Respuesta :
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 [tex]h_{1}[/tex] and [tex]V_{1}[/tex] at the temperature of 40° C.
[tex]h_{1}[/tex] = [tex]h_{f}[/tex] = 167.53 KJ/Kg
[tex]V_{1}[/tex] = [tex]V_{f}[/tex] = 0.001008 [tex]m^{3}[/tex]/ Kg
For [tex]P_{1}[/tex] and [tex]P_{2}[/tex] the interpret value at the table of saturated water temperature is
[tex]P_{1}=P_{\text {Sat at } 40}[/tex] = 7.385 K Pa
[tex]P_{2}=P_{\text {Sat at } 300}[/tex] = 8.588 K Pa
The specific work input is expressed:
[tex]w_{\mathrm{p,in}}=v_{1}\left(P_{2}-P_{1}\right)[/tex]
= 0.001008 [tex]m^{3}[/tex]/ Kg(8.588 K Pa - 7.385 K Pa)
[tex]w_{p,in}[/tex] = 8.65 KJ/Kg
Enthalpy steam is expressed in state 2.
[tex]h_{2}[/tex] = [tex]h_{1}[/tex] + [tex]w_{p,in}[/tex]
[tex]h_{2}[/tex] = 167.53 KJ/Kg + 8.65 KJ/Kg
[tex]h_{2}[/tex] = 176.18 KJ/Kg
The table of saturated water temperature is referred and the interpret value is [tex]s_{3}[/tex] and [tex]h_{3}[/tex] at the temperature 300° C.
[tex]s_{3}[/tex] = [tex]s_{g}[/tex] = 5.7059 KJ/Kg.K
[tex]h_{3}[/tex] = [tex]h_{g}[/tex] = 2749.6 KJ/Kg
For [tex]S_{f}[/tex] and [tex]S_{fg}[/tex] the interpret value at the table of saturated water temperature is
[tex]S_{f}[/tex] = 0.5724 KJ/Kg.K
[tex]S_{fg}[/tex] = 7.6832 KJ/Kg.K
At last the process of heat rejection, the steam quality is expressed.
[tex]x_{4}=\frac{s_{4}-s_{f}}{s_{fg}}[/tex]
[tex]x_{4}=\frac{5.7059-0.5724}{7.6832}[/tex]
[tex]x_{4}[/tex] = 0.6681
The table of saturated water temperature is referred and the interpret value is [tex]h_{fg}[/tex] at the temperature of 40° C as 2406 KJ/Kg.
Enthalpy steam is expressed in state 4.
[tex]h_{4}[/tex] = [tex]h_{f}[/tex] + [tex]x_{4}[/tex][tex]h_{fg}[/tex]
[tex]h_{4}[/tex] = 167.53 KJ/Kg + 0.6681(2406 KJ/Kg)
[tex]h_{4}[/tex] = 1775.1 KJ/Kg
For the Rankine cycle, the specific heat input is expressed.
[tex]q_{in}[/tex] = [tex]h_{3}[/tex] - [tex]h_{2}[/tex]
[tex]q_{in}[/tex] = 2749.6 - 176.18
[tex]q_{in}[/tex] = 2573.4 KJ/Kg
For the Rankine cycle, the specific heat output is expressed.
[tex]q_{out}[/tex] = [tex]h_{4}[/tex] - [tex]h_{1}[/tex]
[tex]q_{out}[/tex] = 1775.1 - 167.53
[tex]q_{out}[/tex] = 1607.57 KJ/Kg
The specific net work output is expressed.
[tex]w_{T,out}[/tex] = [tex]h_{3}[/tex] - [tex]h_{4}[/tex]
[tex]w_{T,out}[/tex] = 2749.6 - 1775.1
[tex]w_{T,out}[/tex] = 974.5 KJ/Kg
The thermal efficiency is expressed.
[tex]\eta_{\mathrm{th}}=1-\frac{q_{out}}{q_{\mathrm{in}}}[/tex]
[tex]\eta_{\mathrm{th}}=1-\frac{1,607.6 \mathrm{kJ} / \mathrm{kg}}{2,573.4 \mathrm{kJ} / \mathrm{kg}}[/tex]
=1 - 0.6246
= 0.3753
Hence, the thermal efficiency is 37.5%
