CHE 303 (Winter 99) Answers to problem set #7



CHE 302 (Fall 2001)

Answers to problem set[pic]#9

21.3 QH = 768.2 Btu/lb, QC = 435.85 Btu/lb, Wnet = 332.35 Btu/lb

(S12 = 0.8386 Btu/lb(R, (S12,sur = - 0.8386 Btu/lb(R

(S23 = 0, (S23,sur = 0

(S34 = - 0.8386 Btu/lb(R, (S34,sur = 0.8386 Btu/lb(R

(S41 = 0, (S41,sur = 0

thermal efficiency = 0.4326

21.6 a) Wnet = 646.2 kJ/kg, thermal efficiency = 0.429

b) Wnet = 552.9 kJ/kg, thermal efficiency = 0.367

21.9 Wnet = 1473.5 kJ/kg, thermal efficiency = 0.4128

flow rate 33.93 kg/s

21.9 Wnet = 2377.2 kJ/kg, thermal efficiency = 0.5028

22.10 a) Qevap = 507.42 Btu/lb, Wnet = 91.08 Btu/lb, COPref = 5.5711

(S12 = 0,

(S23 = - 1.0102 Btu/lb(R

(S34 = 0,

(S41 = 1.0102 Btu/lb(R

b) Qevap = 501.1 Btu/lb, Wnet = 97.4 Btu/lb, COPref = 5.1448

(S12 = 0,

(S23 = - 1.0102 Btu/lb(R

(S34 = 0.0143 Btu/lb(R,

(S41 = 0.9959 Btu/lb(R

22.11 a) Qevap = 107.21 kJ/kg, Wnet = 50.62 kJ/kg, COPref = 2.118

(S12 = 0.03326 kJ/kg(K,

(S23 = - 0.5035 kJ/kg(K

(S34 = 0,

(S41 = 0.4702 kJ/kg(K

b) Qevap = 105.06 kJ/kg, Wnet = 52.77 kJ/kg, COPref = 1.991

(S12 = 0.03326 kJ/kg(K,

(S23 = - 0.5035 kJ/kg(K

(S34 = 0.0094 kJ/kg(K,

(S41 = 0.4608 kJ/kg(K

b) Qevap = 95.27 kJ/kg, Wnet = 62.56 kJ/kg, COPref = 1.523

(S12 = 0.03326 kJ/kg(K,

(S23 = - 0.5035 kJ/kg(K

(S34 = 0.0524 kJ/kg(K,

(S41 = 0.4178 kJ/kg(K

Answers to problem set[pic]#8

19.3 thermal efficiency = 0.525

(S12 = 4.7619 kW/K, (S12,sur = - 4.7619 kW/K

(S23 = 0, (S23,sur = 0

(S34 = - 4.7619 kW/K, (S34,sur = 4.7619 kW/K

(S41 = 0, (S41,sur = 0

(Scycle = 0, (Soverall,sur = 0

19.6 COP = 9.2188

(S12 = 33.898 kJ/K, (S12,sur = - 33.898 kJ/K

(S23 = 0, (S23,sur = 0

(S34 = - 33.898 kJ/K, (S34,sur = 33.898 kJ/K

(S41 = 0, (S41,sur = 0

(Scycle = 0, (Soverall,sur = 0

19.11 thermal efficiency = 0.5417

(S12 = 1.2329 Btu/R, (S12,sur = - 1.2329 Btu/R

(S23 = 0.2403 Btu/R, (S23,sur = 0

(S34 = - 1.4732 Btu/R, (S34,sur = 1.4732 Btu/R

(S41 = 0, (S41,sur = 0

(Scycle = 0, (Soverall,sur = 0.2403 Btu/R

19.14 COP = 9.2188

(S12 = 3.4221 kJ/K, (S12,sur = - 3.4221 kJ/K

(S23 = - 0.0322 kJ/K, (S23,sur = 0

(S34 = - 3.3898 kJ/K, (S34,sur = 3.3898 kJ/K

(S41 = 0, (S41,sur = 0

(Scycle = 0, (Soverall,sur = - 0.0322 kJ/K, impossible

19.19 b) 580.4 J/K

19.14

(S13 = 18.019 J/K, (S13,sur = - 18.019 J/K

(S31 = - 18.019 J/K, (S23,sur = 18.333 J/K

(Soverall,sur = 0, (Soverall,sur = 0.314 kJ/K, impossible

Answers to problem set[pic]#7

17.1 a) 0.35 b) 0.25 c) 0.3044

17.3 a) 0.314 b) 0.272

17.4 a) 3.857 b) 4.0 c) 2.286

17.8 a) 1.947 b) 2.522

17.18 a) See Fig. 17-10 b) 0.4344 c) 1.302

17.20 a) See Fig. 17-10 b) 1.9767 c) 0.336

17.21 a) See Fig. 17-10 b) 2.633 c) 0.380

