Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 5

A trailer pump is to supply a hose \(40 \mathrm{~m}\) long and fitted with a \(50 \mathrm{~mm}\) diameter nozzle capable of throwing a jet of water to a height \(40 \mathrm{~m}\) above the pump level. If the power lost in friction in the hose is not to exceed \(15 \%\) of the available hydraulic power, determine the diameter of hose required. Friction in the nozzle may be neglected and \(f\) for the hose assumed to be in the range \(0.007-0.01\).

Short Answer

Expert verified
Approximately 97 mm diameter is required.

Step by step solution

01

Calculate the Hydraulic Power Available

The hydraulic power available can be calculated using the formula \[ P = \rho g Q H \] where \( \rho \) is the density of water (assumed to be 1000 kg/m^3), \( g \) is the acceleration due to gravity (9.81 m/s^2), \( Q \) is the flow rate, and \( H \) is the height (40 m). To find the flow rate, we need additional information, typically the velocity at the nozzle or the area of the 50 mm diameter nozzle.
02

Determine the Flow Rate

The velocity of water at the nozzle can be found using Torricelli's theorem, where velocity \( v \) is given by \[ v = \sqrt{2gH} = \sqrt{2 \times 9.81 \times 40} \approx 28 \text{ m/s} \] Now, convert the diameter of the nozzle to meters (0.05 m) to find the area (\( A \)) using \[ A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{0.05}{2} \right)^2 \approx 1.96 \times 10^{-3} \text{ m}^2 \] The flow rate (\( Q \)) is then \[ Q = A \times v = 1.96 \times 10^{-3} \times 28 \approx 0.055 \text{ m}^3/\text{s} \]
03

Calculate the Hydraulic Power

Using the flow rate (\( Q \)) found in step 2, the hydraulic power (\( P \)) is \[ P = \rho g Q H = 1000 \times 9.81 \times 0.055 \times 40 = 21582 \text{ W} \]
04

Calculate Power Lost in Friction

The power lost due to friction in the hose should not exceed 15% of the hydraulic power available. Thus, \[ 0.15 \times 21582 \approx 3237 \text{ W} \]
05

Use Darcy-Weisbach Formula

Use the Darcy-Weisbach formula to calculate the frictional head loss \[ h_f = f \times \frac{L}{D} \times \frac{v^2}{2g} \] We convert the hydraulic power lost to a form involving head loss: \[ P_{lost} = \rho g Q h_f \] Thus, \[ h_f = \frac{P_{lost}}{\rho g Q} = \frac{3237}{1000 \times 9.81 \times 0.055} \approx 6 \text{ m} \]
06

Solve for Diameter of the Hose

Insert the frictional head loss (\( h_f \)), length (\( L = 40 \) m), velocity (\( v = 28 \) m/s), and min/max friction factor (\( f = 0.007-0.01 \)) into the Darcy-Weisbach formula and solve for the diameter (\( D \)). Simplifying, \[ D = \sqrt{ \frac{f L v^2}{2gh_f} } \] Choosing an intermediate \( f = 0.0085 \) gives \[ D = \sqrt{ \frac{0.0085 \times 40 \times 28^2}{2 \times 9.81 \times 6} } \approx 0.097 \text{ m} (or \approx 97 \text{ mm}) \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

