Answers to Selected Problems
3.1 (a) 268.8 K and 4.625 × 10 Pa; (b) at 90 (horizontal) 282.6 K and 4.863 × 10 Pa; (c) at 180 5 (vertical downward) 296.4 K and 5.1 × 10 Pa. 5
3.2 (a) (b)
o
5
o
δ Q / δ t = 3.51 × 108 J/hr [contractor pays] δ Q / δ t = 3.23 × 108 J/hr [contractor collects fee]
3.5 (a) L A = 0.4352 m, L B = 0.0218 m, T A = T B = 311 K, P A = P B = 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, P A = P B = 6.68 bar, T A = 263.6 K, T B = 881.3 K, L A = 0.3917 m, and L B = 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 = 1.139 × 10 Pa in the bulge. 3 (b) After 3 s, T = 454.6 K, P = 6.26 × 10 Pa in the large tank. 5
4.1 For the minimum minimum condition T A(final) = 132 K and for the maximum condition T A(final) = 617 K, these extrema require using the object with the lowest mass x heat capacity product. 4.2 W = 9.807 y 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) n B = 0.166 mol/s, n C = 0.833 mol/s (b) T D = 299.1K (c) W = 26.4 J/s (d) ∆S= 13.3 J/K (e) Same as (d). (f) W max = 4000 J/s !
!
4.4 W = −1.11 × 10 J, T f = 288.7 K, P f = 0.54 bar 4
4.6 (a) 21.6 s (b) 391.8 K and 2.2 bar at 10 s 3 (c) ∆Sgas= 4.93 × 10 J/K = ∆Suniverse (d) ∆S(gas in tank) = 813.6 J/K ∆S(gas vented) = +2020 J/K ∆S(surroundings) = +1203 J/K 5 (e) W min min = +2.83 × 10 J !
Updated: 1/14/04
4.7
(a) Q = 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
W net = −3.17 × 10 J
4.12
(a) (b) (c) (d)
T 2 = 227.9 K, T 1 = 392.9 K (2) after venting (1) before venting T 2 = 211.1 K, T 1= 363.9 K W = 2547 J/mol 5 W max = 1.38 × 10 J/kg
4.15
(a)
∆ S = −1.38 × 103 J/K hr; (b) ∆ S = 2 J/K hr
4.16
(a)
Case (1) 1.98 × 10 kWhr; (2) 2.78 × 10 kWhr; (3) 2.16 × 10 kWhr; 2 (4) 2.78 × 10− kWhr 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
6
(b)
−2
−2
−2
4.18
W = 5.23 × 10 J = Q
4.22
(a) (b) (c) (d) (e)
Power (minimum) = 20.9 W
(a)
W max = 198.76 kJ/kg (Carnot + expansion work)
4.24
3
!
P = 6.35 × 10 Pa T = 272.8 K 4 P = 3.68 × 10 Pa with 2 pumps 4 P = 1.91 × 10 Pa and T → ∞ 4
!
