Question

A coal-burning steam power plant produces a net power of 300 MW with an overall thermal efficiency of 32 percent. The actual gravimetric air–fuel ratio in the furnace is calculated to be 12 kg air/kg fuel. The heating value of the coal is 28,000 kJ/kg. Determine (a) the amount of coal consumed during a 24-hour period and (b) the rate of air flowing through the furnace. Answers: (a) 2.89 × 106 kg, (b) 402 kg/s

Answer 1

a) 2.89x10⁶ kg

b) 402 kg/s

Step-by-step explanation:

Let's assume that the power plant operates steadily (without changing in the state) and that the kinetic and the potential energy changes are zero.

a) The rate of the amount of heat inputs to the power plant, is the net power divided by the thermal efficiency (32% = 0.32):

`Q^.` = 300/0.32

`Q^.` = 937.5 MW

The rate is the amount of heat divided by the time, so the amount of heat is the rate multiplied by the time (24h = 86,400s):

Q = 937.5 * 86,400

Q = 8.1x10⁷ MJ

The mass of the coal is the heat divided by its heating value (28,000 kJ/kg = 28 MJ/kg)

m = 8.1x10⁷/28

m = 2.89x10⁶ kg

b) The rate of air/fuel is 12 kg/kg, so the mass of air (ma) is:

ma/m = 12

ma/2.89x10⁶ = 12

ma = 3.47x10⁷ kg

During 24 h(86,400s) the rate of air is:

`ma^.` = 3.47x10⁷/86,400s

`ma^.` = 402 kg/s