeze available for sale during October (at cost) Antifreeze inventory on Oct. 31 (at cost) Other expenses Earned surplus before income taxes as of Sept.30
$ 3,000 700 400 15,100 20,000 11,000 1,200 800
31. Prepare an income statement for the month of October to determine the net income is for the month of October. ANSWER: AC Antifreeze Retailing Company Income Statement As of October Income
Sales
$
Current Assets Accounts Receivable Cash in Bank
15,100 Antifreeze available for sale Earned Surplus before income
20,200
1,700
3,100
800 Total Gross Income Deductions Antifreeze inventory on Oct.31 Salaries 3,000 Delivery Rent 400 Other Expenses 1,200
Government Bonds Inventory Prepaid Rent
36,100
11,600
3,000 1,600
300
700
Total 9,700
Net Income
Long-term Debts Current Liabilities Departments due within 1 year
$
19,200
1,600
1,000
32. From the data above, determine also the total gross income of the AC Antifreeze Company. ANSWER: From the income statement; Gross Income = $36,100
Accounts Payable Social Security Taxes payable
2,300 240 Total
5,140
The following information applies to MADSteel Company on a given date: Long-term debts Debts due within 1 year Accounts payable Machinery and equipment (at cost) Cash in bank Prepaid rent Government bonds Social security taxes payable Reserve for depreciation Reserve for expansion Inventory Accounts receivable
34. Determine the current asset for MADSteel Company
$ 1,600 1,000 2,300 10,000 3,100 300 3,000 240 600 1,200 1,600 1,700
ANSWER: Current Asset = $9, 700 35. Determine the current liabilities of MADSteel Company ANSWER: Current Liabilities= $3, 540 36. Determine the Quick ratio for MADSteel Company SOLUTION:
Quick Ratio = Current assets – Inventory Current liabilities = $9,700 – $1, 600 $3, 540 Quick Ratio = 2.29
37. Determine the current ratio of MADSteel Company
33. Determine the cash asset for the MADSteel Company at the given date.
SOLUTION:
ANSWER: Cash Asset = $6,100
Current ratio =
current assets Current liabilities
Cash Assets Cash in Bank Government Bonds
$ 3,100 3,000 Total
6,100
= $9, 700 $3, 540
Current ratio = 2.74 38. A reactor of special design is the major item of equipment in a small chemical plant. The initial cost of a completely installed reactor is $60,000, and the salvage value at the end of the useful life is estimated to be $10,000. Excluding depreciation costs for the reactor, the total annual expenses for the plant are $100,000. How many years of useful life should be estimated for the reactor if 12 % of the total annual expenses for the plant are due to the cost for the reactor depreciation? The straight-line method for determining depreciation should be used. GIVEN: Vo = $60,000 Vs = $10,000 Total expenses annually exc. Depreciation = $ 100,000 i = 0.12 REQUIRED: n using SLM SOLUTION: d=
Vo
Vs
40. A piece of equipment is originally costing $40,000 was put into use 12 years ago. At the time the equipment was put into use, the service life was estimated to be 20 years and the salvage and scrap value at the end of the service life were assumed to be zero. On this basis, the straight-line depreciation fund was set up. The equipment can now be sold for $10,000, and a more advanced model can be installed for $55,000. Assuming the depreciation fund is available for use, how much new capital must be supplied to make the purchase? GIVEN: Vo = $40,000 Vs = 0 a = 12 n = 20 REQUIRED: New Capital needed SOLUTION:
d=
n
x = total annual expenses inc dep’n/yr
x = $ 13,636.36 = d
60 ,000
d=
Vs
n
40,000
0
20
$2,000 yr
Va = Vo – ad
x = 100,000 + 0.12x
13,636.36 =
Vo
d=
10 ,000
n
