Question

In an air - standard Carnot cycle, heat is transferred to the working fluid at 1150 K, and heat is rejected at 300 K. The heat transfer to the working fluid at 1150 K is 120 kj/kg. The maximum pressure in the cycle P, is 16.5 MPa. Assuming constant specific heat of air, determine the cycle efficiency and pressure at different points of cycle.

Answer 1

Answer:

Explanation:

From the given information:

The efficiency can be calculated by using the formula:

$E= 1 - (\dfrac{T_2}{T_1}) \\ \\ \\ E = 1- (\dfrac{300}{1150}) \\ \\ \\ =0.739$

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, $W_1 = (R_{air} \times T_1 ) In( \dfrac{P_1}{P_2}) ---(1)$

If the equation is rewritten, we can have the following:

$In (\dfrac{P_1}{P_2}) = \dfrac{120}{(0.287 \times 1150)} \\ \\ In (\dfrac{P_1}{P_2}) = 0.364$

By solving the above equation;

$P_2 = 11.471 MPa$

However, at state 2 → 3, there is an adiabatic process.

Thus,

$P_2^{1-\gamma} T_2^{\gamma} = P_3^{1-\gamma} T_3^{\gamma}$

Specific heat rate is denoted by $\gamma$

Thus,

$P_3 = P_2\Big ( \dfrac{T_2}{T_3}\Big)^\dfrac{\gamma}{1-\gamma}}$

Thus;

recall that:

$\dfrac{\gamma}{1-\gamma}} = \dfrac{1.4}{1-1.4} \\ \\ = \dfrac{1.4}{-0.4} \\ \\ = -3.5$

Thus,

$P_3 = \dfrac{11.471 }{(\dfrac{1150}{300})^{-3.5}} \\ \\ = 0.104 \ MPa$

Finally from state 4 - state 1, we have an isentropic or adiabatic process;

As such:

$P_2^{1-\gamma} T_2^{\gamma} = P_4^{1-\gamma} T_4^{\gamma}$

Thus,

$P_4 = P_1(\dfrac{T_1 }{T_4})^{\dfrac{\gamma}{1-\gamma}}$

$P_4 = \dfrac{16.5 \ MPa }{(\dfrac{1150}{300})^{-3.5}} \\ \\ = 0.15 \ MPa$