The Carnot Cycle

• Idealized thermodynamic cycle consisting of four reversible processes (any substance):

Reversible isothermal expansion (1-2, TH=constant) Reversible adiabatic expansion (2-3, Q=0, THTL) Reversible isothermal compression (3-4, TL=constant) Reversible adiabatic compression (4-1, Q=0, TLTH)

1-2 2-3 3-4 4-1

The Carnot Cycle-2Work done by gas = PdV, area under the process curve 1-2-3.

1

2

3

32

1

Work done on gas = PdV, area under the process curve 3-4-1

subtract

Net work1

2

34

dV>0 from 1-2-3PdV>0

Since dV<0PdV<0

The Carnot Principles

• The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs. th, irrev < th, rev

• The efficiencies of all reversible heat engines operating between the same two reservoirs are the same. (th, rev)A= (th, rev)B

• Both Can be demonstrated using the second law (K-P statement and C-statement). Therefore, the Carnot heat engine defines the maximum efficiency any practical heat engine can reach up to.

• Thermal efficiency th=Wnet/QH=1-(QL/QH)=f(TL,TH) and it can be shown that th=1-(QL/QH)=1-(TL/TH). This is called the Carnot efficiency.

• For a typical steam power plant operating between TH=800 K (boiler) and TL=300 K(cooling tower), the maximum achievable efficiency is 62.5%.

ExampleLet us analyze an ideal gas undergoing a Carnot cycle between two temperatures TH and TL.

1 to 2, isothermal expansion, U12 = 0QH = Q12 = W12 = PdV = mRTHln(V2/V1)

2 to 3, adiabatic expansion, Q23 = 0(TL/TH) = (V2/V3)k-1 (1)

3 to 4, isothermal compression, U34 = 0QL = Q34 = W34 = - mRTLln(V4/V3)

4 to 1, adiabatic compression, Q41 = 0(TL/TH) = (V1/V4)k-1 (2)

From (1) & (2), (V2/V3) = (V1/V4) and (V2/V1) = (V3/V4)th = 1-(QL/QH )= 1-(TL/TH) since ln(V2/V1) = ln(V4/V3)

It has been proven that th = 1-(QL/QH )= 1-(TL/TH) for all Carnot engines since the Carnot efficiency is independent of the working substance.

Carnot Efficiency

A Carnot heat engine operating between a high-temperature source at 900 K and reject heat to a low-temperature reservoir at 300 K. (a) Determine the thermal efficiency of the engine. (b) If the temperature of the high-temperature source is decreased incrementally, how is the thermal efficiency changes with the temperature.

th

L

H

th H

H

th H

L

T

T

K

TT

K

TT

1 1300

9000 667 66 7%

300

1300

900

1900

. .

( )

( )

( )

( )

Fixed T and lowering T

The higher the temperature, the higher the "quality"

of the energy: More work can be done

Fixed T and increasing T

L H

H L

200 400 600 800 10000

0.2

0.4

0.6

0.8

1

Temperature (TH)

Eff

icie

ncy

Th( )T

T

200 400 600 800 10000

0.2

0.4

0.6

0.8

1

Temperature (TL)

Eff

icie

ncy

TH( )TL

TL

Lower TH

Increase TL

Carnot Efficiency

• Similarly, the higher the temperature of the low-temperature sink, the more difficult for a heat engine to transfer heat into it, thus, lower thermal efficiency also. That is why low-temperature reservoirs such as rivers and lakes are popular for this reason.

•To increase the thermal efficiency of a gas power turbine, one would like to increase the temperature of the combustion chamber. However, that sometimes conflict with other design requirements. Example: turbine blades can not withstand the high temperature gas, thus leads to early fatigue. Solutions: better material research and/or innovative cooling design.

• Work is in general more valuable compared to heat since the work can convert to heat almost 100% but not the other way around. Heat becomes useless when it is transferred to a low-temperature source because the thermal efficiency will be very low according to th=1-(TL/TH). This is why there is little incentive to extract the massive thermal energy stored in the oceans and lakes.

