AIR COMPRESSOR
1). An air compressor takes air at 100 Kpa and discharges to 600 Kpa. If the volume flow of discharge is 1.2 m 3/sec, determine the capacity of air compressor. A.
4.32 m3/sec
C. 6.85 m3/sec
B.
3.33 m3/sec
D. 7.42 m3/sec
SOLUTION: P1 V1n = P2 V2n n = 1 .4 (for standard ai r) 100(V1)1.4 = (600)(1.2)1.4
V1 = 4.315 m3/sec
2). The discharge pressure of an air compressor is 5 times the suction pressure. If volume flow at suction is 0.1 m3/sec, what is the compressor power assuming n = 1.35 and suction pressure is 98 Kpa? A.
21.67 KW
C. 25.87 KW
B.
19.57 KW
D. 10.45 KW
SOLUTION: W=(n /n -1 )P1V1[( P2P1)n/n-1-1] W=(1.35/(1.35 –1)) (98)(0.1)[ (5P1 / P1) (1.35 –1)/1.35 -1 ]
W=19.57 kw
3). A 10 Hp motor is use to drive an air compressor. The compressor efficiency is 75%. Determine the compressor work. A.7.6 KW B.5.6 KW C.5.0 KW D. 6.5 KW
SOLUTION: Pm= ( Brake Brake or motor Power) ec=Wc/Pm Wc= 10HpxO.746x0.75 10HpxO.746x0.75
W = 5.59 KW
4). The initial condition of air in an air compressor is 98 Kpa and 27°C and discharg air at 400 Kpa. The bore and stroke are 355 mm and 381 mm, respectively with percent clearance of 5% running at 300 rpm. Find the volume of air at suction. A.
600 m3/hr
C. 620 m3/hr
B.
610 m3/hr
D. 630 m3/hr
SOLUTION: Nv = 1 + c – c[ (P2/P1) 1/n] Nv = 1 + 0.05 - 0.05(400/98)1/1.4 Nv = 0.913
VD = π/4 D2 L N VD = π/4 (0.355)2(0.381 )(300/60) VD = 0.1885 m3/sec
V1 = 0.1885(0.913) V1 = 0.17215 m3/sec
V1 = 619.75 m 3 /hr
5). An air compressor has a suction volume of 0.25 m 3/sec at 97 Kpa and discharges to 650 Kpa. How much power is saved by the compressor if there are two stages? A.
8.27 KW
C. 3.86 KW
B.
6.54 KW
D. 10.0 KW
SOLUTION: For single stage: W = (n/ n -1)P1V1 [P2 /P1) n/n-1-1] W= (1.4/1.4-1) (97)(0.25) [(650/97)1.4/1.4-1 -1] W = 61.28 KW
For double stage...x2 stage...x2 Px=(P1xP2)0.5 Px=(97x650)0.5
Px=251.1 Kpa For two stages: W = 2(n/ n -1)P1V1 [P2 /P1) n/n-1-1] W= 2(1.4/1.4-1) (97)(0.25) [(251.1/97)1.4/1.4-1 -1] W = 53 KW
Power Saved = 61.28 - 53 Power Saved = 8.27 KW
6). A 355 mm x 381 mm air compressor has a piston displacement of 0.1885 m 3/sec, Determine the operating speed of the compressor. A.
250 rpm
C. 350 rpm
B.
300 rpm
D. 600 rpm
SOLUTION: VD = (π/4) D2 L N
0.1885=(π/4)0.3552 (0.381)N N = 5 rev/see x 60sec/min
N = 300 rpm
7). The suction condition of an air compressor is 98 Kpa, 27°C and 0.2 m 3/sec. If surrounding air is 100 Kpa and 20°C, determine the free air capacity in m3/sec. A.
0.15
C. 0.25
B.
0.19
D. 0.23
SOLUTION: PoVo/To= PsVs /Ts
100(Vo)/(20 + 273) = 98(0.2)/(27 + 273)
Vo = 0.1914 m3/sec
8). Determine the percent clearance of an air compressor having 87% volumetric efficiency and compressor air pressure to be thrice the suction pressure. A.
5%
C. 15%
B.
7%
D. 11%
SOLUTION: Nv= 1 + c - c(P2/P1) 1/n 0.87 == 1 + c - c( 3P1/P1) 1/1.4
c = 10.91%
9). The compressor work of an air compressor is 100 KW. If the piston speed is 15 m 3/min, determine the mean effective pressure. A.200 Kpa
C. 400 Kpa
B. 300 Kpa
D. 600 Kpa
SOLUTION: W = Pm X VD 100 = Pm(15/60)
Pm = 400 Kpa
10). A double acting air compressor has 16 in. x 7 in., 600 rpm has what volume displacement? A. 688 ft3/min
C. 488 ft3/min
B. 755 ft3/min
D. 977 ft3/min
SOLUTION:
Vo == 2[(π/4 )D2 L N] VO == 2[(π/4) (16/12)2 (7/12) (600)]
Vo == 977.38 ft 3/min
11). A two-stage air compressor has a suction pressure of 14 psi and discharge pressure of 130 psig. What is the intercooler pressure in Kpag. A.
209 Kpag
C. 477 Kpag
B.
600 Kpag
D. 300 Kpag
SOLUTION: P2 == 130 + 14.7 P2 == 147.5 psi a
Px = (P1x P2)0.5 Px = (14x144.7) 0.5 Px = 45 psi x 101.325/14.7 Px = 310.24 Kpaa - 101.325
Px = 208.91 Kpag
12). A two stage air compressor has an intercooler Pressure of 3 kg/cm 2. What is the discharge pressure if suction pressure is 1 kg/cm 2 A.
3 kg/cm2
C. 12 kg/cm2
B.
9 kg/cm2
D. 15 kg/cm2
SOLUTION: Px = (P1x P2)0.5 3 = (1xP2) 0.5
P2 = 9 kg/cm2
13). The piston speed of an air compressor is 140 m/min and has a volume displacement of 0.2 m3/sec. Determine the diameter of compressor cylinder. A.
500 mm
C. 467 mm
B.
358 mm
D. 246 mm
SOLUTION: VD = (π/4) D2 (LN) Piston Speed = 2 (LN) 140 = 2 (LN)
(LN) = 70 m/min
0.2 = (π/4) D2 x70 D=[0.2x(4/π) /70]0.5 D=0.46719 m
D= 467.19 mm
14). An air compressor piston displacement is 5000 cm 3 when operates at 900 rpm and volumetric efficiency of 75%. Determine the mass flow of air at standard density. A.365.3 kg/hr
C. 386.4 kg/hr
B. 243.5 kg/hr
D. 465.2 kg/hr
SOLUTION: Nv = VA/VD 0.75 = VA/5000 VA =5000x0.75 VA =3 750 cm3 VA =3 750 cm3(900 rpm) VA = 3 375 000 cm3/min
ρ = 1.2 kg/m3 (at standard)
m = 1.2(3 375 000/1003) m = 4.05 kg/min
m = 243 kg/hr
15). A two-stage compressor air at 100 Kpa and 22°C discharges to 690 Kpa. If intercooler intake is 105°C, determine the value of n. A.1.400
C. 1.345
B. 1.358
D. 1.288
SOLUTION: Px = (P1xP2 )0.5
Px = (100x690) 0.5 Px = 262.68 Kpa
(Tx/ T1)=(Px / P1) n-1/n
(105 + 273)/ (22 + 273) = (262.68/100) n-1/n ln(1.281) = (n-1/n) ln (2.6268)
n - 1= 0.2564n
n = 1.345
16). The piston displacement of a double acting compressor running at 300 rpm is 0.4 m3/sec. If bore and stroke are unity, determine the length of stroke. A.
350 mm
C. 371 mm
B.
380 mm
D. 400 mm
SOLUTION:
VD = 2[( π/4) D2 L N)] L = D (for unity)
0.4 = 2[(π/4) (D)2(D)(300/60)] D = 0.37067 m
L=D
L = 370.67 mm
17). An air compressor takes air at 97 Kpa at the rate of 0.5 m3/sec and discharge at 500 Kpa. If power input to the compressor is 120 KW, determine the heat loss in the compressor. A.
26.85 KW
C. 30.45 KW
B.
18.55 KW
D. 22.36 KW
SOLUTION: W = (n/ n -1)P1V1 [P2 /P1) n/n-1-1]
W = 1.4(97)(0.5) [(500/97)1.4/0.4 - 1]/1.4-1 W = 101.45 KW
Heat Loss = 120 - 101.45
Heat Loss = 18.55 KW
18). A single acting air compressor has a volumetric efficiency of 87%, operates at 50 rpm. It takes in air at 100 Kpa and 30°C and discharges it at 600 Kpa. The air handled is 6 m 3/min measured at discharge condition. If compression is isentropic, find mean effective pressure in Kpa A.
182
C. 198
B.
973
D. 204
SOLUTION: (P1 V1)k = (P2 V2)k 100(V 1)1.4 = 600(6)1.4 V1 = 21.58 m3/min
VD = 21.58/0.87 VD = 24.8 m3/min
W = (n/ n -1)P1V1 [P2 /P1) n/n-1-1] W = 1.4(100)(21.58) [(600/100)1.4/0.4 - 1]/1.4-1 W = 5049 KJ/min
W =Pm xVD 5,049= Pm X 24.8
Pm = 203.6 Kpa
19). A single acting reciprocating air compressor has a clearance volume of 10%. Air is received at 90 Kpa and 29.3°C and is discharged at 600 Kpa. The compression and expansion are polytropic with n = 1.28. The pressure drop i s 5 Kpa at suction port and 10 Kpa at the discharge port. The compressor piston displacement is 500 cm3 when operating at 900 rpm. Determine the mass of compressed air in kg/hr A.
