Consider a cylinder containing air at 1200 kPa and 350 C and then, the air is expanded
to 140 kPa with a reversible adiabatic process. Calculate the speci�c work (kJ=Kg)
done by the gas. Assume calorically perfect gas.
To calculate the specific work done by the gas in the reversible adiabatic process, we can use the equation:
W = C_v * (T2 - T1)
Where:
W = Specific work done (kJ/kg)
C_v = Specific heat at constant volume (kJ/kg·K)
T2 = Final temperature (K)
T1 = Initial temperature (K)
To find the specific heat at constant volume, C_v, for a calorically perfect gas, we can use the equation:
C_v = R / (γ - 1)
Where:
R = Specific gas constant for the gas (kJ/kg·K)
γ = Specific heat ratio or adiabatic index of the gas
Given:
Initial pressure, P1 = 1200 kPa
Initial temperature, T1 = 350°C = 623.15 K
Final pressure, P2 = 140 kPa
Reversible adiabatic process
First, we need to find the specific heat ratio, γ. For a calorically perfect gas, γ depends on the gas itself. For air, the value of γ is approximately 1.4.
Next, we can calculate the specific heat at constant volume, C_v, using the gas constant for air, R = 0.287 kJ/kg·K:
C_v = R / (γ - 1)
C_v = 0.287 / (1.4 - 1)
C_v ≈ 0.287 kJ/kg·K
Now we have all the necessary values to calculate the specific work done, W:
W = C_v * (T2 - T1)
W = 0.287 * (T2 - T1)
To find the final temperature, T2, we can use the relationship between pressure, volume, and temperature for a reversible adiabatic process:
(P2 * V2^γ) / (P1 * V1^γ) = T2 / T1
Since the volume remains constant (V2 = V1), we can simplify the equation to:
(P2 / P1) = T2 / T1
Rearranging the equation to solve for T2:
T2 = (P2 / P1) * T1
Substituting the given values:
T2 = (140 kPa / 1200 kPa) * 623.15 K
T2 ≈ 72.46 K
Finally, we can calculate the specific work done by substituting the values into the equation:
W = 0.287 * (T2 - T1)
W = 0.287 * (72.46 - 623.15)
W ≈ -152.44 kJ/kg (negative sign indicates work done by the gas)
Therefore, the specific work done by the gas is approximately -152.44 kJ/kg.