n = 3.67 n = 4 years 39. The initial installed cost for a new piece of equipment is $10,000, and its scrap value at the end of its useful life is estimated to be $2,000. The useful life is estimated to be 10 years. After the equipment has been in use for 4 years, it is sold for $7,000. The company which originally owned the equipment employs the straight-line method for determining depreciation costs. If the company had used an alternative method for determining depreciation cost, the asset (or book) value for the piece of equipment at the end of 4 years would have been $5240. The total income-tax rate for the company is 34% of all gross earnings. Capital-gains taxes amount to 34% of the gain. How much net saving after taxes would the company have achieved by using the alternative (in this case, reducing-balance) depreciation method instead of the straight-line depreciation method? GIVEN: Vo = $10,000 Vs = $2,000 V4 DBM = $5,240 n = 10 years V4 SLM = $7,000 Tax = 0.34 REQUIRED: Net saving after taxes SOLUTION: SLM: $7,000 – 7,000 (0.34) = $4,620 DDB: $5,240 – 5,240 (0.34) = $3,458.40 Net saving after taxes = $4,620 - $3,458.40 = $1161.60
Va = 40,000 - 8 ( 2000 ) Va = $24,000 New Capital = 55,000 – (24,000 + 10,000) New Capital = $21,000 41. The original investment for an asset was $10,000, and the asset was assumed to have a service life of 12 years with $2,000 salvage value at the end of the service life. After the asset has been in use for 5 years, the remaining service life and the final salvage value are reestimated at 10 years and $ 1,000, respectively. Under these conditions, what is the depreciation cost during the sixth year of the total life is straight-line depreciation is used? GIVEN: Vo = $10,000 Vs = $2,000 a=5 n = 12 REQUIRED: d during the sixth day SOLUTION: d= d=
Vo Vs n
10,000
d=
2,000
12
$666.67 yr
Va = Vo – ad
V
5
= 10,000 - 5 (666.67) = $6666.65
After 5 years of use:
6666 .65 1,000 10
da =
da = $566.65/yr
Vo = $40,000 Vs = 0 n = 10 REQUIRED: da & % paid off SOLUTION: SYDM
A piece of equipment having a negligible salvage and scrap value is estimated to have a service life of 10 years. The original cost of the equipment was $40,000. Determine the following: 42. Based on the above data, determine the depreciation charge for the fifth year if doubledeclining balance depreciation is used. GIVEN: Vo = $40,000 Vs = 0 n = 10 REQUIRED: da & % paid off Solution: da = 40,000 ( 1-2/10) 5-1 2/10
V1 = 40,000 – 4,363.64 = $ 35,636.36 V2 = 35,636.36 - 4,363.64 = $ 31,272.72 V3 = 31,272.72 - 4,363.64 = $ 26,909.08 V4 = 26,909.08 - 4,363.64 = $ 22,545.94 V5 = 22,545.94 - 4,363.64 = $ 18,181.8 % paid off =
18,181.80 40,000
= 45.45%
% = 45.45
da = $ 3,276.80 43. The depreciation charge for the fifth year if sum-of-the-years-digits depreciation is used. GIVEN: Vo = $40,000 Vs = 0 n = 10 REQUIRED: da & % paid off SOLUTION: SYDM 10
da = 2
5
10 10
1
1
( 40,000)
da = $ 4,363.64 44. The percent of the original investment paid off in the first half of the service life using the double-declining balance method. GIVEN: Vo = $40,000 Vs = 0 n = 10 REQUIRED: da & % paid off SOLUTION:
V = 40,000 1 5
5
% paid off =
2
= $13,107.20
10
13,107.2 40,000
= 32.768%
% = 32.8 45. The percent of the original investment paid off in the first half of the service life using the sumof-the-years-digits method. GIVEN:
46. The original cost of the property is $30,000, and it is depreciated by a 6 percent sinking-fund method. What is the annual depreciation charge if the book value of the property after 10 years is the same as if it had depreciated at $2,500/year by the straight-line method? GIVEN: Vo = $30,000 i = 0.06 n = 10 years REQUIRED: da = $2,500/yr da using SFM if Va SLM = Va SFM SOLUTION: Using SLM: Va = 30,000 – 10 (2,500) Va = $ 5,000 Using SFM: 30,000 –5,000 = R (1+ 0.06) 10 - 1 / (0.06) R = $ 1,896.70 / 10 yr da = $ 189.67/yr 47. A concern has a total income of $1 million/year, and all expenses except depreciation amount to $600,000/year. At the start of the first year of the concern’s operation, a composite account of all depreciable items show a value of $850,000, and the overall service life is estimated to be 20 years. The total salvage value at the end of the service life is estimated to be $50,000. Thirty percent of all profits before taxed must be paid out as income taxes. What would be the reduction in income-tax charges for the first year of operation if the sum-of-the-years-digits methods were used for depreciation accounting instead of the straight-line method? GIVEN: Total income = $ 1 million/yr Annual expenses exc dep’n = $600,000 For the 1 st yr of operation: y = $ 850,000 n = 20 yr Vs = $ 50,000 Income tax rate = 0.45