16.76 C. 98.33
B.
20.45 D. 28.23
SOLUTION:
VD = (π/4) D2 L) N VD = (500)(900) VD = 450,000 cm3/min VD = 0.45 m3/min
P1 = 90 - 5 P1 = 85 kpa
P2=600+10 P2 = 610 kpa
Nv = 1 + c - c(P2/P1) 1/n Nv = 1 + 0.10 - 0.10(610/85)1/1.28 Nv = 0.633684
V1 = 0.45(0.633684) V1 = 0.285 m3/min
m = PV/RT m = 85(0.285)/(0.287)(29.3 + 273) m = 0.2792 kg/min
m = 16.76 kg/hr
20). A single acting air compressor operates at 150 rpm with an initial condition of air at 97.9 Kpa and 27°C and discharges the air at 379 Kpa to a cylindrical tank. The bore and stroke are 355 mm and 381 mm, respectively, with 5% clearance. If the surrounding air is at 100 Kpa and 20°C while the compression and expansion process are PV 1.3 = C, determine free air capacity, m3/sec A.
0.0818
C. 1.23
B.
2.13
D. 4.23
SOLUTION:
VD = (π/4) D2 L N VD = (π/4) (0.355)2 (0.381) (150/60) VD = 0.094278 m3/sec
Nv = 1 + c - c(P2/P1) 1/n Nv = 1 + 0.05 - 0.05(379/97.9) 1/1.3 Nv = 0.908
V1 = 0.908(0.094278) V1 = 0.085604 m3/sec
Solving for free air capacity: PFVF / TF = P1V1/T1 100(VF)/ (20 + 273) = 97.9(0.085604)/(27 + 273)
VF = 0.081851 m 3/sec
PUMPS
1). A double suction centrifugal pumps delivers 70 ft3/sec of water at a head of 12 m and running at 1250 rpm. What is the specific speed of the pump? A.
5014 rpm
C. 2345 rpm
B.
6453 rpm
D. 9968 rpm
SOLUTION: Ns = N(Q)0.5/H 3/ 4 Ns = 374
Q = 70/ ft3 /sec x 7.481 gal/ft3 x 60sec/1 min Q = 15710.10 gal/min
h = 12x3.281 h = 39.37 ft
Ns =1250(15710.10)0.5/(39.37)3/4
Ns = 9968.4 rpm
2). The pump centerline of a centrifugal pump is located 2.5 m above from the high tide level. The sea. water varies two meters from high tide to low tide level. If friction loss at the suction is 0.8 m, determine the total suction head. A.
5.30 m
C. 6.30 m
B.
2.30 m
D. 8.23 m
SOLUTION: hs = total suction head hs = 2.5 + 2 + 0.8
hs = 5.3 m
3). A centrifugal pump requires 40 ft head to deliver water from low level to higher level. If pump speed is 1600 rpm, determine the impeller diameter of the centrifugal pump. A.
185 mm
C. 154 mm
B.
160 mm
D. 176 mm
SOLUTION: v = (2gh )0.5 v = (2(9.81(40x3.28)0.5 v = 15.466 m/sec
v=πDN 15.466 = πD(1600/60) D = 0.18461 m
D = 184.61 mmØ
4). The suction pressure of a pump reads 2 in. of mercury vacuum and discharge pressure reads 130 psi is use to deliver 100 gpm of water with specific volume of 0.0163 ft3/lb. Determine the pump work. A.
4.6 KW
C. 7.4 KW
B.
5.7 KW
D. 8.4 KW
SOLUTION: P1 = 2 in Hg x 101.325/29.92 P1 = 6.773 Kpa P2 = 130 psi x 101.325/14.7 P2 = 896.071 Kpa
dw = 1/υ dw = 1/.0163 dw = 61.35 Ib/ft3 x 9.81/62.4 dw = 9.645 KN/m3
h=(P2 - P1)/ dw h = (896.071 - (-6.773))/ 9.645 h = 93.075 m
Q = 100 gal/min x 3.785 Ii/ gal x 1 m3/1 OOOli x 1/60 Q = 0.006308 m3/sec
P = wQ h P = 9.645(0.006308)(93.075)
P = 5.69 KW
5). A pump is to deliver 150 gpm of water at ahead of 120 m. If pump efficiency is 70%, what is the horsepower rating of motor required to drive the pump? A.
40.44 Hp
C. 38.44 Hp
B.
25.66 Hp
D. 21.33 Hp
SOLUTION: Wp = w Q h Wp = 9.81 (150gal/min x 0.003785m3/1gal x 1/60)(120) W p = 11 .139 KW
BP = 11.139/0.7 BP = 15.913 KW
BP = 21.33 hp
6). A motor is used to drive a pump having an efficiency of 85% and 70% respectively What is the combined efficiency of pump and motor? A.
59.50%
C. 62.50%
B.
61.50%
D. 65.50%
SOLUTION: ecombined=ep(em) ecombined = 0.85(0.7)
ecombined = 59.50%
7). In a boiler feed pump, the enthalpy at the entrance is 765 KJ/kg. If pump has a head of 900 m, what is the exit enthalpy of the pump. A. 897 KJ/kg
C. 774 KJ/kg
B. 465 KJ/kg
D. 864 KJ/kg
SOLUTION: m(h2 - h1) = m x h x 0.00981 h2 - 765 = 900 x 0.00981
h2 = 773.83 KJ/kg
8). A submersible pump delivers 350 gpm of water to a height of 5 ft from the ground. The pump were installed 120 ft below the ground level and a draw down of 8 ft during the operation. If water level is 25 ft above the pump, determine the pump power. A.
7.13 KW
C.
7.24 KW
B.
4.86 KW
D.
864 KW
SOLUTION: h = 5 + 120 - (25 - 8) h = 108/3.281 h = 32.916 m Q = 350 gal/min x 0.003785m3/gal x 1 min/60sec Q = 0.02246 m3/sec
Wp = dw Q h Wp = 9.81(0.02246)(32.916)
Wp = 7.25 KW
9). Determine the number of stages needed for a centrifugal pump if it is used to deliver 400 gal/min of water and pump power of 15 Hp. Each impeller develops a head of 38 ft. A. 6
C. 8
B.4
D.7
SOLUTION: Wp = dw Q h
15 x 0.746 = 9.81 (400 gal/min x 0.00785m3/gal x 1/60)h
h = 45.20 m x 3.281ft/m h = 148.317 ft
Number of stages = 148.317/38 Number of stages = 3.903 stages
Number of stages = 4 stages
10). A boiler feed pump receives 50 Ii/see of water with specific volume of 0.00112 m3/kg at ahead of 750 m. What is the power output of driving motor if pump efficiency is 65 %? A.
505.32 KW
C.
785.56 KW
B.
643.54 KW
D.
356.45 KW
SOLUTION: Wp = dw Q h Wp = (1/0.00112 x 0.00981)(0.050)(750) Wp = 328.46 KW BP = 328.46/0.65
BP = 505.32 KW
11). What power can a boiler feed pump can deliver a mass of 35 kg/s water at a head of 500 m? A.
356.56 KW
C.
456.64 KW
B.
354.54 KW
D.
171.67 KW
SOLUTION: P = m x h x 0.00981 P = 35 X 500 x 0.00981
P = 171.675 KW
12). A pump running at 100 rpm delivers water against a head of 30 m. If pump speed will increased to 120 rpm, what is the increase in head? A.43.2 m
C. 34.6 m
B. 13.2 m
D. 56.3 m
SOLUTION: h2 / h1=N2 2 / N12 h2 / 30=120 2 / 1002
h2 = 43.2 m
Increased = 43.2 - 30
Increased = 13.2 m
13). A pump is used to deliver 50 Ii/see of sea water at a speed of 120 rpm. If speed will increased to 135 rpm, determine the increase in pump capacity. A.56.25 Ii/see B. 34.56 Ii/see
C. 87.54 Ii/see D. 6.260 Ii/see
SOLUTION: Q 2/Q 1 = N2/N1 Q 2 = 135(50)/120 Q 2 = 56.25 L/sec Increased = 56.25 - 50
Increased = 6.25 L/sec
14). A 15 KW motor running at 350 rpm is used to drive a pump. If speed will changed to 370 rpm, what is the increase in power? A.
2.72 KW
C. 56.45 KW
B.
17.72 KW
D. 5.67 KW
SOLUTION: P2 /P1=N2 3 / N13
P2 = (370)3( 15)/(350)3 P2 = 17.72 KW Increased = 17.72 - 15
Increased = 2.72 KW 15). A certain pump is used to deliver 150 gpm of water having a density of 61.2 Ib/ft3. The suction and discharge gage reads 4 in Hg vacuum and 25 psi, respectively. The discharge gage is 2 ft above the suction gage. What is the brake power of the motor if pump efficiancy is 75%? A.
3.24 Hp
C.
5.45 Hp
B.
2.67 Hp
D.
6.89 Hp
SOLUTION: BP = Qdw h/ ep h= Pd- Ps + z Ps = - 4 in Hg x 14.7/29.92 Ps = -1.965 psi Pd = 25 psi
h = [25 - (-1.965)](144)/ 61.2 + 2 h = 65.45 ft
BP = Qdwh/ep BP =(61.2)(150/7.481)(65.45)/33,000(0.75)
BP = 3.24 Hp
16). The discharge pipe of a pump is 400 mm in diameter delivers 0.5 m3/s of water to a building which maintains a pressure of 100 Kpa at a height of 30 m. above the reservoir. If equivalent friction head is 2 m, what power must be furnished by the pump? A.
211 KW
C. 340 KW.
B.
