Answer to Question #156901 in Physics for Arif Saleem

Question #156901

a steam power plant with a power output of 250 MW consumes coal at a rate of 60 tons/h. If the heating value of the coal is 30,000 kj/kg, determine the overall efficiency of this plant


1
Expert's answer
2021-01-21T09:56:49-0500

By definition, the efficiency is the ratio of the production PprodP_{prod} and consumption PconsP_{cons} powers:


η=PprodPcons\eta = \dfrac{P_{prod}}{P_{cons}}

The production power is Pprod=250MWP_{prod} = 250MW. The consumption power is:


Pcons=qmP_{cons} = qm

where q=30000kJkg=30MJkgq = 30000\dfrac{kJ}{kg} = 30\dfrac{MJ}{kg} is the heating value of the coil, and m=60tonsh=601000kg3600s=1509kgsm = 60\dfrac{tons}{h} = 60\cdot \dfrac{1000kg}{3600s} = \dfrac{150}{9}\dfrac{kg}{s} is the rate of coil consumption. Thus, obtain;


η=Pprodqmη=250MW30MJkg1509kgs=0.5\eta = \dfrac{P_{prod}}{qm}\\ \eta = \dfrac{250MW}{30\dfrac{MJ}{kg}\cdot \dfrac{150}{9}\dfrac{kg}{s}} = 0.5

Answer. 0.5.


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