The figure below shows a 4-stage cycle on a PV diagram for an engine that operat

Question

The figure below shows a 4-stage cycle on a PV diagram for an engine that operates with 0.0564 mol of an ideal diatomic gas. The engine operates at a frequency of 30 Hz.

stage i: adiabatc compression

stage ii: isochoric increase in pressure

stage iii adiabatic expansion

stage iv isochoric reduction in pressure

When answering the following questions do not include the sign of the work/heat.

a) What is the amount of heat extracted from the hot reservoir during one cycle?

QH =   (3.s.f)

b) What is the amount of heat added to the cold reservoir during one cycle?

QC =   (3.s.f)

c) What is the amount of work done ON the gas during the adiabatic compression?

Won =   (3.s.f)

d) What is the amount of work done BY the gas during the adiabatic expansion?

Wby =   (3.s.f)

e) What is the thermal efficiency of the cycle?

=   (3.s.f)

f) What is the power output of the engine?

P =   (3.s.f)

Explanation / Answer

given 4 stages
stage 1 : adiabatic compression
Stage 2 : isochoric increase in pressure
Stage 3 : adiabatic expansion
Stage 4 : isochoric reduction in pressure

a. step 2 includes heat absorption from the hot reservoir
   as it is isochoric process, work done by gas W = 0
   hence change in internal energy E = heat extracted from the hot reservoir H
   E = H
   now, E = nCv*dT [ dT is change in temperature, Cv is molar heat capacity at constant volume and n is number of moles of the gas]
   given, n = 0.0564 mol
   ideal diatomic gas, gamma = 7/5
   now, Cp/Cv = gamma
   Cp - Cv = R
   so, Cp = Cv*gamma
   Cv(gamma - 1) = R
   Cv = R/(gamma - 1) = 5R/2

   now, for step 2
   Tf - Ti = dT = 1241 - 497 K = 744 K
   hence E = 0.0564*5*8.31*744/2 = 871.75224 J
   so heat absorbed from reservoir = 871.75224 J

b. heat is rejected to the cold reservoir in step iv
   as this step is isochoric too
   H = E = nCvdT
   dT = 800 - 320 = 480 K

   so, Heat rejected to cold reservoir, H = 0.0564*5*8.31*480/2 =562.4208 J

c. during adiabatic compression ( step 1), work done on gas
   W = PiVi^gamma[Vi^(1 - gamma) - Vf^(1 - gamma) ]/[1 - gamma]
   gamma = 7/5 = 1.4
   Pi = 100,000 Pa
   Pf = 466,000 Pa
   Vi = 1.5*10^-3 m^3
   Vf = 0.5*10^-3 m^3
   W = 206.942 J

d. work done by gas in adiabatic expansion ( step 3)
   W = PiVi^gamma[Vi^(1 - gamma) - Vf^(1 - gamma) ]/[1 - gamma]
   gamma = 7/5 = 1.4
   Pi = 1163,000 Pa
   Pf = 250,000 Pa
   Vi = 0.5*10^-3 m^3
   Vf = 1.5*10^-3 m^3
   W = 516.9622 J
e. thermal efficiency = work done / heat absorbed = 0.593

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