Answer:
Explanation:
From the given information:
The efficiency can be calculated by using the formula:
[tex]E= 1 - (\dfrac{T_2}{T_1}) \\ \\ \\ E = 1- (\dfrac{300}{1150}) \\ \\ \\ =0.739[/tex]
Using the isothermal condition, the process from state 1 → 2 is as follows:
Heat transferred Q₁ = 120 kJ/kg
Workdone W₁ = 120 kJ/kg
However, [tex]W_1 = (R_{air} \times T_1 ) In( \dfrac{P_1}{P_2}) ---(1)[/tex]
If the equation is rewritten, we can have the following:
[tex]In (\dfrac{P_1}{P_2}) = \dfrac{120}{(0.287 \times 1150)} \\ \\ In (\dfrac{P_1}{P_2}) = 0.364[/tex]
By solving the above equation;
[tex]P_2 = 11.471 MPa[/tex]
However, at state 2 → 3, there is an adiabatic process.
Thus,
[tex]P_2^{1-\gamma} T_2^{\gamma} = P_3^{1-\gamma} T_3^{\gamma}[/tex]
Specific heat rate is denoted by [tex]\gamma[/tex]
Thus,
[tex]P_3 = P_2\Big ( \dfrac{T_2}{T_3}\Big)^\dfrac{\gamma}{1-\gamma}}[/tex]
Thus;
recall that:
[tex]\dfrac{\gamma}{1-\gamma}} = \dfrac{1.4}{1-1.4} \\ \\ = \dfrac{1.4}{-0.4} \\ \\ = -3.5[/tex]
Thus,
[tex]P_3 = \dfrac{11.471 }{(\dfrac{1150}{300})^{-3.5}} \\ \\ = 0.104 \ MPa[/tex]
Finally from state 4 - state 1, we have an isentropic or adiabatic process;
As such:
[tex]P_2^{1-\gamma} T_2^{\gamma} = P_4^{1-\gamma} T_4^{\gamma}[/tex]
Thus,
[tex]P_4 = P_1(\dfrac{T_1 }{T_4})^{\dfrac{\gamma}{1-\gamma}}[/tex]
[tex]P_4 = \dfrac{16.5 \ MPa }{(\dfrac{1150}{300})^{-3.5}} \\ \\ = 0.15 \ MPa[/tex]
