Question

By efficiency, we generally mean the ratio of the desired output to the required input. That is, efficiency is a measure of what is accomplished compared to what it costs. One important energy efficiency term for power conversion devices that operate in a cycle is thermal efficiency, ηth.

It is defined as the ratio of net work output (which is desired) to the heat input (which is costly).
Hence ካ = net work output/heat input. This definition of thermal efficiency leads to a value between zero and one.
A steam power plant generates 700,000 lbm/hr of steam in its boiler and has an output of 95,000 kW of power. The plant consumes 72,800 lbm/hr of coal, which has a "heating value" (the amount of thermal energy liberated when the coal is burned in air) of 13,000 Btu/lbm-coal.
a. Determine the overall plant thermal efficiency.
b. If the energy added to the steam from the combustion of coal in the steam-generating unit (the boiler) is 1140 Btu/lbm-steam, what fraction of the energy released from the coal is added to the steam? (This could be called the "boiler efficiency".
c. Energy transferred to the steam that doesn't result in power production (the 95 MW referred to above) has to be "discarded", often by using cooling water from a lake or river. Determine the amount of such heat rejection, in kW.

Answer 1

The answers to the question is as follows

a. 0.3425 or 34.25% efficiency

b. 0.8432 ≅ 0.84

c. 138,871 Kilowatts ≈ 140 kW

Step-by-step explanation:

a.) Overall efficiency =η,overall =
`(Heat equivalent of electric power)/(Heat of coal cumbustion)` i.e (heat amounting to the quantity of produced electric power) / (Heat produced by coal combustion)

The heat equivalent of electric power → 95,000 kw ≡ 324,153,410 BTUs per Hour

heat produced by coal combustion = 72,800 ibm/hr which produces 13000 Btu/ibm-coal hence the amount of heat produced by the coal =

72,800 ibm/hr × 13,000 Btu/ibm-coal = 946400000 Btu/hr

Therefore overall plant efficiency = 324153410/946400000 = 0.3425

which is 34.25% efficiency

b.) Energy added to the steam from combustion of coal = 1140 Btu/lbm-steam

Steam power plant generates = 700,000 lbm/hr

Energy added to the steam per hour from the coal = 700000×1140 = 798000000 Btu/hr,

Fraction of the energy released from the coal that is added to the steam =798000000/946400000 = 0.8432 of the energy released from the coal is added to the steam

c.) The amount of heat rejection is given by

Energy in steam - heat equivalent of energy of power output

798000000 Btu/hr -324,153,410 Btu/Hr = 473846590 Btu/Hr or 138,871 Kilowatts