A power plant taps steam superheated by geothermal energy to 475 K (the temperature of the hot reservoir) and uses the steam to do work in turning the turbine of an electric generator. The steam is then converted back into water in a condenser at 308 K (the temperature of the cold reservoir), after which the water is pumped back down into the earth where it is heated again. The output power (work per unit time) of the plant is 82,500 kilowatts.(a) Determine the maximum efficiency at which this plant can operate.(b) Determine the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours.

Question

07-11-2019
Physics

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A power plant taps steam superheated by geothermal energy to 475 K (the temperature of the hot reservoir) and uses the steam to do work in turning the turbine of an electric generator. The steam is then converted back into water in a condenser at 308 K (the temperature of the cold reservoir), after which the water is pumped back down into the earth where it is heated again. The output power (work per unit time) of the plant is 82,500 kilowatts.(a) Determine the maximum efficiency at which this plant can operate.(b) Determine the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours.

Respuesta :

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matthewadeyinka10 matthewadeyinka10 · Answer 1 · 2019-11-07T22:36:37+01:00

Answer:

Thermal Efficiency, η = [tex]\frac{W₀}{Q₁}[/tex] . . . . . . . . . . . . . . . . Eqn 1

where W₀ = Work Output = Q₁ - Q₀ =82500KW . . . . . . . . . . . . . . . Eqn 2

Q₁ = Heat Supplied/Input = mC(ΔT₁)

Q₁ = Heat Rejected/Output = mC(ΔT₀) . . . . . . . . . . . . . . . . . . . . . . . . Eqn 3

Note: From Carnot's theorem, for any engine working between these two temperatures (T₀/T₁), The maximum attainable efficiency is the Carnot efficiency given as follows;

Therefore, η = 1 - [tex]\frac{Q₀}{Q₁}[/tex] = 1 - [tex]\frac{T₀}{T₁}[/tex]

Remember, T₁ = 475K and T₀ = 308K

η = 1 - (308/475) = 1 - 0.648 = 0.352

Hence, the maximum efficiency at which this plant can operate = 35%

2. To determine the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours.

Remember from Eqn 1, Q₁ = W/η,

Therefore, Q₁= 82500/0.35

Q₁=235,714KW,

So, from Eqn 2, Q₀ = 235714 - 82500

Q₀ = 153214KW (KJ/s) (Released Heat)

In t =24 hours, we can then use this to determine the minimum amount of heat rejected qₓ (KiloJoule), = Q₀t (Remember, you have to convert the time, t, unit to seconds)

= 153214 x t (KiloJoule)

qₓ = 153214 x 24 x 3600 (KiloJoule)

qₓ = 13238 MegaJoule

Respuesta :

Therefore, the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours is 13238 MJoule

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