Question

In a steam power plant, power output from the turbine is 2162 kW and required pumping power is 10 kW. If efficiency of Rankine cycle on which power plant is working is 24%, then the value of mass flow rate of coal used in boiler will be kg/s. Assume calorific value of coal as 29 MJ/kg and ${\eta }_{combustion}=0.95$

• A. 0.325

Answer 1

The correct option is A 0.325
$W_{turbine}=2162kW$ $W_{pump}=10kW$

${\eta }_{rankine}=24\%=0.24$ $CV=29\times {10}^{3}kJ/kg$

${\eta }_{rankine}=\frac{W_{net}}{Q_{input}}=\frac{W_T-W_P}{Q_{boiler}}$

$Q_{boiler}=\frac{2162-10}{0.24}=8966.667kW$

Now let, m${}_f$ = Mass flow rate of fuel in kg/s.

$Q_{boiler}=(m_f\times C{V}_{fuel})\times {\eta }_{combustion}$

$8966.667=m_f\times 29\times {10}^3\times 0.95$

$m_f=\frac{8966.667}{29000\times 0.95}=0.325kg/s$