Question

Liquid water is to be compressed by a pump whose isentropic efficiency is 75 percent from \(0.2 \mathrm{MPa}\) to \(5 \mathrm{MPa}\) at a rate of \(0.15 \mathrm{m}^{3} / \mathrm{min} .\) The required power input to this pump is \((a) 4.8 \mathrm{kW}\) \((b) 6.4 \mathrm{kW}\) \((c) 9.0 \mathrm{kW}\) \((d) 16.0 \mathrm{kW}\) \((e) 12 \mathrm{kW}\)

Short answer

Answer: (d) 16.0 kW

Step-by-step solution

Calculate the isentropic work done

From the conservation of energy, we know that the change in specific enthalpy is equal to the work done. Assuming the process to be isentropic and for incompressible liquid, we have: \(\Delta h = h_{2} - h_{1} = v(P_{2} - P_{1})\) where \(h_{1}\) and \(h_{2}\) are the specific enthalpies at the initial and final states, \(v\) is the specific volume of water, and \(P_{1}\) and \(P_{2}\) are the initial and final pressures. Given: \(v \approx 0.001 \mathrm{m}^{3} / \mathrm{kg}\) (specific volume of liquid water) \(P_{1} = 0.2 \mathrm{MPa}\) \(P_{2} = 5 \mathrm{MPa}\) We can now calculate the change in specific enthalpy: \(\Delta h = (0.001 \mathrm{m}^{3} / \mathrm{kg})(5 \mathrm{MPa} - 0.2 \mathrm{MPa}) = (0.001 \mathrm{m}^{3} / \mathrm{kg}) (4.8 \mathrm{MPa})\) Converting to kJ/kg: \(\Delta h = 4.8 \mathrm{kJ} / \mathrm{kg}\)

Use the isentropic efficiency to find the actual work done

We know the isentropic efficiency of the pump is 75%, so we can write: \(isentropic \, efficiency = \frac{Ideal \, work \, done}{Actual \, work \, done} = 0.75\) Hence, the actual work done, denoted as \(W_{actual}\), is: \(W_{actual} = \frac{Ideal \, work \, done}{isentropic \, efficiency} = \frac{4.8 \mathrm{kJ/kg}}{0.75} = 6.4 \mathrm{kJ/kg}\)

Convert the work done to power input using the given flow rate

The flow rate is given as \(0.15 \mathrm{m}^{3} / \mathrm{min}\). We can convert it to mass flow rate using the specific volume of water: \(Mass \, flow \, rate = \frac{0.15 \mathrm{m}^{3} / \mathrm{min}}{0.001 \mathrm{m}^{3} / \mathrm{kg}} = 150 \mathrm{kg/min}\) Now, we can find the power input by multiplying the actual work done with the mass flow rate: \(Power \, input = W_{actual} \times Mass \, flow \, rate = 6.4 \mathrm{kJ/kg} \times 150 \mathrm{kg/min}\) Since \(1 \, kW = 1 \, kJ/s\), we can convert the power input to kW by dividing the product by \(60\,s/min\): \(Power \, input = \frac{6.4 \mathrm{kJ/kg} \times 150 \mathrm{kg/min}}{60\, s/min} = 16.0 \mathrm{kW}\)

Choose the correct answer among the given options

The calculated power input is \(16.0 \, kW\), which corresponds to option (d). Therefore, the correct answer is: \((d) \, 16.0 \, kW\)

Key concepts

Isentropic Work Calculation

Understanding the isentropic work calculation in a pump involves grasping how work relates to the change in fluid's state without any entropy change (hence isentropic). The key to solving problems like the textbook exercise is to identify the correct physical properties to use. For incompressible fluids, such as liquid water, the specific volume (\(v\)) does not change significantly with pressure, simplifying our calculation.

To calculate the work done by the pump, you first find the change in specific enthalpy (\(\Delta h\)) using the formula \(\Delta h = h_{2} - h_{1} = v(P_{2} - P_{1})\), where \(h_1\) and \(h_2\) represent the specific enthalpies at the initial and final pressures \(P_1\) and \(P_2\). Since this is an isentropic process, the isentropic work done is equal to the change in specific enthalpy.

Enthalpy Change

Enthalpy change (\(\Delta h\)) signifies the total energy change of the fluid as work is done on it without any heat transfer to the environment. In our textbook case, we deal with water, an almost incompressible fluid; thus, the change in volume is negligible, making our \(\Delta h\) calculation straightforward. By multiplying the specific volume (\(v\)) by the difference in pressures (\(P_{2} - P_{1}\)), you can find the enthalpy change directly, translated into work done per unit mass. Remember to convert pressure into the same units as specific volume for consistency; here, we convert megapascals to kilojoules per kilogram, the standard units for enthalpy change.

Specific Volume

Specific volume is an intrinsic property of substances, representing the volume occupied by a unit mass of a substance (\(\mathrm{m}^3/\mathrm{kg}\)). In the context of our pump operation question, we used the specific volume (\(v\)) of liquid water, which we approximated as a constant value due to the incompressible nature of liquids under high pressure. This allows for a simplified approach when making calculations for the work done during compression. The specific volume is critical as it directly influences the enthalpy change and the work done during the isentropic process.

Mass Flow Rate

Mass flow rate is the amount of mass flowing through a cross-section per unit time (\(\mathrm{kg}/\mathrm{min}\)). Calculating mass flow rate in our exercise was essential to find the pump's power input. To get from specific volume to mass flow rate, we applied the formula\(Mass \, flow \, rate = \frac{Volume \, flow \, rate}{Specific \, volume}\), thus transforming volume flow (given in cubic meters per minute) to mass flow rate (how much mass passes in one minute). This conversion is crucial because the actual work we calculate should be considered per unit mass, and applying it to the entire mass flow gives us the power required.

Power Input Calculation

Power input calculation is ultimately what determines the energy requirement for the pump to operate at given conditions. After calculating the actual work per unit mass and knowing the mass flow rate, we determine the power input by multiplying them together. The unit conversion to kilowatts (\(kW\)) is the last but a crucial step, ensuring we match our final answer to the common units of power. The formula\(Power \, input = W_{actual} \times Mass \, flow \, rate\) helps us conclude that a higher flow rate or more work per unit mass would proportionally increase the power input necessary for the pump to function, as seen in the textbook problem's solution.