Question

A steam power plant produces 500 MW of electricity with an overall thermal efficiency of 35%. Determine (a) the rate at which heat is supplied to the boiler and (b) the waste heat that is rejected by the plant. (c) If the heating value of coal (heat that is released when 1 kg of coal is burned) is 30 MJ/kg, determine the rate of consumption of coal in tons(US)/day. Assume that 100% of heat released goes to the cycle. (d) What-if Scenario: What would the fuel consumption rate be if the thermal efficiency were to increase to 40%?

Answer 1

Answer:

Explanation:

500 MW = 500 X 10⁶ J / s

If R be the rate of heat supplied and 35 % is the efficiency of conversion

R x .35 = 500 X 10⁶

R =  500 X 10⁶ / .35

= 1428.57 x 10⁶ J / s

b )

Rate of heat going waste

= .65 x 1428.57 x 10⁶

= 928 .57 x 10⁶ J / s

c )

Heat content of coal = 30 x 10⁶ J /kg

1428.57 x 10⁶ J of heat will come from

= 1428.57 x 10⁶ J  / 30 x 10⁶ J /kg

= 47.62 kg per second

= 47.62 x 24 x 60 x 60 =  4114368 kg in one day

= 4114368 / 907.185 ton per day  

= 4535.31 ton per day

d ) Fuel consumption rate  will decrease by 12.5  % if thermal efficiency is increased to 40 % .  

Answer 2

The rate at which the heat is supplied to the boiler is 1428.57 J/s. Thermal efficiency is the capacity of a plant to utilize the maximum amount of heat supplied to the plant.

What is thermal efficiency?

Thermal efficiency is the capacity of a plant to utilize the maximum amount of heat supplied to the plant.

$\eta = \dfrac WQ\times 100$

Where,

$\eta$ - thermal efficiency = 35%

$W$ - work done (Energy produced) = 500 MW

$Q$ - energy (heat) absorbed = ?

Put the values in the formula,

$Q = \dfrac {500}{35}\times 100\\\\Q = 1428.57 \rm \ J/s$

Therefore, the rate at which the heat is supplied to the boiler is 1428.57 J/s.

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