Answers to Selected Problems

Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Chapter 15
Chapter 16
Chapter 17
Chapter 18
Chapter 19

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3.1
(a) 268.8 K and 4.625105 Pa
(b) at 90o (horizontal) 282.6 K and 4.863105 Pa
(c) at 180o (vertical downward) 296.4 K and 5.1105 Pa
 

3.2
(a) Q/t = 3.51108 J/hr [contractor pays]
(b) Q/t = 3.23108 J/hr [contractor collects fee]
 

3.5
(a) LA = 0.4352 m, LB = 0.0218 m, TA = TB = 311 K, PA = PB = 7 bar
(b) Same as (a)
(c) There is no real solution given the constraints, we must specify more about the expansion (or compression) of A or B. For example, if A expands adiabatically, PA = PB = 6.68 bar, TA = 263.6 K, TB = 881.3 K, LA = 0.3917 m, and LB = 0.0655 m.
 

  3.7
Many variations are possible. If we assume gas in the cylinder is ideal and that it expands adiabatically, then (a) total time t = 0.9 s, (b) height h(max) = 4.78 m, by reducing tube length to 1.64 m.
 

3.12
(a) After 6 s, T = 349.5 K, P = 1.139105 Pa in the bulge.
(b) After 3 s, T = 454.6 K, P = 6.26103 Pa in the large tank.


4.1
For the minimum condition TA(final) = 132 K and for the maximum condition TA(final) =617 K, these extrema require using the object with the lowest mass x heat capacity product.
 

4.2
W = 9.807y in J/kg evaporated, for typical values of ambient dry bulb and wet bulb temperatures of 300 K and 280 K respectively, y = 16.8 km!
 

4.3
(a) B = 0.166 mol/s,-C = 0.833 mol/s
(b) TD = 299.1K
(c) W = -26.4 J/s
(d) S= 13.3 J/K
(e) Same as (d).
(f) = -4000 J/s
 

4.4 W = - 1.11104 J, Tf = 288.7 K, Pf = 0.54 bar
 

4.6
(a) 21.6 s
(b) 391.8 K and 2.2 bar at 10 s
(c) Sgas= 4.93103 J/K = Suniverse
(d) S(gas in tank) = -813.6 J/K,
-----S(gas vented) = +2020 J/K,
-----S(surroundings) = +1203 J/K
(e) Wmin = +2.83105 J
 

4.7
(a) = 2400 W, T = 327.7 K, and Q = 22,200 J after 10 s
(b) T = 1750 K at t = 57.7 s with all stored work consumed.
 

4.9
(a) Wnet = - 3.17106 J
 

4.12
(a) T2 = 227.9 K, T1 = 392.9 K (2) after venting (1) before venting
(b) T2 = 211.1 K, T1= 363.9 K
(c) W = 2547 J/mol
(d) Wmax = 1.38105 J/kg
 

4.13
(a) S = -1.38103 J/K hr
(b) S = 2 J/K hr
 

4.16
(a) Case (1) 1.9810-2 kWhr; (2) 2.7810-2 kWhr; (3) 2.1610-2 kWhr; (4) 2.7810-2 kWhr
(b) Case (1) 420 K, 42 J/K; (2) 300 K, 0 J/K; (3) 300 K, 73.9 J/K; (4) 300 K, 0 J/K
 

4.18 W = 5.23103 J = -Q

4.22
(a) P = 6.35104 Pa
(b) T = 272.8 K
(c) P = 3.68104 Pa with 2 pumps
(d) P = 1.91104 Pa and T
(e) Power (minimum) = 20.9 W
 

4.24
(a) Wmax= -198.76 kJ/kg (Carnot + expansion work)
(b) Wp = ideal pump work (estimated to include PE for lifting water and for the compression of gas space in storage tanks)
 

4.30
(a)  = 6.38106 J/s
(b)= 0.90 = (actual power)/(maximum power)
(c) no, the maximum power outputs are the same


5.2
(a) 
(b) 

5.4
(a) 
(b) 
(c) U
(d) 
 

5.7
(a) 
(b) 
 

5.10 Tf = 1087 K
 

5.12 
 

5.21 Vc RT = (Cp/Cv) RT » 330 m/s (use mass units for R with molecular weight) distance at 2 s » 660 m
 

5.22
(a) 
 

5.23
(b) No, the ratio as shown in Problem 5.17 is expressed in terms of PVT properties and an isentropic derivative which in turn requires non-PVT property information specifically related to the temperature dependence of energy stored in U.
 
