/*! 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 35 A pump has an impeller diameter ... [FREE SOLUTION] | 91Ó°ÊÓ

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A pump has an impeller diameter of \(450 \mathrm{~mm}\), and at its most efficient operating point, it delivers water at a flow rate of \(650 \mathrm{~L} / \mathrm{s}\) with an added head of \(9.5 \mathrm{~m}\). The specific speed of the pump is \(1.5,\) and the shaft power delivered by the motor is \(80 \mathrm{~kW}\). (a) Estimate the shutoff head of the pump. (b) Estimate the efficiency of the pump at its best operating point. Assume water at \(20^{\circ} \mathrm{C}\).

Short Answer

Expert verified
(a) The shutoff head is estimated at 9.5 m, same under assumption. (b) The pump efficiency at its best operating point is 75.44%.

Step by step solution

01

Convert Units

First, we need to convert all quantities to coherent SI units. The flow rate is given in liters per second, which is already in SI units: \( 650 \mathrm{~L/s} = 0.65 \mathrm{~m^3/s} \). The impeller diameter is given in millimeters, which we convert to meters: \( 450 \mathrm{~mm} = 0.45 \mathrm{~m} \). Lastly, the shaft power is given in kilowatts, which is also in SI units. So, \( 80 \mathrm{~kW} = 80,000 \mathrm{~W} \).
02

Calculate Hydraulic Power

The hydraulic power \( P_h \) can be calculated using the formula: \[ P_h = \rho g Q H \] where \( \rho \) is the water density \(1000 \mathrm{~kg/m^3}\) , \( g \) is the acceleration due to gravity \( 9.81 \mathrm{~m/s^2} \), \( Q \) is the flow rate \(0.65 \mathrm{~m^3/s}\), and \( H \) is the head \(9.5 \mathrm{~m}\). Substituting these values in: \[ P_h = 1000 \times 9.81 \times 0.65 \times 9.5 \approx 60,354.75 \mathrm{~W} \].
03

Estimate Efficiency

The efficiency \( \eta \) of the pump is given by the ratio of the hydraulic power to the shaft power. Thus, \[ \eta = \frac{P_h}{P_s} = \frac{60,354.75}{80,000} \approx 0.7544 \text{ or } 75.44\% \]. This is the efficiency of the pump at its best operating point.
04

Estimate the Shutoff Head

The shutoff head is estimated by using the specific speed equation for a pump, which in SI units for a radial flow pump is: \[ N_s = \frac{n \sqrt{Q}}{H^{3/4}} \] Rearranging for \( H_{shutoff} \) when \( Q = 0 \) yields an increase of % in head because of reduced flow: \[ H_{shutoff} \approx H \left(\frac{0.97}{1.5}\right)^{2/3} = 9.5 \]. This typically helps to narrowly overestimate, but specific measurements depend on empirical valve fitting per pump design.

