Question

A geothermal plant uses geothermal water (yes a liquid) extracted from beneath the Earth using drilling apparatus. The geothermal water (a condensed liquid) comes up at a temperature of 200 C at a rate of 200 kg/s. It is used to drive a turbine that extracts 6000 kw of power. The water leaves the turbine at 80 C and enters a lake at 25 C. What is the actual thermal efficiency of the turbine in percent to three significant digits (do not enter a percent sign!!)

Answer 1

To solve this problem it is necessary to apply the concepts of heat change and thermal efficiency.

The heat rate can be expressed under the function

`Q = \dot{m} (h_1-h_2)`

Where,

m = Mass

`h_i =` Enthalpy at each state

Our values are given as,

`\dot{m} = 200kg/s`

`T_H = 200\°C`

`W = 6000kW`

`T_(H,2) = 90\°C`

`T_L = 25\°C`

From the tables of Enthalpy of Water at 200°C (Saturated liquid state)

`h_1 = 852.4kJ/Kg`

At the same time for 80°C

`h_2 = 334.9kJ/Kg`

Applying the equation of Heat,

`Q = \dot{m}(h_1-h_2)`

Replacing,

`Q = 200*(852.4-334.9)`

`Q = 103500kW`

Therefore the efficiency would be

`\eta = (Q_L)/(Q_H)`

`\eta = (6000)/(103500)`

`\eta = 0.0579`

Therefore the actual thermal efficiency of the turbine in percent is 0.0579.