 

5.24
(a) 
(b) 
(c) 
(d)
(e) 
 

5.25
(a) 
(b) 
(c) Yes, using the Gibbs-Duhem relationship:


6.2

thus,  at equilibrium
 

6.3
- from chemical equilibrium criteria, intersections define equilibrium states.
 

6.4
(a) 
----
-----
-----
 

6.5
(a)
(b) T = 366.5 K, PA,1 = PA,2 = 2 bar, PB,2 = 1 bar
      NB,1 = 0 NB,2 = 1 mole NA,1 + NA,2 = 4 moles
(c) P1 = 2 bar, N1 = 5 moles; V1 = 2.5 RT/105 = 0.0752 m3 with T = 367 K and no gas on side 2.
 

6.7
Fraction evaporated 0.227



7.1
if ANN = 0 then AVV = 0 = ANN /V 2
 

7.3
T = 55.1 K
 

7.6
for  = 2.5
 
xA  xB
0.1 0.286, 0.614
0.2
0.4 (critical pt.)
0.3
no solution
0.4 0.2, 0.4 (critical pts.)
0.5
0.175, 0.325
0.6
0.085, 0.28

7.11


8.4
At 365.8 K, 16.5 bar

 

8.5
At T = 319.4 K
 
Pr KH/KS Z = PV/RT
0.1 0.100 0.967
0.2 0.194 0.932
0.4 0.369 0.862
1.0 0.798 0.598
4.0 0.426 0.547
10.0 -0.296 1.108

 

8.9
Using an isenthalpic expansion across an insulated value to 1 bar, 0.236 kg of dry ice could
be produced per kg of CO2 drawn from the cylinder.  For an isentropic expansion, 0.41 kg dry ice/kg of CO2 is possible.
 
 

8.10
(a) Tf = 379.3 K at 48.3 bar
(b) W = - 9.1107 J
 
 

8.14
TB,f|256 to 257 K, PB,f|21.5 bar, xg = 0.63 (fraction vapor)
 
 

8.15
(a) Power = 508 kW
(b) Power = 516 kW
 
 

8.30
- = 90.9 kW and- = 68.3 kW
 
 

8.31

Yes, the vdW EOS gives at the Zeno condition (Z = 1) a straight line: Tr = T/Tc = 27/8 - (9/8) r



9.1
(a) -31.36 kJ/mole water added to acid at the start; -61.1 kJ/mole and added to water
at the start
(b) maximum heat load occurs at the start
(c) total Q = - 28 kJ/mol acid = -11.2 kJ/mol of solution same for both cases
 

9.2
(b) Q = - 3.86 105 J
(c)

(d)
 

9.3
mole % NH3 = 51.37 %
 

9.6
W = 326 J
 

9.7

w = water and s = salt (NaCl)
 

9.8
(a) Wmin = -1069 J (PREOS)
(b) Wmin = -1812 J (ideal gas mixture)
 

9.9

 

9.11
 
x salt (wt fraction)
0.01 7.2 8.5
0.05 42.2 43.9
0.10 90.5 91.1
0.15 156.6 142.0
0.20 242.0 197.0
0.25 355.0 257.0

9.13


 

9.14
(a) 
(b) At -85oC (188 K), there are two phases in equilibrium at xEtOH = 0.07 and 0.35.
 

9.20
= 675 W

 

9.21
(a) 1.32105 W
(b) See Section 14.4 for details, information provided in part (a) is not sufficient, ideal gas state heat capacities and a mixture PVTN EOS are needed.
 

9.27


10.2
H - Ho = PV - RT - a/V
 

10.3




 

10.4

10.6 



 

10.7
b = 
a = 
 

10.9
(a) 
(b) 
(c) for T/Tc = 1.00001,

 

10.10


11.4
(a) 
(b)
(c) for only positive values of , a large number of values 
exist as a function of xi that yield extrema, e.g.
for x1 = 0.2
 
 
0 3.236
0.500 1.699
1.000 1.000
1.500 0.586
2.000 0.209
2.118 0

 

11.5
for 
for 
 

11.7
(c) fraction of solution precipitated = 0.0067
 

11.8
for 2-suffix Margules Tc = w/2 k
 

11.9
for quasi-chemical model Tc = w/2.23 k


12.3
(a) selected values of are given below for LiCl
 
molality -(Meissner)  -(Pitzer)
0.1 0.9651 0.9653
0.01 0.9032 0.9044
0.1 0.7850 0.7927
1.0 0.7636 0.7758
5.0 1.9586 2.0222
10.0 7.9948 11.0827

(b) 
given that s = 9 mol/kg for pure LiCl in H2O. Using the Meissner model for  predictions for both pure and mixed electrolyte.
        Ksp(9)2 (~6)2 = 2916 for pure LiCl at 25oC

        Iterating s*=2.4 and  3H for mixed LiCl and HCl at 25oC
(c) Difficult to say without more information on how much Na2SO4 influences for LiCl.  For example is the effect on  large enough to compensate for the Cl- common ion effect?
 

