Question

Air is compressed steadily and adiabatically from \(17^{\circ} \mathrm{C}\) and \(90 \mathrm{kPa}\) to \(200^{\circ} \mathrm{C}\) and \(400 \mathrm{kPa} .\) Assuming constant specific heats for air at room temperature, the isentropic efficiency of the compressor is \((a) 0.76\) \((b) 0.94\) \((c) 0.86\) \((d) 0.84\) \((e) 1.00\)

Short answer

Answer: (a) 0.76

Step-by-step solution

Recall the adiabatic relations for an ideal gas

From the ideal gas law, we know that \(PV^{\gamma}= \text{constant}\) and \(T V^{\gamma-1}=\text{constant}\) for an adiabatic (isentropic) process, where P is pressure, V is volume, T is temperature, and \(\gamma\) is the ratio of specific heats (\(c_p/c_v\)). For air, we can assume \(\gamma = 1.4\).

Calculate the actual work required for compression

We first need to find the actual work required to compress the air adiabatically. Since the process is adiabatic, we have: \(W_{actual} = c_p (T_2-T_1)\) where \(T_1\) and \(T_2\) are the initial and final temperatures (in Kelvin), and \(c_p\) is the specific heat at constant pressure (approx \(1005\,\mathrm{J/kg\cdot K}\) for air at room temperature). Converting the temperature to Kelvin: \(T_1 = 17^{\circ} \mathrm{C}+273.15 = 290.15 \,\mathrm{K}\) \(T_2 = 200^{\circ} \mathrm{C}+273.15 = 473.15 \,\mathrm{K}\) Now, we calculate the actual work required: \(W_{actual} = 1005 \,\mathrm{J/kg\cdot K}\times (473.15 \,\mathrm{K} - 290.15 \,\mathrm{K}) = 183768.25\,\mathrm{J/kg}\)

Calculate the ideal work required in an isentropic process

Using the adiabatic relations, we can calculate the ideal work required: \(W_{ideal} = \dfrac{R(T_2'-T_1)}{1-\gamma}\) We can find the ideal final temperature \(T_2'\), given by: \(\left(\dfrac{P_1}{P_2}\right)^{\frac{\gamma-1}{\gamma}} = \dfrac{T_2'}{T_1} \Longrightarrow T_2' = T_1 \left(\dfrac{P_1}{P_2}\right)^{\frac{\gamma-1}{\gamma}}\) The gas constant for air, \(R = 287 \,\mathrm{J/kg\cdot K}\). Now we find \(T_2'\): \(T_2' = 290.15 \,\mathrm{K} \times \left(\dfrac{90\,\mathrm{kPa}}{400\,\mathrm{kPa}}\right)^{0.4} = 194.86 \,\mathrm{K}\) Now, we calculate the ideal work required in an isentropic process: \(W_{ideal} = \dfrac{287 \,\mathrm{J/kg\cdot K}(194.86 \,\mathrm{K} - 290.15 \,\mathrm{K})}{1-1.4} = 139251.56 \,\mathrm{J/kg}\)

Calculate the isentropic efficiency

The isentropic efficiency of the compressor can be calculated as the ratio of the ideal work to the actual work: \(\eta_{isentropic} = \dfrac{W_{ideal}}{W_{actual}} = \dfrac{139251.56 \,\mathrm{J/kg}}{183768.25 \,\mathrm{J/kg}} = 0.7576\) So, the isentropic efficiency of the compressor, when rounding up, is approximately 0.76. Thus, the correct answer is: \((a) 0.76\).