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Key Concepts

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

Hydraulic Power
Hydraulic power is a measure of the energy transferred by the pump to the fluid it is moving. It is directly determined by factors like the flow rate and the pressure increase that the pump impart. The formula to calculate hydraulic power, \( P_h \), is given by:\[ P_h = \rho g Q H \]where:
  • \( \rho \) is the fluid density, which for water at \(20^{\circ} \mathrm{C}\), is approximately \(1000 \, \mathrm{kg/m^3}\).
  • \( g \) is the acceleration due to gravity, roughly \(9.81 \, \mathrm{m/s^2}\).
  • \( Q \) is the volumetric flow rate, often converted to cubic meters per second (\(0.65 \, \mathrm{m^3/s}\) in this case).
  • \( H \) is the head, or the height through which the pump moves the water, measured in meters (\(9.5 \, \mathrm{m}\) here).
This formula tells us that the hydraulic power depends heavily on both how much water is moved and how much energy is added to the water through the pumping action. Understanding hydraulic power is crucial as it influences pump selection and energy usage when designing fluid systems.
Impeller Diameter
The impeller diameter is a key dimension in pump design and significantly affects pump performance, such as flow rate and the energy added to the fluid. In our exercise, the impeller diameter is \(450 \, \mathrm{mm}\), or \(0.45 \, \mathrm{m}\) when converted to meters for consistency with SI units.An impeller, which is the rotating component, works like a turbine, generating fluid flow and pressure from rotational motion. Key points about the impeller diameter include:
  • A larger diameter generally means that more energy can be transferred to the fluid per revolution, potentially increasing the pump's capacity to move fluid.
  • The impeller diameter impacts characteristics like the head produced and the efficiency of the pump. As seen in the exercise, it operates alongside factors such as fluid density and gravity to determine the hydraulic power output.
  • A properly selected impeller size ensures optimal pump performance and avoids issues like cavitation or excessive energy consumption.
Understanding the role of impeller diameter helps in choosing the right pump for a given application, ensuring efficiency and reliability.
Specific Speed
Specific speed is a dimensionless parameter that helps to characterize the performance of pumps at their best efficiency point. It allows pump engineers to compare different pumps regardless of their sizes or designs.The specific speed, \(N_s\), is calculated as:\[ N_s = \frac{n \sqrt{Q}}{H^{3/4}} \]where:
  • \( n \) is the rotational speed of the pump in revolutions per minute.
  • \( Q \) is the flow rate, proportional to the pump output.
  • \( H \) is the head, representing the energy increase imparted to the fluid.
In our exercise, the specific speed value of \(1.5\) indicates the design characteristics of the pump and suggests its application for medium flow and head requirements. The specific speed helps in identifying which impeller design is most suitable for a particular application by categorizing pumps into radial, mixed, or axial flow types. It guides not only the selection of pumps but also performance expectations in real-world scenarios.
Shutoff Head
The shutoff head refers to the maximum head that a pump can achieve when the discharge valve is completely closed, meaning the flow rate is zero. It gives insight into the pump's capabilities and potential maximum lift height.Determining shutoff head involves using specific speed and operating conditions. Although complex, a simplified estimation using adjusted head formulas gives a rough idea. From typical empirical data, we approach shutoff head calculation as:\[ H_{shutoff} \approx H \left(\frac{0.97}{1.5}\right)^{2/3} \]where the specific speed reduces the effective operating head, adjusted further by empirical constants.
  • Understanding the shutoff head is useful for ensuring pump safety and effectiveness under varying conditions.
  • It can help avoid operational issues like undue pressure on the pump casing or excessive energy consumption.
In practical terms, shutoff head should always be considered when designing systems where flow rates might vary or need to be stopped without damaging the pump. This parameter ensures longevity and reduces the likelihood of unexpected system failures.

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Most popular questions from this chapter

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.

Affinity laws are typically used to identify homologous values of flow rate, head, and power between geometrically similar pumps. Develop an affinity law for relating homologous values of torque.

A proposed hydropower plant is to be located at a site with an available head of \(9 \mathrm{~m}\), and it is desired to obtain a (shaft) power of \(35 \mathrm{MW}\) from this site. The axial-flow turbine units under consideration operate at an angular speed of \(150 \mathrm{rpm}\), have a specific speed of \(5,\) and have an estimated maximum efficiency of \(80 \%\). (a) How may of these units are required? (b) What total flow rate must be available to generate the desired power? Assume water at \(20^{\circ} \mathrm{C}\).

Water at \(20^{\circ} \mathrm{C}\) is to be pumped out of a reservoir at a rate of \(20 \mathrm{~L} / \mathrm{s}\) through a vertical 150 -mm-diameter ductile iron pipeline with an estimated roughness height of \(0.3 \mathrm{~mm}\). Appurtenances installed in the intake pipe are expected to contribute to a total local head loss coefficient of 12 . The pump being considered for installation has a required net positive suction head of \(5.5 \mathrm{~m}\). (a) What is the maximum elevation of the pump relative to the water surface in the reservoir? (b) If the total head loss coefficient can be reduced to 1.2 , how much higher can the pump be placed? Assume standard atmospheric conditions.

A large centrifugal fan generates a flow of \(7 \mathrm{~m}^{3} / \mathrm{s}\) with a motor speed of \(1140 \mathrm{rpm}\). A smaller geometrically similar fan has a motor speed of \(1725 \mathrm{rpm}\), operates at the same efficiency as the larger fan, and generates the same pressure increase. What flow rate is generated by the smaller fan?

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