A coal-fired generating station has an efficiency of 38% and produces an
average electrical output of 1000 MWe.
(a) How much coal does it burn per second? Assume the coal is bituminous.(b) How much heat (in joules) is released to the environment per second?
(c) How many liters of water at 20°C could the heat in part (b) boil (at STP) per second?

Question

Quizplus · Accepted Answer

The Answer is (a) 1000 MWe output at 38% efficiency requires 1000 MW)/(0.38) = 2632 MW thermal energy from the combustion of coal. Using an energy content of coal of 31 MJ/kg then the mass of coal required per second will be

( 2632 \mathrm { MJ } / \mathrm { s } ) / ( 31 \mathrm { MJ } / \mathrm { kg } ) = 84.9 \mathrm {~kg} / \mathrm { s }

(b) In 1 second the total energy is

2632 \mathrm { MJ }

and the electrical energy is

1000 \mathrm { MJ }

so the waste heat will be

2632 \mathrm { MJ } - 1000 \mathrm { MJ } = 1632 \mathrm { MJ }

.
(c)

1 \mathrm {~kg}

of water has a heat capacity of

4180 \mathrm {~J} / { } ^ { \circ } \mathrm { C }

and a latent heat of vaporization of

2.26

MJ. To heat water from

20 ^ { \circ } \mathrm { C }

to

100 ^ { \circ } \mathrm { C }

will require

\left( 4180 \mathrm {~J} / \mathrm { kg } ^ { \circ } \mathrm { C } ight) imes \left( 100 ^ { \circ } \mathrm { C } - 20 ^ { \circ } \mathrm { C } ight) = 0.334 \mathrm { MJ } / \mathrm { kg }

So to heat water from

20 ^ { \circ } \mathrm { C }

too

100 ^ { \circ } \mathrm { C }

and boil it requires

( 0.334 \mathrm { MJ } / \mathrm { kg } ) + ( 2.26 \mathrm { MJ } / \mathrm { kg } ) = 2.59 \mathrm { MJ } / \mathrm { kg }

Therefore,

1632 \mathrm { MJ } / \mathrm { s }

will boil

( 1632 \mathrm { MJ } / \mathrm { s } ) / ( 2.59 \mathrm { MJ } / \mathrm { kg } ) = 630 \mathrm {~kg} / \mathrm { s } ext { or } 630 \mathrm {~L} / \mathrm { s }

A Coal-Fired Generating Station Has an Efficiency of 38% and Produces