480 KW
D. 240 KW
SOLUTION: Q = AxV
0.5 = (π/4 x( 0.400)2) V
V = 3.9788 m/s
h=(v2/2g) +P/ w + h +z h = (3.9788)2/2(9.81) + 100/9.81 + (30 + 2) h = 43m
Wp = w Q h Wp = 9.81 (0.50){43)
Wp = 210.92 KW
17). A centrifugal pump is designed for 1800 rpm and head of 61 m. Determine the speed if impeller diameter is reduced from 305 mm to 254 mm. A.
1000 rpm
C. 1500 rpm
B.
1250 rpm
D. 1600 rpm
SOLUTION: h2/ h1= D2 2/D12 h2 = (254)2 (61)/(305)2 h2 = 42.30 m h2 /h1=N22/N12 42.3/61 = (N2)2/1800
N2 = 1499 rpm
18). Water from a reservoir A 10 m elevation is drawn by a motor driven pump to an upper reservoir B at 72 m elevation. Suction and discharge head loss are 0.15 m; respectively. For discharge rate of 15 L/sec, find the power input to the motor if overall efficiency is 65%. A.
12.65 KW
B.
23.54 KW
C.
14.17KW
D.
45.35 KW
SOLUTION: Z=72m-10 m Z=62 m
Wp = wQ h h = (72-10)+0.15 h = 62.15 m Wp = 9.81(0.015){62.15) Wp = 9.145 KW
Power input = 9.145/0.65
Power input = 14.07 KW
19). The elevation of suction reservoir is 5 m above the pump centerline and delivers to 85 m elevation tank which maintain 150 Kpa. If 1.5 m 3/sec of water is used to deliver a total head of 3m, determine the power needed by the pump. A.
1446 KW
C. 4675 KW
B.
2567 KW
D. 3456 KW
SOLUTION: h = (85 - 5) + 3 + 150/9.81 h = 98.29 m
P = wQ h P = 9.81 (1.5){98.29)
P = 1446.34 KW 20). Water from a reservoir is pumped over a hill through a pipe 900 mm in diameter and a pressure of one kg/cm2 is maintained at the pipe discharge where the pipe is 85 m from the pump centerline. The pump have a positive suction head of 5 m. Pumping rate of the pump at 1000 rpm is 1.5 m3/sec. Friction losses is equivalent to 3 m of head loss. What amount of energy must be furnished by the pump in KW? A. 1372kw
C. 1234kw
B. 1523kw
D. 1723kw
SOLUTION: Pd = 1 kg/cm2 Z1= 85m Z2=5m
Vd = Q/A Vd = 1.5/(π /4)(0.9)2 Vd = 2.358 m/sec
Pd = 1 kg/cm2 x 101.325/1.033 Pd = 98.088 Kpa Ps = 0 (open to atmosphere)
h = (Zd - Zs ) + (Pd -Ps)/w +Vd2/2g+ (hfs+ hfd) h = (85 – 5)+(98.088 - 0)/9.81 + (2.358)2/2(9.81) + 3
h = 93.28 m Water Power = w Q h Water Power= 9.81(1.5)(93.28)
Water Power = 1372.6 KW
GAS TURBINE
1). Air enters the compressor of a gas turbine at 100 Kpa and 300 0K with a volume flow rate of 5 m3/sec. The compressor pressure ratio is 10 and its isentropic efficiency is 85%. At the inlet to the turbine, the pressure is 950 Kpa and the temperature is 1400oK. The turbine has an isentropic efficiency of 88% and the exit pressure is 100 Kpa. On the basis of air standard analysis, what is the thermal efficiency of the cycle in percent? A.
42.06
C.
31.89
B.
60.20
D.
25.15
Solution:
Wc' = 1621/0.85
Solving for the mass flow rate:
Wc' = 1907 KW
PV = mRT
Wt = mCp(T3 – T4)
100(5) = m(0.287)(300)
Wt= 5.81 (1.0)(1400 - 736)
m = 5.81 kg/s
Wt = 3858 kW
Solving for T2:
Wt' = 3858(0.88)
T2/T1= (P2/P1 )k-1/k
Wt' = 3395 KW
T2/300 = (10/1)1.4-1/1.4 T2= 579°K
Wn' = Wt' – Wc' Wn' = 3395 - 1907
Solving for T4:
Wn' = 1488 kW
T3/T4= (P3/P4)k-1/k 1400/T4 = (950/100)1.4-1/1.4
Q A = mCp(T3 – T2)
T4 = 736°K
Q A = 5.81(1.0)(1400-579) Q A= 4770 kW
Wc = mCp(T2 – T1) Wc = 5.81 (1.0)(579 - 300)
Efficiency = 1488/4770
et= 31.19%
Wc = 1621 KW
2). Air is drawn into a gas turbine working on the constant pressure cycle at 1 bar 21⁰C and compressed to 5.7 bar. The temperature at the end of heat supply is 680⁰C. Taking expansion and compression to be adiabatic where C v = 0.718 kJ/kg, Cp = 1.055kJ/kg-k, calculate the heat energy supplied per kg at constant pressure. A.
472 kJ/kg
C.
501 kJ/kg
B.
389 kJ/kg
D.
489kJ/kg
Solution: Heat Energy Supplied:
T2 /(21+273) = (5.7/1)1.4-1/1.4
Q A = mCp(T3 – T2) T2 = 483.41 K Solving for T2 : T2 / T1 = (P2 / P1)k-1/k
T3 = 680 +273 =953 K
Q A = 1.005 (953 – 483.41)
Q A = 471.94 kJ/kg
Thus;
3). There are required 2200 kW net from a gas turbine unit for pumping of crude oil. Air enters the compressor section at 100kpa, 280 K, the pressure ratio r p = 10. The turbine section receives the hot gases at 1 100 k. Assume the closed Brayton cycle and determine the required air fl ow. A.
7.91%
C.
8.11%
B.
7.16%
D.
8.91%
Solution:
= (1)(1.0062)(1100 – 540.6)
Pnet =m Wnet
= 562.87 kJ/Kg
Solving for Wnet :
Q A = mCp(T4 – T1) =(1)(1.0062)(569.74 – 280)
T2 /T1 = (P2/P1) k-1/k
=291.54kJ/kg
T2 /280 = (10)1.4-1/1.4 T2 = 540.60 K
Wnet = 562.87 - 291.54 = 271.33 kJ/kg
T4 /T3 = (P4/P3) k-1/k T4 /1100 = (1/10)1.4-1/1.4
Thus;
T4 = 569.74 K
2200 = m (271.33)
Q A = mCp(T3 – T2)
m = 8.11kg/s
4). The intake of the compressor of an air –standard Brayton cycle is 35 000 ft3/min at 14 psia and
95⁰F. The compression ratio is 4 and the temperature at the turbine inlet is 1500⁰F. The exit pressure of the turbine is 14 psia. Determine the mean effective pressure. A.
25.06 psi
C.
25.06 psi
B.
28.05 psi
D.
38.05 psi
V1 = 14.68 ft3 /lb
Solution: Pm = Wn / Vd
V2 = V1 /rk Solving for Wn and Vd :
= 14.68 / 5
m = (P1V1) / (RT1)
= 2.94 ft3 /lb
= (14)(144)(35 000) / (53.34)(95 + 460) m = 2383.49 lb/min
P2 = P1(rk)k = 14(5)1.4
T2 = T1rk k-1
= 133.26 psi
= (95 + 460)(4) 1.4 -1 T2 = 966.31 K
V3 = V2 (T3 / T2) = 2.94 {(1500+460) / 966.31} 3
V1 = 35 000 ft /min /2383.49 lb/min
= 5.96 ft3 /lb
V4 = V3 (P3 / P4)1/k
Wt = m Cp (T3 – T4)
= 5.96 (133.26) 1/1.4
= (1)(0.24)(966.31 – 555)
= 29.80 ft3 /lb
= 98.71 Btu/lb
T4 = T3 (V3/V4)k-1
Wn = 223.30 – 98.71
= (1500+460) (5.96/29.80) 1.4-1
= 124.59 Btu/lb
= 1029.60 ⁰R Then; Wc = m Cp (T2 – T1)
Pm = 124.59(778) / (29.80 – 2.94)144
= (1)(0.24)(1960 – 1029.60)
Pm = 25.06 psi
= 223.30 Btu/lb
5). Calculate the work done per kg of gas expanding from 6.33 kg abs to 1.05 kg/cm2 abs. in a gas
turbine of 82% internal efficiency. Initial temperature, 750 ⁰C; k = 1.34,M=29. A. 349 kJ/kg
C. 249 kJ/kg
B. 425 kJ/kg
D. 525 kJ/kg
Solution:
Solving for the temperature after
Let;
expansion:
Wta = actual turbine work Wti = ideal turbine work
T3/ T4 = (P3 / P4) k-1/k
Wta = eWti
(750+273) / T4 = (6.33 /1.05) 1.34 -1/1.34 Solving for, Wti :
T4 = 648.50 K
Cp = kCv = 1.34Cv
Then;
Cp – Cv = R
Wti = m Cp (T3 – T4)
Cp – Cv = 8.314 / 29 Cp – Cv = 0.288
= 1(1.135)(1023 – 648.50) Eq.2
From eq. 1 and 2
= 425.06 kJ/kg Thus; Wta= 0.82(425.06)
Cp = 1.135
Wta = 348.55kJ/kg
6). Kerosene is the fuel of a gas turbine plant : fuel – air ratio, mf = 0.012, T3 = 972K, pressure ratio, rp=4.5, exhaust to atmosphere. Find the available energy in kJ per kg air flow. Assume k=1.34 and Cp=1.13. A.
352.64kJ/kg
C.
252.64kJ/kg
B.
