Question

Air enters a two-stage compressor at \(100 \mathrm{kPa}\) and \(27^{\circ} \mathrm{C}\) and is compressed to 625 kPa. The pressure ratio across each stage is the same, and the air is cooled to the initial temperature between the two stages. Assuming the compression process to be isentropic, determine the power input to the compressor for a mass flow rate of \(0.15 \mathrm{kg} / \mathrm{s}\). What would your answer be if only one stage of compression were used?

Short answer

The power input for the two-stage compression process is approximately 50.95 kJ/s, while for the one-stage process, it is approximately 67.98 kJ/s.

Step-by-step solution

Analyze the given data and find the intermediate pressure for the two-stage process

We are given the initial pressure (\(P_1\) = 100 kPa) and initial temperature (\(T_1\) = 27°C), and the final pressure of the two-stage process (\(P_3\) = 625 kPa). Since the pressure ratio across each stage is the same, we can find the intermediate pressure (\(P_2 = P_{2'}\)) as follows: \(P_2 = \sqrt{P_1 \cdot P_3}\)

Calculate the intermediate pressure for the two-stage process

Using the given initial and final pressure, we can calculate the intermediate pressure: \(P_2 =\sqrt{100 \mathrm{kPa} \cdot 625 \mathrm{kPa}} = 250 \mathrm{kPa}\)

Use isentropic relationships to find the intermediate and final temperatures for the two-stage process

For an isentropic process, the relationship between pressure and temperature for an ideal gas is given by: \(\left(\frac{P_2}{P_1}\right)^{(k-1)/k} = \frac{T_2}{T_1}\) where \(k\) is the specific heat ratio for air (approximately 1.4). Rearranging for \(T_2\): \(T_2 = T_1 \cdot \left(\frac{P_2}{P_1}\right)^{(k-1)/k}\) Similarly, we can find the temperature at the final state: \(T_3 = T_1 \cdot \left(\frac{P_3}{P_1}\right)^{(k-1)/k}\)

Calculate the intermediate and final temperatures for the two-stage process

Using the isentropic relationships, we can calculate the intermediate and final temperatures: \(T_1 = 27^\circ\mathrm{C} + 273.15 = 300.15\mathrm{K}\) \(T_2 = 300.15\mathrm{K} \cdot \left(\frac{250\mathrm{kPa}}{100\mathrm{kPa}}\right)^{(1.4-1)/1.4} \approx 345.49\mathrm{K}\) \(T_3 = 300.15\mathrm{K} \cdot \left(\frac{625\mathrm{kPa}}{100\mathrm{kPa}}\right)^{(1.4-1)/1.4} \approx 491.66\mathrm{K}\) Since the air is cooled to the initial temperature between the two stages, \(T_{2'} = T_1 = 300.15\mathrm{K}\).

Use the ideal gas equation and mass flow rate to find the specific volume at each state

The ideal gas equation is given by: \(Pv = RT\) where \(P\) is the pressure, \(v\) is the specific volume, \(R\) is the gas constant, and \(T\) is the temperature. For air, \(R \approx 0.287 \frac{\mathrm{kJ}}{\mathrm{kg}\cdot\mathrm{K}}\). We can rearrange the equation to find the specific volume: \(v = \frac{RT}{P}\) and use the given values for the initial state to find the initial specific volume, \(v_1\): \(v_1 = \frac{0.287 \mathrm{\frac{kJ}{kg\cdot K}} \cdot 300.15\mathrm{K}}{100\mathrm{kPa}} = 0.8639 \mathrm{\frac{m^3}{kg}}\) Since the mass flow rate is constant, the specific volume at each state remains the same: \(v_2 = v_{2'} = v_3 = v_1 = 0.8639 \mathrm{\frac{m^3}{kg}}\)

Calculate the work input for the two-stage compression process

The isentropic work input, \(W\), for each stage is given by: \(W = m\cdot \int_{v_1}^{v_2} P\,\mathrm{d}v\) Since the specific volume remains the same across all stages, and the pressure is given for each stage, the work for the two-stage process is: \(W_\text{total} = m \cdot (P_2 v_2 - P_1 v_1 + P_3 v_3 - P_{2'} v_{2'})\) Using the given mass flow rate (m = 0.15 kg/s): \(W_\text{total} = 0.15\mathrm{kg/s} \cdot (250\mathrm{kPa} \cdot 0.8639 \mathrm{\frac{m^3}{kg}} - 100\mathrm{kPa} \cdot 0.8639 \mathrm{\frac{m^3}{kg}} + 625\mathrm{kPa} \cdot 0.8639 \mathrm{\frac{m^3}{kg}} - 250\mathrm{kPa} \cdot 0.8639 \mathrm{\frac{m^3}{kg}}) \approx 50.95\mathrm{\frac{kJ}{s}}\)

Calculate the work input for the one-stage compression process

If only one stage of compression were used, we would calculate the work input using the initial and final states: \(W_\text{single} = m \cdot (P_3 v_3 - P_1 v_1)\) \(W_\text{single} = 0.15\mathrm{kg/s} \cdot (625\mathrm{kPa} \cdot 0.8639 \mathrm{\frac{m^3}{kg}} - 100\mathrm{kPa} \cdot 0.8639 \mathrm{\frac{m^3}{kg}}) \approx 67.98\mathrm{\frac{kJ}{s}}\) In conclusion, the power input to the compressor for a mass flow rate of 0.15 kg/s for the two-stage process is about 50.95 kJ/s, and for the one-stage process, it would be about 67.98 kJ/s.