Access full Solution Manual only here Solutions to Chemical and Engineering Thermodynamics, 5 th ed
Chapter 4
http://www.book4me.xyz/solution-manual-for-chemical-biochemical-and-engineering-thermodynamics-sandler/
4 4.1
System = Ball (1) + Water (2)
(a)
Energy balance: M1U1 f + M2U 2f − M1U1i − M2U 2i = 0
⇒ M1CV,1cT1 f − T1i h + M2 CV,2 cT2f − T 2i h = 0 ; also T1 f − T 2f . Thus f
T =
M1CV,1T1i + M2 CV,2 T 2i M1CV,1 + M2 C V,2
=
5 × 103 × 0.5 × 75 + 12 × 103 × 4.2 × 5 5 × 103 × 0.5 + 12 × 103 × 4.2
= 8.31° C [Note: Since only ∆T s are involved, ° C were used instead of K)]. dT T (b) For solids and liquids we have (eqn. 4.4-6). 4.4-6). That ∆S = M C P = MC P ln 2 for the case in T T 1
z
which C P is a constant. Thus Ball: ∆S = 5 × 10 1 03 g × 0.5
J g ⋅ K
31 + 273.15 U J R 8.31 = −53161 . V K T 75 + 273.15 W
× lnS
= −53161 . s Water: ∆S = 12 × 103 g × 4.2
J g ⋅ K
31 + 273.15 U J R 8.31 = +59622 . V K T 5 + 273.15 W
× lnS
and ∆S (Ball + Water ) = 596.22 − 531.61
4.2
J
= 64.61
J
K K Note that the system system Ball + Water is isolated. isolated. Therefore J ∆S = S gen = 6461 . K Energy balance on the combined system of casting and the oil bath
c
h
c
h
M cCV,c T f − Tci + Mo CV,o T f − T oi = 0 since there is a common final temperature.
20 kg × 0.5
kJ
kJ
dT f − 450iK + 150 kg × 2.6 kg ⋅ K dT f − 450iK = 0 kg ⋅ K
This has the solution T f = 60oC = 313.15 K Since the final temperature is known, the change in entropy of this system can be calculated 273.15 + 60 I F 273.15 + 60I . kJ from ∆S = = 20 × 0.5 × lnF H 273.15 + 450K + 150 × 2.6 × lnH 273.15 + 50K = 4135 K
4.3
Closed system energy and entropy balances dU
dV
dt
dt
;
dS
Q
+ S gen ; dt T dS Thus, in general Q = T − TS gen and dt = Q + Ws − P
=
Solutions to Chemical and Engineering Thermodynamics, 5 th ed
W s =
dU
−Q+ P
dV
dt
dt
=
Chapter 4
dU
dS dV − T + TSgen + P dt dt dt
dU
dS dV − T +P dt dt dt dU dV (a) System at constant U & & V ⇒ = 0 and =0 dt dt
Reversible work: WsRev = WsRev c S gen = 0h =
WscSgen =
dS
(b) System at constant S & & P ⇒
dt
Rev Rev
h = WS
0
= −T
dP
= 0 and
dS dt
= 0⇒ P
dt
dV dt
=
d dt
( PV )
so that
c
h
dU
Ws S gen = 0 = W Srev =
dt
+
d dt
( PV ) =
d dt
(U + PV ) =
dH dt
4.4
700 bar, 600oC
10 bar, T = = ?
