/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 Nine pump units are placed in pa... [FREE SOLUTION] | 91Ó°ÊÓ

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Nine pump units are placed in parallel at a pump station. Each unit has a power demand of \(40 \mathrm{~kW}\) and adds \(35 \mathrm{~m}\) of head under optimal conditions. The best efficiency of each unit is \(60 \%\). The liquid being pumped is water at \(20^{\circ} \mathrm{C}\). When all units are operating under optimal conditions, what is the flow rate delivered by the pump station?

Short Answer

Expert verified
The total flow rate is approximately \(0.6291 \text{ m}^3/\text{s}\).

Step by step solution

01

Calculate Individual Pump Efficiency

The efficiency of each pump is given as \(60\%\), which means the useful power output is \(60\%\) of the power input.
02

Compute the Hydraulic Power Output

The hydraulic power output for a single pump can be calculated using the formula:\[ P_{hydraulic} = \rho \cdot g \cdot Q \cdot H \]where \(\rho\) is the density of water \(1000 \text{ kg/m}^3\), \(g\) is the acceleration due to gravity \(9.81 \text{ m/s}^2\), \(Q\) is the flow rate, and \(H\) is the head. We need to find \(Q\).Given:- Power input \(P_{input} = 40 \text{ kW} = 40,000 \text{ W}\)- Efficiency \(\eta = 60\% = 0.60\)The useful output power \(P_{output}\) is \(\eta \cdot P_{input} = 0.60 \times 40,000 = 24,000 \text{ W}\).
03

Solve for Flow Rate of a Single Pump

Rearrange the hydraulic power equation to solve for \(Q\):\[ Q = \frac{P_{hydraulic}}{\rho \cdot g \cdot H} \]Substitute the known values:\[ Q = \frac{24,000}{1000 \times 9.81 \times 35} \approx 0.0699 \text{ m}^3/\text{s} \]Thus, the flow rate for a single pump is approximately \(0.0699 \text{ m}^3/\text{s}\).
04

Calculate Total Flow Rate for All Pumps

Since all nine pumps are operating in parallel, the total flow rate \(Q_{total}\) is nine times the flow rate of a single pump:\[ Q_{total} = 9 \times 0.0699 = 0.6291 \text{ m}^3/\text{s} \]Therefore, the total flow rate delivered by the pump station is approximately \(0.6291 \text{ m}^3/\text{s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

pump efficiency
Pump efficiency is a measure of how well a pump converts the energy put into it (usually electrical energy) into the energy needed to move fluids. It is limited by mechanical and fluid dynamics losses. Efficiency is often expressed as a percentage. The equation for pump efficiency is given by:\[\eta = \frac{P_{output}}{P_{input}}\]Where \(P_{output}\) is the hydraulic power delivered by the pump and \(P_{input}\) is the power consumed by the pump. In the exercise, the efficiency is 60%, meaning only 60% of the input energy turns into useful work to move water, while the rest is lost as heat or other inefficiencies. Understanding pump efficiency helps in selecting the right pump for your needs and improving energy cost savings.
flow rate calculation
The flow rate is the volume of fluid that moves through a pump in a given time. It's typically expressed in cubic meters per second (m³/s). To compute the flow rate for a pump, we use the relationship between hydraulic power, fluid density, gravitational acceleration, flow rate, and the head added by the pump:\[Q = \frac{P_{hydraulic}}{\rho \cdot g \cdot H}\]Where \(\rho\) is the fluid density (for water, it's 1000 kg/m³), \(g\) is the acceleration due to gravity (approximately 9.81 m/s²), \(H\) is the head added by the pump, and \(P_{hydraulic}\) is the hydraulic power output. By rearranging the formula to solve for \(Q\), we determine how much fluid is being pumped by a single unit. This is crucial for knowing if a pump can meet operational demands efficiently.
hydraulic power
Hydraulic power is the effective power used to lift or move the fluid within a system. It is different from the power input because it only accounts for the useful work done by the pump. The formula for calculating hydraulic power is:\[P_{hydraulic} = \rho \cdot g \cdot Q \cdot H\]Hydraulic power depends on the flow rate \(Q\), fluid density \(\rho\), gravity \(g\), and head \(H\). It is typically in watts (W) when using SI units. In pump systems, maximizing hydraulic power while minimizing energy loss is a key goal, leading to more efficient systems. Achieving higher hydraulic power without increasing energy input involves optimizing system design and operation.
parallel pump operation
In parallel pump operation, multiple pumps work together to increase the total flow rate. This configuration is useful when a single pump cannot provide the required flow rate or when flexibility in operation is needed. The primary benefit of parallel operation is that the flow rate is additive, meaning if each pump delivers a flow rate \(Q\), the total flow rate is the sum of the individual flow rates:\[Q_{total} = n \times Q\]where \( n \) is the number of pumps in operation. Parallel systems offer redundancy, ensuring that if one pump fails, others can continue operation. However, care must be taken in system design to avoid issues such as uneven load distribution or increased chance of cavitation. Understanding the dynamics of pumps in parallel helps in designing systems that are both reliable and efficient in meeting fluid transport demands.

