Question

Air at \(27^{\circ} \mathrm{C}\) and 5 atm is throttled by a valve to 1 atm. If the valve is adiabatic and the change in kinetic energy is negligible, the exit temperature of air will be \((a) 10^{\circ} \mathrm{C}\) \((b) 15^{\circ} \mathrm{C}\) \((c) 20^{\circ} \mathrm{C}\) \((d) 23^{\circ} \mathrm{C}\) \((e) 27^{\circ} \mathrm{C}\)

Short answer

Answer: (e) 27° Celsius.

Step-by-step solution

Write down the given information.

The air entering the valve is initially at: - Temperature, \(T_1 = 27^{\circ} \mathrm{C} = 27+273.15 = 300.15 \mathrm{K}\) - Pressure, \(P_1 = 5 \mathrm{atm}\) The pressure at the exit of the valve: - Pressure, \(P_2 = 1 \mathrm{atm}\)

Understand the constant enthalpy process.

Since the process is adiabatic (no heat exchange) and involves negligible change in kinetic energy, it can be assumed to be a constant enthalpy process. This means that the enthalpy of the air entering the valve, \(h_1\), is equal to the enthalpy of the air leaving the valve, \(h_2\).

Find the relationship between temperature, pressure, and enthalpy for air.

For an ideal gas, the relationship between temperature, pressure and specific enthalpy can be represented as: \(h = C_p T\) Where: \(h\) is the specific enthalpy, \(C_p\) is the specific heat capacity at constant pressure, and \(T\) is the temperature. Since the enthalpy remains constant during the process and \(C_p\) is also a constant for the ideal gas, we can write: \(C_p T_1 = C_p T_2\)

Solve for exit temperature \(T_2\) .

We can now find the temperature at the exit of the valve, as the initial temperature \(T_1\) is given. From the previous step, we know \(C_p T_1 = C_p T_2\). Dividing both sides by \(C_p\) and rearranging, we get: \(T_2 = T_1\) Therefore, the exit temperature of air: \(T_2 = 300.15 \mathrm{K}\) Converting the temperature back to Celsius: \(T_2 (^{\circ} \mathrm{C}) = 300.15 - 273.15 = 27^{\circ} \mathrm{C}\) The exit temperature of air after the throttling process is \(27^{\circ} \mathrm{C}\). So, the correct answer is: \((e) 27^{\circ} \mathrm{C}\)