4.14 - An Introduction to Thermal Physics

Question

A heat pump is an electrical device that heats a building by pumping heat in from the cold outside. In other words, it's the same as a refrigerator, but its purpose is to warm the hot reservoir rather than to cool the cold reservoir (even though it does both). Let us define the following standard symbols, all taken to be positive by convention:
$$\begin{gathered} T_h=\text{ temperature inside building } \\ {T}_c=\text{ temperature outside } \\ {Q}_h=\text{ heat pumped into building in }1\text{ day } \\ {Q}_c=\text{ heat taken from outdoors in }1\text{ day } \\ W=\text{ electrical energy used by heat pump in }1\text{ day } \end{gathered}$$
(a) Explain why the "coefficient of performance" (COP) for a heat pump should be defined as Q_h / W.
(b) What relation among Q_h , Q_c, and W is implied by energy conservation alone? Will energy conservation permit the COP to be greater than 1 ?
(c) Use the second law of thermodynamics to derive an upper limit on the COP, in terms of the temperatures T_h and T_c alone.
(d) Explain why a heat pump is better than an electric furnace, which simply converts electrical work directly into heat. (Include some numerical estimates.)

Step-by-Step Solution

Answer

a) $\mathrm{COP}=\frac{{\mathrm{Q}}_{\mathrm{h}}}{\mathrm{W}}$

b) $\mathrm{COP}=\frac{{\mathrm{Q}}_{\mathrm{h}}}{{\mathrm{Q}}_{\mathrm{h}}-{\mathrm{Q}}_{\mathrm{c}}}.$COP of the heat pump will always be greater than one

c) Upper limit of COP is $\mathrm{COP}\le \frac{{\mathrm{T}}_{\mathrm{h}}}{{\mathrm{T}}_{\mathrm{h}}-{\mathrm{T}}_{\mathrm{c}}}$

d) Heat pump is better.

Part(a): Step 1 - Given Information

Given a heat pump , where W is power taken by the pump and Q_h= is Power output in heat.

Find the formula for COP

Part(a): Step 2: Explanation

COP is a dimensionless quantity, which gives performance of heat pump

 Assume a heat pump that is used for air cooling having a coefficient of performance as 2.

This will mean that 2 KW of cooling power is achieved for each KW of power consumed

So the expression for the COP is 

$\mathrm{COP}=\frac{{\mathrm{Q}}_{\mathrm{h}}}{\mathrm{W}}$

Part(b) : Step 1 : Given information

$\mathrm{COP}=\frac{{\mathrm{Q}}_{\mathrm{h}}}{\mathrm{W}}$

Find  relationship among Q_h , Q_c, and W is implied by energy conservation alone? 

Explain if COP can be greater than 1 ?

Part(b): Step 2 : Explanation

Coefficient of performance is ratio which gives efficiency of pump, by simple division of output by input.
We can directly relate the electrical energy used by heat pump in 1 day to the heat taken from the outdoors in 1 day and heat pumped into the building in one day.

We the expression of COP changed to 

$\mathrm{COP}=\frac{{\mathrm{Q}}_{\mathrm{h}}}{{\mathrm{Q}}_{\mathrm{h}}-{\mathrm{Q}}_{\mathrm{c}}}$

Simplify this by dividing with Q_h, we get

$$\begin{gathered} \text{ COP }=\frac{\frac{Q_h}{Q_h}}{\frac{Q_h}{Q_h}\frac{Q_c}{Q_h}} \\ COP=\frac{1}{1-\frac{Q_c}{Q_h}} \\ COP\times \left(1-\frac{Q_c}{Q_h}\right)=1...............................\left(1\right) \end{gathered}$$

From the equation (1) we can say that COP of the heat pump will always be greater than one.

Part(c): Step 1 : Given information

$\mathrm{COP}=\frac{{\mathrm{Q}}_{\mathrm{h}}}{{\mathrm{Q}}_{\mathrm{h}}-{\mathrm{Q}}_{\mathrm{c}}}$

Find the upper limit of COP in terms of T_h and T_c.

Part(c): Step 2: Explanation

Entropy of the system is related to heat and temperature:

$$\begin{gathered} S_h=\frac{Q_h}{T_h} \\ {S}_c=\frac{Q_c}{T_c} \\ \because S_c\le S_h \\ \frac{Q_c}{T_c}\le \frac{Q_h}{T_h} \\ {Q}_c\times T_h\le Q_h\times T_c \\ \frac{Q_c}{Q_h}\le \frac{T_c}{T_h} \end{gathered}$$

Now substitute this value in COP $=\frac{1}{1-\frac{Q_c}{Q_h}}$, we get

$\mathrm{COP}\le \frac{1}{1-\frac{{\mathrm{T}}_{\mathrm{c}}}{{\mathrm{T}}_{\mathrm{h}}}}$

Simplify by multiplying numerator and denominator with T_h , we get.

$\mathrm{COP}\le \frac{{\mathrm{T}}_{\mathrm{h}}}{{\mathrm{T}}_{\mathrm{h}}-{\mathrm{T}}_{\mathrm{c}}}$

Part(d): Step 1: Given information

Heat pump and Furnace 

Explain why a heat pump is better than an electric furnace

Part(d): Step 2 : Explanation

Lets assume a heat pump working in the temperature ranges of 300 K and 278 K.
Calculate the COP:
$\mathrm{COP}=\frac{300 \mathrm{K}}{300 \mathrm{K}-278 \mathrm{K}}=13.63$
Now , compare the COP for:
Electric Furnace:
COP_{Electric Furnace} =1
Heat Pump:
COP_{Heat Pump} =13.63
This clearly shows that the efficiency of the heat pump is more than electric furnace.

So heat pump is better.