0.5 kg of air (ideal gas) executes a Carnot power cycle having a thermal efficiency of 50 per cent. The heat transfer to the air during the isothermal expansion is 40 kJ. At the beginning of the isothermal expansion the pressure is 7 bar and the volume is 0.12 m3.

Question

Determine:
(i) The maximum and minimum temperatures for the cycle in K;
(ii) The volume at the end of isothermal expansion in m3; and
(iii) The heat transfer for each of the four processes in kJ.
For air Cv = 0.721 kJ/kg K, and Cp = 1.008 kJ/kg K.

Answer 1

To solve this problem, we can use the formulas for the thermal efficiency of a Carnot cycle and the specific heat capacities of air.

(i) The thermal efficiency of a Carnot cycle is given by:

η = 1 - (Tmin / Tmax)

where η is the thermal efficiency, Tmin is the minimum temperature, and Tmax is the maximum temperature. We are given that the thermal efficiency is 50% (or 0.5).

0.5 = 1 - (Tmin / Tmax)

Solving for Tmin, we get:

Tmin = (1 - 0.5) * Tmax
Tmin = 0.5 * Tmax

(ii) During the isothermal expansion, the heat transfer is given as 40 kJ. The formula for heat transfer during an isothermal process is:

Q = n * R * T * ln(Vf / Vi)

where Q is the heat transfer, n is the number of moles, R is the gas constant, T is the temperature, Vf is the final volume, and Vi is the initial volume. We are given that the initial volume is 0.12 m^3 and the heat transfer is 40 kJ. Since we are given the mass of air (0.5 kg), we can convert it to moles using the molar mass of air (28.97 kg/kmol) to get:

n = 0.5 / (28.97 * 10^-3) = 17.25 kmol

Substituting the values into the equation, we can solve for the final volume:

40 kJ = 17.25 * R * T * ln(Vf / 0.12)

(iii) The heat transfers for the four processes in a Carnot power cycle are:

- Heat transfer during isothermal expansion: Q_expansion = 40 kJ
- Heat transfer during adiabatic expansion: Q_ab_expansion = 0 kJ (adiabatic process)
- Heat transfer during isothermal compression: Q_compression = -40 kJ (opposite of heat transfer during expansion)
- Heat transfer during adiabatic compression: Q_ab_compression = 0 kJ (adiabatic process)

Now, let's calculate the values.

(i) Using the equation for thermal efficiency, we can find the maximum temperature (Tmax):

0.5 = 1 - (0.5 * Tmax)

Solving for Tmax:

0.5 * Tmax = 1 - 0.5
0.5 * Tmax = 0.5
Tmax = 1 K

The maximum temperature for the cycle is 1 K.

To find the minimum temperature (Tmin):

Tmin = 0.5 * Tmax
Tmin = 0.5 * 1
Tmin = 0.5 K

The minimum temperature for the cycle is 0.5 K.

(ii) To find the final volume (Vf) during isothermal expansion, we solve the equation for heat transfer during isothermal expansion:

40 kJ = 17.25 * R * T * ln(Vf / 0.12)

Rearranging the equation:

ln(Vf / 0.12) = (40 kJ) / (17.25 * R * T)

Since we know that R = Cp - Cv, we can use the values for Cp and Cv to calculate R:

R = 1.008 kJ/kg K - 0.721 kJ/kg K = 0.287 kJ/kg K

Substituting the values:

ln(Vf / 0.12) = (40 kJ) / (17.25 * 0.287 kJ/kg K * T)

ln(Vf / 0.12) = (40) / (17.25 * 0.287 * T)

ln(Vf / 0.12) = 0.8755 / T

Taking the exponential of both sides:

Vf / 0.12 = e^(0.8755 / T)

Vf = 0.12 * e^(0.8755 / T)

(iii) The heat transfers for each process:

- Heat transfer during isothermal expansion: Q_expansion = 40 kJ (given)
- Heat transfer during adiabatic expansion: Q_ab_expansion = 0 kJ (adiabatic process)
- Heat transfer during isothermal compression: Q_compression = -40 kJ (opposite of heat transfer during expansion)
- Heat transfer during adiabatic compression: Q_ab_compression = 0 kJ (adiabatic process)

Therefore, the heat transfer for each process is:

Q_expansion = 40 kJ
Q_ab_expansion = 0 kJ
Q_compression = -40 kJ
Q_ab_compression = 0 kJ