17.22 a) See Fig. 17-21 for PV diagram, TP diagram is similar to Fig. 17-10 except that it is in one-phase region b) 2.633 c) 0.380

Answers to problem set[pic]#1

2.9 a) slug b) slug = lbf(s2/ft c) gc = 1 d) 2.857 ft/s2

e) 11,247 ft(lbf at NYC, 2100 ft(lbf on the moon

3.4 Total internal energy, entropy, and kinetic energy are extensive. These variables are intensive when expressed as specific quantities.

3.7 T = T(T,P) makes no sense, as the same variable cannot be both an independent and dependent variable.

T(V,P,U) violates the statement that any homogeneous, pure substance may be described in terms of any two independent variables.

3.9 a,b) 1672.6 cm3/g c) 292 cm3/g

d) 1.63 ft3/lb e) –2106 kJ/kg

f) – 805 Btu/lb

3.10 a) Ideal gas: P, V, and T are single-values functions.

b) P and T are single-values functions. V could possess three roots, the largest root is the gas volume.

6.14 a) 1.441(106 kg/hr b) 1.184(106 kg/hr

6.20 Analysis cannot be trusted since disagreement between C and H balances is over 25%.

Answers to problem set[pic]#2

[pic]

[pic]

e)

[pic]

4.12 a) 2,490.1 kJ/kg b) 1,115.6 Btu/lb

5.6 a) pN2 = 49.20 kPa, pO2 = 64.60 kPa, pHe = 86.15 kPa

b) vN2 = 44.22 L, vO2 = 58.06 L, vHe = 77.47 L

5.8 a) 2mvx b) vx/V1/3, Nvx/V1/3

c) Nvx/V1/3(2mvx d) P = [pic]

5.9 a) Ideal gas: 24.497 L/mol, van de Waals for He: 24.485 L/mol

b) Ideal gas: 24.497 L/mol, van de Waals for CO2: 24.335 L/mol

c) Ideal gas should give better prediction for He since He molecule is much more symmetrical and smaller than CO2 molecule.

5.12 a) PV3 – (bP + RT)V2 + aV – ab = 0

b)

[pic]

c) Case 1: above Tc

3 roots, one real, which is the gas volume, and two imaginary

Case 2: at Tc

3 equal real roots, which are the gas (and liquid) volume

Case 3: below Tc

3 unequal real roots, largest root is gas volume, smallest root is liquid volume, intermediate root has no physical meaning

6.10 6,000 L high-octane gasoline, 4,000 L straight-run gasoline

6.11 15.8 mol distillate

Answers to problem set[pic]#3

6.22 Coal required = 1.591(109 kg, hydrogen required = 1.764(109 m3

7.8 (Part 1) (a) Isolated, (b) Open. Q = 0. W = 0, (c) Open. Q = 0. W = 0

(Part 2) Assume no heat transfer from the water to the surroundings except through the hot plate. (a) Closed. Q = 0. W = (negative) electric work, (b) Closed. Q = negative amount. W = 0, (c) Closed. Q = negative amount. W = 0 (d) Open. Q = 0. W = 0 (e) Closed. Q = negative amount. W = (positive) electric work, (f) Open. Q = 0. W = (positive) electric work.

(Part 3) (a) Open. Q = 0. W = 0, (b) Open. Q = negative amount. W = 0, (c) Open. Q = 0. W = (negative) PV works, (d) Open. Q = negative amount. W = (negative) PV works.

