Question

A particular power plant operates with a heat-source reservoir at 350 oC and a heat-sink reservoir at 30 oC. It has a thermal efficiency equal to 55% of the Carnot-engine thermal efficiency for the same temperatures a). what is the thermal efficiency of the plant? b). To what temperature must the heat-source reservoir be raised to increase the thermal efficiency of the plant to 35%?

Answer 1

Answer

given,

heat-source reservoir at 350°C

heat-sink reservoir at 30°C

thermal efficiency = 55 %

To calculate the thermal efficiency

$\eta_{carnot}=1 - \dfrac{T_L}{T_H}$

$\eta_{carnot}=1 - \dfrac{30+273}{350+273}$

$\eta_{carnot}=1 - \dfrac{303}{623}$

$\eta_{carnot}=0.5136$

thermal efficiency of the plant = 55 % of Carnot engine

$\eta_{plant}=0.55\ \eta_{carnot}$

                  =$0.55\times 0.5136$

                  =0.282

$\eta_{plant}=28.2\%$

calculation of thermal efficiency with raised heat source temperature

$\eta_{carnot}=1 - \dfrac{T_L}{T_H}$

$\eta_{carnot}=1 - \dfrac{30+273}{T_H}$

thermal efficiency of the plant = 55 % of Carnot engine

$\eta_{plant}=0.55\ \eta_{carnot}$

$0.35=0.55\times (1 - \dfrac{30+273}{T_H})$

$0.35=0.55\times (1 - \dfrac{303}{T_H})$

T_H = 833.25 K

T_H = 560°C

Answer 2

The thermal efficiency of the plant is 28.2%. The ratio of the heat used by the plant and the total heat energy in the fuel that has been given to the plant.

What is the thermal efficiency of a plant?

It can be defined as the ratio of the heat used by the plant and the total heat energy in the fuel that has been given to the plant.

$\eta = \dfrac {W_T}{Q_T}\times 100$

Where,

$W_t$ - Work done by the engine = 30°C = 303 K

$Q_T$ - the amount of heat in the fuel = 350°C = 623 K

Put the values in the formula,

$\eta_c = \dfrac {303}{623}\times 100\\\\\eta_c = 55\%$

Thus the thermal efficiency of the plant is,

$\eta_p = 55\times 0.5136\\\\\eta_p = 28.2\%$

Therefore, the thermal efficiency of the plant is 28.2%.

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