Heat Pumps and Refrigerators

2. Heat Pumps and Refrigerators

a. Coefficient of Performance Heat pumps and refrigerators are used to transfer heat through work input and are characterized by a coefficient of performance COP (= heat transfer ÷ work input) instead of an

efficiency. For a heat pump COP H =Q H /Work input = Q out /(Q in –Q out ),

(71) and for a refrigerator COP R =Q L /Work input = Q in /(Q in –Q out ).

For a Carnot heat pump Q out /Q in =Q H /Q L =T H /T L . Therefore,

(73) Likewise, for Carnot refrigerators

COP H =T H /(T H –T L ).

(74) Figure 14a contains an energy band diagram for a heat pump and refrigerator that in-

COP R =T L /(T H –T L ).

teracts with fixed–temperature thermal energy reservoirs. Figure 14b illustrates the corre- sponding availability. Both the Carnot COPs → ∞ as T H →T L , and approach either zero (in case of COP R ) or unity (in case of COP H ) as the difference (T H –T L ) becomes very large. The

availability COP COP avail = |W cyc,opt |/|W cyc |, where

(75) W cyc,opt =W cyc,min =W cyc _ – To σ cyc .

i. Example 9

A Carnot heat pump delivers heat to a house maintained at a 25ºC temperature in a 0ºC ambient. The temperature of the evaporator in the heat pump is –10ºC, while the condenser temperature is 35ºC so that external irreversibilities exist. Determine: The COP based on the evaporator and the condenser temperatures. The COP based on the house and the ambient temperatures. The minimum work input that is required. The availability efficiency. The irreversibility.

Solution Using the relation

COP Carnot = |Q condenser |/|Work Input| = |Q condenser |/(|Q condenser | – |Q evaporator |),

(A) Q condenser /Q evaporator =T condenser /T evaporator .

(B)

W sh =837

Figure 13: Exergy band diagram a steam power plant. Therefore,

COP Carnot =T condenser /(T condenser –T evaporator ) = (308)/(308 – 263) = 6.844. In the absence of external irreversibilities, the evaporator and ambient temperatures

should be identical, as should the condenser and the house temperatures. Therefore COP Carnot,ideal = |Q house |/|Work Input| = T house /(T house –T ambient ) = 11.92, and W Carnot = |Q house |/COP= 1/11.92 = 0.0839 kJ per kJ of heat pumped into the house. Consider the generalized availability equation

cv =∑ m ˙ i ψ i + ∑ j = 1 Q ˙ Rj , ( 1 − T 0 / T Rj , ) −∑ m inlets ˙ exits e ψ e − dE ( cv − TS 0 cv )/ dt .

For a steady state cyclical process with one inlet and exit, d/dt = 0, ˙ mm i = ˙ e (steady), and ψ ˙ i = ψ ˙ e (cyclical). Therefore,

W cyc,min = |Q house |(1– T 0 /T house )/T house = |Q house |/COP Carnot , and (C) W cyc,min = 1/11.92 = 0.0839 kJ per kJ of heat pumped in.

The availability COP COP avail = 0.08389/0.146 = 0.57. We considered an internally reversible process with external irreversibilities existing at the evaporator and condenser due to the temperature differences between these res- ervoirs and the ambient and the house, respectively. In that case W cyc = |Q house |/COP = 1/6.844 = 0.146 kJ per kJ of heat pumped into the house. The overall irreversibility associated with every kJ of heat that is pumped into the en- vironment is

I=W cyc,min –W cyc =T 0 σ cyc = –0.08389 – (–0.146) =

0.06211 kJ for every kJ of heat pumped into the house. Since the ambient temperature is 0ºC, 0.06211 kJ for every kJ of heat pumped into the house. Since the ambient temperature is 0ºC,

(a)

Q 0 Q opt,0

(b)

actual

Figure 14 : (a) Energy, and (b) exergy band diagrams for a heat pump. σ –1

cyc = 0.0621/273 = 0.00023 kJ K for every kJ of heat pumped into the house.

Remarks The overall irreversibility can also be obtained by considering entropy balance equa- tion for the system by including the thermal reservoirs at 25ºC and 0ºC, i.e., Entropy change in the isolated system = Entropy change in the house (at temperature

T H ) + Entropy change in the ambient (at temperature T L ) + Entropy change in the control volume of interest during the cyclical process due to internal irreversibilities. Therefore,

σ = ∆S H + ∆S L +0= ∆S H + ∆S L . Based on each unit heat transferred to the house, the two entropy changes are

∆S H =Q H /T H = 1/298 = 0.00336 kJ K –1 , and ∆S L =–Q L /T L . Since W cyc = 0.146 kJ per kJ of heat pumped into the house,

Q L = 1 – 0.146 = 0.854 kJ per kJ of heat pumped into the house, ∆S –1

L = – 0.854/273 = –0.00313 kJ K , and

σ = 0.00336 – 0.00313 = 0.00023 kJ per kJ of heat pumped into the house. This is the same answer as that obtained in the solution.