452.64kJ/kg
D.
552.64kJ/kg
Solution:
The available Energy, Q:
Q= (1+mf ) Cp (T3 – T4)
T4 = 663.63K
Solving for T4 :
Then;
T3 /T4 = (P3 /P4) k-1/k
Q = (1+0.012) (1.13)(972 – 663.63)
972/T4 = (4.5)1.34-1/1.34
Q = 352.64 kJ/kg
7). A gas turbine power plant operating on the Brayton cycle delivers 15 MW to a standby electric generator. What is the mass flow rate and the volume flow rate of air if the minimum and maximum pressure are 100 kPa and 500kPa respectively and temperatures of 20⁰C and 1000⁰C. A.
31.97 kg/s , 26.88 m3/s
C.
41.97 kg/s, 26.88 m3/s
B.
36.98 kg/s, 28.99 m3/s
D.
46.98 kg/s, 28.99 m3/s
Solution: PV = mRT
Wt = m Cp (T3 – T4) 15000 = m (1)(1273 – 803.75)
Solving for m:
m = 31.97 kg/s
T3/T4= (P3/P4)k-1/k (100+273)/T4 = (500/100)1.4-1/1.4 T4 = 803.75°K
Then; 100V = 31.97 (0.287)(20+273)
V = 26.88 m 3/s
8). In an air – standard Brayton cycle the inlet temperature and pressure are 20⁰C and 101.325 kPa. The turbine inlet conditio ns are 1200kPa and 900⁰C. Determine the air flow rate if the turbine produces 12 MW. A.
21.41 kg/s
C.
19.25 kg/s
B.
20.20 kg/s
D.
18.10 kg/s
Solution:
(900+273)/T4 = (1200/101.325)1.4-1/1.4 T4 = 578.89°K
Wt = m Cp (T3 – T4) Solving for T4: T3/T4= (P3/P4)k-1/k
Then; 12000 =m(1)(1173 – 578.89)
m = 20.20 kg/s
9). The net power output of an air-standard Brayton cycle is 200 KW. Air enters the compressor
at 32⁰C and leaves the high-temperatures heat exchanger at 800⁰C. What is the mass flow rate of air if it leaves the turbine at 350⁰C ? A.
0.57 kg/s
C.
0.77 kg/s
B.
0.67 kg/s
D.
0.87 kg/s
Solution :
Wn = Wt - Wc Wc = m Cp (T2 – T1) = m (1) (525.30 – 305)
Solving for Wt and Wc :
=220.3 m T1 = 32 + 273 Wt = m Cp (T3 – T4)
= 305K
= m (1)(1073 – 623) T3 = 800 +273 =107
=450 m
T4 = 350+1073 = 623K Then; T2/T1 = T3/T4
200 = 450m – 220.3m
T2/305 = 1073/623
m = 0.87kg/s
T2 = 525.30K 10). In an air-standard Brayton cycle air enters the compressor at 101.325 kPa and 27⁰C.
Determine the network if the maximum temperature is 1000⁰C and the pressure ratio is 9. A.
456.88 kJ/kg
C.
331.47kJ/kg
B.
421.56kJ/kg
D.
301.74kJ/kg
(1000+273)/T4 = (9)1.4-1/1.4
Solution:
T4 = 679.50K Wn = Wt - Wc Solving for Wt and Wc :
Wc = mCp(T2 – T1) Wc = 1 (562.03 - 300) Wc = 262.03kJ/kg
T2 / T1 = (P2 / P1)k-1/k
Wt = mCp(T3 – T4) Wt= 1 (1273 – 679.50)
1.4-1/1.4
T2 /(27+273) = (9)
Wt = 593.05kJ/kg
T2 = 562.03 K
Wn = 262.03+593.05
Wn =331.47 kJ/kg k-1/k
T3/T4= (P3/P4)
11). In a gas turbine unit, air enters the combustion chamber at 550 kpa, 227C and 43 m/s. the products of combustion leave the combustor at 511kPa, 1004C and 140 m/s. Liquid fuel enters with heating value of 43,000 kJ/kg. For fuel-air ration of 0.0229, what is the combustor efficiency of the unit in percent? A. 64%
C. 78%
B. 92%
D. 102%
Solution: Comb Eff. = Heat Absorbed / Heat supplied by Fuel Heat supplied by fuel = mr Q h = 0.0229 (43,000) =984.7 kJ/kg air Heat Absorbed = Cp (t2-t1) + 0.5(V22 = V12) = 1 (1004 - 227) + 0.5(1402 - 432)/1000 = 785.88 kJ/kg air Then ec = 785.88 / 984.7
ec = 0.7981 = 79.81 %
12). Products of combustion with k of 1.35, 556 K, molecular weight M = 29, are moving within an exhaust pipe at 174 m/s; 1.12 kg/cm 2 abs. static pressure. Find the total pressure and temperature. A. 1.61 kg/cm2 ,580 K
C. 1.13 kg/cm2 ,570 K
B. 1.61 kg/cm2 ,570 K
D. 1.13 kg/cm2 ,580 K
Let;
Ptotal = total pressure Ttotal = total temperature Ptotal = P(1+((k-1)/2)M2)(k-1) /k
Solving for the Mach number, M: R = (8.314 kJ/ kg mol - K) / (29 kg/ kg-mol)
= 0.287 kJ/kg. K = 29.26 kg.m/kg.K
M = V/√(kgRT)
= (174 m/s)/√(1.35(9.81 m/s2)(29.26 kg.m/kg.K)(556K)) = 0.375 Then; Ptotal = (1.12 kg/cm2)(1+ 0.5(1.35 – 1)(0.375)2)(1.35 -1)/1.35 = 1.13kg/cm2
Solving for the total temperature: T total = T [ 1 + (k – 1 / 2) ] M2 = 556 [ 1 + (1.35 – 1/ 2) (0.375)2 ] Thus;
T total = 569.68 K
13). An ideal gas turbine operates with a pressure ratio of 10 and the energy input in the high temperature heat exchanger is 300 kW. Calculate the air flow rate for a temperature limts of 30OC and 12000C . A.
0.25 kg/s
C. 0.41 kg/s
B.
0.34 kg/s
D. 0.51 kg/s Q A = mCp ( T3 – T2 )
Solving for T2 : T2/T1 = (P2/P1)
k-1/k
T2/ 30 + 273 = (10/ )
1.4-1/1.4
T2 = 585 K Then; 300 = m (1) (1473 - 585) Thus;
m = 0.34 kg/s
14). The compressor inlet air air temperature an a gas turbine plant is 990C. Calculate the compressor air exit temperature if it requires 400 kJ/kg of work. A. 4990C
C. 550C
B. 4000C
D. 599C
Solution: WC = m Cp (T2 – T1) 400 = 1(1) [ T2 – (99 = 273)]
Thus;
T2 = 4990C
15). What is the efficiency of the compressor in a gas turbine plant if the compressor power is 300 Kw. Power input is 400 kW. A. 75%
C. 85%
B. 80%
D. 70%
Solution: ec = 300/400 thus;
ec = 0.75 or 75%
16). A gas turbine working on air standard Brayton cycle has air enter into the compressor at atmospheric condition and 22 oC. The pressure ratio is 9 and the maximum temperature in the cycle is 1077oC. Compute for the cycle efficiency per kg of air in percent. A. 44.85%
C. 41.65%
B. 43.92%
D. 46.62%
Solution: ec = 1 – 1/rp(k-1/k) ec = 1 – 1/(9)(1.4-1/1.4)
ec = 0.4662 or 46.62%
17). What is the thermal efficiency of an air-standard Brayton cycle if the pressure ratio is 10. A. 48.21%
C. 45.36%
B. 50.16%
D. 42.44%
Solution: e = 1 - 1/rp(k-1/k) e = 1 - 1/(10)(1.4-1/1.4)
e = 48.21% 18). In an air-standard Brayton cycle, the compressor receives air at 101.325 kPa, 21 oC and it leaves at 600 kPa at the rate of 4 kg/s. Determine the turbine work if the temperature of the air entering the turbine is 1000oC. A. 3000 kW
C. 2028 kW
B. 2701 kW
D. 3500 kW
Solution: Wt = mCp(T3-T4) Solving for T4: T4/T3 = (P4/P3)k-1/k (T4/1000+273) = (101.325/600)1.4-1/1.4 T4 = 765.83 K t4 = 492.83 oC Thus; Wt = (4)(1)(1000-492.83)
Wt = 2028 kW 19). A gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The air enters the turbine at 120 psia and 2000 R and leaves at 15 psia an d 1200 R. Heat is rejected to the surroundings at a rate of 6400 Btu/s, and air flows through the cycle at a rate of 40 l bm/s. Assuming the turbine to be isentropic and the compressor to have an isentropic efficiency of 80 percent, determine the net power output of the plant. Account for the variation of specific heats with temperature. Solution: T3 = 2000 R; h3 = 504.71 Btu/lbm T4 = 1200 R; h4 = 291.3 Btu/lbm rp = P2/P1 = 120/15 =8
Q out = m(h4 – h1); h1 = 291.3 – 6400/40 = 131.3 Btu/lbm
Pr1 = 1.474
Pr2 = (P2/P1)(Pr1) = (8)(1.474) = 11.79 h2s = 238.07 Btu/lbm
Wcin = m(h2s - h1)/ec = (40)(238.07 – 131.3)/(0.8) = 5339 Btu/s WTout = m(h3 - h4) = (40)(504.71 – 291.3) = 8536 Btu/s Wnet out = WTout - WCin = 8536 – 5339 = 3197 Btu/s = 3373 kW 20). For what compressor efficiency will the gas-turbine power plant in Problem 19 produce zero net work? Solution: ec = (h2s - h1)/ (h3 - h4) ec = (238.07 – 131.3)/(504.71 – 291.3)
ec = 0.5 or 50%
HEAT TRANSFER
1). A 15 cm thick wall has a thermal conductivity of 5 W/m-oK. If inside .md outside surface temperature of the wall are 200°C and 30°C, lespectively. Determine the heat transmitted. A. 5.67 KW/m2
C. 8.87 KW/m2
B. 4.68 KW/m2
D. 6.87 KW/m2
SOLUTION: Q = kA (t2-t1)/X Q = 5A(200 - 30)/0.15 Q/A = 5666.67 W/m2
Q/A = 5.6667 KW/m2 2). Two walls of cold storage plant are composed of an insulating material (k = 0.25 KJ/hr -m-°C), 100 mm thick at the outer layer and material (k = 3.5 KJ/hr-m-°C), 15 cm thick at inner layer. If the surface temperature at the cold side is 30°C and hot side is 250°C, find the heat transmitted per square meter. A. 0.138 KW/m2
C. 0.025 KW/m2
B. 0.450 KW/m2
D. 0.065 KW/m2
SOLUTION: Q = k1A( t2-t1) /X1 Q= k2A (t2-t3) /X2 Q/A= (t1-t3)/[(X1/k1)+(X2/k2)] Q/A = 250 – 30)/[(0.15/ 13.5) + (0.100 /0.25)]
Q/A = 0.138 KW/m2 3). Sea water for cooling enters a condenser at 27 OC and leaves at 37 OC. The condenser temperature is 45°C, what is the log mean temperature difference? A. 50.56°C
C. 37.82°C
B. 12.33°C
D. 80.54°C
SOLUTION:
θmean= (θmax- θmin)/ln (θmax/ θmin) θmax = 45°C - 27°C θmax = 18°C θmin = 45°C - 37°C
θmin = 8°C θmean= (θmax- θmin)/ln (θmax/ θmin) θmean= (18- 8)/ln (18/ 8) θmean= 12.33 OC
4). Determine the thermal conductivity of a material that is use in a 2 m2 test pannel, 25 mm thick with a temperature difference of 10.8 oFbetween the surfaces. During the 5 hours test period, the heat transmitted is 200 KJ. A.