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Most popular questions from this chapter

If the performance curve of a certain pump model is given by $$ h_{\mathrm{p}}=30-0.05 Q^{2} $$ where \(h_{\mathrm{p}}\) is the head added in \(\mathrm{m}\) and \(Q\) is the flow rate in \(\mathrm{L} / \mathrm{s}\), what is the performance curve of a pump system containing \(n\) of these pumps in series? What is the performance curve of a pump system containing \(n\) of these pumps in parallel?

A site being considered for hydropower development has an available head of \(800 \mathrm{~m}\) and a reliable flow rate of \(2.5 \mathrm{~m}^{3} / \mathrm{s}\). A generator operating at a speed of \(600 \mathrm{rpm}\) is feasible, and the expected hydraulic efficiency of a turbine unit at this site is \(95 \%\). What type of turbine should be considered for use?

A pump with a rotary speed of 1725 rpm delivers \(25 \mathrm{~L} / \mathrm{s}\) at its most efficient operating point. Under this condition, the inflow velocity is normal to the inflow surface of the impeller, the component of the velocity normal to the outflow surface of the impeller is \(4 \mathrm{~m} / \mathrm{s}\), and the efficiency of the pump is \(80 \%\). The width of the impeller at the outflow surface is \(15 \mathrm{~mm}\), and the blade angle at the outflow surface is \(50^{\circ}\). (a) Estimate the head added by the pump. (b) Use the affinity laws to estimate the head added and the flow rate delivered by the pump when the rotational speed is changed to \(1140 \mathrm{rpm}\).

Water at \(15^{\circ} \mathrm{C}\) is pumped at a rate of \(20 \mathrm{~L} / \mathrm{s}\) using a \(5 \mathrm{~kW}\) pump. If the efficiency of the pump is \(80 \%,\) what is the head added to the water as it passes through the pump?

Tests on a pump under standard atmospheric conditions show that when water at \(20^{\circ} \mathrm{C}\) is pumped at \(60 \mathrm{~L} / \mathrm{s}\) and the head added by the pump is \(40 \mathrm{~m},\) cavitation occurs when the pressure head plus velocity head on the suction side of the pump is \(3.9 \mathrm{~m}\). (a) Determine the required net positive suction head and the cavitation number of the pump. (b) If this same pump is operated on a mountain under the same flow rate and added head condition but the temperature of the water is \(5^{\circ} \mathrm{C}\) and the atmospheric pressure is \(90 \mathrm{kPa}\), by how much must the elevation of the pump above the sump reservoir be reduced compared with the test condition? Assume that the friction loss in the suction pipe remains approximately the same and that the sump reservoir is open to the atmosphere in both cases.

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