REQUIRED: Reduction in income taxes if SYDM is used instead SLM SOLUTION: Using SYDM: da = 2 ( 850,000 –50,000 ) = $ 76,190.48 Gross Earnings = Total income – total expense GE = $ 1,000,000 – ( $600,000 + 76,190.48 ) = $ 323,809.52 Net Profit = $ 323,809.52 (1 - 0.45) = $ 178,095.24 Income tax = 0.45 (323,809.52) = $ 145,714.28 Using SLM: d= d=
Vo
percent sinking-fund method were used to determine depreciation costs instead of straight-line method? GIVEN: Vo = $50,000,000 Vs = 0 n = 10 REQUIRED: Annual profits increase SOLUTION: Using SFM:
Vs
1 1
Va = $28034.35
n
Using SLM
850 ,000
Vo
50 ,000
d=
20 d=
5 1 0.05 !0 1 0.05
Va = 50,000 - (50,000-0)
$40,000 yr
Vs
n
d = $ 25,000 inc = $28034.3 - $25,000 inc = $ 3.034.35
Basis: 1 yr GE = $ 1,000,000 – (600,000 + 40,000) = $ 360,000 Net Profit = $ 360,000 (1 - 0.45) = $ 198,000 Income Tax = 0.45 ($ 360,000) = $ 162,000 Reduction in net profit = $198,000 - $178,095.24 = $19,904.76 Reduction in income tax = $162,000 - $ 145.714.28 = $16,285.72 Reduction in income tax = $16,285.72 48. The total value of anew plant is $2 million. A certificate of necessity has been obtained permitting a write-off of 60 percent of the initial value ay 5 years. The balance of the plant requires a write-off period of 15 years. Using the straight-line method and assuming negligible salvage and scrap value, determine the total depreciation cost during the first year. GIVEN: Vo = $2,000,000 Vs = 0 n = 15 a=5 Vs = negligible REQUIRED: d after 1 year SOLUTION: Vo new = 2,000,000 – 2,000,000 ( 0.6) = $ 800,000 d= d=
Vo
51. In order to make it worthwhile to purchase a new piece of equipment, the annual depreciation costs for the equipment cannot exceed $3,000 at any time. The original cost of the equipment is $30,000, and it has a zero salvage and scrap value. Determine the length of service life necessary if the equipment is depreciated by the sum=of-the-years-digits method by the straight-line method. GIVEN: da = $ 3,000 Vo = $30,000 Vs = 0 REQUIRED: n SOLUTION: Using SYDM: a=1 da = 2 ( n-1+1)/ n (n+1) x ( Vo –Vs ) 3000n2 + 3000 n = 60000n n = 19 years 52. Referring to the previous number, determine the length of service life necessary if the equipment is depreciated by the straight-line method. GIVEN: da = $ 3,000 Vo = $30,000 Vs = 0 REQUIRED: n SOLUTION: Using SLM:
a=1 d=
Vo
Vs
n
Vs
3000 = 30000n
n
80,000
n= 10 years
0
15
d = $ 53, 333. 33/ yr 50. A profit-producing property has an initial value of $50,000, a service life of 10 years, and zero salvage and scraps value. By how much would annual profits before taxes be increased if a 5
A materials-testing machine was purchased for $20,000 and was to be used for 5 years with an expected residual salvage value of $5,000. Graph the annual depreciation charges and year-end book values obtained by using: 53. By using Straight-line depreciation GIVEN:
Vo = $20,000 Vs = $5,000
Vo = $20,000 Vs = $5,000
n=5 REQUIRED: da & Va SOLUTION:
n=5 REQUIRED: da & Va SOLUTION: DDBM
SLM d= d=
Vo
Vs = 2 5,000 = 0.5 20,000 Vo
Vs
f = 2
n
20,000
5,000
5
= $ 3,000 Va
Va1 = Vo – ad = $17,000 Va2 = $14,000 Va3 = $11,000 Va4 = $ 8,000 Va5 = $ 5,000 54. By using Sum-of-digits depreciation GIVEN: Vo = $20,000 Vs = $5,000 n=5 REQUIRED: da & Va SOLUTION: n
d=
n
(Vo-Vs) t
nt = 15 d1 =
5 15 (20,000-5, 000) = $ 5.000
d2 = $ 4,000 d3 = $ 3,000 d4 = $ 2,000 d5 = $ 1,000 Va1 = Vo – d = $ 15,000 Va2 = $ 11,000 Va3 = $ 8,000 Va4 = $ 6,000 Va5 = $ 5,000 55. By using Double-declining balance depreciation GIVEN:
1
= Vo 1
f
a
= 20,000
1
1
0.5 = $ 10,000
