Question

A 65000 Kw steam power station uses coal of calorific value 15000 kcal per kg if the coal consumption per kwh is 0.5 kg and the load factor of the station is 40% calculate overall efficiency and coal consumption per day​

Answer 1

Step-by-step explanation:

A 65000 Kw steam power station uses coal of calorific value 15000 kcal per kg if the coal consumption per kwh is 0.5 kg and the load factor of the station is 40% calculate overall efficiency and coal consumption per dayA 65000 Kw steam power station uses coal of calorific value 15000 kcal per kg if the coal consumption per kwh is 0.5 kg and the load factor of the station is 40% calculate overall efficiency and coal consumption per day.

Main Answer

A 65000 Kw steam power station uses coal of calorific value 15000 kcal per kg if the coal consumption per kwh is 0.5 kg and the load factor of the station is 40% calculate overall efficiency and coal consumption per day

Answer 2

Answer:

(i) overall efficiency=28.7% (ii) coal comsumption per day= 312 tons

Step-by-step explanation:

(ii)units generated per day = Max output× load factor× hours in a day

units generated per day = 65000×0.4×24=624000kwh

coal consumption per day = Units generated per day×coal consumption per kwh

coal consumption per day = 624000×0.5= 312000kg = 312 tons

(i)

units generated per day= max output× hours in a day

units generated per day= 65000×24 = 1560000kwh

coal consumption per day= 312000 kg

over all efficiency = electrical heat equivalent/ heat of combustion

overall efficiency = 156000×860/312000×15000

overall efficiency = 0.2866

approx efficiency 28.7%