MAE 204 Thermodynamics Homework #8: Problem Solutions and Analysis

This document provides detailed solutions to MAE 204 Homework #8, focusing on thermodynamics principles. The solutions cover three problems: the first involves an insulated piston-cylinder device containing R-134a undergoing an isentropic expansion, requiring the determination of the final temperature and work done. The second problem analyzes a Carnot heat-engine cycle using water as the working fluid, calculating heat added, work output, and heat rejection. The third problem explores the maximum thermal efficiency of a Carnot cycle, the coefficient of performance (COP) of a heat pump, and the work output of a reversible heat engine, and COP of a refrigerator. The solutions utilize thermodynamic properties, energy principles, and cycle analysis to arrive at the answers, with clear steps and explanations for each problem.

Answer 1
1. Assumptions are- no heat transfer, no kinetic and potential energy change, reversible
process. For an isentropic process. S1 =S2
At P1 = 0.8MPa and saturated vapor condition the ν1 = 0.0256m3/Kg (specific
volume)
U1 = 246.76KJ/Kg (internal energy), s1 = 0.91865KJ/KgK
To find out mass. V/ν1=m. Therefore. 0.05/0.025621 = 1.952Kg
At P2 = 0.4MPa and S2=S1,
dryness fraction x2 = (s2-sf)/sfg = (0.91835-0.2476)/0.67929 = 0.984
based on this dryness fraction internal energy at state 2 is U2 = uf+x2*Ufg
U2 = 63.62+0.984*171.45 = 232.91KJ/Kg
Temperature at his internal energy = 8.9C
Based on energy principle, Energy in-Energy out = energy change
-W = ΔU = m(u2-u1)
W = m(u1-u2) = 1.952*(246.76-232.91) – 27.09Kj
Answer 2
2. The cycle is shown below
Process 1-2 is constant temperature process

Process 2-3 is isentropic process
Process 3-4 is constant temperature process
Process 4-1 is isentropic process
State 1 = saturated liquid at 250C. h1 = 1085KJ/Kg, s1 = 2.7815 Kj/KgK
State 2 = saturate vapor = at 250C. h2 = 2801KJ/Kg, s2 = 6.069Kj/KgK
Heat added = h2-h1 = 2801-1085 = 1716Kj/Kg
Work output = heat in *efficiency = 1716*0.3 = 514.8Kj/Kg
h3 =h2-workout = 2801-514.8=2286.2Kj/Kg
State 3 S3 = S2 = 6.069KJ/KgK. h3 = 2286.2Kj/Kg
x=Quality (Water,h=2286.2,s=6.069) = 0.809. It is saturate mixture
The corresponding temperature = 119C
State 4. Enthalpy h4 = 995.8Kj/Kg (at 119.3C and s1)
Heat rejection = h3-h4 = 2286.2-995.8 =1290.4KJ/Kg
x=Quality (Water,h=995.8,s=2.7815) = 0.2243. It is saturate mixture
Answer 3
3. (a). The maximum thermal efficiency is calculated based on Carnot cycle. According
to this
Efficiency = 1-(T2/T1) = 1− 273+33
273+ 400 = 0.545 = 54.5%
(b). Cop of a heat pump = T 2
T 2−T 1 = 330
330−250 =4.125. Based on the carnot heat
pupm the maximum COP is 4.125. It cannot get 5COP
©. Reversible heat engine working between same temperature limit has same
efficiency.
Efficiency = Work out/heat in =
0.3 = work/100. Work = 30KW
(d). COP of a refrigerator = TL/(TH-TL)
4 = 273/(TH-273). TH = 341.25K
References
 Cengel, Y., & Boles, M. (2015). Thermodynamics. New York: McGraw-Hill
Education.

 Black, W., & Hartley, J. (1996). Thermodynamics. New York: HarperCollins College
Publishers.