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Prove directly (by calculating the heat taken in and the heat expelled) that a Carnot engine using an ideal gas as the working substance has an efficiency of1-tcth

Short Answer

Expert verified

Hence we proved that

Carnot engine using an ideal gas as the working substance has the efficiency of 1-TcTh.

Step by step solution

01

To Prove

Carnot engine using an ideal gas as the working substance has the efficiency of 1-TcTh

02

Explanation

The Carnot cycle begins with the isothermal expansion of 1 mol of gas, which changes its state from (P1,V1,Th)to(P2,V2,Tc).The heat absorbed QHby the gas from the source at constant temperature

This given by:

Qh=W1=RThlogeV2V1---(1)

The Carnot cycle's second stage is the adiabatic expansion of 1 mol of gas taking its state fromP2,V2,Thto P3,V3,TcThe work done W2by the gas is given by:

W2=RTh-Tcγ-1---(2)

The Carnot cycle's third stage involves isothermal compression of 1 mol of gas taking its state fromP3,V3,TctoP4,V4,Thby the gas to the sink at constant temperature Tcis given by

Qc=W3=RTclogeV3V4---(3)

The fourth stage of the Carnot cycle is adiabatic compression, which involves compressing 1 mol of gas to its original condition.

P4,V4,Tcto P1,V1,ThThe work done W4on the gas is given by:

W4=RTh-Tcγ-1---(4)

The efficiency of Carnot engine is given by

η=output workheat suppliedη=Qh−QcQh=1−QcQh−−−(5)

Substitute (1) and (3) in (5)

η=1-RTclogeV3V4RThlogeV2V1---(6)

03

Further Continuation to the proof

For an adiabatic expansion:

TVγ-1=Constant

In the second stage, for an adiabatic expansion:

ThV2γ−1=TcV3γ−1TcTh=V2r−1V3r−1=V2V3r−1V2V3=TcTh1r−1−−−−−−(7)

In the fourth stage, for an adiabatic compression:

TcV4γ−1=ThV1γ−1T2T1=V1γ−1V4γ−1=V1V4γ−1V1V4=TcTh1γ−1−−−(8)

On comparing equation (7) and (8)

V1V4=V2V3⇒V3V4=V2V1−−−(9)

Now (9) in (6)

η=1-TcTh

Hence proved.

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Most popular questions from this chapter

Consider an ideal Hampson-Linde cycle in which no heat is lost to the environment.

(a) Argue that the combination of the throttling valve and the heat exchanger is a constant-enthalpy device, so that the total enthalpy of the fluid coming out of this combination is the same as the enthalpy of the fluid going in.

(b) Let xbe the fraction of the fluid that liquefies on each pass through the cycle. Show that

x=Hout-HinHout-Hliq,

where Hinis the enthalpy of each mole of compressed gas that goes into the heat exchanger, Houtis the enthalpy of each mole of low-pressure gas that comes out of the heat exchanger, and Hliqis the enthalpy of each mole of liquid produced.

(c) Use the data in Table 4.5to calculate the fraction of nitrogen liquefied on each pass through a Hampson-Linde cycle operating between 1 bar and 100 bars, with an input temperature of 300K. Assume that the heat exchanger works perfectly, so the temperature of the low-pressure gas coming out of it is the same as the temperature of the high-pressure gas going in. Repeat the calculation for an input temperature of 200K.

Can you cool off your kitchen by leaving the refrigerator door open? Explain.

A small scale steam engine might operate between the temperatures 20°Cand 300°C, with a maximum steam pressure of 10bars. Calculate the efficiency of a Rankine cycle with these parameters.

To get more than an infinitesimal amount of work out of a Carnot engine, we would have to keep the temperature of its working substance below that of the hot reservoir and above that of the cold reservoir by non-infinitesimal amounts. Consider, then, a Carnot cycle in which the working substance is at temperatureThwas it absorbs heat from the hot reservoir, and at temperatureTcwas it expels heat to the cold reservoir. Under most circumstances the rates of heat transfer will be directly proportional to the temperature differences:

QhΔt=KTh-ThwandQcΔt=KTcw-Tc

I've assumed here for simplicity that the constants of proportionality Kare the same for both of these processes. Let us also assume that both processes take the

same amount of time, so theΔt''s are the same in both of these equations.*

aAssuming that no new entropy is created during the cycle except during the two heat transfer processes, derive an equation that relates the four temperaturesTh,Tc,Thwand Tcw

bAssuming that the time required for the two adiabatic steps is negligible, write down an expression for the power (work per unit time) output of this engine. Use the first and second laws to write the power entirely in terms of the four temperatures (and the constant K), then eliminateTcwusing the result of part a.

cWhen the cost of building an engine is much greater than the cost of fuel (as is often the case), it is desirable to optimize the engine for maximum power output, not maximum efficiency. Show that, for fixed Thand Tc, the expression you found in part bhas a maximum value at Thw=12Th+ThTc. (Hint: You'll have to solve a quadratic equation.) Find the correspondingTcw.

dShow that the efficiency of this engine is 1-Tc/ThEvaluate this efficiency numerically for a typical coal-fired steam turbine with Th=600°CandTc=25°C, and compare to the ideal Carnot efficiency for this temperature range. Which value is closer to the actual efficiency, about 40%, of a real coal-burning power plant?

At a power plant that produces 1 GW109 watts) of electricity, the steam turbines take in steam at a temperature of 500o, and the waste heat is expelled into the environment at 20o
(a) What is the maximum possible efficiency of this plant?
(b) Suppose you develop a new material for making pipes and turbines, which allows the maximum steam temperature to be raised to 600o. Roughly how much money can you make in a year by installing your improved hardware, if you sell the additional electricity for 5 cents per kilowatt-hour? (Assume that the amount of fuel consumed at the plant is unchanged.)

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