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Q. 4.2

Expert-verifiedFound in: Page 134

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

At a power plant that produces 1 GW10^{9} watts) of electricity, the steam turbines take in steam at a temperature of 500^{o}, and the waste heat is expelled into the environment at 20^{o}(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 600^{o}. 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

Temp at cold reservoir = 20^{o }C and

Temp at hot reservoir = 500^{o}

Convert temp in K.

T_{c} =20^{o }C = 273+20= 193 K

T_{h }=500^{o}C = 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%

The temperature of the cold reservoir = 20^{o}The temperature of the hot reservoir = 600^{o}The power plant produces 10^{9} watts of electricity that is 10^{9} J/s.Rate of additional electricity is 5 cents /kwh.

T_{h}=600^{o}=873 K and T_{c}=20^{o}=293 K

Percentage of improved efficiency

$1-\frac{293}{873}(100\%)=66.437\%$

With the improved efficiency the amount of electricity produced is

$P=\frac{{10}^{9}\times 66.44}{62.1}\phantom{\rule{0ex}{0ex}}=1.0699\times {10}^{9}\mathrm{W}$

${P}_{\text{additional}}=0.0699\times {10}^{9}\mathrm{W}\phantom{\rule{0ex}{0ex}}=6.{9910}^{7}\mathrm{W}\phantom{\rule{0ex}{0ex}}=6.{9910}^{4}\mathrm{kW}$

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

$=\frac{{10}^{4}\times 6.99\times 5}{3600}\text{cents}\phantom{\rule{0ex}{0ex}}=97.083\text{cents}$

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)

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