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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.)

Short Answer

Expert verified

a) 62.1%

b) $ 30 million

Step by step solution

01

Part(a)- Step 1- Given 

Temp at cold reservoir = 20oC and

Temp at hot reservoir = 500o


02

Pat(a)-Step 2- Explanation

Convert temp in K.

Tc =20oC = 273+20= 193 K

Th=500oC = 273+500=773 K

The efficiency of power plant is given as

e1-TcTh

Substitute the values we get

e1-293773=0.621

In percentage it is 62.1%

03

Part(b)-Step 1- Given information

The temperature of the cold reservoir = 20o
The temperature of the hot reservoir = 600o
The power plant produces 109 watts of electricity that is 109 J/s.
Rate of additional electricity is 5 cents /kwh.

04

Part(b)-Step 2 - Explanation

Th=600o=873 K and Tc=20o=293 K

Percentage of improved efficiency =1-293873(100%)=66.437%= 66.44% ( rounding)

With the improved efficiency the amount of electricity produced is

P=109×66.4462.1=1.0699×109W

So additional electricity produced is

Padditional=0.0699×109W=6.99107W=6.99104kW

If sold for 5 cents per kilo-watt-hour then the money made in one second is:

role="math" localid="1647250244806" =104×6.99×53600cents=97.083cents

Money made in one year is

=97.083 x 3600 x 24 x 365 cents

=3,06,16,09,488 cents

covert in $

=$ 3,06,16,094.88

= $ 30 Million ( approx)

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Most popular questions from this chapter

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?

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.

In an absorption refrigerator, the energy driving the process is supplied not as work, but as heat from a gas flame. (Such refrigerators commonly use propane as fuel, and are used in locations where electricity is unavailable.* ) Let us define the following symbols, all taken to be positive by definition:
Qf= heat input from flame
Qc= heat extracted from inside refrigerator
Qr= waste heat expelled to room
Tf= temperature of flame
Tc= temperature inside refrigerator
Tr= room temperature

(a) Explain why the "coefficient of performance" (COP) for an absorption refrigerator should be defined as Qc / Qf.
(b) What relation among Qf, Qc, and Qr is implied by energy conservation alone? Will energy conservation permit the COP to be greater than 1 ?
(c) Use the second law of thermodynamics to derive an upper limit on the COP, in terms of the temperatures Tf, Tc, and Tr alone.

In table 4.1, why does the entropy of water increase with increasing temperature, while the entropy of steam decreases with increasing temperature?

In a real turbine, the entropy of the steam will increase somewhat. How will this affect the percentages of liquid and gas at point4in the cycle? How will the efficiency be affected?

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