Question

A 5-kilowatt pump operating at steady state draws in liquid water at 1 bar, \(15^{\circ} \mathrm{C}\) and delivers it at 5 bar at an elevation \(6 \mathrm{~m}\) above the inlet. There is no significant change in velocity between the inlet and exit, and the local acceleration of gravity is \(9.8 \mathrm{~m} / \mathrm{s}^{2} .\) Would it be possible to pump \(7.5 \mathrm{~m}^{3}\) in \(10 \mathrm{~min}\) or less? Explain.

Short answer

No, the pump does not have sufficient power to pump 7.5 m³ in 10 minutes.

Step-by-step solution

- Determine the Flow Rate

Calculate the required flow rate to pump 7.5 cubic meters of water in 10 minutes. Convert the time into seconds: \[ 10 \text{ minutes} = 10 \times 60 = 600 \text{ seconds} \] Required flow rate: \[ \dot{V} = \frac{7.5 \text{ m}^3}{600 \text{ s}} = 0.0125 \text{ m}^3/ \text{s} \]

- Calculate the Work Done by the Pump

The power required by the pump is given as 5 kW. The mechanical work done by the pump per second is:\[ W = 5 \text{ kW} = 5000 \text{ J/s} \]

- Calculate the Change in Potential Energy per Second

Calculate the energy needed to raise the water by 6 meters per second. Volume flow rate, \( \dot{V} \), is 0.0125 m³/s, and gravity, \( g \), is 9.8 m/s². Density of water, \( \rho \), is 1000 kg/m³. Potential energy per second: \[ \rho \dot{V} g h = 1000 \times 0.0125 \times 9.8 \times 6 = 735 \text{ J/s} \]

- Calculate the Energy to Increase Pressure

The energy required to increase the pressure from 1 bar to 5 bar, given that 1 bar = 100 kPa = 100,000 N/m². This energy is: \[ W_p = \dot{V} \Delta P \] Pressure change, \( \Delta P \), is 400,000 N/m²: \[ W_p = 0.0125 \times 400,000 = 5000 \text{ J/s} \]

- Check the Total Energy Required by the Pump

Total energy required per second: The sum of potential energy and pressure energy: \[ 735 + 5000 = 5735 \text{ J/s} \]

- Compare with Available Power

Available power from the pump is 5000 J/s. Since 5735 J/s > 5000 J/s, it implies the pump cannot maintain the flow rate needed to deliver 7.5 m³ in 10 minutes.

Key concepts

Pump Analysis

Analyzing the performance of a pump involves examining its capacity to move fluid from one location to another. For this exercise, we look into a pump operating at steady state, which means the properties like pressure and flow rate remain constant over time. The pump draws water at a specific pressure and temperature and delivers it at a higher elevation and pressure. By checking if a 5-kilowatt pump can move 7.5 cubic meters of water within 10 minutes, we delve into understanding its mechanical workload and limitations. Important factors in pump analysis include flow rates, energy requirements for overcoming pressure differences, and elevation changes, all while considering constraints like the power capacity of the pump.

Energy Balance

In engineering thermodynamics, energy balance is crucial for analyzing systems such as pumps. Energy balance involves accounting for all energy inputs and outputs to ensure they match. Calculating the work done by the pump per second (5 kW or 5000 J/s) and comparing it with the sum of the energies needed for lifting the water and increasing the pressure (5735 J/s) indicates an imbalance. Since the required energy (5735 J/s) exceeds the supplied energy (5000 J/s), the pump cannot achieve the desired task within the given constraints. This example underscores the importance of balancing energy in engineering tasks to predict performance accurately.

Flow Rate Calculation

Flow rate, represented as \( \dot{V} \), measures the volume of fluid passing through a point per unit time. It is critical in determining if a pump can handle a specified load. Here, we need to pump 7.5 cubic meters of water in 10 minutes. Converting minutes to seconds (600 seconds), the flow rate calculation yields \( 0.0125 \, m^3/s \). This value helps ascertain the pump's ability to maintain an adequate flow rate under given conditions. An accurate flow rate is essential to ensuring that the required volume of water is moved efficiently and within the time frame specified.

Potential Energy Change

Potential energy change involves the energy required to move a fluid to a higher elevation. Here, we calculate it using the formula \( \rho \dot{V} g h \), where \( \rho \) is the density of water (1000 kg/m³), \( \dot{V} \) is the flow rate, \( g \) is the gravitational constant (9.8 m/s²), and \( h \) is the height difference (6 meters). The calculation shows us that 735 J/s is needed to lift the water by 6 meters. Understanding this concept is vital as it helps in planning and optimizing the energy needs of fluid-moving equipment, ensuring all potential energy changes are correctly accounted for.

Pressure Energy

Pressure energy refers to the energy required to increase the pressure of the fluid in the system. For our pump, raising the pressure from 1 bar (100 kPa) to 5 bar (500 kPa) creates a pressure change, \( \Delta P \), of 400,000 N/m². Using \( W_p = \dot{V} \Delta P \), we find the required energy per second is 5000 J/s. This determines whether the pump can meet the pressure requirements set for the job. By understanding the necessary pressure energy, engineers can design and calibrate pumps to fulfill specific operational criteria efficiently and effectively, ensuring that systems operate within their intended performance range.