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Chapter 8: Problem 26

A 1: 4 scale model of a water pump is operated at a speed of \(4500 \mathrm{rpm}\). At its best efficiency point, the efficiency of the model pump is \(84 \%,\) and the model delivers a flow rate of \(0.7 \mathrm{~m}^{3} / \mathrm{s}\) with an added head of \(4.9 \mathrm{~m}\). If the full-scale pump has a rotational speed of \(120 \mathrm{rpm}\), what is the flow rate and head delivered by the fullscale pump operating at its most efficient point? What is the power requirement of the full-scale pump at its best efficiency point? Assume water at \(20^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Full-scale pump delivers 1.196 m鲁/s with a head of 0.0556 m and requires 315.29 W.

Step by step solution

01

Understand the Similarity Laws

In scaled model testing, the affinity laws (or similarity laws) are used to relate the performance parameters of the model to the full-scale version. The key parameters are the flow rate, head, and speed. The affinity laws for flow rate, head, and power are as follows:1. Flow rate: \( Q_2 = Q_1 \left( \frac{N_2}{N_1} \right) \left( \frac{D_2}{D_1} \right)^3 \)2. Head: \( H_2 = H_1 \left( \frac{N_2}{N_1} \right)^2 \left( \frac{D_2}{D_1} \right)^2 \)3. Power: \( P_2 = P_1 \left( \frac{N_2}{N_1} \right)^3 \left( \frac{D_2}{D_1} \right)^5 \)Where \( Q \) is the flow rate, \( H \) is the head, \( P \) is the power, \( N \) is the rotational speed, and \( D \) is the diameter.
02

Identify Given Values and Ratios

Given values:- Model speed ( \( N_1 \): \( 4500 \) rpm- Full-scale speed (\( N_2 \)): \( 120 \) rpm- Model flow rate (\( Q_1 \)): \( 0.7 \ \mathrm{m}^{3}/\mathrm{s} \)- Model head (\( H_1 \)): \( 4.9 \ \mathrm{m} \)- Scale ratio: \( \frac{1}{4} \) (meaning \( \frac{D_2}{D_1} = 4 \))
03

Calculate Flow Rate for Full-Scale Pump

Use the affinity law for flow rate:\[ Q_2 = Q_1 \left( \frac{N_2}{N_1} \right) \left( \frac{D_2}{D_1} \right)^3 \]Substitute known values:\[ Q_2 = 0.7 \left( \frac{120}{4500} \right) \left( 4 \right)^3 \]Calculate:\[ Q_2 = 0.7 \times 0.0267 \times 64 = 1.196 \ \mathrm{m}^{3}/\mathrm{s} \]
04

Calculate Head for Full-Scale Pump

Use the affinity law for head:\[ H_2 = H_1 \left( \frac{N_2}{N_1} \right)^2 \left( \frac{D_2}{D_1} \right)^2 \]Substitute known values:\[ H_2 = 4.9 \left( \frac{120}{4500} \right)^2 \left( 4 \right)^2 \]Calculate:\[ H_2 = 4.9 \times 0.000711 \times 16 = 0.0556 \ \mathrm{m} \]
05

Calculate Power Requirement for Full-Scale Pump

First, calculate the model's power using:\[ P_1 = \frac{\rho \cdot Q_1 \cdot g \cdot H_1}{\eta} \]Where \( \rho = 1000 \ \mathrm{kg}/\mathrm{m}^3 \) (density of water), \( g = 9.81 \ \mathrm{m/s}^2 \), \( \eta = 0.84 \):\[ P_1 = \frac{1000 \cdot 0.7 \cdot 9.81 \cdot 4.9}{0.84} = 39960 \ \mathrm{W} \]Now, use the affinity law for power:\[ P_2 = P_1 \left( \frac{N_2}{N_1} \right)^3 \left( \frac{D_2}{D_1} \right)^5 \]Substitute known values:\[ P_2 = 39960 \left( \frac{120}{4500} \right)^3 \left( 4 \right)^5 \]Calculate:\[ P_2 = 39960 \times 0.0000077 \times 1024 = 315.29 \ \mathrm{W} \]
06

Summary of Calculations

For the full-scale pump:- Flow rate is \(1.196 \ \mathrm{m}^{3}/\mathrm{s}\)- Head is \(0.0556 \ \mathrm{m}\)- Power requirement is \(315.29 \ \mathrm{W}\)

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Key Concepts

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

Scale Model Testing
Scale model testing involves creating a smaller version of a mechanical system, like a pump, to predict how a full-size version will perform. This method is cost-effective and helps engineers test various scenarios without the expense of building full-scale prototypes. By using a scale model, engineers can simulate conditions the large pump will face and gather data on its expected performance.