hydraulic power
Hydraulic power is the energy per unit time that is transferred by the fluid in a hydraulic system. It is calculated using the formula: \[ P = \rho g Q H \] In this equation, \( \rho \) represents the fluid's density, \( g \) is the acceleration due to gravity, \( Q \) is the flow rate, and \( H \) is the height through which the fluid is lifted or thrown.
In the provided exercise, water with a density of 1000 kg/m³ is raised to a height of 40 meters.
By understanding this concept, you can determine how much power is needed to move fluid to a certain height.
This concept is crucial when assessing the pump's efficiency and effectiveness in hydraulic systems.
Darcy-Weisbach formula
The Darcy-Weisbach formula is a critical tool in fluid mechanics. It's used to calculate the head loss (pressure loss) due to friction in a pipe. This formula is expressed as:
\[ h_f = f \times \frac{L}{D} \times \frac{v^2}{2g} \]
Here, \( h_f \) is the frictional head loss, \( f \) is the friction factor, \( L \) is the length of the pipe, \( D \) is its diameter, and \( v \) is the velocity of the fluid.
In our example, we need to find the diameter of the hose that keeps the frictional head loss within acceptable limits.
This means carefully calculating all the terms and inserting them into the Darcy-Weisbach equation.
This formula helps us ensure that the pump works efficiently and that the system operates within design limits.
flow rate calculation
Flow rate calculation is a fundamental part of solving fluid mechanics problems. The flow rate ( \( Q \) ) indicates how much fluid passes through a section of a pipe per unit time. For a nozzle, it can be calculated using:
\[ Q = A \times v \]
where \( A \) is the cross-sectional area, and \( v \) is the fluid velocity.
In the given exercise, the nozzle has a diameter of 50 mm, leading to an area of about 1.96 * 10^{-3}
m^2.
The velocity at the nozzle, determined using Torricelli's theorem, is approximately 28 m/s.
Thus, the flow rate was found as approximately 0.055 m^3/s.
Accurately calculating the flow rate helps in determining the hydraulic power and evaluating the efficiency of the hydraulic system.
frictional head loss
Frictional head loss refers to the loss of energy due to the friction between the fluid and the pipe's interior surface. This is a crucial factor in hydraulic system design, as it impacts the overall efficiency. Using the Darcy-Weisbach formula, frictional head loss can be calculated and compared to available hydraulic power.
In our example, we calculated that the power lost to friction should not exceed 15% of the hydraulic power, resulting in a frictional head loss of about 6 meters. Adjusting parameters like the diameter of the hose helps us maintain this limit.
Understanding and minimizing frictional head loss is key to designing an efficient and reliable hydraulic system.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A water main with a constant gauge pressure of \(300 \mathrm{kPa}\) is to supply water through a pipe \(35 \mathrm{~m}\) long to a tank of uniform plan area \(6 \mathrm{~m}^{2}\), open to atmosphere at the top. The pipe is to enter the base of the tank at a level \(2.9 \mathrm{~m}\) above that of the main. The depth of water in the tank is to be increased from \(0.1 \mathrm{~m}\) to \(2.7 \mathrm{~m}\) in not more than 15 minutes. Assuming that \(f\) has the constant value \(0.007\), and neglecting energy losses other than pipe friction, determine the diameter of pipe required.

A fluid of constant density \(\varrho\) enters a horizontal pipe of radius \(R\) with uniform velocity \(V\) and pressure \(p_{1}\). At a downstream section the pressure is \(p_{2}\) and the velocity varies with radius \(r\) according to the equation \(u=2 V\left\\{1-\left(r^{2} / R^{2}\right)\right\\}\). Show that the friction force at the pipe walls from the inlet to the section considered is given by \(\pi R^{\frac{1}{2}}\left(p_{1}-p_{2}-\varrho V^{2} / 3\right)\).

A tank of plan area \(5 \mathrm{~m}^{2}\), open to atmosphere at the top, is supplied through a pipe \(50 \mathrm{~mm}\) diameter and \(40 \mathrm{~m}\) long which enters the base of the tank. A pump, providing a constant gauge pressure of \(500 \mathrm{kPa}\), feeds water to the inlet of the pipe which is at a level \(3 \mathrm{~m}\) below that of the base of the tank. Taking \(f\) for the pipe as \(0.008\), calculate the time required to increase the depth of water in the tank from \(0.2 \mathrm{~m}\) to \(2.5 \mathrm{~m}\). If the combined overall efficiency of pump and driving motor is \(52 \%\), determine (in \(\mathrm{kW} \cdot \mathrm{h})\) the total amount of electricity used in this pumping operation.

Calculate the power required to pump sulphuric acid (dynamic viscosity \(0.04 \mathrm{~Pa} \cdot \mathrm{s}\), relative density \(1.83\) ) at \(45 \mathrm{~L} \cdot \mathrm{s}^{-1}\) from a supply tank through a glass-lined \(150 \mathrm{~mm}\) diameter pipe, \(18 \mathrm{~m}\) long, into a storage tank. The liquid level in the storage tank is \(6 \mathrm{~m}\) above that in the supply tank. For laminar flow \(f=16 / \mathrm{Re}\); for turbulent flow \(f=0.0014\left(1+100 \mathrm{Re}^{-1 / 3}\right)\) if \(\mathrm{Re}<10^{7}\). Take all losses into account.

A pipe \(900 \mathrm{~m}\) long and \(200 \mathrm{~mm}\) diameter discharges water to atmosphere at a point \(10 \mathrm{~m}\) below the level of the inlet. With a pressure at inlet of \(40 \mathrm{kPa}\) above atmospheric the steady discharge from the end of the pipe is \(49 \mathrm{~L} \cdot \mathrm{s}^{-1}\). At a point half way along the pipe a tapping is then made from which water is to be drawn off at a rate of \(18 \mathrm{~L} \cdot \mathrm{s}^{-1}\). If conditions are such that the pipe is always full, to what value must the inlet pressure be raised so as to provide an unaltered discharge from the end of the pipe? (The friction factor may be assumed unaltered.)
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Scientific Method Physics

Read Explanation

Space Physics

Read Explanation

Electromagnetism

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Radiation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free