12.8
Some representative values for 

molality (NaCl or LiCl)(CuCl2)
0.1                                 0.96                                         0.55
1.0                                 0.30                                         0.18
10.0                               0.096                                       0.055

At T = 300 o22.3 so multiply above values by (22.3 573)/(78.5)(298) = 0.74
 

12.10
I = 0.546 or m = I / 3 = 0.182 molal


13.2
Using Joback’s method with Tb = 225.5K (exp. value)
 -Tc = 364 K, Pc = 46.7 bar, and = 0.1606
- = -2.83 + 0.269 T = 3.1510-4 T 2 + 4.210-8 T 3
 

13.3
Tc = 663.5 K, Pc = 32.21 bar, Vc = 429.5 cm3/mol, and Tb = 450.38 K
Using Eq. (13-18) for vapor pressure:

 

13.4
For pure caffeine, Tb640 K, Tc = 872 K, Vc = 488.5 cm3/mol, and Pc = 41.46 bar
Using the PR EOS for density estimate for 5 mol % caffeine CO2 mixture at 80 bar, 310 K = 1151 kg/m3. The vapor pressure of liquid caffeine is approximated by Eq. (13-18)



14.2
Using the PR EOS,




 

14.6
(a) 
(b) 
(c) Using the RKEOS and a suitable mixing rule, amix and bmix parameters can be calculated and used to calculate all PVTN properties. With an ideal-gas state Cp all derived properties needed for the cycle calculation can be obtained using a departure function approach. Then,

 

14.8
(a) 
(b)  (about 150 times smaller!)
 

14.12
(a) and (b) 
      so CTI’s claim of 1 kWhr production violates the 2nd Law limit
(c) All heat transfer and work production and utilization steps have some irreversibility in practical systems.
 

14.14
(a) 
(b) no fundamental laws or concepts are violated
(c)  about 17% more area for heat exchange is required for the GH process
(d) 

NB: If an electric motor drive is used for the feed pump in the conventional Rankine system, then the net output would be reduced possibly accounting for the discrepancy.


15.1

 

15.3
x (Naphthalene) = 0.196
 

15.6
(c) q (critical quality) = 
(e) O2: q = 0.54 and H2: q = 0.36
 

15.12
xKCl = 0.413 and T = 624 K
 

15.13

 
 

15.15
Anesthetic pressure of CCl4- 4.910-3 bar.
 
 

15.19
(a) 
(b) 
(c)
     (i) , solution is regular
     (ii), less structural ordering in mixture versus pure state
     (iii), more structural ordering in mixture versus pure state, implies 


16.9
(a) yA = 0.764 yB = 0.236
(b) not in equilibrium, estimate yB assuming equilibrium
 

16.12

 

16.14

 

16.16
using expansion valve
using expansion turbine
Consider net energy, entropy, and work flows using 1st and 2nd Law concepts to show process is not feasible as described.
 

16.20
(a) P = 21.9 bar
(b) T = 51.5 K
(c) 
 

16.22
T, K       Cp (effective), J/gK
293                         4.69
313                         7.20
333                         8.29
353                         6.36
373                         3.64


17.3
(a) ; at A: L1-L2-H-G and at B: L1-H-I-G
(b) L1 = 0 and M1 = 0 use P-explicit EOS to determine required Aijk and Aij derivatives
(c) 
(d) 
in this case
 

17.6
d ln P/d(1/T) = 


18.1

 

18.4
J/kg mol
 

18.5
Ti = 269 K = -4oC (winter!)
 

18.6
Tbath = 317 K, efficiency = 0.065
 
 

18.8
z = depth = 764 m
 

18.9
(tube bottom) = 0.501, if 4/7
 

18.11

 
 

18.14
ytritium = 0.065 and ydeuterium = 0.538 at rim
 

18.16
h = 252 m, for equilibrium ocean, fresh water cannot rise above a level 252 m below sea surface. For the well-mixed case, process becomes feasible if Z = depth > 9850 m (32,000 ft.).


19.1

 

19.2

 

19.5

 

19.7
xbutanol = 0.436
 

19.8
(a) 0.046 J/m2
(b) 296 K


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