Steady-state balance equations dM = 0 = M1 + M 2 dt
dU
0 dV 0 = M1 H1 + M2 H 2 = 0 = M1H1 + M2 H 2 + Q 0 + Ws − P
dt
dt
or H1 = H 2
Drawing a line of constant enthalpy on Mollier Diagram we find, at P = 10 bar, T ≅ ≅ 308° C At 700 bar and 600° C
At 10 bar and 308° C
3
V = 0.003973 m kg
V ≈ 0.2618 m3 kg
H = 3063 kJ kJ kg
H ≈ 3063 kJ kg
S = 5.522 kJ kg K
S = 7.145 kJ kg K
Also dS
⇒ Sgen = M1c S 2 − S1h or
0
dt
Q T
= 0 = M1S1 + M2 S 2 +
S gen
+ S gen = 0
M 1
4.5
= S2 − S 1 = 7.145 − 5.522 = 1.623
1
System
2
Ws
Energy balance ∆U = cU 2 f − U2i h + cU1f − U1i h = Q
adiabatic
+ W S −
z
constant
PdV volume
kJ kg ⋅ K
Solutions to Chemical and Engineering Thermodynamics, 5 th ed
c
h
c
h
Chapter 4
c
h c
h
Ws = MC p T2 f − T2i + MCp T1 f − T1i = MCp T2 f − T2i + T1 f − T1i
but T1 f = T2f = T f ⇒
W s
f
i
i
= 2T − T1 − T 2
MC P
Entropy balance adiabatic
∆S = c S2 f − S2i h + cS1f − S1i h =
Q 0 dt +Sgen T
z
0 for maximum work
cS2 f
− S2i h + cS1f − S1i h = 0 = MCP l n
R T2 f T 1 f U V=0; T T2i T 1i W
or lnS
T 2 f T 2i
+ MC P ln
T 1 f T 1i
T1 f T2f = T1i T2i ; but T1 f = T2f = T f
2
⇒ cT f h = cT1i T 2i h or T f = T1i T 2i and W s MC P
4.6
= 2T f − T1i − T2i = 2 T1i T2i − T1i − T2i
1 bar
10 bar
290 K
575 K
(a) Entropy change per mole of gas ∆ S = C P ln
T 2 T 1
− R ln
P2
eqn. (4.4-3)
P1
575 J 10 J ln − 8314 . ln = 09118 . mol K 290 mol K 1 mol K (b) System = contents of turbine (steady-state system) dN Mass balance = 0 = N1 + N 2 ⇒ − N 2 = N1 = N dt dV 0 dU 0 Energy balance + Ws − P = 0 = N1 H 1 + N2 H 2 + Q dt dt Ws = N a H 2 − H1 f = NCP a T2 − T1 f Thus ∆S = = 29.3
J
W s
W =
N
= CP aT2 − T 1f = 29.3
J mol K
× (575 − 290)K
J
= 83505 .
mol (c) In Illustration Illustration 4.5-1, 4.5-1, W = = 78348 . J mol because of irreversiblitities ( ∆S ≠ ≠ 0) , more work is done on the gas here. What happens to this this additional energy energy input? It appears as an increase of the internal energy (temperature) of the gas.
4.7
Heat loss from metal block dU dT = C P =Q dt dt
Solutions to Chemical and Engineering Thermodynamics, 5 th ed
−W =
C P
T − T 2 T
RQ = heat out of metal S−Q = heat into heat engine T
Q(−1)
dT aT − T 2 f dt
T
T 2
t
z
z F H 1 − T I K dT
0
LF NH
T 2
= −W ⇒ − Wdt = CP
T 1
−W = CP aT2 − T1f − CP ⋅ T2 l n −W = CPT 2 MG1 −
T 2 T 1
L N
= CP MaT2 − T1f − T2 l n
T 2 O T 1 PQ
T 1 I T O − − ln 2 P J T 2 K T 1 Q
T 2
Q=
Chapter 4
F
T 1 I
z C dT = C aT − T f = C T H G 1 − T J K P
P
2
1
P 2
2
T 1
Alternate way to solve the problem
T 2
System is the metal block + heat engine (closed) dU dT E.B.: = C P = Q + W dt dt
S.B.:
dS dt
Q
=
T 2
Q = T 2
+Sgen
dS
0 for maximum work
;
dU
= T 2
dS
+ W ; dU = CPdT ; dS =
C P
dt dt dt T dU dS C P T F I W = − T 2 = CP dT − T2 dT = C P 1 − 2 dT H dt dt T T K
dT
T 2
T 2
1
1
F T I F T I W = z Wdt = z CP 1 − 2 dT = C P z 1 − 2 dT H T K H T K T T T T O LF T I − W = CP aT2 − T1 f − T2 CP ln 2 = CP T 2 MG1 − 1 J − ln 2 P T 1 NH T 2 K T 1 Q
4.8
This problem is not well posed since since we do not know exactly exactly what is happening. There are several several possibilities: (1) Water contact is is very short so neither neither stream changes T very much. In this case we have the the Carnot efficiency = η =
−W Q
=
Thigh − T low T high
=
22 27 + 273
=
22 300
= 0.07 0733 = 7.33 33%
(2) Both warm surface water water (27° (27°C) and cold deep water (5° (5 °C) enter work producing device, and they leave at a common temperature.