Va2 = $ 5,000 Va3 = $ 2,500 Va4 = $ 1,250 Va5 = $ 625 56. An asset with an original cost of $10,000 and no salvage value has a depreciation charge of $2381 during its second year of service when depreciated by the sum-of-digits method. What is its expected useful life? GIVEN: Vo = 120,000 Vs = 0 a=2 da = $2381/yr REQUIRED: n SOLUTION: n = 2(n-a+1)/(n (n+1) x ( Vo-Vs) 2381n2 + 2381n = 2n –2 (10,000) 2381 n2 - 17619n + 20,000 = 0 By quadratic formula: n = 5.9998 n = 6 years 57. An electronic balance costs P90, 000 and has an estimated salvage value of P8, 000 at the end of its 10 years lifetime. What would be the book value after 3 years, using straight-line method in solving for the depreciation? GIVEN: Co = P90, 000 CL = P8, 000 L = 10 n=3 REQUIRED: d using SLM SOLUTION: C
d= d=
D
3
90,000
O
C
L
L
8,000
10
= P 8,200
= n(d) = 3(8,200) = P24,600
C = C - D = 90,000 - 24,000 O 3 3
C = P 65,200 3
A broadcasting corporation purchased equipment for P53, 000 and paid P1, 5000 for freight and delivery charges top the job sites. The equipment has a normal life of 10 year with a trade-in value of P5, 000 against the purchase of anew equipment at the end of the life. 58. referring to the problem above, determine the annual depreciation by straight-line method. GIVEN: Co = P53, 000 + 1,500 = 54,500 CL = P5, 000 L = 10 REQUIRED: da using SLM SOLUTION: C
d= d=
54,500
O
C
SOLUTION:
CO
d=
F A
CL
=
, 12% , 16 D
=
F
P54,000 42.7533
= P 1,263
A,6.5%,10 F
12
=d
A,
12 % , 16 1263 (24.1331)
d12 = P30, 480
L
L
60,000 6,000
C12 = Co - d 12 = 60,000 – 30,480
5,000
C12= P29, 520
10
d = P4, 950 59. From the preceding number, determine annual depreciation by sinking fund method. Assuming interest 6 ½% compounded annually. GIVEN: Co = P53,000 + 1,500 = 54,500 CL = P5, 000 L = 10 REQUIRED: da using SFM SOLUTION: P49,500 CO CL d= = 54,500 5,000 = 13.3846 F F , 6.5% , 10 A,6.5%,10 A d= P 3,668 A firm brought equipment for P56, 000. Other expenses including installation amounted to P4, 000. The equipment is expected to have a life of 16 years with a salvage value of 10% of the original cost. 60. Determine the book value at the end of 12 years by SLM: GIVEN: Co = P56,000 + 4,000 = P60,000 CL = 0.1 C o L = 10 i = 0.12 REQUIRED: C12 using SLM SOLUTION: C
d=
O
C
L
L
d12 = P3, 375 (12) = P 40,500 C12 = Co - d 12 = 60,000 – 40,500 C12= P19, 500 61. Determine the book value at the end of 12 years by SFM: GIVEN: Co = P56, 000 + 4,000 = P60, 000 C L = 0.1 C o L = 10 i = 0.12 REQUIRED: C12 using SFM
For numbers 62 – 64. A certain type of machine losses 10% of its value each year. The machine cost P2, 000 originally. Make cut a schedule showing the following: 62. By yearly depreciation GIVEN: Co = P2, 000 dep’n = 10% REQUIRED: C5 SOLUTION: Year Book value at value the beginning 1
Dep’n (10%)
Total dep’n
Book
at the end
P 2,000
P 200
P 200
P 1,800 2
1,800
180
380
1,620
162
542
1,458
145.8
687.8
1,620 3 1,458 4 1,312 5
1,312.20
63. The total depreciation after 5 years. GIVEN: Co = P2, 000 ANSWER: P819.12
131.22
819.12
1,180.98
dep’n=10%REQUIRED: C5
64. Estimate the book value at the end year for 5 years. Answer from the above table: P1, 180.98 65. Determine the rate of depreciation, the total depreciation up to the end of 8 th year and book value at the end of 8 years for an asset that costs P15,000 new and has an estimated scrap value of P2,000 at the end of 10 years by DBM. GIVEN:
Co = P15, 000 CL = P2, 000 REQUIRED: d8 & V8 SOLUTION: k=1-
n
Cs=
L
CL = C O
10
2,000 15,000 = 0.1825 or 18.25%
C8 = co
2 L
8
1
2
L
= 15,000
2 =
10
2
10
= P2, 517
D9 = Co - C 8 = 15,000 – 2,517 D9 = P 12,483
5
=
n Co
CL =
L
5 220 ,000 20 ,000
10
= P100, 000
C5 = Co – D5 = 220,000 – 100,000 = P 120,000 C5 = P 120,000 68. Refer to the previous problem. What is the book value of the calciner after 5 years of use? Assume a scrap value of P22, 000 for textbook for DBM. GIVEN: Co = P220, 000 L = 10 n=5 REQUIRED: D5 & SOLUTION:
C5
