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Q. 4.2

Expert-verified
Found in: Page 134

### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

# 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.)

a) Maximum possible efficiency of the plant = 62.1%

b) Money can be made in a year $30 million See the step by step solution ### Step by Step Solution ## Part(a)- Step 1- Given Information Temp at cold reservoir = 20o C and Temp at hot reservoir = 500o ## Pat(a)-Step 2- Explanation Convert temp in K. Tc =20o C = 273+20= 193 K Th =500oC = 273+500=773 K The efficiency of power plant is given as $\mathrm{e}\le 1-\frac{{\mathrm{T}}_{\mathrm{c}}}{{\mathrm{T}}_{\mathrm{h}}}\phantom{\rule{0ex}{0ex}}Substitutevalues,\phantom{\rule{0ex}{0ex}}\mathrm{e}\le 1-\frac{293}{773}\phantom{\rule{0ex}{0ex}}\mathrm{e}\le 0.621$ Maximum efficiency in percentage is 62.1% ## Part(b)-Step 1- Given information The temperature of the cold reservoir = 20oThe temperature of the hot reservoir = 600oThe power plant produces 109 watts of electricity that is 109 J/s.Rate of additional electricity is 5 cents /kwh. ## Part(b)-Step 2 - Explanation Th=600o=873 K and Tc=20o=293 K Percentage of improved efficiency $1-\frac{293}{873}\left(100%\right)=66.437%$ With the improved efficiency the amount of electricity produced is $P=\frac{{10}^{9}×66.44}{62.1}\phantom{\rule{0ex}{0ex}}=1.0699×{10}^{9}\mathrm{W}$ ${P}_{\text{additional}}=0.0699×{10}^{9}\mathrm{W}\phantom{\rule{0ex}{0ex}}=6.{9910}^{7}\mathrm{W}\phantom{\rule{0ex}{0ex}}=6.{9910}^{4}\mathrm{kW}$ If sold for 5 cents per kilo-watt-hour then the money made in one second is: $=\frac{{10}^{4}×6.99×5}{3600}\text{cents}\phantom{\rule{0ex}{0ex}}=97.083\text{cents}$ 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)