0.045 W/m-oK
C. 0.023 W/m-oK
B.
0.560 W/m-oK
D. 0.370 W/m-oK
SOLUTION: ∆OC= 5/9 (∆OF) ∆OC = 5/9 (10.8) ∆OC = 6°C Q = kA(t2-t1) /x 200/5x3600 = k(2)(6)/0.025 k = 2.3148 X 105 KW/m-o k
k = 0.023148 W/m- o k 5). A steam pipe having a surface temperature of 200 °C passes through a room where the temperature is 27 OC. The outside diameter of pipe is 80 mm and emissivity factor is 0.8. Calculate the radiated heat loss for 3 m pipe length. A.
1434.47 W
C. 2756.57 W
B.
3746.35 W
D. 3546.45 W
SOLUTION: Ao=πDL Ao = π(0.08)(3) Ao = 0.7539 m2 Solving for heat due to radiation: Q R = 20,408.4 x 10-8σAoFe(T14 - T24) J/hr T1 = 200 + 273 T1 = 473°K T2 = 27 + 273 • T2 = 3000K
Q R = 20,408.4 x 10-8(0.8)(0.7539) [(473)4 - (300)4] Q R = 5164079.866 J/hr x 1hr/3600sec
Q R = 1434.47 W 6). A counter flow heat exchanger is designed to heat fuel oil from 30°C to 90°C while the heating fluid enters at 140°C and leaves at 105°C. Determine the arithmetic mean temperature difference. A.72.5OC
B. 62.5 OC
C . 45.5 OC
D.67.5 OC
SOLUTION: AMTD = (θmax +θmin )/2
θmax = 105 – 30 θmax =75 °C θmin =140 - 90 θmin = 50°C AMTD = (75 +50 )/2
AMTD = 62.5 °C 7). A heat exchanger has an overall coefficient of heat transfer of 0.50 KW/m2-oK. Heat loss is 11 KW and the mean temperature difference is 15°C. What is the heat transfer area in ft2? A.
51.80 ft2
C. 56.80 ft2
B.
37.30 ft2
D. 15.80 ft2
SOLUTION:
Q = UA θm 11 = 0.50 (A )15 A = 1 .467 m2 x (3.281)2ft2/ m2
A = 15.79 ft 2
8). Brine enters a circulating brine cooler at the rate of 60 m3/hr at -lO°Cand leaves at -16°C. Specific heat of brine is 1.072 KJ/kg-OK and specifiC gravity of 1.1. Determine the tons of refrigeration.
A.
53.5 TR
C.
33.5 TR
B.
65.3 TR
D.
44.5 TR
SOLUTION: Density of brine = 1.1 (1 000 kg/m3) Density of brine = 1100 kg/m Mass flow ratr= 1100(60)/3600 m = 18.33 kg/sec
Q = m Cp ∆t Q = 18.33(1.072) [-10 -(-16)] Q = 117.92 KW TR = 117.92/3.516
TR = 33.54 TR 9). A heat exchanger has a hot gas temperature of 300°C and surface conductance on hot side is 200 W /m2-oK. If heat transmitted is 1000 W/m2, what is the surface temperature on the wall at hot side? A. 295°C
C. 234°C
B. 465°C
D. 354°C
SOLUTION:
Q = A (ho) (∆t) Q/A = (ho) (t2-t1) 1000 = 200 (300 -t1)
t1 = 295°C 10). A heat exchanger is to be designed for the following specifications: Hot gas temperature, 1145°C Cold gas temperature, 45°C Unit surface conductance on the hot side, 230 W/m 2 -OK Unit surface conductance on the cold side,290 W/m2-oK Thermal conductivity of the metal wall, 115 W/m-oK Find the maximum thickness of metal wall-between the hot gas and the cold gas, so that the maximum temperature of the wall does not exceed 545°C. A.
10 mm
C. 20 mm
B.
30 mm
D. 40 mm
SOLUTION: Hot side Q = h1 A (∆t) Q/A = h1 (t1 - t2) Q/A = 230(1145 - 545) Q/A =138000 Cold side Q = h2 A (∆t) Q/A = h2 (t2 – t3) Q/A = 290(545 - 45) Q/A =138000 138000 = (1145 - 45)/[(1/ 230) + (x/ 115) + (1/ 290)] x = 0.020115 m
x=20.115mm 11). An uninsulated steam pipe passes through a room in which the air and walls are at 250C. The outside diameter of the pipe is 70 mm, and its surface temperature and emissivity are 200°C and 0.80, respectively. If the coefficient associated with free convection heat transfer from the surface to the air is 15 W/m2-oK, what is the rate of heat loss from the surface per unit length of pipe? A. 998 w/m2
C. 762 w/m2
B. 872 w/m2
D. 422 w/m2
SOLUTION: Qc = heat transmitted by convection Qc = ho Ao (t 1 - t2)
Qc = 15[π (0.07) L](200 - 25) Qc /L = 577.268 W/m Q R = heat transmitted by radiation Q R = 20,408.4 X 10-8 A Fe(T14 - T24) ,J/hr
where: Ao = π(0.07)L T1 = 200 + 273
T1 = 473°k T2 = 25 + 273 T2 = 298°k
Q R = 20,408.4 X 10-8 [π(0.07)L (0.8)[(473)4 - (298)4] Q R /L = 1,514,032 J/hr x 1/3600 Q R/L = 420.564 W/m
Q/L= Qc + QR Q/L = 577.268 + 420.564
Q/L = 997.832 W/m 12). A turbo-charged, 16 cylinder, Vee-type diesel engine has an air consumption 0f 3,000 kg/hr per cylinder at rated load and speed. This air is drawn in through a filter by a centrifugal compressor directly connected to the exhaust gas turbine. The temperature of the air from the compressor is 145°C and a counter flow air cooler reduces the air temperature to 45°C before it goes to the engine suction header. Cooling water enters air cooler at 30°C and leaves at 38°C. Calculate the log mean temperature difference. A.
47°C
C.
34°C
B.
87°C
D.
65°C
θmean = (θmax- θmin)/ln (θmax/ θmin) (θmax =145-38 Θmax =107 °C θmin =45-30 θmin =15 °C θmean = (107- 15)/ln (107/ 15) θmean = 46.82 = 47 oC
13). A turbo-charged, 16 cylinder, Vee-type diesel engine has an air consumption of 3,000 kg/hr per cylinder at rated load and speed. This air is drawn in through a filter by a centrifugal compressor directly connected to the exhaust gas turbine. The temperature of the air from the
compressor is 145°C and a counter flow air cooler reduces the air temperature to 45°C before it goes to the engine suction header. Cooling water enters air cooler at 30°C and leaves at 38°C. Calculate the quantity of cooling water in m3/hr required to cool the total air requirements of the engine at rated load and speed. A.
144
C. 123
B.
132
D. 124
SOLUTION: Heat gained by water = heat lost by air
(m cp ∆t)water = (m cp ∆t)air mw(4.187)(38 - 30) = 48,000(1.0)(145 - 45) mw = 143,301 kg/hr
Ave. temp of water = (30 + 38)/2 Average temperature = 34°C From steam table, Vf at 34°C = 0.0010 056 m3/kg Volume flow = 143,301x0.001 0056)
Volume flow = 144.1 m3/hr 14). An oil heater heats 100 kg per minute of oil from 35°C to 100°C in a counter flow heat exchanger. The average specific heat of the oil is 2.5 KJ/kg-oC. Exhaust gases used for heating enter the heater with an average specific heat of 1 KJ/kg-OC, a mass flow rate of 250 kg/min and an initial temperature of 200°C. The overall heat transfer coefficient is 75 W 1m2 -oC. Determine heating surface in square meters A.
