A simple ideal Rankine cycle which uses water as the working fluid operates its condenser at 40°C and its boiler at 250°C. Calculate the work produced by the turbine, the heat supplied in the boiler, and the thermal efficiency of this cycle when the steam enters the turbine without any superheating

The thermal efficiency is 37.5%.

Explanation:

The T-s diagram with steam was attached below.

The table of saturated water temperature is referred and the interpret value is and at the temperature of 40° C.

= = 167.53 KJ/Kg

= = 0.001008 / Kg

For and the interpret value at the table of saturated water temperature is

= 7.385 K Pa

= 8.588 K Pa

The specific work input is expressed:

= 0.001008 / Kg(8.588 K Pa - 7.385 K Pa)

= 8.65 KJ/Kg

Enthalpy steam is expressed in state 2.

= +

= 167.53 KJ/Kg + 8.65 KJ/Kg

= 176.18 KJ/Kg

The table of saturated water temperature is referred and the interpret value is and at the temperature 300° C.

= = 5.7059 KJ/Kg.K

= = 2749.6 KJ/Kg

For and the interpret value at the table of saturated water temperature is

= 0.5724 KJ/Kg.K

= 7.6832 KJ/Kg.K

At last the process of heat rejection, the steam quality is expressed.

= 0.6681

The table of saturated water temperature is referred and the interpret value is at the temperature of 40° C as 2406 KJ/Kg.

Enthalpy steam is expressed in state 4.

= +

= 167.53 KJ/Kg + 0.6681(2406 KJ/Kg)

= 1775.1 KJ/Kg

For the Rankine cycle, the specific heat input is expressed.

= -

= 2749.6 - 176.18

= 2573.4 KJ/Kg

For the Rankine cycle, the specific heat output is expressed.

= -

= 1775.1 - 167.53

= 1607.57 KJ/Kg

The specific net work output is expressed.

= -

= 2749.6 -  1775.1

= 974.5 KJ/Kg

The thermal efficiency is expressed.

=1 - 0.6246

= 0.3753

Hence, the thermal efficiency is 37.5%

answer:

a

explanation: