Answer to Question #188583 in Molecular Physics | Thermodynamics for Ahmad

Question

a steam plant uses 3.045 tonne of coal per hour .the steam generated is fed to a turbine whose output is 4.1MW .if the calorific value is 28 MJ/kg,determine the thermal efficiency of the plant

Answer 1

To be given in question

Time (t)=1hour=3600sec

Power output$(P_{o})$ =4.1MW =4.1$\times10^6watt$

$P_{o}=4.1\times10^6watt$

Mass(m)=3.045tonne

=3.045$\times10^3kg$

Heat of combustion=28MJ/kg =$28\times10^6watt$

Total heat given by the coil =

Mass of coil$\times$ heat combustion of coil

Heat input=$3.045\times10^3\times28\times10^6J$ =8.526$\times10^{10}J$

Power input =$\frac{heat input}{time}$

$P_{I}=\frac{8.526\times10^{10}}{3600}$ =$23.69\times10^{6}watt$

$P_{I}=23.69\times10^6watt$

$Efficiency=\frac{output power}{inputpower}$

Put value

Efficiency=$\frac{4.1\times10^6}{23.69\times10^6}$ =0.1730

Percentage efficiency (%)=$0.1730\times100$

Percentage efficiency =17.30%