16.11
C.13.11 ~
B.
63.11
D. 36.11
SOLUTION: Heat gained by oil = Heat lost by gas
(m cp ∆t)Oil = (m cp ∆t)gas 100(2.5)(100 - 35) = 250(1 )(200 - to) to = 135°C Heat transferred = (100/60)(2.5)(100 - 35)
Heat transferred = 270.83 KW
∆tA = 135 - 35 ∆tA = 100°C ∆tB = 200 - 100 ∆ts = 100°C Since ∆tA = ∆tB, use the average value ∆tm = (100 + 100)/2 ∆tm = 100°C
Q = U A ∆tm 270.83 = (0.075) (A) (100)
A = 36.11 m 2
15). A surface condenser serving a 50,000 KW steam turbo-generator unit receives exhaust steam at the rate of 196,000 kg/hr. Vacuum in condenser is 702 mm Hg. Sea water for cooling enters at 29.5°C and leaves at 37.5°C. For steam turbine condenser, manufacturers consider 950 Btu/lb of steam turbine condensed as heat given up to cooling water. Calculate logarithmic mean temperature difference in oF. A.10
C. 12
B. 14
D. 16
SOLUTION: Condenser pressure = 101.325 - (702 x 101.325/760) Condenser pressure = 7.733 Kpa From steam table, at 0.007733 Mpa, tsat = 40.86°C
∆tA = 40.86 - 29.5 ∆tA = 11.36°C ∆tB = 40.86 - 37.5 ∆tB = 3.36°C θmean=11.36 - 3.36 In(11 .36/ 3.36) θmean = 6.56°C
θmean =6.56 °C x 1.8°F/°C θmean = 11.82 °F
16). A surface condenser serving a 50,000 KW steam turbo-generator unit receives exhaust steam at the rate of 196,000 kg/hr. Vacuum in condenser is 702 mm Hg. Sea water for cooling enters at 29.5°C and leaves at 37.5°C. For steam turbine condenser design, manufacturers consider 950 Btu/lb of steam turbine condensed as heat given up to cooling water. Calculate the required quantity of cooling water in cubic meters per hour. A. 10,374
C. 11,345
B. 12,445
D. 13,509
.
SOLUTION: Heat absorbed by cooling water = 950(196,000)(2.205) Heat absorbed by cooling water = 410,571,000 Btu/hr
Average SG of sea water is 1.03 and Cp of gas is 0.93 Btu/lb-oF
m cp ∆t = 410,571,000 m(0.93)(37.5 - 29.5)= 410,571,000 m = 30,657,930.11 Ibs/hr
V = 30,657,930.11/(62.5)(1.03) V = 477,003 ft3/hr V = 477,003/35.31
V = 13,509 m3/hr
17). Calculate the energy transfer rate across 6" wall of firebrick with a temperature differences across the wall of 50°C. The thermal conductivity of the firebrick is 0.65 Btu/hr-ft-oF.
A.
285 W/m2
C.
112W/m2
B.
369W/m2
D.
429W/m2
SOLUTION: Q = kA(ta- tb) /x
ta - tb = 50(9/5) ta - tb = 90°F Q/A = k (ta- tb) /x Q/A = 0.65(90)/(6/12) Q/A = 117 Btu/hr-ft2
Q/A = 117 Btu/hr-ft2 x 1055J/Btu x 1 hr/3600sec x 10.76 ft2/m2
Q/A = 369 W/m 2
18). Water is flowing in a pipe with radius of 25.4 cm at a velocity of 5 m/sec at the temperature in the pipe. The density and viscosity of the water are as follows: density = 997.9 kg/sec ; viscosity = 1.131 Pa-s. What is the Reynolds Number for this situation? A.
2241
C. 3100
B.
96.2
D. 1140
SOLUTION:
R = DV/ υ where: D= 2(25.4) D= 50.8 cm D = 0.508 m
Vo = velocity Vo = 5 m/sec
υ = kinematic viscosity υ = μ/ρ υ = 1.131/997.9 υ = 0.0011334 m2/sec Re=0.508(5)/ 0.0011334
Re = 2241 19). The hot combustion gases of a furnace are separated from the ambient air and its surrounding, which are at 25°C, by a brick wall 0.15 m thick. The brick has a thermal conductivity of 1.2 W/m-oK and surface emissivity of 0.8. Under steady state conditions and outer surface temperature of 100°C is measured. Free convection heat transfer to the air adjoining this surface is characterized by a convection coefficient of 20 W/m2-oK. What is the brick inner surface temperature in OC? A.
623.7°C
C. 461.4°C
B.
352.5 °C
D. 256.3°C
SOLUTION: Qc =convection heat transfer Qc = Aho (t1 - t2) Qc /A = 20(100-25)
Qc /A= 1500 W/m2 . Q= radiated heat loss Qr/A= 20,408.4 x 10-8 Fe (T14 -T24),J/hr-m2 Qr/A = 20,408.4 x 10-8 (0.8)[(100+273)4 - (25+273)4] Qr/A = 1,872,793 J/hr-m2 x (1/3600) Qr/A = 520 W/m2
Q /A = Qc/A + Qr/A Q /A =1500 + 520 Q /A =2020 W/m2
Q /A = k(ta - tb)/X 2020 = 1.2(ta - 100)/0.15
ta =352.5°C 20). Steam initially saturated at 2.05 Mpa, passes through a 10.10 cm standard steel pipe for a total distance of 152 m. The steam line is insulated with a 5.08 cm thickness of 85% magnesia. For an ambient temperature of 22°C, what is the quality of the steam which arises at its destination if the mass flow rate is 0.125 kg steam per second? Properties of steam: Pressure 2.05 Mpa; Temperature =213.67 °C Enthalpy : hf= 914.52 hfg = 1885.5 hg =2800.0 k for 85% magnesia=0.069 W/m-oK ; ho for still air= 9.36 W/m-oK A.
93%
C. 84%
B.
98%
D. 76%
SOLUTION: Heat transferred= change in enthalpy of steam
Q(ΣR)=ti -to Q[(ln (r2/r1)/2πkL) +1/ hoAo]= ti -to Q=ms(h1 - h2) h = hf + xhfg r1 = 10.10/2
r1 = 5.05 cm r2 = 5.08 + 5.05 r2 = 10.13 cm
Ao=2πrL Ao = 2π(0.1013)(152) Ao= 96.746 m2 Conduction :
Q =2πkL (ti – t1)/In(r2/r1) Convection: Q =hoAo(t1-to) Combined conduction and convection: ti-to= Q(ΣR) ti-to =Q[(ln (r2/r1)/2πkL) +1/ hoAo]
(213.67-22)=Q[(In(0.1013/0.0505)/2π(0.069)(152)+1/96.746(9.36)] Q =16,427.4 W Q =16.427 KW
Q=ms(h1 - h2) 16.4274=0.125(2800 - h2) h2 = 2668.6
h = hf + xhfg 2668.6=914.52 + x(1885.5)
x = 93%
COOLING TOWER
1. Determine the approximate amount of air to be handled and the quantity of make-up water required by a cooling tower that is to cool 12.62 lps from 36 ℃ to 31℃. Atmospheric conditions are 35℃ DB and 25 ℃ WB. Assume that air leaves the tower at 32 ℃ DB and 90% RH. Properties of air entering the tower: h=80.38 kJ/kg d.a. and W=0.0177 kg/kg d.a. Properties of air leaving the tower: h=102.0 kJ/kg d.a. and W=0.0274 kg/kg d.a. a.12.22 kg/s, 0.119 kg/s
c. 12.22 kg/s, 1.19 kg/s
b.12.22 kg/s, 0.911 kg/s
d. 12.22 kg/s, 1.91 kg/s
Solution: mu=(W2-W1)ma Q A=mw Cpw(Δt)
Q R=ma(h2-h1)
Q R=Q A ma(h2-h1) =mw Cpw(Δt) ma(102-80.38)=12.62(4.187)(36-32) ma= 12.87 kg/sec
mu=(0.0274-0.0177)12.87
mu= 0.125 kg/s
2. The change of enthalpy of air in a cooling tower is 81.42 kJ/kg d.a. and the mass flow rate of air is 206 kg/min. water enters the tower at the rate of 190 lpm and 46ºC. Determine the exit temperature of water. a. 25ºC
c. 24ºC
b. 24.92ºC
d. 42ºC
c. Q A=mw Cpw(Δt) Q R=ma(h2-h1) Q R=Q A ma(h2-h1) =mw Cpw(Δt) 206(81.42)=190(46-Tb)
Tb= 42.27ºC 3. The change of temperature of water entering the cooling tower and the WB temperature of surrounding air is 23ºC, and the efficiency of the tower is 65%. If the mass flow rate of the water is 15 kg/s. determine the heat carried away by the air, in kW. a. 983.93 kW
c. 938. 93 kW
b. 993.83 kW
d. 939. 83 kW
Solution: Q R= mwCpw (Ta - Tb) Q A=ma(h2-h1) Q R= Q A mwCpw (Ta - Tb) =ma(h2-h1)
e= (Ta-Tb)/(Ta-WB1) 0.65= (Ta-Tb)/(23) (Ta-Tb)= 14.95 ºC
Q R= mwCpw (Ta - Tb) Q R= 15(4.187)(14.95)
Q R = 938.93 Kw
4. The amount of water carried by air in a cooling tower is 6.8 kg/min. The change in humidity ratio in the tower outlet and inlet is 0.025 kg/kg d.a. Determine the volume flow rate of air needed if the specific volume is 0.8123 m3 /kg d.a. a. 221 m3/min
c. 221 m3/min
b. 122 m3/min
d. 212 m3/min
Solution: Va=ma x υa mu=(W2-W1)ma 6.8=(0.025) ma ma = 272 kg/min
Va=ma x υa Va = 272 x0.8123
Va = 221 m3/min 5. The approach and efficiency of a cooling tower are 10ºC and 65%, respectively. If the temperature of water leaving the tower is 27ºC, what is the temperature of water entering the tower? a. 45.57 ºC
c. 47.55ºC
b. 55.47 ºC
d. 54.75ºC
Solution: t4=27, Approach=10oC e= (T3-T4)/(T3-Wb1)
Approach= t4-wb1=10 ºC Wb1=27 ºC -10 ºC =17 ºC
e= (T3-T4)/(T3-Wb1) 0.65 = (T3-27)/(T3-17) 0.65 (T3-17) = (T3-27) 0.65T3- 0.65(17) = (T3-27)
T3=[27 -0.65(17)]/0.35
T3= 45.57 ºC
6. In an induced draft cooling tower the circulating flow of water is 126 kg/s, entering the tower at 38OC (h=159.21 kJ/kg) and leaves at 27OC (h=113.25).The inlet air conditions are27 OC DB and 15OC WB (h=42.0kJ/kg d.a.; W=0.0056 kg/kg d.a.) while exit condition is 32 OC DB and 90%RH (h=102.0 kJ/kg d.a., W=0.0275 kg/kg d.a.). How much is the make-up water required. a. 2.118 kg/s
c. 8.112 kg/s
b. 1.828 kg/s
d. 2.811 kg/s
Solution: mu=(W2-W1)ma
Q A=mw Cpw(Δt) Q R=ma(h2-h1) Q R=Q A ma(h2-h1) =mw (ha-hb) ma(102.0-42.0)=126(159.21-113.25) ma = 96.516 kg/sec mu = 96.516(0.0275-0.0056)
mu = 2.1137 kg/s 7. Water at 55OC is cooled in a cooling tower having an efficiency of 65%. Atmospheric air is 32OC and 70% RH (WB=27.4 OC). The heat rejected from the condenser is 2,300,000 kJ/hr. Find the pump capacity in liters per second to circulate the cold water.
