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

$\eta = \frac{Q_L}{Q_H}$

$\eta = \frac{6000}{103500}$

$\eta = 0.0579$

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