T H T O T L
Solutions to Chemical and Engineering Thermodynamics, 5 th ed
Chapter 4
http://www.book4me.xyz/solution-manual-for-chemical-biochemical-and-engineering-thermodynamics-sandler/ dM
M.B.:
= 0 = M H + M L + M0 ⇒ M0 = −b M H + M L g
dt dU
E.B.:
= 0 = M H HH + M L HL + M0 H0 + W = 0
dt
b g = M H c H0 − HH h + M L c H0 − H L h = M H CP aT0 − TH f + M LCP aT0 − T L f
W = − M H HH − M L HL + M H + M L H0
Q = 0 = M H SH + M L SL + M0 S 0 + T dt
dS
S.B.:
0
b
+Sgen
g
0
M H SH + M L SL − M H + ML S 0 = 0
c
h
c
h
M H SH − S0 + M L SL − S0 = 0 ⇒ M H C P ln
F T H I M F T L I M H G T 0 J K H G T 0 J K
H
M H
T0 = TH
T H T 0
+ M LC P ln
T L T 0
=0
L
= 1 or TH M H TL M L = T 0 M H + M L
b M H + ML g
ML
T L
b M H + M L g
From this can calculate T 0 . Then
a
f
a
f
W = M H CP T0 − TH + M L CP T0 − TL
This can be used for any flow rate ratio. (3) Suppose very large amount of surface water water is contacted with a small small amount of deep water, i.e., i.e., M H >> M L . Then T0 ~ T H
a
f
a
f
a
f
W = M H CP TH − TH + ML CP TH − TL ~ ML CP TH − TL
(4) Suppose very large amount of deep water water is contacted with a small small amount of surface water, i.e., i.e., M H << M L , T0 ~ T L .
a
f
a
f
a
f
W = M H CP TL − TH + M LCP TL − TL ~ MH CP TL − TH
4.9
System = contents of the turbine. This is a steady-state, adiabatic, adiabatic, constant volume system. dM (a) Mass balance = 0 = M1 + M 2 or M 2 = − M 1 dt Energy balance dV constant dU adiabatic +Ws − P = 0 = M1H1 + M2 H 2 + Q volume dt dt Entropy balance
dS
Q +Sgen T
0, by problem statement
c
Thus M 2 = − M 1 = −4500 kg h
= 0 = M1S1 + M2 S 2 +
dt
h
M.B.
WS = − M1 H1 − H2
E.B.
S2 = S 1
S.B.
Solutions to Chemical and Engineering Thermodynamics, 5 th ed
State 1
T 1 = 500° C P1 = 60 bar
State 2
P2 = 10 bar S2 = S 1 = 68803 .
W s = 4500
kJ
kg
h
Chapter 4
Steam → Tables
H 1 = 34222 . kJ kJ kg
Steam → Tables
T 2 ≅ 2404 . °C
S 1 = 68803 . kJ kJ kg
H 2 ≈ 29205 . kJ kJ kg
kgK
× (2920.5 − 3422.2)
kJ kg
= − 2257650
kJ h
= − 627.1 kW
(b) Same exit pressure a P2 = 10 bar f , and still adiabatic ⇒ Ws = − M1c H1 − H2 h .
Here, however,
c
Ws = 0.8Ws ( Part a ) = 0.8 −2.258 × 106
h
kJ
= 4500c H 2 − 3422.2h
h Steam → Tables
⇒ H 2 = 30208 . kJ kg
P = 10 bar
kJ
h T 2 ≅ 2867 . K
S 2 ≈ 70677 . kJ kg K
Thus
c
h
Sgen = − M1 S1 − S2 = −4500
kg
× (6.8803 − 7.0677)
kJ
= 843.3
kJ
h kg K K⋅h (c) Flow across valve is a Joule-Thompson Joule-Thompson (isenthalpic (isenthalpic expansion) ... See Illustration Illustration 3.4-1. Thus, Hinto valve = H out of valve , and the inlet conditions to the turbine are
H1 = Hout of valve = H into valve = 34222 . kJ kJ kg
P1 = 30 bar
Steam → Tables
T 1 ≈ 4848 . °C S 1 ≈ 71874 . kJ kg K
Flow across turbine is isentropic, as in part (a) Steam → Tables
S2 = S 1 = 71874 . kJ kg K
P2 = 10 bar
T 2 ≅ 3181 . °C H 2 ≈ 30904 . kJ kJ kg
kJ kJ × (3090.4 − 3422.2) = − 1.493 × 106 = − 414.8 kW h kg h Since compression compression is isentropic, isentropic, and and gas is ideal ideal with constant heat capacity, we have W s = 4500
kg
4.10
F T 2 I = F P2 I R C = H G T 1 J K H G P1 J K
P
F P I R C So that T2 = T 1G 2 J H P1 K
P
8.3 14 14 36.8 F 3 × 106 I = 29815 . G = 32675 . K . H 2 × 106 J K
Problem 3.31, that Ws = NCP aT2 − T1 f
Now using, from solution to
Solutions to Chemical and Engineering Thermodynamics, 5 th ed
W s = 125
kg
s
×
1 mol 16 g
× 36.8
J mol K
Chapter 4
× (326.75− 298.15)K ×
1000 g kg
6
= 8.23 × 10 J s The load on the gas cooler is, from Problem 3.31,
a
f
Q = NC p T3 − T2
=
125 kg kg s × 1000 g kg
× 36.8
16 g mol
J mol K
× (298.15 − 326.75) K
= − 8.23 × 10+6 J s
4.11
(a) This is a Joule-Thomson expansion ⇒ H (70 bar, T = ?) = H(1.0133 bar, T = 400° C) ≈ H(1 bar, T = 400° C)
= 32782 . kJ kJ kg and T = 447° C , S = 6619 661 . 9 kJ kg K
(b) If turbine is adiabatic adiabatic and reversible reversible c S gen = 0h , then Sout = S in = 6.619 kJ kg K and P = 1013 .