69. Refer to the previous problem. What is the book value of the calciner after 5 years of use? Assume a scrap value of P20, 000 for textbook for DDBM. GIVEN: Co = P220, 000 L = 10 n=5 REQUIRED: D5 & C5 SOLUTION: 5
5
2 2 C = Co 1 L = P220,000 1 10 5
Cs = P 72, 090
Year
67. Mr. Dim bought a calciner for P220, 000 and used it for 10 years, the life span of the equipment. What is the book value of the calciner after 5 years of use? Assume a scrap value of P20, 000 for SLM; GIVEN: Co = P220, 000 L = 10 n=5 REQUIRED: D5 & C5 SOLUTION: SLM
D
(220,000( 22,000/220,000)^(5/10)= P 69,570
70. A structure costs P12, 000 new. It is estimated to have a life of 5 years with a salvage value at the end of life of P1, 000. Determine the book value at the end of three years. GIVEN: Co= P12, 000 C L =P1, 000 L=5 REQUIRED: Va SOLUTION: Co - Cl = 12,000 - 1,000 = P 11,000
= 0.2 or 20 % 8
1
Cs = P 69, 570.00
C9 = Co (1 – k) 8 = P 2,992 D8 = Co - C 8 = 15,000 – 2,992 D8 = P 12,008 66. Determine the rate of depreciation, the total depreciation up to the end of 8 th year and book value at the end of 8 years for an asset that costs P15, 000 new and has an estimated scrap value of P2, 000 at the end of 10 years by DDBM. GIVEN: Co = P15, 000 CL = P2, 000 d8 & V8 REQUIRED: SOLUTION:
d = R
CL L Co = Co
1 2 3 4 5
Year in reverse order
Dep’n during the year
5 4 3 2 1
5/15 (11,000) = 3,664 4/15 (11,000) = 2,933 15 (11,000) = 2,200 2/15 (11,000) = 1,467 1/15 (11,000) = 733
Book value P 8,333 5,400 3,200 1,733 1,000
71. Operator A produces 120 spindle/hr on a lathe. His hourly rate is $1.80. Operator B, using an identical lathe, is able to produce 150 identically units/hr. The overhead charge for a lathe is fixed at $2.50/hr. Determine operator B’s hourly rate so that his cost per piece is identical to A’s. SOLUTION: Cost per unit for operator A = 1.80 + 2.50 = $0.0358 120 Let x = hourly rate of operator B. Then; x + 2.50 = $0.0358 150 x = $2.88 72. In a type 1 warehouse, initial cost will be $24, 000.This warehouse has adequate capacity for the near future, but 12 years from now an addition will be required that costs $15, 000. A type 2 warehouse costs $34, 000. This type has the same capacity as the type 1 warehouse with its addition. What will be the present cost of the type 1 warehouse? Which of these should be built, assuming that depreciation is negligible and that the interest rate is 7%? GIVEN: Present cost = $15, 000
REQUIRED: SOLUTION:
Present worth factor, i = 7% n = 12 present cost of warehouse 1
4 i = 0.051
P = S/ (1+i) n = 15, 000 x 0.444 = $6660
Six equal payments of $3, 000 are required; their worth at the beginning of the 16 th year is P = R [ ( 1+ i ) n – 1] / i (1 + i ) n = 3000 [ (1+0.051) 6 – 1] / 0.051 ( 1+0.051) 6 P = $15, 210
the present cost of type 1 is seen to be 24, 000 + 6660 = $30, 660. Since this is smaller than the type 2 building, it is more economical to build 73. Determine the equal (year-end) payments that will be available for the next four years if we invest $4, 000 at 6%. SOLUTION: R = $4, 000 0.06 (1 + 0.06) 4 = $4, 000 (0.28859) (1 + 0.06) 4 - 1 R = $1154.36 74. A new snow removal machine costs $50, 000. The new machine will operate at a reputed savings of $400 per day over the present equipment in terms of time and efficiency. If interest is at 5% and the machine’s life is assumed to be 10 years with zero salvage, how many days per year must the machine be used to make the investment economical? SOLUTION: Assume straight-line depreciation and no salvage value. Annual depreciation = 50, 000 – 0 = $5, 000 10 Average annual unearned interest = ½ [50, 000 x 0.05 + 50, 000 (0.05)] = $1375 10 Annual cost = $6375
