Question

Air enters a compressor at ambient conditions of \(100 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\) at a rate of \(4.5 \mathrm{m}^{3} / \mathrm{s}\) with a low velocity, and exits at \(900 \mathrm{kPa}, 60^{\circ} \mathrm{C},\) and \(80 \mathrm{m} / \mathrm{s}\). The compressor is cooled by cooling water that experiences a temperature rise of \(10^{\circ} \mathrm{C}\). The isothermal efficiency of the compressor is 70 percent. Determine \((a)\) the actual and reversible power inputs, \((b)\) the second-law efficiency, and \((c)\) the mass flow rate of the cooling water.

Short answer

Answer: The second-law efficiency for the given air compressor is 90.47 percent and the mass flow rate of the cooling water is 108.4 kg/s.

Step-by-step solution

Calculate the Actual Power Input

First, let's find the actual power input (\(W_{actual}\)) using the given isothermal efficiency. Isothermal efficiency (\(\eta_{iso}\)) is defined as the ratio of the work done in an isentropic process to the actual work done. The isothermal efficiency is given as 70 percent. \(\eta_{iso} = \frac{W_{isentropic}}{W_{actual}}\) We can calculate the isentropic work as: \(W_{isentropic} = \int_{P_1}^{P_2} \frac{V}{n}dP\) Where \(V\) is the volume flow rate, \(n\) is the polytropic index, \(P_1\) and \(P_2\) are the initial and final pressures, respectively. For an isentropic process, \(n = 1\) (isothermal process). In this case, we have \(V = 4.5 m^3/s\) and \(P_1 = 100 kPa,\) and \(P_2 = 900 kPa\). Calculating the isentropic work: \(W_{isentropic} = \int_{100}^{900} \frac{4.5}{1}dP = 4.5(P_2 - P_1) = 4.5(900 - 100) = 3600 \thinspace \mathrm{kW}\) Now, solving for the actual power input: \(W_{actual} = \frac{W_{isentropic}}{\eta_{iso}} = \frac{3600}{0.7} = 5142.86 \thinspace \mathrm{kW}\)

Calculate the Reversible Power Input

To find the reversible power input (\(W_{rev}\)), we can use the entropy balance and enthalpy difference. First, we calculate the change in entropy between inlet and outlet: \(\Delta S = S_2 - S_1 = m(C_p \ln{\frac{T_2}{T_1}} - R \ln{\frac{P_2}{P_1}})\) Where \(m\) is the mass flow rate of air, \(C_p\) is the specific heat at constant pressure, \(T_1\) and \(T_2\) are the initial and final temperatures, and R is the specific gas constant. Considering an ideal gas, the mass flow rate can be calculated as: \(m = V \cdot \frac{P_1}{R T_1} = \frac{4.5 \times 100}{287\times293.15} =15.33 \thinspace \mathrm{kg/s}\) Now, we calculate the change in entropy: \(\Delta S = 15.33(1.005 \ln{\frac{333.15}{293.15}} - 0.287 \ln{\frac{900}{100}}) = -1.848 \thinspace \mathrm{kW/K}\) Next, we can determine the reversible work: \(W_{rev} = W_{actual} - T_0 \Delta S\) Where \(T_0\) is the ambient temperature. \(W_{rev} = 5142.86 - 293.15 \times (-1.848) = 5142.86 + 541.46 = 5684.32 \thinspace \mathrm{kW}\)

Calculate the Second-law Efficiency

Now, we can calculate the second-law efficiency (\(\eta_{II}\)) using the actual power input and the reversible power input. \(\eta_{II} = \frac{W_{actual}}{W_{rev}} = \frac{5142.86}{5684.32} = 0.9047\) The second-law efficiency is 90.47 percent.

Calculate the Mass Flow Rate of the Cooling Water

Finally, we will find the mass flow rate of the cooling water (\(m_{cw}\)) using the energy balance. Considering the energy balance on the compressor: \(W_{actual} = \dot{Q}_{out} + \dot{m}(h_2 - h_1)\) Where \(\dot{Q}_{out}\) is the heat transfer rate from the compressor to the cooling water, \(\dot{m}\) is the mass flow rate of air, and \(h_1\) and \(h_2\) are the enthalpies at the inlet and outlet, respectively. Calculating the enthalpy difference: \(\Delta h = h_2 - h_1 = C_p(T_2 - T_1) = 1.005(60 - 20) = 40.20 \thinspace \mathrm{kJ/kg}\) Now, calculating the heat transfer rate: \(\dot{Q}_{out} = W_{actual} - \dot{m}\Delta h = 5142.86 - 15.33\times40.20 = 5142.86 - 615.86 = 4527 \thinspace \mathrm{kW}\) The heat transfer rate from the compressor to the cooling water is 4527 kW. Considering the energy balance on the cooling water: \(\dot{Q}_{out} = m_{cw} C_{p,cw} \Delta T_{cw}\) Where \(C_{p,cw}\) is the specific heat at constant pressure for the cooling water, and \(\Delta T_{cw}\) is the temperature difference of the cooling water. Solving for the mass flow rate of the cooling water: \(m_{cw} = \frac{\dot{Q}_{out}}{C_{p,cw} \Delta T_{cw}} = \frac{4527}{4.18 \times 10} = 108.4 \thinspace \mathrm{kg/s}\) The mass flow rate of the cooling water is 108.4 kg/s.