Engineering Thermodynamics: Carnot Heat Engine Essay

1. A food freezer is to produce a 5 kW cooling effect and its COP is 1.3. How many kW of power will this refrigerator require for operation?
2. A refrigerator is used for cooling food in a grocery store is to produce a 10,000 kJ/h cooling effect, and it has a coefficient of performance of 1.35. How many kilowatts of power will this refrigerator require to operate?
3. You were convinced by a cunning sales person to buy a new heat pump with a COP of 2.5. Your flat is to be maintained at 23 C and loses 60 000 kJ/h while you and your super-computer generates 4 kW together

i) Determine the required power input to the heat pump.

4. An inventor claims to have developed a heat engine that receives 700 kJ of heat from a source at 500 K and produces 300 kJ of net work while rejecting the waste heat to a sink at 290 K. Is this a reasonable claim? Why?

5. A completely reversible heat pump has a COP of 1.6 and a sink temperature of 300 K. Calculate

  • the temperature of the source and (b) the rate of heat transfer to the sink when 1.5 kW of power is supplied to this heat pump.



a) Assuming a steady state operation for the refrigerator.
The coefficient of performance for a freezer is given as COPref =
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
The refrigerator requires to operate.
b) Ql= 10,000kJ/h COPref =1.35
the work input W=
Converting W to kJ/s but 1kJ/s=1kiloWatt
Therefore the power required for the refrigerator to operate is 2.0576 kW.
2. Heat pumps
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
Therefore W==5.06668kW
The power input in the heat pump is 5.06668kW.
ii.COPhp= =
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 = ??h = 0.022kg/s ? 276.45kJ/kg = 6.0819kJ/s
Which is equal to 6.0819kW
COPhp= =
The rate of heat absorption form the atmosphere Ql is Qh – W
Ql= 6.0819-1.2
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,
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=
The sink temperature Tl=300K
Thot – 1.6Thot=480
0.6Thot=480 T?hot= 800K
Temperature of the source Thot is 800K
Power supplied to the heat pump, Qh= 1.5kW
COPhp= work input Win=
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
S1=8.6893 +? (8.9049-8.6893)=8.73242kJ/kgK
The entropy Sg at 50 kPa and 160 0C
S2= 7.9413+? (8.1592-7.9413)=7.98488kJ/kgK
The entropy S at 35 kPa and 160 0C
S=7.98488+? (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=
The specific enthalpy h=hf+xhfg
=561.43+ (0.2394)2163.5
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
Power produced by the turbine = ?5kg/s
7. Entropy
Assuming air is a perfect gas, the adiabatic process obeys the law pv?=constant
Therefore, where ? 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;
Isentropic efficiency= isentropic efficiency =84%
Isentropic work =Cv (T2 –T?1) Cv =0.718kJ/kgK
Pv = ?RT v=2.4m3/s
Isentropic work = 0.718?(530-290)
Actual work output=isentropic efficiency? isentropic work
Power required to drive the compressor =? ? actual work output


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

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