7.9 (a) (1) A quantity of work is done on the mass, thereby increasing its PE by 200 kJ. (2) Assuming frictionless, W(push) = 0. Assuming friction, then W(push) will be transferred to the surroundings as heat. In either case, the total energy of the mass will not change. (3) Just before toppling the mass possesses PE = 200 kJ. At the split instant before impact, the PE is transformed in KE = 200 kJ. (4) Upon impact, KE is transformed into (U = 200 kJ. The increase in U of the mass manifests itself as a temperature increase. (5) Given time, a quantity of heat, Q = 200 kJ is transferred from the mass to the surroundings and at this point the temperature of the mass and surroundings will be equal.

(b) Nothing.

(c) Overall nothing. Locally, at some point on Earth's surface 200 kJ of work was performed, and somewhere else on Earth's surface 200 kJ of heat was absorbed.

7.9 (a) Qnet = 40 J. Wnet = 0.

(b) Qnet = 0. Wnet = 40J.

(c) Qnet = 450 Btu. Wnet = -1000 Btu.

(d) Qnet = 950 kJ. Wnet = -1000 kJ.

(e) Qnet = 7000 kJ. Wnet = -7000 kJ.

9.6 (a) (U = 467.58 Btu/lb. (H = 472.88 Btu/lb. Saturation conditions.

(b) (U = 958.9 Btu/lb. (H = 1044.4 Btu/lb. Saturation conditions.

9.7 Data from table B-6

(a) H = 399.65 kJ/kg. U = 270.05 kJ/kg.

(b) H = 528.9 kJ/kg. U = 365.3 kJ/kg.

(c) H = 333.8 kJ/kg. U = 238.1 kJ/kg.

9.9 (a) (U = 0. (H = 12.0 kJ

(b) (U = 0. (H = 0.573 Btu

(c) Appear reasonable as (H's are small.

9.12 (a) WAB = - 5.429 Btu

(b) QC = - 60.571 Btu

Answers to problem set[pic]#4

9.14 (a) –4,000 J

(b) –20,000 J

(c) 21,000 J

(d) [pic] 11,000 J

9.15 -3,475 J, -94.4oC

9.20 Steam with 15.25% quality

9.21 80,000 kJ

9.23 330oC

10.14 (a) –5,000 J

(b) –10,900 J

(c) 2,000 J

10.16 107.5 kJ

10.20 -5,612 J/mol

Answers to problem set[pic]#5

10.21 (a) The two forms may co-exist in equilibrium at 25oC

(b) The input of energy required to transform one mole of the rhombic form to one mole of the monoclinic form.

(c) 326.4 J/mol

(d) Both forms cannot be stable at 60oC and 1 atm. Given T and P, enthalpy consideration is not sufficient to determine which form is stable. The Gibbs free energy must be considered.

11.8 (a) Net work = 1122.8 kJ/kg, heat load on the condenser = 2243.2 kJ/kg

(b) 89.06 kg/s

(c) Summer net work = 1061.3 kJ/kg, mass flow 94.22 kg/s, winter is about 6% more efficient

11.10 (a) 404.3oK

(b) 379.6oK

(c) 442.7oK

11.13 1.13(107 Btu/hr

12.8 (a) –46,385 Btu/lbmol

(b) 180 kJ/mol

13.5 (a) Acetylene is the limiting reactant

(b) 333%

13.10

|Species |CO2 |H2O |O2 |N2 |

|Mol % |1.96 |3.92 |16.67 |77.45 |

13.12 19 moles air to natural gas

Answers to problem set[pic]#6

13.14 0.221

14.6 (a) 141 kJ/mol supplied to reactor

(b) 268 kJ/mol supplied to reactor

15.1d 1274oC

16.7 probability = 1.0867(10-50 , percentage of all possible combination = 1.0867(10-48

16.8 probability = 6.30(10-12

16.8 (a) 16.13 and 16.14 are reversible

(b) 16.11 and 16.12 are irreversible

(c) 16.11 and 16.12 are spontaneous

(d) 16.11, 16.12, (16.13 and 16.14), 16.15

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