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.

## Answer:

### 1.Refrigeration

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

W=

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 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

W=

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

A.

i.COPhp=2.5

Qloss=60,000kJ/h

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

COPhp=

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

=4.8819kW

3. Carnot heat engine

i.COPhp=2.5

Qloss=60,000kJ/h

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

COPhp=

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

=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

Ql=

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

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

Ql=

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

1.6=

1.6Thot-480=Thot

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

The sink temperature Tl=300K

1.6=

1.6Thot-480=Thot

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

=1079.3719kJ/kg

The temperature is 133.520C

6. Entropy

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

=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

Power produced by the turbine = ?5kg/s

=14,117.8945kW

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;

P2=

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)

=172.32kJ/kg

Actual work output=isentropic efficiency? isentropic work

=0.84?172.32=144.7488kJ/kg

Power required to drive the compressor =? ? actual work output

=144.7488kJ/kg

=417.3976kJ/s

=417.3976kW

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

=14,117.8945kW

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;

P2=

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)

=172.32kJ/kg

Actual work output=isentropic efficiency? isentropic work

=0.84?172.32=144.7488kJ/kg

Power required to drive the compressor =? ? actual work output

=144.7488kJ/kg

=417.3976kJ/s

=417.3976kW

## References

Fermi, E. (2012). Thermodynamics. Massachusetts: Courier Corporation.

Hans J. Kreuzer, I. T. (2010). Thermodynamics. Singapore: World Scientific.