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.
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.
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
The power input in the heat pump is 5.06668kW.
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
The rate of heat absorption form the atmosphere Ql is Qh – W
3. Carnot heat engine
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
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
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
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
The temperature is 133.520C
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
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.