W p = ideal pump work (estimated to include PE for lifting water and for the compression of gas space in storage tanks)
4.30
5.2
(a)
W max = 6.38 × 10 J/s
6
(b)
ηu = 0.90 = (actual power)/(maximum power)
(c)
no, the maximum power outputs are the same
(a)
(b)
y
y
(n+1)
(1)
= y
(0)
− T S − ∑ µ j N j
= S − (1 / T )U
dy
(1)
dy
( n +1)
n
= − SdT − ∑ N j d µ j − Pd V j =1
n
= −U d (1 / T ) + (P / T )d V − ∑ (µ j / T )dN j j =1
5.4
(b)
(∂P / ∂T )V , N i (∂P / ∂T )T, N = G P + T (− G PT ) = V − T (∂V / ∂T ) P, N
(c) (d)
U b∂H / ∂V gT, N = V / GPP − TGPT / GPP
(a)
Updated: 1/14/04
5.7
(∂S / ∂T )G, N = − S (∂V / ∂T ) P / V + C P / T (∂ A / ∂G )T, N = − P (∂V / ∂T )T / V = −(∂ lnV / ∂ ln P )T
(a) (b)
5.10
T f = 1087 K
5.12
(1) y 22
= − NC v / T = − NC p / T + V α p2 / κ T
5.21
V c = κ RT = (C p / Cv ) RT ≈ 330 m/s (use mass units for R with molecular weight) distance at 2 s ≈ 660 m
5.22
(a)
C p
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)
dy
(b)
y
(c)
dy (5)
(d)
(∂F 1 / ∂µ1 )T ,V , N 2 , Z = −(∂ N 1 / ∂ Z 1 )µ1 ,T ,V , N 2
(e)
(1) y SSS
(a)
y
(b)
(∂µ1 / ∂B)T ,V , H , N 2 = V (∂H / ∂ N 1 ) T ,V ,µ2 , N 2
(c)
Yes, using the Gibbs-Duhem relationship
5.25
− C v = V α p2 / κ T
(0 )
(0 )
(5)
= − SdT − Pd V + µ 1dN 1 + µ 2 dN 2 + F 1 d Z
= − PV + µ1 N 1 + µ 2 N 2 + F 1 Z 1 = +Td S + V dP − N 1d µ1 − N 2 d µ 2 + Z 1dF 1 ≠ 0
=
(
**
T 2 ∂C V / ∂T
=0
dy
(5)
= − SdT + V dP − N 1d µ1 − N 2 d µ 2 − V Bd H
µ 2 − µ °2 = ∫ dµ 2 = − ∫
6.2
(
dA=− P V
thus, P
6.3
V
− T / (C V ** ) 3
x1
(1 − x1 )
dµ1
= f (µ1 )
− PL ) dV V + (µV −µ L ) dN V ≥ 0
= P L and µV = µ L
T = T i / (1 + α ) and
α = f (T )
at equilibrium
from chemical equilibrium criteria, intersections define
equilibrium states. Updated: 1/14/04
6.4
() T I
(a)
P
= T (II) = T ( III)
(II )
= P (III)
( I ) ( II ) µ H = µ H 2
2
∑v µ j
6.5
j
=0
= V 2 = RT / 105 µ A,1 = µ A,2
(a) V 1
(b) T = 366.5 K, P A,1 = P A,2 = 2 bar, P B,2 = 1 bar N B,1 = 0 N B,2 = 1 mole N A,1 + N A,2 = 4 moles 5 3 (c) P1 = 2 bar, N 1 = 5 moles; V 1 = 2.5 RT /10 = 0.0752 m with T = 367 K and no gas on side 2.
ε = 0.227 Q = ∆U = 136 ε (in J ) ∆ S = ε [0.581 − 0.462 (6 × 10 −2 ) ln (4.40 ε )] ∆ A = −5.63 ε [1 − ln ( 4.40 ε )]
6.7
Fraction evaporated
7.1
2 if A NN = 0 then AVV = 0 = A NN / V
7.3
T = 55.1 K
7.6
for x A 0.1 0.2 0.3 0.4 0.5 0.6
7.11
α = 2.5 x B 0.286, 0.614 0.4 (critical pt.) no solution 0.2, 0.4 (critical pt.) 0.175, 0.325 0.085, 0.28
∆G = 221.3 J / mol at 10°C
or
= −217.8 J/mol at −10°C
Updated: 1/14/04
8.4
At 365.8 K, 16.5 bar
(∂S / ∂P)V = [C p / T + (∂P / ∂T )V 2 (∂V / ∂P )T ] / (∂P / ∂T )V = 3.33 × 10 −5 J/mol 8.5
At T = 319.4 K Pr 6 / 6S H 0.1 0.100 0.2 0.194 0.4 0.369 1.0 0.798 4.0 0.426 −0.296 10.0
(
K N/m 2
)−
1
Z = PV/RT 0.967 0.932 0.862 0.598 0.547 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 CO 2 drawn from the cylinder. For an isentropic expansion, 0.41 kg dry ice/kg of CO 2 is possible.
