A heat engine uses the closed cycle shown in the diagram below. ( Figure 1 ) The
ID: 2256697 • Letter: A
Question
A heat engine uses the closed cycle shown in the diagram below. (Figure 1) The working substance is nmoles of monatomic ideal gas. Find the efficiency ? of such a cycle. Use the values for pressure and volume shown in the diagram, and assume that the process between points 1 and 3 is isothermal. Find the efficiency
? of the heat engine.
A heat engine uses the closed cycle shown in the diagram below. (Figure 1) The working substance is n moles of monatomic ideal gas. Find the efficiency ? of such a cycle. Use the values for pressure and volume shown in the diagram, and assume that the process between points 1 and 3 is isothermal. Find the efficiency? of the heat engine.Explanation / Answer
Effeciency of a heat engine is given by the ratio of work done by the engine to the heat energy entering the engine (system).
When the cycle moves from 1 to 2, volume changes under constant pressure. As, compression takes place so the work done by the engine is negative. No, heat transfer takes place in this process. Q12=0
W12 = P*(V2-V1)
=- 200 * (900-300) kPa cm^3
= -120000 kPa cm^3
= -120000* 1000/1000000 J
=-120 J
From 2 to 3 as there is no change in volume so work done is 0.
W23= 0
But from 2 to 3 heat is taken by the engine, which is equal to the change in internal energy of the gas.As the gas is monoatomic so specific heat capacity at constant volume= Cv= 3/2*R
Q23= nCv(T3-T2)= 1.5*n*R*(T3-T2)
using ideal gas equation, Q23= 1.5*(P3*V3 - P2*V2)
= 1.5*300*400 kPa cm^3 = 1.5*300*400*1000/10^6 J = 180 J
From 3 to 1 there is isothermal expansion, so work done by the system is
W31 = nRTln(V2/V1)
As it is ideal gas so, the P1V1=P2V2=nRT
W31= P1*V1 *ln(V2/V1)
= 900*200*ln(900/300) kPa cm^3
= 180000*(1000/10^6)*1.0986 J
= 180* 1.0986 J = 197.748 J
The heat supplied to the engine from 3 to 1 is equal to W31 as entire heat is tranformed into work(as there is no change in internal energy during isothermic process).
Q31 = W31 = 197.748 J
Effeciency = ( W12 + W23 + W31)/ (Q12+Q23+Q31)
= ( -120+0+197.748)/(0+180+197.748)
= 77.748/377.748 = 0.206 = 20.6 %