a.
8.57 L/s
c. 7.58 L/s
b.
5.87 L/s
d. 6.57 L/s
SOLUTION: Q R=Q A=mwCP ∆T
e= ∆T/TWi-WBA 0.65= ∆T/55-27.4 ∆T= 17.94 OC
(2,300,000 kJ/hr)hr/3600 s= mw 4.187(17.94) mw = 8.57 kg/s 8.57 kg/s( 1 li/kg)= 8.57 li/s 8. An atmospheric cooling tower is to cool the jacket water of a four stroke 800 kW Diesel generator set. The tower efficiency is 60% at a temperature approach of 10OC. If the ambient air has a relative humidity of 70% and DB of 32 OC (WB=27.45OC), determine the cooling water supplied to the diesel engine in liters per hour. Generator efficiency is 97%, useful work=30%, and cooling loss=25%. a.
39,800 L/hr
c. 38,800 L/hr
b.
45,800 L/hr
d. 40,800 L/hr
SOLUTION: Q R=Q A=mwCP ∆T
e= ∆T/TWi-WBA Approach=TWO-WBA 10 OC = TWO-27.45 OC TWO =37.45 OC 0.60= TWi -TWO /TWi - WBA 0.60= TWi -37.45 /TWi – 27.45 TWi = 52.45 OC
∆T= TWi -TWO ∆T = 52.45 – 37.45 ∆T = 15 OC Q R= (800kW)0.25/0.97(0.30) Q R = 61.8556 kW
Q R = Q A = mwCP ∆T 687.285 kW = mw(4.187)(15) mw = 10.94 kg/s kg/s( 1 li/kg)= 10.94 li/s = 39384 li/hr 9. Fifty gallons of water per minute enters the tower at 46OC. Atmospheric air at 16OC DB and
55% RH (υ=0.828 m3/kg, W=0.0056 kg/kg d.a.) enters at 2.85 m 3/s and leaves at 32 OC saturated (W=0.0308 kg/kg d.a.). Determine the volume of water that leaves the tower (water that falls to the basin). a.
3.10 L/s
c. 5.10 L/s
b.
4.10 L/s
d. 6.10 L/s
SOLUTION: m3-m4 = mA (W2-W1) m3= (50 gal/min)3.785 li/gal(1 kg/li)/60 s m3 = 3.154 kg/s0 mA= (2.85 m3/s)/(0.828 m3/kg) mA= 3.44 kg/s
m3-m4 = 3.44 (0.0308 – 0.0056) m4 = 3.067 kg/s
V4 = 3.067 (1 li/kg)
V4 = 3.067 li/s 10. A 250,000 kg/hr of water at 35°C enters a cooling tower where it is to be cooled to 17.5°C. the energy is to be exchanged with atmospheric air entering the units at 15°C and leaving the unit at 30°C. The air enters at 30% RH and leaves at 85% RH. If all process are assumed to occur at atmospheric pressure, determine the percentage of total water flow that is made up water. a. 2.22%
b. 3.33%
c. 4.44%
d. 1.11%
Solution : Percentage make-up water = Mass of make-up water / 250 000 Solving for mass of make-up water:
At 15⁰C and 30% RH: h1= 23.02 kJ/kg W1 = 0.0033 kg/kg
At 30⁰C and 85% RH : h2 = 89.01 kJ/kg w2 =0.0233 kg/kg Heat loss by water = heat gained by air MwCpwt = ma (h2- h1) 250000 (4.187)(35-17.5) = ma(89.01-23.2) ma = 277 589.41 kg/hr Then ;the mass of make-up water, m s : m5 = ma (w2-w1) m5= 277 589.41(0.0233-0.0033) m = 5 551.79 kg/hr thus, the percentage make-up water, %make-up = 5 551.79 / 250 000
= 0.0222 or 2.22%
FANS AND BLOWERS
1). Calculate the required motor capacity needed to drive a forced –draft fan serving a stoker fired boiler using coal as fuel. Combustion data includes the following : Atmospheric air
101.3 kPa ; 20°C
Weight of fuel burned per hour
10 tons
Ultimate analysis of fuel : C = 78%
S = 1%
H = 3%
A = 8%
O = 3%
M = 7%
Excess air
30%
Fuel bed and air heater resistance
18 cm WG
Fan efficiency
60%
A.87.84 KW
B. 82.87 KW
C.84.87 KW
D. 88.72 KW
Pmotor= Pair/efan Solving for Pair Theoretical air required for the combustion of coal. Wt = 11.5 C + 34.5 (H – O/8) + 4.3 S = 11.5 ( 0.78) + 34.5 (0.03 – 0.03/8) + 4.3 (0.01) = 9.92 kg air / kg fuel Actual weight of air supplied into the boiler : Wa= ( 1 + e ) Wt Wa = ( 1 + 0.30 ) ( 9.92 kg air/ kg fuel ) [ 10 (1000) kg fuel / hr ] Wa= 128,942 kg air / hr Volume of air demanded by the boiler from the forced draft fan : Q = 128.942/1.2 Q = 107,451.77 m3/hr Q = 29.85 m3/s Then, Pair= [ 1.2 (0.00981) ] (29.85) [ (0.18)(1000/1.2)] = 52.71 kW Thus, Pmotor= 52.71/0.60
Pmotor= 87.84 kW 2). A boiler requires 75,000 m3 /hr of standard air. What is the motor power if it can deliver a total pressure of 145 mm of water gage. The mechanical efficiency of fan is 64%. A. 40.30 KW
B. 46.30 KW
C. 42.45 KW
Pmotor= Pair/efan Solving for Pair : Pair = γ Q h
D. 43.69 KW
Where : h = 0.145 ( 1000/1.2) h = 120.83 m then, Pair = [ ( 1.2 )(0.00981) ] (75000/3600)(120.83) Pair = 29.63 kW Thus ; Pmotor = 29.63/0.64
Pmotor = 46.30 kW 3). A 40 in. diameter fan rated at 160,000 cfm standard air at 16 i n. and pressure is operating at 1200 rpm. Solve for the specific speed.
a.386,845.18 rpm c.384,845.18 rpm
b.380,125.20 rpm d.392,865.28 rpm
Ns = specific speed Ns = N(Q) 0.5/h 3/4 Ns = 1200(160,000)0.5 /(4/3)3/4
Ns = 386,845.18 rp
4). The motor power needed to drive the fan is 75 kW and the volume flow rate delivered by fan is 23 3 /s and 20 cm water gage. The density of air is 1.2 kg/3. What is the fan efficiency? a. 60%
b.62%
c. 64%
d. 65%
efan= Pair /Pmotor Solving for Pair: Pair = [ ( 1.2 )(0.00981)(23) ][ 0.20 (1000/1.2)] Pair = 45.126 kW ℎ,
efan = 45.126/75 efan = 0.60168
efan = . %
5). In a certain installation, a fan when driven by a 7.5 HP motor at a speed of 600rpm delivers 510 3 of air per minute at a total pressure of 5cm WC. If in the same installation, 6.5cm WC pressure is required. What power and motor speed will the fan be driven? a. 564.11 rpm, 11.12HP b. 684.11 rpm, 10.12HP c. 684.11 rpm, 11.12HP d. 584.11 rpm, 10.12HP
Fan drive speed : (N2/N1)2 = h2/h1 (N2/600)2 = 6.5/5.0 (N2/600)2 = 6.5/5.0 N2=684.11 rpm Motor Power required : P2/P1= (N2/N1)3 P2/7.5= (684.11/600)3
P2=11.12hp
6). A tabular air heater is installed in line with the boiler, and the fan is now required to supply heated air for combustion at 90°C. What drive power is required and the new total pressure that this fan will operate if it is going to deliver the same volume of heated air at 1200 rpm? a. 3.28 cm WG, 4.10HP
c. 3.28 cm WG, 5.50 HP
b.
