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Sagot :
The overall efficiency (in %) of the plant is 30% if the steam power plant with a power output of 150 MW consumes coal at a rate of 60 tons/h.
What is efficiency?
Efficiency is the capacity to achieve something or get the desired outcome without wasting resources, time, money, energy, or effort.
It is given that:
A steam power plant with a power output of 150 MW consumes coal at a rate of 60 tonnes per hour.
As we know,
Thermal efficiency Te = (W/Q)x100
m(coal) = 60 tons/h = 50/3 kg/s
The heating value of the coal is 30,000 kj/k = 30 MJ/kg
Q = (50/3)30
Q = 500 MW
Te = (150/500)100
Te = 30%
Thus, the overall efficiency (in %) of the plant is 30% if the steam power plant with a power output of 150 MW consumes coal at a rate of 60 tons/h.
The question is incomplete.
The complete question is:
A steam power plant with a power output of 150 MW consumes coal at a rate of 60 tonnes per hour. If the heating value of coal is 30,000 KJ/kg, determine the overall efficiency (in %) of the plant.
Learn more about efficiency here:
https://brainly.com/question/13828557
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