Carnot CycleHot Reservoir

T1

Cold Reservoir T2

C

Q1

Q2

W

Volume

Pressure

•

•

a

b

•d

T1

Q1

Carnot Cycle

Q2

VnRT

P 1=

V

constP

.=

T2•c

Q=0Q=0

VnRT

P 2=

Volume

Pressure

•

•

a

b

•d

T1

Q1

Q2

VnRT

P 1=

V

constP

.=

T2•c

Q=0Q=0

VnRT

P 2=

W

Carnot Cycle

From a to b: isothermal, so that U = 0 and Q = - W

Thus, Q1 = +nRT1ln(Vb/Va) (+ve quantity)

Carnot Cycle

Similarly, from c to d: isothermal, so that U = 0 and Q = - W

Thus, Q2 = +nRT2ln(Vd/Vc) = -nRT2ln(Vc/Vd) (-ve)

From b to c: adiabatic, Q = 0, so that TV-1 is constant.Thus, T1Vb

-1 = T2Vc-1 or 1

2

1

b

c

V

V

T

T

Similarly, d to a: adiabatic, Q = 0, so that TV-1 is constant.Thus, T2Vd

-1 = T1Va-1 or

1

2

1

a

d

V

V

T

T

Carnot Cycle

We see that:11

2

1

a

d

b

c

V

V

V

V

T

T

a

b

d

c

V

V

V

V

Which means that

Now also:)/ln()/ln(

)/ln()/ln(

2

1

2

1

2

1

dc

ab

dc

ab

VVTVVT

VVnRTVVnRT

QQ

This is an important result. Temperature can be defined (on the absolute (Kelvin) scale) in terms of the heat flows in a Carnot Cycle.

But as the volume ratios are equal: 2

1

2

1

TT

QQ

What’s Special about a Carnot Cycle?

(1) Heat is transferred to/from only two reservoirs at fixed temperatures, T1 and T2 - not at a variety of temperatures.

(2) Heat transfer is the most efficient possible because the temperature of the working substance equals the temperature of the reservoirs. No heat is wasted in flowing from hot to cold.

(3) The cycle uses an adiabatic process to raise and lower the temperature of the working substance. No heat is wasted in heating up the working substance.

(4) Carnot cycles are reversible. Not all cycles are!

What’s Special about a Carnot Cycle?

(5) The Carnot theorem states that the Carnot cycle (or any reversible cycle) is the most efficient cycle possible. The Carnot cycle defines an upper limit to the efficiency of a cycle.

• Where T1 and T2 are the temperatures of the hot and cold reservoirs, respectively, in degrees Kelvin.

As T2 > 0, c is always <1.

• Recall that for any cycle, the efficiency of a heat engine is given as:

1

2

11==QQ

QW

E

• For an engine using a Carnot cycle, the efficiency is also equal to:

1

21=TT

C

Problem

1. A Carnot engine using 0.020 mol of an ideal gas operates between reservoirs at 1000.0 K and 300.0 K. The engine takes in 25 J of heat from the hot reservoir per cycle. Find the work done by the engine during each of the two isothermalsteps in the cycle.

2. Consider a Carnot heat engine that operates between a heat source at 750 K and a heat sink at 300 K. The work output of the engine is used to drive a Carnot refrigerator that removes heat from the cooled space at −15°C at a rate of 400 kJ/min and rejects it to the same heat sink of the engine which is at 300 K. Determine (a) the rate of heat supplied to the heat engine and (b) the total rate of heat rejection to the environment.

4. A Carnot heat-pump works with the saturated liquid–vapor mixture of refrigerant-134a and the net power input to the cycle is 7 kW. The refrigerant mass flow rate is 0.264 kg/s. Temperature measurement with thermocouples show that the maximum temperature (in Kelvin) in the cycle is 1.25 times the minimum temperature (in Kelvin). If the refrigerant changes from saturated vapor to saturated liquid during the heat rejection process, what is the ratio of the maximum to minimum pressures in the cycle?

3. Consider a Carnot heat engine with thermal efficiency of th operating between a hot reservoir with temperature TH and a cold reservoir of temperature TL . It is possible to change the temperature of the hot reservoir, but the temperature on the cold reservoir is fixed. If it is desired to double the thermal efficiency of this engine, what should the temperature of the hot reservoir be?