77. The engineer has 15 years to accumulate this fund. His quarterly deposits are determined by u sing the sinking factor. There are 15 x 4 or 60 interest payments to be made at a quarterly interest rate of 0.0125 SOLUTION: R = Si/[(1+i)n – 1 ] Using n = 60 R = $15, 210 ( 1/88.5745) R = $172.00 where 88.5745 represents the amount of annuity. 78. A low carbon steel machine part, costing $350 installed, lasts 6 years when operating in a corrosive atmosphere. An identically shaped part, but treated for corrosion resistance, would cost $650 installed. How long would the corrosion resistance part have to last to be at least as good investment as the untreated part? Assume money is worth 7%. SOLUTION: R=P [ i(1+i)n ] [ (1+i)n - 1] = 350 x 0.2098 R = $73.50 per year required $73.50 = $650 x c.r.f. c.r.f. = 73.50/ 650.00 = 0.113
To invest in the machine, the yearly savings must at least be equal to $6375. The number of days m the machine must be used is therefore
for
n = 14 n=x
crf = 0.114 crf =
0.113 m = 6375 / 400 = 15.9 or 16 days
n = 15
crf = 0.109
crf =
0.109 75. Based on the sinking fund method and using the data in the previous problem, what number of days must the machine be used if the amount to be accumulated in 10 years is $50, 000? SOLUTION: R = S i / [(1+i)n – 1 ] For i = 5% and n = 10
Differenc e = 15 -14 = 1.00
0.005
0.004
0.004/0.005 = (15 – x)/ 1.00 x = 14.20 years
i/[(1+i)n – 1 ] = 0.0795 therefore
R = 50, 000 x 0.0795 R = $3975
m = 3975/400 = 9.9 or 10 days 76. A consulting engineer decides to set up an educational fund for his son that will provide $3000 per year for 6 years starting in 16 years. The best interest rate he can expect to get is 5% compounded quarterly. He wants to accumulate the necessary capital by making quarterly deposits until his son starts college. What will be his needed quarterly deposit? SOLUTION: i = (1+ 0.05 )4 - 1
79. The total cost of a cast product consists of (1) the raw material cost that is directly proportional to the weight, of the casting, (2) the machining cost that varies inversely as the weight, and (3) overhead cost that remains constant per unit produced regardless of weight. Find the weight giving the minimum cost per casting. SOLUTION: Let Ct = total cost Cw = cost based on weight Cm = machining cost Co = overhead unit cost W = weight
Cw = k1W (direct proportion of raw material cost to weight) Cm = k2/W (inverse proportion of machining cost to weight) Co = Co (constant value)
which may be considered to balance each other). If a fixed amount of evaporation is to be obtained and each units to have 1, 000 ft 2 of heating surface with a service life of five years, the annual fixed costs Cf would be (using cost data of $25, 000 for a single evaporator of 5, 000 ft 2, employing the 0.6 factor, and neglecting the interest).
The minimum total cost can be determined by differentiating cost with respect to weight. dCt/dW = k 1 – k2/W
Cf =( 1, 000) 0.6 25, 000 N 5, 000 5
…….
dollars per year
the minimum cost can be ascertained by equating the right hand side of the above eqn to 0. W = ($ x lb) 1/2 = lb ( $/lb)1/2 Therefore, minimum cost occurs when: W = (k2 /k1)1/2 lb
where N is the number of effects. Because of the steam economy in multiple-effect operation, the direct costs for steam will decrease and the total of all annual direct costs, C D, has been established for this type of operation as CD = 65, 000 N -0.95 dollars SOLUTION: Ct = 1, 900 N = 65, 000 N -0.95 dCt = 1, 900 – 61, 800 N -0.95 = 0 dN N = 5.95 or 6
80. Based on the previous problem, what is the minimum total cost? SOLUTION: Ct = [k1(k2 / k1)1/2 = k2 (k1 / k2) ½ + Co]
Differentiating:
81. Methyl alcohol condensed at 148 F is to be cooled to 100 F for storage at a rate of 10, 000gal/hr by water available at 75 F in a countercurrent heat exchanger. The over-all heat transfer coefficient is constant and estimated at 200 Btu/ft 2-hr-F. Heat exchanger annual costs including operation are estimated at $2 per ft 2 including depreciation. The cooler is to operate 5, 000 hr/year, and the value of heat utilized is estimated at $5x10 -7 per Btu. What is the estimated optimum cost of the heat exchanger if the cost for surface is $9 per ft 2?
