Answer to Question #274414 in Molecular Physics | Thermodynamics for roseann

Question

A steam power plant burns coal to produce a 600 MW net power with an overall thermal efficiency of 45 percent. The actual air-fuel ratio in the furnace is calculated to be 12 kg air/kg fuel. The heating value of the coal is 30,000 kJ/kg. 

(i)  What is the rate and amount of coal consumed during a 24-hour period?

(ii) What is the rate of air flowing through the furnace? 

Answer 1

Solution;

(a)

Rate and amount of coal consumed in 24-hour period;

$P_{in}=\frac{P}{\eta}=\frac{600MW}{0.45}=1333.3MW$

$Q_{in}=P_{in}\Delta t=1333.33×(24×3600s)=11.52×10^7MJ$

Amount of coal;

$m_c=\frac{Q_in}{q_{HV}}=\frac{11.52×10^7MJ}{30MJ/kg}=3.84×10^6kg$

Rate ;

$\dot{m}=\frac m{\Delta t}=\frac{3.84×10^6kg}{24×3600}=44.44kg/s$

(b)

Rate of air flowing through the furnace;

$\dot{m_{air}}=(AF)\dot{m_c}=12×44.44=533.33kg/s$