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.
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.
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
To solve this problem, we can use the formulas for the Carnot power cycle and the specific heat capacities of air.
(i) To find the maximum and minimum temperatures for the cycle, we need to use the thermal efficiency formula for the Carnot cycle:
η = 1 - Tc/Th
Given that the thermal efficiency (η) is 50%, we can rearrange the formula to find the ratio of the hot to cold temperatures:
Tc/Th = 1 - η = 1 - 0.5 = 0.5
The minimum temperature (Tc) occurs during the isothermal compression, and the maximum temperature (Th) occurs during the isothermal expansion.
From the given data, we know that during the isothermal expansion, the heat transfer to air (Qh) is 40 kJ. We can use the formula for isothermal expansion to find the maximum temperature:
Qh = m * Cp * ΔT
Where m is the mass of air, Cp is the specific heat capacity at constant pressure, and ΔT is the change in temperature during isothermal expansion.
We don't know ΔT, but we can find it using the ideal gas law:
P * V = m * R * T
Where P is the pressure, V is the volume, m is the mass, R is the specific gas constant, and T is the temperature.
Since this is isothermal expansion, we can set the initial and final pressures equal:
P_initial * V_initial = P_final * V_final
P_initial = 7 bar = 700 kPa
V_initial = 0.12 m^3
P_final is not given, but we can find it using the ideal gas law:
P_final = P_initial * (V_initial / V_final)
And since ΔT = T_final - T_initial, we can rewrite the equation as:
P_final = P_initial * (V_initial / (V_initial + ΔV)) where ΔV = V_final - V_initial
P_final = P_initial * (V_initial / (V_initial + ΔV)) = (700 kPa) * (0.12 m^3 / (0.12 m^3 + ΔV))
Now we can substitute this value of P_final into the formula for isothermal expansion:
Qh = m * Cp * ΔT
40 kJ = (0.5 kg) * (1.008 kJ/kg K) * ΔT
Solving for ΔT:
ΔT = 40 kJ / (0.5 kg * 1.008 kJ/kg K) = 79.36 K
Therefore, the maximum temperature (Th) is:
Th = T_initial + ΔT = T_initial + 79.36 K
Now we can use the ratio of temperatures to find Tc:
Tc = 0.5 * Th
(ii) To find the volume at the end of isothermal expansion, we can use the given data:
V_initial = 0.12 m^3
From the ideal gas law, we already found the equation for P_final:
P_final = (700 kPa) * (0.12 m^3 / (0.12 m^3 + ΔV))
We know that P_final is the pressure at the end of isothermal expansion, which means it is the same as the pressure at the beginning of isothermal compression. Therefore, we can use the same formula for P_final as follows:
P_initial = (700 kPa) * (0.12 m^3 / (0.12 m^3 + ΔV))
Solving for ΔV:
ΔV = 0.12 m^3 * (700 kPa / P_initial - 1)
Finally, we can find V_final:
V_final = V_initial + ΔV = 0.12 m^3 + 0.12 m^3 * (700 kPa / P_initial - 1)
(iii) To find the heat transfer for each of the four processes, we can use the formulas for each process in the Carnot cycle:
1. Isothermal Expansion: Qh = m * Cp * ΔT
2. Adiabatic Expansion: Qc = 0 (no heat transfer)
3. Isothermal Compression: Qc = m * Cp * ΔT
4. Adiabatic Compression: Qh = 0 (no heat transfer)
From our calculations, we already know Qh for the isothermal expansion is 40 kJ. Using the given specific heat capacity of air, we can calculate Qc for the isothermal compression using the formula:
Qc = m * Cp * ΔT = (0.5 kg) * (1.008 kJ/kg K) * ΔT
Since the adiabatic processes have no heat transfer, Qc for the adiabatic expansion and Qh for the adiabatic compression are both zero.
Therefore, the heat transfer for each process is:
1. Isothermal Expansion: Qh = 40 kJ
2. Adiabatic Expansion: Qc = 0
3. Isothermal Compression: Qc = (0.5 kg) * (1.008 kJ/kg K) * ΔT
4. Adiabatic Compression: Qh = 0
To find the value of Qc for the isothermal compression, we need to calculate ΔT using the formula:
ΔT = Th - T_initial = Th - (T_initial + ΔTc)
Where ΔTc is the change in temperature during isothermal compression.
From the given ratio of temperatures, we can find T_initial:
T_initial = Th - ΔTc = (T_initial + ΔT) - ΔTc = Th - ΔTc + ΔT - ΔTc
Simplifying this equation:
2ΔTc = ΔT
ΔTc = ΔT/2 = 79.36 K / 2 = 39.68 K
Now we can calculate Qc for the isothermal compression:
Qc = (0.5 kg) * (1.008 kJ/kg K) * ΔTc
Therefore, the heat transfer for each process is:
1. Isothermal Expansion: Qh = 40 kJ
2. Adiabatic Expansion: Qc = 0
3. Isothermal Compression: Qc = (0.5 kg) * (1.008 kJ/kg K) * 39.68 K
4. Adiabatic Compression: Qh = 0
I hope this helps! Let me know if you have any further questions.