Question

Thermal power plants consists of two 50 MW units, each running at 6000 hours and one 20MW units runs at 3000 hours per year. Energy produced by the plant is 840 x 106kWh per year. Find Plant load factor and plant use factor.​

Answer 1

Answer:

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Answer 2

Given:

Energy Produced per Year= 840× 10∧6 kWh

2 units of 50 MW running at 6000 hours per year.

1 unit of 20 MW running at 3000 hours per year.

To Find:

(i) Plant load factor

Solution:

First, we need to calculate Average Load and Maximum Demand.

a). Average Load

The energy produced per year = Average Load × Total number of hours per year.

∴ 840 × 10∧6 = Average Load × [365 days/ year × 24 hours/ day]

∴ Average Load = $\frac{840 * 10^6 }{365 * 24 }$

∴ Average Load = 95.890 × 10∧3  kw = 95.89 mw.

b). Maximum Demand

Plant capacity = [50 × 2] + 20.

∴ Plant capacity = 120 MW.

Assuming plant capacity = Maximum demand.

∴ Maximum demand = 120 MW.

Now, we can calculate Plant Load Factor.

(i) Plant Load Factor

Load factor = $\frac{Average Load}{Maximum Demand}$

= $\frac{95.89}{120}$

∴ Load factor = 0.7990 = 79.90%