bar. This suggests that that a two-phase mixture is leaving leaving the turbine S V = 73594 . kJ kg K
Let x = fraction vapor
S L = 13026 . kJ kg K
)(1.3026) = 6.619 kJ kg K or x = 08778 Then x(7.3594) + (1 − x)( . .
Therefore the enthalpy of
fluid leaving turbine is H = 0.8788 ×
H
+ (1 − 0.8778) ×
2675.5 V
(sat’d, 1 bar )
= 2399.6
417. 46 L
H (sat’d, 1 bar )
Energy balance dV 0 0 = M in Hin + Mout H out + Q 0 +Ws − P
dt
but M in = − M out
W s
⇒−
M in
= 3278.2 − 2399.6 = 878.6
kJ kg
(c) Saturated vapor at at 1 bar S = 7.3594 kJ kg K ; H = 2675.5 kJ kg
−
W s M in
= 3278.2 − 2675.5 = 602.7 kJ kJ kg Actual
Efficiency (% ) = S gen
602.7 × 100 8786 .
= 686% .
M in
= 7.35 3594 − 6.61 619 = 0. 74 740 kJ Kh
kJ kg
Solutions to Chemical and Engineering Thermodynamics, 5 th ed
(d)
Chapter 4
0 = M1 + M2 ⇒ M2 = − M 1
W
Steam
Water
70 bar
1 bar
447° C
25° C
0 = M1c H1 − H2 h + Ws + Q − P
0 = M1c S1 − S 2 h +
Q
dt
Q
dV
+ S gen
T
Simplifications Simplifications to balance equations dV S gen = 0 (for maximum work); P = 0 (constant volume) dt Q Q = where T 0 = 25° C (all heat transfer at ambient ambient temperature) T T 0
H (sat'd sat'd liq, liq, 25° C) = 104.89
Q
= T0 c S2 − S 1 h ;
M
−W s
−W s
M
kJ kg
; S (sat'd sat'd liq, liq, T = 25° C) = 0.3674
kJ kg K
= H1 − H2 + T0 c S2 − S1 h = c H1 − T0 S1 h − c H2 − T0 S 2 h
max
M
= 3278.2 − 298.15 × 6.61 619 − 104.89 89 − 298.15 × 0.36 3674 max
= 1304.75 75 + 4.65 65 = 1309.4 kJ kJ kg 4.12
Take that that portion portion of the methane methane initially initially in the tank that that is also in the tank finally to be in the system. This system is isentropic S f = S i . (a) The ideal gas solution
F I R C H G J K
P f S f = S i ⇒ Tf = Ti Pi N =
PV RT
⇒ N i =
PV i RT i
p
14 3 6 35 . 8.3 14 = 300F I = 1502 . K H 70 K
= 1964.6 mol; N f =
P f V RT f
∆ N = N f − N i = −17684 . mol (b) Using Figure 2.4-2. 70 bar ≈ 7 MPa, T = = 300 K Vi = 00195 .
m3 kg
28
, so that mi =
0.7 m3 00195 .
35.90 kg × 1000 N i =
Si = 505 . kJ kJ kg K = S f
g mol
m3 kg
g kg
= 1282 mol
= 3590 . kg.
= 196.2 mol
Solutions to Chemical and Engineering Thermodynamics, 5 th ed
Chapter 4
At 3.5 bar = 0.35 MPa and S f = 5.05 kJ kg K ⇒ T ≈ 138 K. Also,
V f = 0192 .
m3 kg
0.7 m3
, so that m f =
0192 .
