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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:
Th=temperature inside buildingTc=temperature outsideQh=heat pumped into building in1dayQc=heat taken from outdoors in1dayW=electrical energy used by heat pump in1day
(a) Explain why the "coefficient of performance" (COP) for a heat pump should be defined as Qh / W.
(b) What relation among Qh , Qc, 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 Th and Tc 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.)

Short Answer

Expert verified

a) COP=QhW

b) COP=QhQh-Qc.COP of the heat pump will always be greater than one

c) Upper limit of COP is COPThTh-Tc

d) Heat pump is better.

Step by step solution

01

Part(a): Step 1 - Given Information

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

Find the formula for COP

02

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

COP=QhW


03

Part(b) : Step 1 : Given information

COP=QhW

Find relationship among Qh , Qc, and W is implied by energy conservation alone?

Explain if COP can be greater than 1 ?

04

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

COP=QhQh-Qc

Simplify this by dividing with Qh, we get

COP=QhQhQhQhQcQhCOP=11-QcQhCOP×1-QcQh=1...............................(1)

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

05

Part(c): Step 1 : Given information

COP=QhQh-Qc

Find the upper limit of COP in terms of Th and Tc.

06

Part(c): Step 2: Explanation

Entropy of the system is related to heat and temperature:

Sh=QhThSc=QcTcScShQcTcQhThQc×ThQh×TcQcQhTcTh

Now substitute this value in COP =11-QcQh, we get

COP11-TcTh

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

COPThTh-Tc

07

Part(d): Step 1: Given information

Heat pump and Furnace

Explain why a heat pump is better than an electric furnace

08

Part(d): Step 2 : Explanation

Lets assume a heat pump working in the temperature ranges of 300 K and 278 K.
Calculate the COP:
COP=300K300K-278K=13.63
Now , compare the COP for:
Electric Furnace:
COPElectric Furnace =1
Heat Pump:
COPHeat Pump =13.63
This clearly shows that the efficiency of the heat pump is more than electric furnace.

So heat pump is better.

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