Important factors to consider in scale model testing are:
  • Scale Ratio: The size difference between the model and the full-scale pump, like 1:4, which indicates the model is one-quarter the size of the actual pump.
  • Similar Materials: Ensuring that the model uses materials that mimic those of the full-sized pump in terms of density and weight characteristics.
  • Environmental Conditions: Simulating the temperature, fluid dynamics, and other environmental factors that the full-scale pump will encounter.
By understanding how these elements work together, we can anticipate how the actual pump will function, leading to better design and efficiency.
Fluid Mechanics
Fluid mechanics is a branch of physics that studies how fluids behave under various forces. This field is crucial when designing systems that involve liquids or gases, such as pumps. Pumps are designed to move liquid through pipes and into desired locations and require a solid understanding of fluid properties.

Key concepts in fluid mechanics include:
  • Viscosity: A fluid's resistance to motion, affecting how it flows in a pump system.
  • Density: The mass per unit volume of the fluid; water at 20掳C is often used as a standard in calculations with a density of 1000 kg/m鲁.
  • Pressure: The force exerted by a fluid per unit area; changes in pressure control how fluids move in pipes and systems.
Grasping these concepts helps engineers design pumps with ideal operating conditions, leading to enhanced performance and efficiency.
Pump Efficiency
Pump efficiency measures how well a pump converts the electrical or mechanical energy input into fluid movement. A highly efficient pump uses less energy to move a given volume of fluid compared to a less efficient one. It's expressed as a percentage, representing the ratio of useful power output to total power input.

Things affecting pump efficiency include:
  • Design: The shape and size of the pump affects how fluid flows through it with minimal resistance.
  • Operating Conditions: Efficiency can vary depending on how a pump is used, such as speed and temperature conditions.
  • Maintenance: Well-maintained pumps run more efficiently since wear and tear are less likely to impede flow.
Understanding pump efficiency allows engineers to optimize design to ensure that energy use is minimized while performance is maximized.
Hydraulic Similarity
When discussing the design and testing of pumps, hydraulic similarity is a concept that refers to maintaining consistent fluid dynamics across different scales of pump systems. It ensures that the flow patterns and forces inside the pump remain consistent, regardless of the size.

Under hydraulic similarity, we consider:
  • Dynamic Similarity: Conditions that ensure flow patterns in the model and actual pump are identical.
  • Geometric Similarity: All dimensions of the model and full-size pump are proportionally scaled.
  • Kinematic Similarity: The velocity of fluid flow in the model matches that in the full-scale system.
Ensuring hydraulic similarity is critical in scale model testing as it helps predict how a full-sized pump will behave under real-world conditions. By using similarity laws, engineers can save time and cost in pump design and testing.

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Most popular questions from this chapter

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?

A pump manufacturer makes three homologous series of pumps with three different specific speeds. The manufacturer documents the performance of each homologous series by nondimensional functional relationships. An engineer identifies the desired homologous series for a particular project as the one that has a best efficiency point with a flow coefficient of \(0.035,\) a head coefficient of \(0.14,\) and a power coefficient of \(0.006 .\) For a particular pump with an impeller diameter of \(600 \mathrm{~mm}\) and a rated rotational speed of \(1140 \mathrm{rpm}\), determine the rated discharge, total dynamic head, brake horsepower, and efficiency of the pump. Assume water at \(20^{\circ} \mathrm{C}\).

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 \(1140 \mathrm{rpm}\).

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 \(20^{\circ} \mathrm{C}\).

Show that if the effective head on a Pelton wheel is \(h_{\mathrm{e}}\), the velocity coefficient of the nozzle is \(C_{\mathrm{v}}\), and the bucket speed of the wheel is \(U\), then the theoretical maximum efficiency is attained by the Pelton when $$ U=\frac{1}{2} C_{\mathrm{v}} \sqrt{2 g h_{\mathrm{e}}} $$
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