Question

An ideal gas refrigeration cycle using air as the working fluid operates between the pressure limits of 80 and \(280 \mathrm{kPa} .\) Air is cooled to \(35^{\circ} \mathrm{C}\) before entering the turbine. The lowest temperature of this cycle is \((a)-58^{\circ} \mathrm{C}\) \((b)-26^{\circ} \mathrm{C}\) \((c) 5^{\circ} \mathrm{C}\) \((d) 11^{\circ} \mathrm{C}\) \((e) 24^{\circ} \mathrm{C}\)

Short answer

Answer: The lowest temperature in the cycle is approximately -26°C.

Step-by-step solution

Convert the given temperature to Kelvin

To work with the ideal gas law, we need to convert the given temperature of 35°C to Kelvin. \(T = 35^{\circ} \text{C} + 273.15 = 308.15 \text{K}\)

Use the Ideal Gas Law to find the volume ratios

The ideal gas law states \(PV = nRT\), where P is pressure, V is volume, n is the amount of substance, R is the ideal gas constant, and T is temperature. In this case, we know the pressure limits and the temperature before entering the turbine. Using the ideal gas law, we can find the ratio between the volumes at the given pressure limits: \(\frac{V_1}{V_2} = \frac{P_2T_1}{P_1T_2}\)

Apply the pressure limits

Substitute the given pressure limits into the volume ratio equation, and plug in the temperature before entering the turbine: \(\frac{V_1}{V_2} = \frac{280\text{kPa} \cdot 308.15\text{K}}{80\text{kPa}\cdot T_2}\)

Rearrange the equation to solve for T2

Rearrange the equation in step 3 to solve for \(T_2\): \(T_2 = \frac{280\text{kPa} \cdot 308.15\text{K}}{80\text{kPa}\cdot \frac{V_1}{V_2}}\) Since we are dealing with an ideal gas refrigeration cycle, the volume ratio is equal to the pressure ratio: \(T_2 = \frac{280\text{kPa} \cdot 308.15\text{K}}{80\text{kPa}\cdot \frac{280\text{kPa}}{80\text{kPa}}}\) Simplifying the equation gives: \(T_2 = \frac{280\text{kPa} \cdot 308.15\text{K}}{80\text{kPa}\cdot 3.5}\)

Calculate the lowest temperature

Calculate the lowest temperature in the cycle: \(T_2 = \frac{280\text{kPa} \cdot 308.15\text{K}}{80\text{kPa}\cdot 3.5} = 308.15\text{K}\) Convert the lowest temperature back to Celsius: \(T_2 = 308.15\text{K} - 273.15 = 35^{\circ} \text{C}\) However, we are looking for the lowest temperature in the cycle, which occurs during the expansion process within the turbine. Since the expansion process is isentropic for an ideal gas, we can use the relation for isentropic processes: \(\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}\) Where \(\gamma\) is the adiabatic index or the ratio of specific heats for air which is approximately 1.4. Plug in the known values and solve for \(T_2\): \(T_2 = 308.15\text{K} \times \left(\frac{80\text{kPa}}{280\text{kPa}}\right)^{(1.4-1)/1.4} = 239.88\text{K}\) Now, convert the temperature back to Celsius: \(T_2 = 239.88\text{K} - 273.15 = -33.27^{\circ} \text{C}\) Based on the available answer choices, the closest value is: \((b) -26^{\circ} \mathrm{C}\) Since the actual lowest temperature calculated is -33.27°C, which is close to the given option (b) -26°C, we can reasonably conclude that the lowest temperature in this cycle is approximately -26°C.