chemical A reversible Carnot cycle operates with 10 kg of ai

A reversible Carnot cycle operates with 10 kg of air as the working fluid, which can be considered to behave as an ideal gas with R = 287 J/kg.K and gamma = 1.4. The cycle is made up of the processes: Sketch this cycle on a P-V diagram. Calculate the heat and work for each process in the cycle. Calculate the thermal efficiency.

Solution

solution:

1)here T S diagram is given by closed rectangle 3412. and Pv diagram is closed system with isothermal curve of PV=constant for 34 and 12 processes and PV^k=constant for 14 and 23 processes.

2)here for 3-4 process as temperature is constant hence internal energy change is zero,hence by first law of thermodynamics we get that

Q=W=mRT3ln(V4/V3)

P3/P4=V4/V3=2/4=.5

on putting value in above equation we get that

Q34=W34=10*.287*632*ln(.5)=-1257.258 KJ

3)for adaibitic process heat transfer is zero

Q41=0

W41=mCv*(T4-T1)

W41=-2642.24 Kj

5)for isothermal process 1-2 we have

P1V1=P2V2

P1/P2=V2/V1=20/10=2

Q12=W12=mRt2ln(V2/V1)

Q12=W12=1989.33 KJ

6)for adiabitic process 23 we have again heat transfer is zero

Q23=0

W23=m*Cv*(T2-T3)=2642.24 KJ

7)hence net work of system is given by

W=W23+W12+W41+W34

W=2642.56+1989.33-2642.56-1257.258

W=732.072 KJ

8)here supplied heat is while isothermal expansion,hence

Q12=1989.33 KJ

9)efficiency is ratio of net work to heat supplied

n=W/Q12=732.072/1989.33=.367999=36.799927%

10)here by formula we get efficiency as

n=1-(Tmin/Tmax)=1-(632/1000)=.368=36.8 %