Q. 4.2 - An Introduction to Thermal Physics

Question

At a power plant that produces 1 GW10⁹ 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.)

Step-by-Step Solution

Answer

a) Maximum possible efficiency of the plant = 62.1%

b) Money can be made in a year $ 30 million

Part(a)- Step 1- Given Information

Temp at cold reservoir = 20^o C and 

Temp at hot reservoir = 500^o

Pat(a)-Step 2- Explanation

Convert temp in K.  

T_c =20^o C = 273+20= 193 K

T_h =500^oC = 273+500=773 K

The efficiency of power plant is given as 

$$\begin{gathered} \mathrm{e}\le 1-\frac{{\mathrm{T}}_{\mathrm{c}}}{{\mathrm{T}}_{\mathrm{h}}} \\ Substitute values, \\ \mathrm{e}\le 1-\frac{293}{773} \\ \mathrm{e}\le 0.621 \end{gathered}$$

Maximum efficiency in percentage is 62.1%

Part(b)-Step 1- Given information

The temperature of the cold reservoir = 20^o
The temperature of the hot reservoir = 600^o
The 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

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

$$\begin{gathered} P=\frac{{10}^9\times 66.44}{62.1} \\ =1.0699\times {10}^9 \mathrm{W} \end{gathered}$$

$$\begin{gathered} P_{\text{additional }}=0.0699\times {10}^9 \mathrm{W} \\ =6.{9910}^7 \mathrm{W} \\ =6.{9910}^4 \mathrm{kW} \end{gathered}$$

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

$$\begin{gathered} =\frac{{10}^4\times 6.99\times 5}{3600}\text{ cents } \\ =97.083\text{ cents } \end{gathered}$$

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)