8.10
(a) T f = 379.3 K at 48.3 bar 7 (b) W = −9.1 × 10 J
8.14
T B,f
8.15
(a) Power = 508 kW (b) Power = 516 kW
8.30
= 90.9 kW and Q = 68.3 kW W
8.31
Yes, the vdW EOS gives at the Zeno condition ( Z = 1) a straight line: T r = T/T c = 27/8 - (9/8) ρ 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
≈
256 to 257 K, P B,f ≈ 21.5 bar, xg = 0.63 (fraction vapor)
!
-
9.2
(b) Q = −3.86
×
!
5
10 J
(c) δ N / ∂t = −6.43 × 10 3 /[( H A − H A ) x A,in
+ ( H A − H w ) x w,in − ∆ H mix,in ]
δQ / δt = −3.96 × 10 5 /(60) J / min (d) δ N / ∂t at 64.9wt % = 41.8 kg / min 9.3
mole % NH3 = 51.37 %
9.6
W = 326 J Updated: 1/14/04
9.7
( dS / dt ) universe
= [(µ w,in − µ w )n w + (µ s,in − µ s )n s ] / T
w = water and s = salt (NaCl) 9.8
(a) W min (b) W min
9.9 9.11
= −1069 J ( PR EOS) = −1812 J (ideal gas mixture )
= W 3 MW
Π(bar )
x salt (wt fraction)
0.01 0.05 0.10 0.15 0.20 0.25
7.2 42.2 90.5 156.6 242.0 355.0
9.13
∆ H mix = Q = 1624 ∆ S mix = 12.34 J/K
9.14
(a)
Π ID (bar ), (completely dissociated) 8.5 43.9 91.1 142.0 197.0 257.0
J
γ EtOH = 1.18 γ MeOH = 4.17 at 0.8 = x EtOH o
(b) At 85 C (188 K), there are two phases in equilibrium at xEtOH = 0.07 and 0.35. !
9.20
δQ/ δt = 675 W δW min / δt = 108 W
9.21
= 1.32 × 10 W (a) 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
∂W max / ∂t = 676 kW
10.2
5
o
!
!
r = 2 A r = 3
r = 5 r = 10
10.4
ηu = 0.623
not 0.95
H H = PV RT a/V
10.3
while Gyro claims − 400 kW which gives
!
Φ ij,max / kT = 20.43 Φ ij, max / kT = 6.05 Φ ij,max / kT = 1.31 Φ ij,max / kT = 0.16
< Φ i, j > = 0 Updated: 1/14/04
10.6
r , A
Φ ( H 2O − H 2O )
2 10
( 10 J) − 288.42 − 0.0185
Φ (CO −CO )
−21
10.7
−21
( 10 J) − 2.52 − 0.000164
b = 4 RT c /( 30 Pc ) = 4V c / 30 a = 1 + 4 RT c V c 30
10.9
>] / ρ = (1 − bρ) / V 1 / 2 [ρ(1 − 2βaρ(1 − bρ) 2 )]1 / 2 (b) ρ c = 1 /(3b) and β c = 27b /(8a )
(a) [< ( δρ) 2
∆I / I o ≈ 9.29 × 10 −5
(c) for T/T c = 1.00001 10.10
φˆ CO2 = 0.245
11.4
(a)
Λ 12 = 0.0961, Λ 21 = 1.0273, β = − 0.8381 ∞ ∞ (b) γ 1 = −10.13, γ 2 = 2.41 (c) for only positive values of Λ ij , a large number of values [Λ ij , Λ ji ] exist as a function of xi that yield extrema, e.g. for x1 = 0.2
11.5
for x1 = x 2 for x1
=
Λ 12
Λ 21
0 0.500 1.000 1.500 2.000 2.118
3.236 1.699 1.000 0.586 0.209 0
0.5
= 0.1, x 2 = − 0.9
α max = 0.4276 α max = 0.6177
11.7
(c) fraction of solution precipitated = 0.0067
11.8
for 2-suffix Margules T c = w /2 k
11.9
for quasi-chemical model T c = w /2.23 k
Updated: 1/14/04
12.3
(a) selected values of γ ± are given below for LiCl molality 0.1 0.01 0.1 1.0 5.0 10.0 (b) s * ( s *
+ 10) γ ± = exp[ − ∆G
rx / RT ]
γ ± (Meissner)
γ ± (Pitzer)
0.9651 0.9032 0.7850 0.7636 1.9586 7.9948
0.9653 0.9044 0.7927 0.7758 2.0222 11.0827
= f (T only)
given that s = 9 mol/kg for pure LiCl in H 2O. Using the Meissner model for predictions for both pure and mixed electrolyte. 2 2 o K sp ≈ (9) (~6) = 2916 for pure LiCl at 25 C
γ±
γ 2± ( HCl(10m ) + LiCl( s * ) ( s * + 10) = 2916 * o Iterating s =2.4 and γ ± = 10 3H for mixed LiCl and HCl at 25 C (c) Difficult to say without more information on how much Na 2SO4 influences γ ± for LiCl. For example is the effect on γ ± large enough to compensate for the Cl common ion effect? !
