Question

A geothermal power plant uses geothermal water extracted at 160°C at a rate of 440 kg/s as the heat source, and produces 22 MW of net power. If the environment temperature is 25°C, determine (a) (0.5 point) the actual thermal efficiency, (b) (0.5 point) the maximum possible thermal efficiency, and (c) (0.5 point) the actual rate of heat rejection from this power plant

Answer 1

Answer:

The actual efficiency is:

$n_{actual} = 2,76 %$

The maximum efficiency possible is:

$n_{max} = 31,18 %$

The rate of heat rejection is:

$Q_{rej}  = 774621,3 kW$

Explanation:

Considering the specific heat of water 4,1813 KJ/kgK the energy that goes in the system is:

$Q_{in} = 440 kg/s * 4,1813 KJ/kgK * 433 K = 796621,3 kW$

The heat energy that goes out in an ideal situation (Tout = 25°C):

$Q_{out} = 440 kg/s * 4,1813 KJ/kgK * 298 K = 548252,1 kW$

The maximum heat that can be obtained is:

$Q_{max}  = Q_{in} - Q_{out} = 248369,2 kW$

So the maximum efficiency possible is:

$n_{max} = \frac{Q_{max} }{Q_{in} } = 31,18 %$

The actual efficiency is:

$n_{actual} = \frac{W_{out} }{Q_{in} } = \frac{22000 kW} }{796621,3 kW } =2,76 %$

Where $W_{out}$ is the obtained work equal to 22 MW

From the balance of energy, the rate of heat rejection is:

$Q_{rej}  = Q_{in} - W_{out} = 774621,3 kW$