d. 3.95 cm WG, 4.10 HP
3.95 cm WG, 5.5HP
Solving for the new head :
h2/h1=ρ2/ρ1
ρ2/ρ1=T1/T2 ρ2/ρ1=(25+273)/(90+273) ρ2/ρ1=0.82 , h2/h1= 0.82
h2=3.28 cm WG
Solving for the new drive power : P2/P1= ρ2/ρ1 P2/P1= 0.82 P2/5= 0.82
P2=4.10 hp ∶ when air is heated, its density decreases and the pressure needed is move the air to
the combustion chamber will be lesser resulting to the dec rease in fan power requirement. 7). A fan has a suction pressure of 5cm water vacuum with air velocity of 5 m/s. The discharge has 20 cm WG and discharge velocity of 10 m/s. Find the head of fan. a. 198.61 m
b. 189.61 m
c.212.15 m
h=hs+hv Where : hs=(hdw-hsw)ρw/ρa hs=(0.20-(-0.05)1000/1.2 hs=208.33 m hv=(Vd2-Vs2)/2g hv=(102-52)/2(9.81) hv=3.82 m Thus, h=208.33+3.82
h=212.15 m
d. 200 m
8). An Air Handling Unit (AHU) for an airconditioning system has a centrifugal fan with backward curved blades mounted on a scroll housing driven by a motor at 750 rpm. The fan delivers 2000 cfm of air against 3 in. WC static pressure (including resistance of ducts, elbows, cooling coils, and outlet grills) and 0.80 in. WC velocity pressure. Calculate the tip speed of the whee l. a. 3100 fpm
b. 3000 fpm
c. 3586 fpm
Solution: The speed of the wheel : V = (2ghv)0.5 Solving for the hv:
ρahva=ρwhvw (0.075)hva=(0.8/1.2)(62.4) hva=55.46 ft of H2O Note :
ρa=0.075 lb/ft3 ρw=62.4 lb/ft3 then; V = (2(32.2)55.46)0.50 V = 59.77 ft/s
V = 3,586 ft/s
d. 3500 fpm
CHIMNEY
1). A boiler uses 2500 kg of coal per hour and air required for combustion is 16 kg per kg of coal. If ash loss is 10%, determine the mass of gas entering the chimney. A.
42,250 kg/hr
C. 85,452 kg/hr
B.
78,300 kg/hr
D. 33,800 kg/hr
Solution: mg = ma + mf - mash A/F = ma/mf 16 = ma/mf ma = 16 mf mg = 16 mf + mf - 0.10 mf mg = 16.9 mf
mg = 16.9(2500) mg = 42,250 kg/hr 2). The flue gas density of chimney is 0.75 kg/m3 and air density of 1.15 kg/m3. If the driving pressure is 0.25 Kpa, determine the height of chimney. A.
54.6 m
C. 74.6 m
B.
63.7 m
D. 68.5 m
Solution: hw = H(da - dg) 0.25 = H(1.15 - 0.75)(0.00981 kN/m3)
H = 63.71 m 3). The actual velocity of gas entering i n a chimney is 8 m/sec. The gas temperature is 5°C and pressure of 98 Kpa with a gas constant of 0.287 KJikg_°K. Determine the chimney diameter if mass of gas is 50,000 kg/hr. A.
1.57 m
C. 3.56 m
B.
1.81 m
D. 1.39 m
SOLUTION: P Vg = mg Rg T 98(Vg) = (50,000/3600)(0.278)(25 + 273) Vg = 12.12 m3/sec Vg = A x v
11.74 =( π/4) D2x (8)
D = 1.39 m φ
4). A coal fired steam boiler uses 3000 kg of coal per hour. Air required for combustion is 15.5 kg per kg of coal at barometric pressure of 98.2 Kpa. The flue gas ha temperature of 285°C and an average molecular weight of 30. Assuming an ash los of 11 % and allowable gas velocity of 7.5 m/sec, find the diameter of chimney. SOLUTION: Amt. of air required =15.5(3000)= 46,500 kg/hr
R = 8.314/M R = 8.314/30 R = 0.277 KJ/kg oK By mass balance: mc , ma + mf = mash + mg 46,500 + 3000 = 0.11 (3000) + mg mg = 49,170 kg/hr PVg = mRgTg 98.2(Vg) = (49,170/3600)(0.277)(285 + 273) Vg = 21.498 m3/sec Let D = diameter of chimney Q = Axv
Q = (π/4) D2)v
21.498 =(π/4) D2)(7.5) D = 1.91 m 5). A power plant situated at an altitude having an ambient air of 96.53 Kpa and 23.88°C. Flue gases at a rate of 5.0 kg/see enter the stack at 200°C and leaves at 160°C. The fl ue gases gravimetric analysis are 18% C02, 7% 02 and 75% N2. Calculate the height of stack necessary for a driving pressure of 0.20 Kpa. SOLUTION: Solving for the molecular weight and gas constant of the flue gas: C02 02 N2 18% 7% 75% 0.18/44 = 0.00409 0.07/32 = 0.00219
0.75/28 = 0.02678 0.03306 Chimney Mg = 1/0.03306 Mg = 30.25 Rg = 8.3143/30.25 Rg = 0.275 Tg = (200 + 160)/2 Tg = 180°C
dg = P/RT dg= 96.53/ (0.275)(180 + 273) dg = 0.775 kg/m3 da = P/RT da = 96.53 /(0.287)(23.88 + 273) da = 1.133 kg/m3 Draft = H(da - dg) 0.20 = H(1.133 - 0.775)(0.00981)
H = 56.95 m
6). A power plant situated at an altitude having an ambient air of 96.53 Kpa and 23.880C. Flue gases at a rate of 5.0 kg/see enter the stack at 200°C and leaves at 1600C. The flue gases gravimetric analysis are 18% C02, 7% 02 and 75% N2. Calculate the diameter of stack in meters for a driving pressure of 0.20 Kpa. SOLUTION: Solving for the molecular weight and gas constant of the flue gas: C02 02 N2 18% 7% 75%
Mg = 1/0.03306 Mg = 30.25
Rg = 8.314/30.25 Rg = 0.275
Tg = 200 + 160 2 Tg = 180°C dg = P/RT dg=96.53 /(0.275)(180 + 273) dg = 0.775 kg/m3 vt = theoretical velocity vt = (2ghw)0.5 weight of flue gas=0.775(0.00981) 0.18/44 = 0.00409 0.07/32 = 0.00219 0.75/28 = 0.02678 0.03306
hw = 26.30 m of flue gas vt ={[2(9.81)(26.3)]0.5 v = 22.716m/sec Actual velocity = 40% vt Actual velocity = 0.40(22.716) Actual velocity = 9.1 m/sec Q = Axv
(5/0.775) = (π/4) D 2)(9.1) D = 0.95 m φ
7). A steam generator with economizer and air heater has an overall draft loss of 21.78 cm of water. If the stack gases are at 177 oC and if the atmosphere is at 101.3 Kp and 26°C, what theoretical height of stack in meters is needed when no draft f an are used? Assume that the gas constant for the flue gases is the same as that for air. A.
565
C. 545
B.
535
D. 550
SOLUTION: da = P/RT =101.325/ (0.287)(26 + 273) da=.180 kg/m3
dg = P/RgTg dg=101.3/ (0.287)(177 + 273) dg = 0.784 kg/m3
Draft = 0.2178(1000) Draft = 217.8 Kg/m2 Draft = H(da - dw) 217.8 = H(1.18-0.784)
H = 550 m 8). If the actual draft required for a furnace is 6.239 cm of water and the frictional l osses in the stack are 15% of the theoretical draft, calculate the required stack height in meters. Assume that the 'flue gas have an average temperature of 149°C and m olecular weight of 30. Assume air temperature of 21°C. A.
215
C.220
B.
230
D. 210
SOLUTION: hw = total draft
hw = 6.239 + 0.15hw hw = 7.34 cm water Chimney hw = 0.0734(9.81) hw = 0.72 Kpa da = P/RT da =101.325/(0.287)(21 + 273) da = 1.2 kg/m3 dg = P/RT d g=101.325/(8.314/30)(149 + 273)
dg = 0.867 kg/m3 hw = H(da - dg) 0.72 = H(1.2 - 0.867)(0.00981)
H = 220 m
9). A steam boiler plant consumes 9,000 kg of coal per hour and produces 20 -kg of dry f1ue gases per kg of coal fired. Outside air temperature is 32°C, average lomperature of the flue gas entering the chimney is 343°C and average temperature of dry fl ue gas in the chimney is 260°C. The gage fluid density is 994.78 kg per m3 Ilnd the theoretical draft of 2.286 cm of H20 at the chimney base is needed when the ilarometric pressure is 760 mm Hg. Determine the height of the chimney. A.
46
C. 40
B.
50
D. 56
SOLUTION: hw = H(da - dg) hw = h x w hw = (0.02286)(994.78) hw = 22.74 kg/m2 P=da R T da=P/RT P = 760 mm Hg P = 101 .325 kpa da =101.325/(0.287)(32 + 273) da = 1.157 kg/m3 dg = P /R T dg=101.325/ (0.287)(260 + 273) dg = 0.662 kg/m3 hw = H(da - dg) 22.74 = H(1.157 - 0.662)
H = 46 m
