In a Carnot cycle, the maximum pressure and temperature are limited to 18 bar and 410°C. The ratio of isentropic compression is 6 and isothermal expansion is 1.5.

Question

Assuming the volume of the air at the beginning of isothermal expansion as 0.18 m3, determine:
(i) The temperature and pressures at main points in the cycle.
(ii) Change in entropy during isothermal expansion.
(iii) Mean thermal efficiency of the cycle.
(iv) Mean effective pressure of the cycle.
(v) The theoretical power if there are 210 working cycles per minute.

Answer 1

(i) To calculate the temperature and pressures at the main points in the cycle, we can use the Carnot cycle equations.

The Carnot cycle consists of four main points: (1) isentropic compression, (2) isothermal compression, (3) isentropic expansion, and (4) isothermal expansion.

Given:
- Maximum pressure (point 1) = 18 bar
- Maximum temperature (point 1) = 410°C
- Ratio of isentropic compression (point 1 to point 2) = 6
- Ratio of isothermal expansion (point 3 to point 4) = 1.5
- Volume at the beginning of isothermal expansion (point 4) = 0.18 m^3

First, let's calculate the pressure and temperature at point 2 (after isentropic compression):
P2/P1 = (V1/V2)^(γ)
6 = (V1/V2)^(γ)
Taking logarithm on both sides:
γ*log(V1/V2) = log(6)
log(V1/V2) = log(6)/γ

Since we are given the ratio of compression (6), we can substitute this value into the equation to solve for V1/V2:
log(V1/V2) = log(6)/γ
log(V1/V2) = log(6)/1.4
V1/V2 = e^(log(6)/1.4)
V1/V2 ≈ 2.792

Next, we can calculate the volume at point 2 (V2):
V1/V2 = 2.792
V2 = V1/2.792
V2 = 0.18 m^3/2.792
V2 ≈ 0.0645 m^3

Using the equation of state for an ideal gas, we can calculate the temperature at point 2 (T2):
P1V1/T1 = P2V2/T2
18 bar * 0.18 m^3 / (410 + 273) K = P2 * 0.0645 m^3 / T2
T2 = P2 * 0.0645 m^3 * (410 + 273) K / (18 bar * 0.18 m^3)
T2 ≈ 328.1 K ≈ 55.9 °C

Now, let's calculate the pressure and temperature at point 3 (after isentropic expansion):
P3 = P2 * (V2/V3)^(γ)
P3 = 18 bar * (0.0645 m^3/0.18 m^3)^(1.4)
P3 ≈ 2.925 bar

Using the same equation of state, we can calculate the temperature at point 3 (T3):
P2V2/T2 = P3V3/T3
18 bar * 0.0645 m^3 / 328.1 K = 2.925 bar * V3 / T3
T3 = 2.925 bar * V3 * 328.1 K / (18 bar * 0.0645 m^3)
T3 ≈ 474.9 K ≈ 201.8 °C

Finally, let's calculate the pressure and temperature at point 4 (after isothermal expansion):
P4 = P3 * (V3/V4)^(γ)
P4 = 2.925 bar * (0.0645 m^3/0.18 m^3)^(1.4)
P4 ≈ 0.332 bar

Using the same equation of state, we can calculate the temperature at point 4 (T4):
P3V3/T3 = P4V4/T4
2.925 bar * 0.0645 m^3 / 474.9 K = 0.332 bar * V4 / T4
T4 = 0.332 bar * V4 * 474.9 K / (2.925 bar * 0.0645 m^3)
T4 ≈ 1535.3 K ≈ 1262.2 °C

Therefore, the temperature and pressures at the main points in the cycle are:
(i) T1 = 410°C, P1 = 18 bar
T2 ≈ 55.9 °C, P2 ≈ 18 bar
T3 ≈ 201.8 °C, P3 ≈ 2.925 bar
T4 ≈ 1262.2 °C, P4 ≈ 0.332 bar

(ii) To calculate the change in entropy during isothermal expansion, we can use the equation:
ΔS = Q / T
Since the expansion is isothermal, the change in entropy is equal to the heat transfer during the expansion divided by the temperature.
Q = P3V3 * ln(V4/V3)
ΔS = P3V3 * ln(V4/V3) / T3

(iii) The mean thermal efficiency of the cycle is given by the equation:
η = 1 - (1/γ)
where γ is the ratio of specific heat capacities.
Given that γ = 1.4, we can substitute this value into the equation to find the mean thermal efficiency of the cycle.

(iv) The mean effective pressure of the cycle is given by the equation:
MEP = (P1 * V1 - P3 * V3) / (V1 - V3)
Given the values of P1, V1, P3, and V3, we can substitute them into the equation to find the mean effective pressure.

(v) The theoretical power of the cycle can be calculated using the equation:
P = MEP * V1 * n
where MEP is the mean effective pressure, V1 is the initial volume, and n is the number of working cycles per minute.
Given the value of MEP, V1, and n, we can substitute them into the equation to find the theoretical power of the cycle.