Question

# Chapter 16 Part 2: Question 16.3 An ideal Carnot engine takes 2000 J of heat from...

Chapter 16 Part 2: Question 16.3

An ideal Carnot engine takes 2000 J of heat from a reservoir at 500 K, does some work, and discards some heat to a reservoir at 350 K. Let’s see what happens if the Carnot engine described is run backward as a refrigerator.What is its performance coefficient?

SOLUTION

SET UP AND SOLVE We know that a refrigerator will act like an engine but with the signs for QH, QC, and Wreversed because the cycle is run backwards. From the equation, QCQH=?TCTH, the heat QC discarded by a Carnot engine that runs between these two temperatures is

QC=?QHTCTH=?(2000J)350K500K=?1400J

Then from the first law, the work W done by the engine is

W=QH+QC=2000J+(?1400J)=600J

Therefore, for our refrigerator, we will use the following values QC=?1400J, QH=2000J, W=600J, with all the signs reversed because the cycle is run backward:

QC=+1400J, QH=?2000J, W=?600J

From the equation, K=QC|W|, the performance coefficient K is

K=QC|W|=1400J600J=2.33

Because the cycle is a Carnot cycle, we may also use the equation Kcarnot=TCTH?TC,

K=TCTH?TC=350K500K?350K=2.33

REFLECT For a Carnot cycle, e and K depend only on the temperatures, as shown by the equations ecarnot=1?TCTH=TH?TCTH and Kcarnot=TCTH?TC, and we don't need to calculate Q and W. For cycles containing irreversible processes, however, these equations are not valid, and more detailed calculations are necessary.

Part A - Practice Problem:

Suppose you want to increase the performance coefficient K, would it be better to decrease TH by 40 K or to increase TC by 40 K ?

Suppose you want to increase the performance coefficient , would it be better to decrease  by 40  or to increase  by 40  ?

 Decrease TH by 40 K . Increase TC by 40 K .

given ideal carnot engine

Qin = 2000 J

Th = 500 K

Tc = 350 K

a. for reverse cycle

Qrej = 2000 J

Th = 500 K

Tc = 350 K

Qext = ?

now

Qext = Qc

Qrej = Qh

(Qc - Qh)/Qh = (Tc - Th)/Th

Qc/2000 - 1 = 350/500 - 1

Qc = 4*350 = 1400 J

Work done W = Qh - Qc = 600 J

K = Qc/W = 1400/600 = 7/3 = 2.33

for carnott cycle, Kcarnott = K = 2.33

A. to increase K

K = Qc/(Qh - Qc) = 1/(Qh/Qc - 1) = 1/(Th/Tc - 1)

k1 = 1/((Th - 40)/Tc - 1) = Tc/(Th - Tc - 40)

k2 = 1/(Th/(Tc + 40) - 1) = (Tc + 40)/(Th - Tc - 40)

hence

k2 > k1

hence

increasing Tc by 40 K will improve perfprmance coeficient more

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