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# Thermodynamics: Efficiency of a Carnot cycle

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I need a calculation of the 'Carnot cycle' and a proof of the 'efficiency expression.

Efficiency expression: e = benefit / cost = W / Q Q = heat absorbed

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https://brainmass.com/physics/internal-energy/thermodynamics-efficiency-carnot-cycle-211954

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Thermal physics(schroeder) : carnot cycle
I need a calculation of the 'Carnot cycle' and a proof of the 'efficiency expression;. DETAILED please

Efficiency expression: e = benefit / cost = W / Q Q = heat absorbed

As we know the Carnot cycle is the combination of four processes
i) Isothermal expansion (A to B)
ii) Adiabatic expansion (B to C)
iii) Isothermal compression (C to D) and
iv) Adiabatic compression (D to A)
The thermodynamic states of the n moles of ideal gas taken as working substance are given as the coordinates.
i) Isothermal expansion (A to B)
The thermodynamic state changes from (P1V1T1) to (P2V2T1) i.e. the pressure and the volume changes while the temperature remains same T1 as that of the source.
The work done in isothermal change at temperature T1, as the volume changes from V1 to V2 is given by
W1 = n R T1 ln (V2/V1) -------------------------- (1)
As there is no change in the temperature of the gas, there will be no change in its internal energy and hence the work done by the gas is equal to the energy absorbed by the gas from the source. Thus the energy absorbed by the gas is given by
Q1 = W1 = n R T1 ln (V2/V1)
ii) Adiabatic expansion (B to C)
The thermodynamic state changes from (P2V2T1) to (P3V3T2) i.e. the pressure and the volume changes as well as the temperature decrease from T1 to T2.
The work done in adiabatic change as temperature increases from T1 to T2 is given by
W2 = --------------------------- (2)
Where  is the ratio of specific heats of the gas, called its adiabatic exponent
As in adiabatic process there is no heat is absorbed or given out by the gas the work done by the gas is equal to the loss in its internal energy and hence its temperature is reducing accordingly hence in this process Q = 0.
iii) Isothermal compression (C to D)
The thermodynamic state changes from (P3V3T2) to (P4V4T2) i.e. the pressure and the volume changes while the temperature remains same T2 as that of the sink.
The work done in isothermal change at temperature T2, as the volume changes from V3 to V4 is given by
W3 = n R T2 ln (V4/V3) -------------------------- (3)
As there is no change in the temperature of the gas, there will be no change in its internal energy and hence the work done on the gas is equal to the energy given out by the gas to the sink. Thus the energy given out by the gas is given by
Q2 = W3 = n R T2 ln (V4/V3)
(as V4 <V3, the work done by the gas will be negative means energy is given out)

iv) Adiabatic compression (D to A)
The thermodynamic state changes from (P4V4T2) to (P1V1T1) i.e. the pressure and the volume changes as well as the temperature increase from T2 to T1.
The work done in adiabatic change as temperature increases from T2 to T1 is given by
W2 = --------------------------- (4)
(here T2 is < T1 hence the work done by the gas is negative means the work is done on the gas to compress it.)
As in adiabatic process there is no heat is absorbed or given out by the gas the work done on the gas is equal to the gain in its internal energy and hence its temperature increases accordingly. Hence in this process Q = 0.
Thus the net work done by the gas in the whole cycle is given by
W = W1 + W2 + W3 + W4
Or W = n R T1 ln (V2/V1) + + n R T2 ln (V4/V3) +
Or W = n R T1 ln (V2/V1) + n R T2 ln (V4/V3) -------------- (5)
Now as for isothermal change the product of pressure and volume remains constant we get for isothermal changes i) and iii)
i) P1V1 = P2V2 and
iii) P3V3 = P4V4
similarly for adiabatic changes ii) and iv) we can write the relation for pressure and the volume as
ii) and
iv)
Multiplying above four equations and solving we get

Or
Or
Substituting this result in equation (5) we get
W = n R T1 ln (V2/V1) + n R T2 ln (V1/V2)
Or W = n R T1 ln (V2/V1) - n R T2 ln (V2/V1)
Or W = n R (T1- T2) ln (V2/V1) ------------------ (6)

Now the engine will be more efficient if the work done by it in every cycle W is more and the heat absorbed from the source is less. Thus the efficiency of the engine is given by
 = W/Q1
Substituting the values from equations (1) and (6) the efficiency of the Carnot cycle is given by
 = [n R (T1- T2) ln (V2/V1)] / [ n R T1 ln (V2/V1)]
Or  = (T1- T2)/ T1
Or
This is the formula for the efficiency of the Carnot cycle. For the result in % multiply it by 100.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

© BrainMass Inc. brainmass.com October 5, 2022, 4:42 pm ad1c9bdddf>
https://brainmass.com/physics/internal-energy/thermodynamics-efficiency-carnot-cycle-211954