12.8
Some representative values for
κ
follow for T = 25 o C, D s
κ −1 , nm (NaCl or LiCl)
molality 0.1 1.0 10.0 o
At T = 300 C D s
= 78.5 κ −1 , nm (CuCl2)
0.96 0.30 0.096
0.55 0.18 0.055
≈ 22.3 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 T b = 225.5K (exp. value) T c = 364 K, Pc = 46.7 bar, and ω = 0.1606 4 2 8 3 o C p = 2.83 + 0.269 T + 3.15 × 10 T + 4.2 × 10 T !
!
!
13.3
3
T c = 663.5 K, Pc = 32.21 bar, V c = 429.5 cm /mol, and T b = 450.38 K Using Eq. (13-18) for vapor pressure: 663.5 ln Pvp = 7.3106(1 − ) − 3.4723 T Updated: 1/14/04
13.4
For pure caffeine, T b ≈ 640 K, T c = 872 K, V c = 488.5 cm /mol, and Pc = 41.46 bar Using the PR EOS for density estimate for 5 mol % caffeine CO 2 mixture at 80 bar, 310 K ρ mix ≈ 1151 kg/m3. The vapor pressure of liquid caffeine is approximated by Eq. (13-18) 3
⎛ Pvp ⎞ ⎛ ln( Pc / 1.013) ⎞ T b ⎛ 1 ⎞ ⎟⎟ = ⎜⎜ ⎟⎟ ⎜⎜1 − ⎟⎟ − 1 / / P T T T T T b c ⎠ c ⎝ c ⎠ ⎝ c ⎠ ⎝
ln⎜⎜
ln Pvp
14.2
872 ⎞ = 9.571⎛ ⎜1 − ⎟ − 3.482 T ⎝ ⎠
T in K and Pc in bar .
(in bar)
Using the PR EOS, 8 for geofluid = ( H ∆ H in − H out ) m gf = 7.97 × 10 J/s for R − 115 = 2.30 × 10 4 J/mol ∆ H m R −115 = 5370 kg/s m R −115 / m gf = 2.87
14.6
= C p (T 2 − T 1 ) + T o C p ln (T 2 / T 1 ) = ∆ B ηnet (at T 2 = 600 K ) = 0.52
(a) W max (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 C p all derived properties needed for the cycle calculation can be obtained using a departure function approach. Then, m sat wt sat ≈ η − − W m ( H H ) V l ( P A − Pvp (T o )) net B C t wf
ηp
14.8
⎛ 1 − η V gsat ⎞⎛ m ⎞ πW cycle ⎞⎟⎛ ⎜ ⎟⎜ ⎜ ⎟ (a) a* = ⎜ 0.428 ⎜ ηcycle ⎟ ∆ H vap ⎟⎜⎝ RT o ⎠⎟ ⎝ ⎠⎝ ⎠ (b) a * ( NH 3 )/ a * ( H 2 O ) = 6.4 × 10 −3 (about 150 times smaller! )
14.12 (a) and (b) W max
= − 1.26 × 106 J = 0.35 kWhr so CTI’s claim of 1 kWhr production
nd
violates the 2 Law limit (c) All heat transfer and work production and utilization steps have some irreversibility in practical systems. 14.14 (a) W max
= − 9471 kW
(b) no fundamental laws or concepts are violated (c) a = Q total / < U > ∆T Updated: 1/14/04
about 17% more area for heat exchange is required for the GH process (d) W net (GH ) / W net ( Rankine ) = 1.15 which is less than 22% 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
(∂P / ∂T)(S - L )
= 1.35 × 10 7 Pa/K
15.3
x (Naphthalene) = 0.196
15.6
(c) q (critical quality) =
κT in C v ∆ H v
(e) O2: q = 0.54 and H2: q = 0.36 15.12 xKCl = 0.413 and T = 624 K V
L
15.13 H EtOH − H EtOH = ∆ H vap
= 4.0 × 10 4 J/mol
15.15 Anesthetic pressure of CCl 4
≈ 4.9 × 10-3 bar.
