Question

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

Answer 1

The thermal efficiency is 37.5%.

Step-by-step explanation:

The T-s diagram with steam was attached below.

The table of saturated water temperature is referred and the interpret value is
`h_(1)` and
`V_(1)` at the temperature of 40° C.

`h_(1)` =
`h_(f)` = 167.53 KJ/Kg

`V_(1)` =
`V_(f)` = 0.001008
`m^(3)`/ Kg

For
`P_(1)` and
`P_(2)` the interpret value at the table of saturated water temperature is

`P_(1)=P_{\text {Sat at } 40}` = 7.385 K Pa

`P_(2)=P_{\text {Sat at } 300}` = 8.588 K Pa

The specific work input is expressed:

`w_{\mathrm{p,in}}=v_(1)\left(P_(2)-P_(1)\right)`

= 0.001008
`m^(3)`/ Kg(8.588 K Pa - 7.385 K Pa)

`w_(p,in)` = 8.65 KJ/Kg

Enthalpy steam is expressed in state 2.

`h_(2)` =
`h_(1)` +
`w_(p,in)`

`h_(2)` = 167.53 KJ/Kg + 8.65 KJ/Kg

`h_(2)` = 176.18 KJ/Kg

The table of saturated water temperature is referred and the interpret value is
`s_(3)` and
`h_(3)` at the temperature 300° C.

`s_(3)` =
`s_(g)` = 5.7059 KJ/Kg.K

`h_(3)` =
`h_(g)` = 2749.6 KJ/Kg

For
`S_(f)` and
`S_(fg)` the interpret value at the table of saturated water temperature is

`S_(f)` = 0.5724 KJ/Kg.K

`S_(fg)` = 7.6832 KJ/Kg.K

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

`x_(4)=(s_(4)-s_(f))/(s_(fg))`

`x_(4)=(5.7059-0.5724)/(7.6832)`

`x_(4)` = 0.6681

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

Enthalpy steam is expressed in state 4.

`h_(4)` =
`h_(f)` +
`x_(4)`
`h_(fg)`

`h_(4)` = 167.53 KJ/Kg + 0.6681(2406 KJ/Kg)

`h_(4)` = 1775.1 KJ/Kg

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

`q_(in)` =
`h_(3)` -
`h_(2)`

`q_(in)` = 2749.6 - 176.18

`q_(in)` = 2573.4 KJ/Kg

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

`q_(out)` =
`h_(4)` -
`h_(1)`

`q_(out)` = 1775.1 - 167.53

`q_(out)` = 1607.57 KJ/Kg

The specific net work output is expressed.

`w_(T,out)` =
`h_(3)` -
`h_(4)`

`w_(T,out)` = 2749.6 - 1775.1

`w_(T,out)` = 974.5 KJ/Kg

The thermal efficiency is expressed.

`\eta_{\mathrm{th}}=1-\frac{q_(out)}{q_{\mathrm{in}}}`

`\eta_{\mathrm{th}}=1-\frac{1,607.6 \mathrm{kJ} / \mathrm{kg}}{2,573.4 \mathrm{kJ} / \mathrm{kg}}`

=1 - 0.6246

= 0.3753

Hence, the thermal efficiency is 37.5%