28
kg
g
3646 . kg × 1000 N f =
= 3.646 kg. kg.
m3
kg
g
= 1302 . mol
mol
∆ N = N f − N i = 130.2 − 1282 = −1151.8 mol 4.13
d S = C
∆S =
dT T
+ R
dV V
eqn. (4.4-1)
b T + cT z LMN(a − R) + bT
2
+ dT 3 +
e 2
T
dV O dT PQ T + Rz V
so that S aT2 , V 2 f − SaT1, V 1f = ( a − R) ln
+
d
3
T 2 T 1
+ baT2 − T 1 f +
c
2
cT22 − T 12 h
e
V 2
2
V 1
cT23 − T 13 h − cT2−2 − T1−2 h + R ln
Now using PV = RT ⇒
a
f
a
f
S T2 , P2 − S T1, P1 = a ln
+
T 2 T 1
V 2 V 1
=
T 2
⋅
P1
T 1 P2
+ baT2 − T 1f +
d
⇒ c
T22 − T 12 h c 2
e
P
T23 − T 13 h − cT2−2 − T1−2 h − R ln 2 c 3 2 P 1
Finally, eliminating T 2 using T2 = T1 P2 V 2 P1V 1 yields
a
f
a
f
S P2 ,V 2 − S P1, V 1 = a ln
+ + −
F P2V 2 I + b + a P2V 2 − P1V 1 f GH PV J 1 1 K R
c 2
a P2V 2 f2 − a P1V 1f 2
3
a P2V 2 f3 − a P1V 1f 3
2 R d 3 R
eR2
2
P
d P2V 2 −2 i − d P1V 1−2 i − R ln P2
1
Solutions to Chemical and Engineering Thermodynamics, 5 th ed
Chapter 4
http://www.book4me.xyz/solution-manual-for-chemical-biochemical-and-engineering-thermodynamics-sandler/ 4.14
System: contents of valve (steady-state, adiabatic, constant volume system)
Mass balance
0 = N1 + N 2
Energy balance
0 dV 0 0 = N1 H1 + N 2 H 2 + Q 0 + Ws − P
dt
⇒ H 1 = H 2
Q T
Entropy balance 0 = N1 S1 + N2 S 2 + S gen +
0
S gen
⇒ ∆ S = S 2 − S 1 =
N (a) Using the Mollier Mollier Diagram for for steam (Fig. (Fig. 3.3-1a) or the Steam Tables Tables T 1 = 600 K P2 = 7 bar
⇒
P1 = 35 bar H 2 = 30453 . J g
T 2 ≈ 293° C S 2 = 7277 . J g K
H1 = H 2 = 30453 . J g . Thus S 1 = 65598 . J g K ; K ; T exit = 293° C
∆S = S2 − S 1 = 0.7 17 17 J g K
(b) For the the ideal ideal gas, H 1 = H 2 ⇒ T1 = T 2 = 600 K ∆ S = S aT2 , P2 f − SaT1 , P1 f = C p ln = − R ln
P2 P1
T 2 T 1
− R ln
P2 P1
= 1338 . J mol K ⇒
∆S = 0743 . J mol K
4.15
From the Steam Tables P = 15538 . MPa V L = 0.001157 m3 / kg
V V = 0.12736 m3 / kg
U L = 850.65 kJ / kg
U V = 2595.3 kJ / kg
H L = 852.45 kJ / kg
H L = 2793.2 kJ / kg
S L = 2.3309 kJ / kg ⋅ K
S V = 6.4323 kJ / kg ⋅ K
o
At 200 C,
∆ H vap = 1940.7 kJ / kg k g ∆S vap = 4.1014 kJ / kg kg ⋅ K (a) Now assuming that there there will be a vapor-liquid mixture mixture in the tank at the end, the properties properties of the steam and water will b e P = 0.4578 MPa MPa
V L = 0.001091 m3 / kg
V V = 0.3928 m3 / kg
U L = 631.68 kJ / kg
U V = 2559.5 kJ / kg
H L = 632.20 kJ / kg
H V = 2746.5 kJ / kg
S L = 1.8418 kJ / kg ⋅ K
S V = 6.8379 kJ / kg ⋅ K
∆ H vap = 2114.3 kJ / kg
∆S vap = 4.9960 kJ / kg ⋅ K
o
At 150 C,
(b) For simplicity simplicity of calculations, calculations, assume assume 1 m3 volume of tank.
