University Engineering Thermodynamics: Problems and Solutions

This document presents solutions to a series of engineering thermodynamics problems. It begins with calculations for refrigeration systems, determining work input and power requirements based on given coefficients of performance and cooling loads. The solutions then move on to heat pump analysis, calculating power input, heat generation, and the coefficient of performance under different conditions. Further, the assignment explores Carnot cycle efficiency and entropy calculations, including determining entropy changes, dryness fraction, and enthalpy values for steam under various pressure and temperature conditions. The document concludes with an analysis of air compression processes, calculating outlet pressure, isentropic work, and actual work output for a compressor, along with the power required to drive it. The solutions utilize thermodynamic tables and formulas to arrive at the answers.

Running Head: Engineering Thermodynamics 1
Engineering Thermodynamics
Student’s name
University affiliation

Engineering Thermodynamics 2
1. Refrigeration
a) Assuming a steady state operation for the refrigerator.
The coefficient of performance for a freezer is given as COPref = Ql
Wnet
Where Ql is the cooling effect energy and,
W is the work input (Fermi, 2012). Therefore we can calculate for w by substituting given
figures in above equation.
Ql= 5kW
COPref = 1.3
W= Ql
COPref = 5
1.3 =3.8462 kW
The refrigerator requires 3.8462 kW to operate.
b) Ql= 10,000kJ/h COPref =1.35
the work input W= Ql
COPref = 10000
1.35 =7,407.4074 kJ /h
Converting W to kJ/s 7,407.4074 kJ
3600 =2.0576 kJ /s but 1kJ/s=1kiloWatt
Therefore the power required for the refrigerator to operate is 2.0576 kW.
2. Heat pumps
A.
i. COPhp=2.5
Qloss=60,000kJ/h 60000
3600 =16.6667 kW
Qgenerated=4kW
The temperature to be maintained Thot is 23 0C = (273+23) =300K
Qh is the heat required to maintain the 230C
Qh= Qloss – Q generated= 16.6667-4= 12.6667kW

Engineering Thermodynamics 3
COPhp= Qh
W
Therefore W= Qh
COPhp =12.6667
2.5 =5.06668kW
The power input in the heat pump is 5.06668kW.
ii. COPhp= Thot
Thot−Tcold = 300
300−(273+1)=13.4545
The best possible COP for the heat pump when the outside temperature is 10C is 13.4545.
B. For an R-134 at 800kPa, the saturation temperature is 31.310C.Therefore, the working
fluid at 400C is superheated. From the thermodynamics tables, the enthalpy of this fluid at
400C h=276.45kJ/kg.
Power consumed W= 1200W=1.2kW
The energy Qh = ṁ×h = 0.022kg/s × 276.45kJ/kg = 6.0819kJ/s
Which is equal to 6.0819kW
COPhp= Qh
W = 6.0819
1.2 =5.06825
The rate of heat absorption form the atmosphere Ql is Qh – W
Ql= 6.0819-1.2
=4.8819kW
3. Carnot heat engine
Energy in supplied Qh=700kJ at a temperature Th=500K
Work input= -300kJ
Sink temperature is 290 K
For a heat pump Qh= W + Ql
Ql=Qh –W =700kJ- 300kJ=400kJ
For a Carnot cycle Ql=TlS and,
Qh=ThS
Therefore Qh
Ql = Th
Tl

Engineering Thermodynamics 4
Ql= Qh×Tl
Th = 700× 290
500 =406 kJ
There Carnot heat engine has to reject 406 kJ to the sink and thus the net-work produced is (700-
406) 294 kJ
This is not a reasonable claim because all adiabatic processes in a Carnot cycle are isothermal
(Hans J. Kreuzer, 2010). This makes the claim of the heat engine to produce 300kJ net-work as
false.
4. Carnot heat pump
a) Given the COPhp=1.6 COPhp= Thot
Thot−Tcold
The sink temperature Tl=300K
1.6= Thot
Thot−300
1.6Thot-480=Thot
Thot – 1.6Thot=480
0.6Thot=480 Thot= 800K
Temperature of the source Thot is 800K
Power supplied to the heat pump, Qh= 1.5kW
COPhp= Qh
W work input Win= 1.5
1.6 =0.9375 kW
The rate of heat of transfer to the sink Ql=Qh –W=1.5kW- 0.9375kW=0.5625kW
5. Entropy
a) The inlet pressure P1=35kPa and temperature T1=1600C
From the thermodynamics tables, Tsat at 35kPa is between 69.090C and 75.860C.This means that
the water vapor is super-heated and double interpolation is applied.
PRESSURE (kPa) ENTROPY
(kJ/kgK)
1500C 1600C 2000C
10 8.6893 8.73242 8.9049
35 8.2652075
50 7.9413 7.98488 8.1592
The entropy Sg at 10kPa and 160 0C

Engineering Thermodynamics 5
S1=8.6893 + 160−150
200−150 × (8.9049-8.6893)=8.73242kJ/kgK
The entropy Sg at 50 kPa and 160 0C
S2= 7.9413+ 160−150
200−150 × (8.1592-7.9413)=7.98488kJ/kgK
The entropy S at 35 kPa and 160 0C
S=7.98488+ 35−10
50−10 × (8.73242-7.98488) =8.2652075kJ /kgK
The outlet pressure P2 300kPa at an entropy 8.2652075kJ /kgK. At this pressure, the
Sg=6.9917kJ/kgK
The dryness cfraction x= S−Sg
Sfg = 8.265207−6.9917
5.32 =0.2394
The specific enthalpy h=hf+xhfg
=561.43+ (0.2394)2163.5
=1079.3719kJ/kg
The temperature is 133.520C
6. Entropy
Inlet pressure 4MPa
The exhaust pressure 50kPa has a Tsat =81.320C and it is superheated at 1000C
Enthalpy at this temperature h=2682.4kJ/kg at it takes up 95%.
Finding 100% of this 100× 2682.4
95 =2823.5789 kJ /kg
Power produced by the turbine = 2823.5789 kJ /kg×5kg/s
=14,117.8945kW

Engineering Thermodynamics 6
7. Entropy
Assuming air is a perfect gas, the adiabatic process obeys the law pvγ=constant
Therefore, T 1
T 2
γ
γ−1 = P 1
P 2 where γ is the isentropic index of air =1.4
P1 is the inlet pressure=100kPa
T1 is the inlet temperature =170C (17+273) = 290K
P2 is the outlet pressure
T2 is the outlet temperature= 2570C (257+273) =530K
The outlet pressure P2, can be calculated using the following formulae;
P2=
P1
, T 1
T 2
γ
γ −1
= 100 kPa
, 290
530
1.4
1.4 −1
=825.2253 kPa
Isentropic efficiency= actual work output
isentropic work isentropic efficiency =84%
Isentropic work =Cv (T2 –T1) Cv =0.718kJ/kgK
Pv = ṁRT v=2.4m3/s
ṁ= pv
RT = 100 ×1000 × 2.4
0.287× 1000 ×290 =2.8836 kg /s
Isentropic work = 0.718×(530-290)
=172.32kJ/kg
Actual work output=isentropic efficiency× isentropic work
=0.84×172.32=144.7488kJ/kg
Power required to drive the compressor =ṁ × actual work output
=2.8836 kg /s ×144.7488kJ/kg
=417.3976kJ/s
=417.3976kW

Engineering Thermodynamics 7
References
Fermi, E. (2012). Thermodynamics. Massachusetts: Courier Corporation.
Hans J. Kreuzer, I. T. (2010). Thermodynamics. Singapore: World Scientific.