83. A capitalized cost for a piece of equipment has been found to be $55, 000. This cost is based on the original cost plus the present value of an indefinite number of renewals. An annual interest rate of 12% was used in determining the capitalized cost. The salvage value of the equipment at the end of the service life was estimated to be 10 years. Under these conditions, what would be the original cost of the equipment? k = $55, 000 i = 12% Vs= 0 N = 10 yrs REQUIRED : Cv SOLUTION: k = CR ( 1 - i) n + Vs (1 + i ) n – 1 55000 = CR ( 1 – 0.12)10 (1 + 0.12 ) 10 – 1 CR = $37, 291.47 GIVEN:
SOLUTION: Since the outlet temperature is fixed; ∆t is fixed at 100 – 75 = 25 F Hourly cost:
2
= $4x10-4 per (ft2-hr) 5, 000
∆t1 = (M/ R) (T1 – t1) = 4/5 x 103 (148 – 75) = 11.7 F U x 25 200 x 25 Outlet temp. for water : 148 – 11.7 = 136 F cp = 0.5 ρ methanol = 0.79g/cc q = 10, 000 x 8.33 x 0.79 (0.5)(148 -100) = 1, 580, 000 Btu/ hr ∆tm = 25 – 11.7 = 17.7 ln (25/11.7) q = UA∆t A = 1, 580, 000 200 x 17.7 A = 446 ft2 446 x 9 = $4, 014 Estimated optimum cost: 82. What is the most economical number of effects to use i n the recovery of black liquor in a paper plant if the following cost data are available? The annual fixed costs increase essentially linearly with each effect (except for condensing, feeding, and other equipment costs for multiple units
k = Cv +
CR (1 + 0.12 ) 10 – 1 Cv = $37, 291.47 84. A heat treating furnace is used to preheat small steel parts. The furnace uses fuel oil consisting $0.04 per gallon, with a heating value of 142, 000Btu/gal. The furnace has a firebrick lining, the outside temperature of which is 1210 F, this is to be covered with insulation costing $300 per 1000 board feet. The air temperature is 110 F. Operations is 7200 hr/yr. Conductivity is 0.028 for insulation in Btu/hr-ft 2-F. Calculate the most economical thickness of insulation. Furnace life is 8 years. Assume negligible temperature drop from insulation to air. GIVEN: Heat treating furnace T1 = 1210 F T2 = 110 F Cost of fuel oil $ = 0.04 / gal Heating value = 142000 Btu/gal Cost of insulation = $300/1000bdft
Operation = 7200 hrs/yr k = 0.028 Btu REQUIRED: Most economical thickness of insulation SOLUTION: Basis: A = 1 ft 1 q = - kA∆T = - (0.028)(1)(110-1210) x q = 30.8 x cost of fuel = 30.8 x 1gal/142000 Btu ($0.04/gal)(7200hr/yr x cost of fuel = 0.06246705 x cost of insulation = 0.06246705 + 0.3 x x dTc = -0.06246705 + 0.3 = 0 2 x x = 5.48 in 85. A batch inorganic chemical operations gives product C from two chemicals A and B according to the following empirical relation: C = 2.8 (AB – AC – 1.2 BC + 0.5C 2 )0.5 where A, B and C are pounds of respective components. The reaction rate is sufficiently high to be neglected, and the time to make any batch is essentially the charging and discharging time, including heating up, which totals 1 hr. If A costs $0.10 per lb and B costs $0.05 per lb, what is the ratio of B to A to give the minimum costs of raw materials per lb of product . what is the cost per lb of C? GIVEN: Reaction: A + B C 1 hr – time per batch A = $0.10/lb B = $ 0.05/ lb REQUIRED: minimum cost of raw materials / lb of product SOLUTION: Ratio of B: A = 0.5 Assume: 1 LB OF a AND 1LB OF b are used in the making of C C = 2.8 ( 1 – C – 1.2C + 0.5C2 ) 0.5 dC = 1.4 ( 1 – C – 1.2C + 0.5 C 2) 0.5
GIVEN: Reaction: A + B C 1 hr – time per batch A = $0.10/lb B = $ 0.05/ lb REQUIRED: cost per lb of C SOLUTION: By ratio and proportion: If A = $0.1 /lb and B = $0.05/lb C = $3.08/ lb 87. Seven million pounds of water per year is to be obtained from 8 percent solids slurry to be filtered on a leaf filter to produce a cake containing 40 percent solids. The area of the filter is 200 ft2. Tests show a value of 2 x 104 for k in pound units. The cake is not washed. The dumping and cleaning time is 3 hr and costs $39 each cycle. Filtration costs are $14 per hr, and inventory charges maybe neglected. What is the cycle time for minimum costs? GIVEN: 7000000 lbwater/yr – can obtained from 8% slurry A filter = 200 ft 2 Dumping and cleaning time = 3 hr Filtration costs = $14/hr REQUIRED: cycle time for minimum cost SOLUTION: Q = A ( k θf) 0.5 = 200 ( 2x10 4)(θf)0.5 7, 000, 000 lb/ yr is also equal to 799.08 lb/hr 799.08 = 200 ( 2x10 4)(θf)0.5 θf = 0.014 hrs
88. Referring to the previous problem, calculate the cycle time for maximum production. GIVEN: 7000000 lbwater/yr – can obtained from 8% slurry A filter = 200 ft 2 Dumping and cleaning time = 3 hr Filtration costs = $14/hr θf = 0.014 hrs REQUIRED: cycle time for maximum production SOLUTION: θtotal = θf + θw + θd θw = 0 ; since the cake is not washed θf = 0.014 hrs θd = 3 hrs θtotal = ( 0+0.014+3) θtotal = 3.04 hrs.
equating to 0 0 = 1.4 (-2.2 +C) 0 = -3.08 +C C = $3.08 86. Based from the previous problem, what is the cost per lb of C?
