Cyclical Integral for a Reversible Heat Engine

2. Cyclical Integral for a Reversible Heat Engine

We will show that for any reversible heat engine involving an ideal gas as the me- dium the cyclical integral

∫ δQ/T = 0. For such a heat engine using the First law δq – δw = du.

If the processes within it are in quasiequilibrium, then δq rev – δw rev = du. Since δw rev = Pdv, and for an ideal gas du = c v0 dT,

δq rev – P dv = c v0 dT, i.e., δq rev /T – (R/v)dv = c v0 dT/T. Integrating over a cycle, we find that the RHS of these relations is zero, i.e., ∫ δq rev /T = ∫R dv/v + ∫c v0 dT/T = 0.

(8) The Carnot cycle (due to Sadi Carnot, 1796–1832) uses ideal gas as its working fluid

and consists of four quasiequilibrium processes , as illustrated in Figure 3 : adiabatic compres- sion, isothermal heat addition Q H from a higher–temperature reservoir at a temperature T H , adiabatic expansion, and heat rejection Q L to a lower–temperature reservoir at T L . From Eq. (8) it is evident that

Q L /Q H =T L /T H (9) for a Carnot cycle. Hence the efficiency of a Carnot cycle is given by the expression

(10) If the P–v–T relationship for ideal gases were of the form P v = R f( Θ) where T =

η=W cycle /Q H =1–Q L /Q H =1–T L /T H .

f( Θ), in that case du = c v0 f´( Θ) d Θ, and ∫ δQ/T = ∫ δQ/f(Θ) = 0. Therefore,

Q L /Q H = f( Θ H )/f( Θ L ), i.e.,

(12) From Eqs. (10) or (12) it is obvious that all Carnot heat engines running between the

Q L /Q H =f( Θ L , Θ H ).

reservoirs at the same high and low temperatures, and using an ideal gas as a medium have the same efficiency. This is known as Carnot’s Second Corollary. Let us now state Carnot’s cor- ollaries:

First : The thermal effi- ciency of an irreversi- ble power cycle is al-

ways less than the thermal efficiency of a

reversible power cycle

T= T H

when each operates between the same two

Q=0

Q=0

reservoirs.

Second : All reversible power cycles with any

T= T L

medium of fluid oper-

ating between the same two thermal reservoirs

must have the same thermal efficiencies.

For instance, a Carnot cycle using steam as a medium

Figure 3. Schematic illustration of a Carnot cycle. must have the same effi- ciency as one using ideal gas as the working fluid if they operate between the same thermal reservoirs. The Kelvin–Planck statement of the Second law is violated if Carnot’s Second Corollary is violated as illustrated in the next example. We will prove this I) for a cycle with ideal gas as medium and II) then with steam as medium. Also in general, for any Carnot cycle operating with any medium

∫ δQ/T = 0.

d. Example 4 An automobile engine consists of an Carnot cycle–based heat engine using steam as

its medium, and operating between a thermal reservoir at 500 K and ambient air at 300 K. Assume its efficiency to be 50%. A similar heat engine which uses ideal gas as its medium is operated in reverse in order to replenish the heat lost by the

higher–temperature reservoir. This tandem operation of steam heat engine and ideal gas heat pump occurs perpetually in the absence of any fuel input to power the auto-

mobile. Is this possible? (See Figure 4 .) Solution The efficiency of a Carnot cycle–based engine using ideal gas as its working fluid is η G = 1 – 300 ÷ 500 = 0.4, i.e., Q L = 0.6 Q H . We are asked to assume that for the steam

engine η s = 0.5, Q L = 0.5 Q H . Since the engines are reversible, the ideal gas–based machine can be operated as a heat pump. Using both the steam heat engine and ideal

gas heat pump in tandem, it is possible to obtain a net energy production (as work) equal to 0.1Q H or 10 kJ if Q H =100 kJ while the entire heat lost by the higher–temperature reservoir is replenished. The combined system within the dashed boundary removes 10 kJ of thermal energy from the lower–temperature reservoir at 300 K and converts it completely into work with 100% efficiency in violation of the Kelvin–Planck statement of the Second law. Clearly, this is impossible.

Remarks An assumption was made that η s > η G . Instead, if we assume η s < η G , the tandem op-

eration of an ideal gas engine and a similar steam heat pump can be hypothesized to prove that this is not possible. Thus, the only realistic scenario occurs when η s = η G . The conclusion from this example is that the efficiency of a Carnot cycle using steam

as its medium is the same as for a cycle employing air as the medium as long as they operate between the same TERs. In other words the Carnot efficiency is independent of the medium used in the cycle or constitutive relation (e.g. ideal gas law) of the me- as its medium is the same as for a cycle employing air as the medium as long as they operate between the same TERs. In other words the Carnot efficiency is independent of the medium used in the cycle or constitutive relation (e.g. ideal gas law) of the me-

e. Example 5 Steam is generated at a temperature of 1000 K. It is possible to transfer 2000 kW

from it to a Carnot heat engine. Calculate the work done if the engine is used in:

A desert where the ambient temperature is 47ºC?

A polar region where the ambient temperature is –13ºC? Solution

Q H = 2000 kW. Therefore, heat rejection from the engine at 47ºC (320 K) is Q L = –Q H T L /T H = –2000 × (320 ÷ 1000) = –640 kW.

W = ∫ δQ = Q H –Q L = 2000 – 640 = 1360 kW, and

η = 1 – 320 ÷ 1000 = 0.68. At the lower temperature of –13ºC (260 K), the heat rejection Q L = –Q H T L /T H =– 2000 × (260 ÷ 1000) = –520 kW, W = 2000 – 520 = 1480 kW, and η = 1 – 260 ÷ 1000 = 0.74.

A larger amount of work is possible with the same thermal input if the temperature of the lower–temperature reservoir is reduced.

f. Example 6 What is the work required to run a Carnot heat pump that provides 2000 kW of ther- mal energy to a 1000 K high-temperature reservoir in a desert that has an ambient

temperature of 47ºC. Solution The COP = Q H /W = T H /(T H –T L ) = 1000 ÷ (1000–320) = 1.47. Therefore, W = 2000 ÷ 1.47 = 1360 kW. This work input equals the output of the heat engine discussed in Example 5 above.