Question

A coal-burning steam power plant produces a net power of 300MW with an overall thermal efficiency of 32 percent. The actual gravimetric air-fuel ration 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 (b) the rate of air flowing through the furnace.

Answer 1

a)the amount of coal consumed during a 24 hour period is 2887500kg

b)the rate of air flowing through the furnace is 401.01kg/s

Step-by-step explanation:

Hello ! To solve this problem we must be clear about several concepts of thermal plants and apply them at each step of the calculations, I will explain all the steps carefully.

1. determine the heat energy entering the plant, this is determined by remembering that the concept of efficiency is the power of the plant's output plant over heat.

`eficiency=(W)/(Q)`

`Q=(W)/(eficiency) =(300000)/(0.32) =937500Kw`

2.

We found the ratio between time, the heat power of the fuel, and the mass of fuel using the problem data

t=time=24h=86400s

`C=28000KJ/KG\\(t)/(c) =(86400s)/(28000kJ/kg) =3.08(kg)/(KW)`

3.

Now we multiply the ratio found in step two by the heat added in step 1, to find the fuel mass

`Mcoal=(937500Kw)(3.08(kg)/(Kw) )=2887500kg`

the amount of coal consumed during a 24 hour period is 2887500kg

4. to find the rate of air flowing through the furnace, we use the gravimetric air-fuel ration

`(mair)/(mcoal) =12\\Mair=12(Mcoal)\\\\Mair=12( 2887500kg)=34650000kg`

finally we divide the air mass for a period of 24h to find the flow

`Mair=(34650000 kg)/(24h) (1h)/(3600s) =401.01kg/s`

the rate of air flowing through the furnace is 401.01kg/s