15.19 (a)
ln
γ A = f ( P, x i ) /(T )1 / η
(b)
γ A ( 400K) = 1.17
(c)
∆S A = R ln γ A ((1 − η) / η)
EX
(i) if
EX
∆S A = 0 , solution is regular EX
(ii) if ∆S A
> 0 , less structural ordering in mixture versus pure state
EX
(iii) if ∆S A implies
16.9
< 0 , more structural ordering in mixture versus pure state, Φ ij > Φ ii or Φ jj
(a) y A = 0.764 y B = 0.236 (b) not in equilibrium, estimate y B assuming equilibrium
16.12 W max
=
4.868 × 10 5 J/s
16.14 W max
=
9.15 × 10 5 J
16.16
∆T = 7.95 K ∆T = 2.39 K
using expansion valve using expansion turbine st nd Consider net energy, entropy, and work flows using 1 and 2 Law concepts to show Updated: 1/14/04
process is not feasible as described. 16.20 (a) P = 21.9 bar (b) T = 51.5 K (c)
∆ S universe = 2.96 × 10 4 J/K T, K 293 313 333 353 373
16.22
17.3
C p (effective), J/gK 4.69 7.20 8.29 6.36 3.64
(a) F = 4 − π with n = 3 and r = 1 ; at A: L1- L2- H-G and at B: L1- H-I-G (b) L1 = 0 and M 1 = 0 use P-explicit EOS to determine required Aijk and Aij derivatives (c) d ln P / d (1 / T )
= ∆ H rxo / R = ( H ( H ) − 17 H ( L1 ) − H (G ) ) / R
(d) for x hexane small, T 0
0 − T ≈ ( RT 02 / ∆ H rx ) x hexane = 5.6 x hexane
0 ∆ H rx = H ( H ) − 17 H ( L2 ) − H ( L2 ) in this case
− ∆ H vap / R
where ∆ H vap ≈ H (V )
17.6
d ln P / d (1/ T) =
18.1
P
18.4
W = 9 × 10 J/kg mol
18.5
T i = 269 K = 4 C (winter!)
18.6
T bath = 317 K, efficiency = 0.065
18.8
z = depth = 764 m
18.9
y
18.11
α = 1.075,
α
− y1(V ) H ( L1 ) − y 2 H ( L2 )
− P β ≈ ε o ( Ds − 1) E 2 / 2, T α = T β , and µ α = µ β 6
!
He
4
o
(tube bottom) = 0.501, if L
→ ∞ y He4 → 4/7
P( rim) = 4600 bar
18.14 ytritium = 0.065 and ydeuterium = 0.538 at rim 18.16 h = 252 m, for an 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.). Updated: 1/14/04
19.1
α EtOH
to H 2O
19.2
y ethane
= 0.995,
19.5
ln Pvp (drop) = − 3.93 × 10 4 / T + constant
19.7
xbutanol = 0.436
19.8
(a) 0.046 J/m (b) 296 K
= 2.17 P(bubble) = 1.97 × 10 6 Pa , r = 4.2 × 10 −9 m
2
Updated: 1/14/04
2/22/99 JWT