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

The Carnot cycle is a theoretical cycle that represents the most efficient heat engine possible. The cycle efficiency can be determined using the equation Efficiency (η) = 1 - (Tc/Th). In this case, the efficiency is 73.9%. The pressure at different points of the cycle can be determined using the ideal gas law and the specific heat capacity equation.

Step-by-step explanation:

The Carnot cycle is a theoretical cycle that represents the most efficient heat engine possible. It consists of four processes: two reversible isothermal processes and two reversible adiabatic processes. To determine the cycle efficiency, we can use the equation:

Efficiency (η) = 1 - (Tc/Th)

where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. In this case, Tc = 300 K and Th = 1150 K. Plugging these values into the equation gives:

η = 1 - (300/1150) = 0.739 = 73.9%

The pressure at different points of the cycle can be determined using the ideal gas law:

PV = mRT

where P is the pressure, V is the volume, m is the mass, R is the specific gas constant, and T is the temperature in Kelvin. Since the working fluid is air and its specific heat is constant, we can use the equation:

C = cp - cv

where C is the specific heat capacity, cp is the heat capacity at constant pressure, and cv is the heat capacity at constant volume. The pressure at each point of the cycle can be calculated using the information given in the problem statement.

Answer 2

Step-by-step explanation:

From the given information:

The efficiency can be calculated by using the formula:

`E= 1 - ((T_2)/(T_1)) \\ \\ \\ E = 1- ((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) * T_1 ) In( (P_1)/(P_2)) ---(1)`

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

`In ((P_1)/(P_2)) = (120)/((0.287 * 1150)) \\ \\ In ((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 ( (T_2)/(T_3)\Big)^(\gamma)/(1-\gamma)}`

Thus;

recall that:

`(\gamma)/(1-\gamma)} = (1.4)/(1-1.4) \\ \\ = (1.4)/(-0.4) \\ \\ = -3.5`

Thus,

`P_3 = (11.471 )/(((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((T_1 )/(T_4))^{(\gamma)/(1-\gamma)}`

`P_4 = (16.5 \ MPa )/(((1150)/(300))^(-3.5)) \\ \\ = 0.15 \ MPa`