For nos. 89- 91. In processing 500 ton/day of ore assaying 50% mineral, 300 tons of concentrate containing 66.7 % are obtained at a cost of sales ( all fixed operating cost are excluded) of $15 per ton concentrate. An investment of $200, 000 of concentrate that will assay 71% mineral. If the plant operates 200 days/year, equipment must pay out in 5 years with interest at 15% and no salvage value and no additional labor or repair costs need to be considered.
89. Based on the stated problem above, calculate the additional cost per ton of concentrate for capital recovery on the new equipment. GIVEN: 500 tons/day of ore with assay of 50 % mineral 300 tons/day of ore with assay of 66.7% mineral fixed cost = $15/ton Operation = 200 days/year Pay out period = 5 years Interests = 15% No salvage value REQUIRED: additional cost per ton concentrate SOLUTION: Let x additional cost of equipment FCI = $15 + x Annual share with 15% interest = 0.15 (15+x) = 2.25 + 0.15x with 5 year payment period 15 + 0.15x 5 Cost of equipment invested: = $200, 000 x / 300 tons/day (200 days/yr) =3.33 x tons of concentrate per year = [500 (0.5) + 300 (0.667) + 300 (0.71) ]tons/day ( 200d/yr) = 19893 / x +1 Cost total/yr = FCI + Cost concentrate + Annual Charges + Investments
Fixed cost = $15/ton Operation = 200 days/year Pay out period = 5 years Interests = 15% No salvage value Selling price = $ 263.13 REQUIRED: efficiency SOLUTION: E = 500(0.5) + 300(0.6667) – 300(0.71) 500(0.5) + 300(0.6667) E = 52.677% 92. A ties on a plant railroad sliding are to be replaced. Untreated ties consisting $ 2.50 installed have a life of 7 years. If created ties have a life of 10 years, what is the maximum installed cost that should be paid for treated ties if money is worth 8 percent? GIVEN: Untreated ties = $2.50 n=7 Created ties = X n = 10 REQUIRED: maximum installed cost SOLUTION: Getting the annual depreciation cost 0.08 =
(2.50/7) – ( x /10) X – 2.50 X = $3.10
Ct = 15 +3.825x +(19893/x+1) 9+ 2.25 +0.15x +3.33x dCt/dx = 3.48x - 19893/ (x+1)2 = 0 3.48 ( x2 +2x +1) – 19893 = 0 by quadratic equation: x = $74.6 90. Determine the selling price in dollars per ton (100% mineral basis) required for which the cost of the n new equipment is j ustified. GIVEN: 500 tons/day of ore with assay of 50 % mineral 300 tons/day of ore with assay of 66.7% mineral Fixed cost = $15/ton Operation = 200 days/year Pay out period = 5 years Interests = 15% No salvage value REQUIRED: selling price SOLUTION: Since x = $74.6 19893 / (74.6 + 1) = $ 263.13 91. What is the % increase in recovery and rejection for the new process based on mineral and gauged? GIVEN: 500 tons/day of ore with assay of 50 % mineral 300 tons/day of ore with assay of 66.7% mineral
93. Powdered coal having a heating value of 13, 500 Btu/ lb is to be compared with fuel oil worth $2.00 per bbl (42 gal) having a heating value of 130,000 Btu/gal as a source of fuel in the processing plant. If the efficiency of the conversion of the fuel is 64% for coal and 72 % for oil, with all other costs being equal, what is the maximum allowable selling price for coal per ton? SOLUTION: Powdered coal Let x be the selling price per ton (X / ton)(ton/ 2000lb) (lb / 13500Btu) (0.64) = 2.37 x 10 -8- /Btu Fuel Oil (2.00/bbl) (bbl/42gal) (gal/130000Btu)(0.72) = 2.67x 10 -7 / Btu Equating both prices: X = $11.13 / ton 94. A steam boiler is purchased on the basis of guaranteed performance. However, initial tests indicate that the opening (income) cost will be P400 more per year than guaranteed. If the expected life is 25 years and money is worth 10 %, what deduction from the purchase price would compensate the buyer for the additional operating cost? SOLUTION A = 400 N = 25 i = 0.10
