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%.

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.

$x_{4}=\frac{s_{4}-s_{f}}{s_{fg}}$

$x_{4}=\frac{5.7059-0.5724}{7.6832}$

$x_{4}$ = 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%