/*! 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 14 A \(675 \mathrm{~mm}\) water mai... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 14

A \(675 \mathrm{~mm}\) water main runs horizontally for \(1500 \mathrm{~m}\) and then branches into two \(450 \mathrm{~mm}\) mains each \(3000 \mathrm{~m}\) long. In one of these branches the whole of the water entering is drawn off at a uniform rate along the length of the pipe. In the other branch one-half of the quantity entering is drawn off at a uniform rate along the length of the pipe. If \(f=0.006\) throughout, calculate the total difference of head between inlet and outlet when the inflow to the system is \(0.28 \mathrm{~m}^{3} \mathrm{~s}^{-1}\). Consider only frictional losses and assume atmospheric pressure at the end of each branch. \([4.41 \mathrm{~m}]\)

Short Answer

Expert verified
Total head loss across the system is approximately 7.395 m.

Step by step solution

01

Calculate the Velocity in the Main Water Pipe

First, find the flow area of the 675 mm main using the formula \( A = \pi \cdot \left(\frac{D}{2}\right)^2 \) where \( D = 0.675 \) m. Compute \( A = \pi \left(\frac{0.675}{2}\right)^2 \approx 0.3572 \, \mathrm{m}^2 \). Next, find the velocity \( v \) using the formula \( Q = A \cdot v \) where \( Q = 0.28 \, \mathrm{m}^3/\mathrm{s} \). Thus, \( v = \frac{0.28}{0.3572} \approx 0.784 \, \mathrm{m/s} \).
02

Calculate Head Loss in the Main Water Pipe

Use the Darcy-Weisbach equation to compute the head loss \( h_f \) due to friction in the 675 mm pipeline: \( h_f = f \cdot \frac{L}{D} \cdot \frac{v^2}{2g} \). Substitute \( f = 0.006 \), \( L = 1500 \, \mathrm{m} \), \( D = 0.675 \, \mathrm{m} \), \( v \approx 0.784 \) m/s, and \( g = 9.81 \, \mathrm{m/s}^2 \) to get:\[ h_f = 0.006 \cdot \frac{1500}{0.675} \cdot \frac{0.784^2}{2 \cdot 9.81} \approx 0.713 \, \mathrm{m} \].
03

Determine Velocity and Head Loss in Each 450 mm Branch

The initial flow splits evenly between two branches. Calculate the area \( A = \pi \cdot \left(\frac{0.450}{2}\right)^2 \approx 0.159 \mathrm{~m}^2 \). Velocity in the branch \( v \) is given by \( v = \frac{Q/2}{A} = \frac{0.14}{0.159} \approx 0.88 \mathrm{~m/s} \). Calculate head loss for each branch using \( h_{f} = f \cdot \frac{L}{D} \cdot \frac{v^2}{2g} \). \( h_{f} = 0.006 \cdot \frac{3000}{0.450} \cdot \frac{0.88^2}{2 \cdot 9.81} \approx 1.909 \mathrm{~m} \).
04

Calculate Additional Head Loss due to Water Extraction

In each of the branches, there is an additional head loss due to the water drawn off. For the first branch (all water drawn off), the additional loss calculated based on linear extraction leads to another head loss of \( 1.909 \, \mathrm{m} \). For the second branch, where only half of the water is drawn off, calculate head loss as half of that in the first branch: \( \Delta h_{draw-off} = 0.955 \, \mathrm{m} \).
05

Combine All Head Losses

The total head loss is the sum of losses in the main and the two branches. For accurate calculation, consider: - Head loss in the main pipeline: \( 0.713 \, \mathrm{m} \)- Head loss in the first branch due to friction and water extraction: \( 1.909 + 1.909 = 3.818 \, \mathrm{m} \)- Head loss in the second branch due to friction and water extraction: \( 1.909 + 0.955 = 2.864 \, \mathrm{m} \)Total head loss across all components: \( 0.713 + 3.818 + 2.864 = 7.395 \, \mathrm{m} \).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Darcy-Weisbach equation
The Darcy-Weisbach equation is integral to calculating head loss in fluid systems, particularly in pipelines where frictional losses are a concern. This equation is crucial in hydraulic engineering to determine how much energy is lost due to the friction of flowing fluids within the pipe walls.
Calculating head loss involves knowing the flow characteristics and the pipe dimensions. The formula is given by:
\[ h_f = f \cdot \frac{L}{D} \cdot \frac{v^2}{2g} \]
where \( h_f \) is the head loss, \( f \) is the friction factor, \( L \) is the length of the pipe, \( D \) is the diameter, \( v \) is the velocity of the flow, and \( g \) is the acceleration due to gravity.
  • This equation helps engineers design and analyze piping systems efficiently.
  • In practice, engineers use charts or empirical equations to find the friction factor \( f \), which depends on the flow regime and pipe roughness.
  • It considers both the physical and operational conditions in pipelines.
Frictional losses in pipes
Frictional losses in pipes are a significant consideration in fluid dynamics and hydraulic engineering.
The flow of fluid through a pipe causes friction against the pipe walls, resulting in a loss of mechanical energy. This loss is directly related to several factors.
  • **Pipe Length and Diameter**: Longer pipes and smaller diameters increase the frictional losses as the contact surface for friction grows.
  • **Flow Velocity**: Higher velocities increase friction since the fluid interacts more intensely with the pipe walls.
  • **Fluid Properties**: Viscosity and density affect how significantly the fluid is slowed down by friction. Additionally, turbulence in the flow can greatly impact friction loss relations. Smooth, steady flows (laminar flow) have different friction behavior than turbulent ones.
The calculation of these losses is critical, as neglecting them can lead to inefficient system designs and energy wastage.
Flow velocity calculations
Calculating flow velocity is a fundamental aspect of hydraulic engineering.
Velocity in a pipeline is determined by the flow rate and the cross-sectional area of the pipe. The general formula is: \[ v = \frac{Q}{A} \] Where \( v \) is the velocity, \( Q \) is the volumetric flow rate, and \( A \) is the cross-sectional area of the pipe.
  • The flow rate \( Q \) is often measured in cubic meters per second.
  • The cross-sectional area \( A \) can be calculated using the pipe diameter \( D \) in the formula: \( A = \pi \left(\frac{D}{2}\right)^2 \).
Correctly calculating the velocity is essential for understanding how the fluid will behave within the system. It aids in predicting the pressure loss and ensuring the system is efficient and well-sized for its purpose.
Hydraulic engineering concepts
Hydraulic engineering encompasses principles that are pivotal in designing and managing systems where fluid flows through channels, such as pipes, open channels, and tunnels.
Focusing on fluid mechanics, hydraulic engineering involves understanding:
  • **Fluid Flow Dynamics**: How fluids move and interact in different conditions.
  • **Pressure and Head Dynamics**: Understanding how fluid pressure changes based on elevation and flow speed.
  • **System Efficiency**: Ensuring that engineered systems use materials and energy efficiently to transport water or other fluids.
Implementing these concepts helps engineers build systems that control and optimize the movement of fluids. Mastery of hydraulic principles enables the design of efficient systems, reducing environmental impact and operation costs.

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 rectangular cross-section tank, \(2 \mathrm{~m} \times 3 \mathrm{~m}\), is filled with water up to a depth of \(2 \mathrm{~m}\). Calculate the time to reduce the volume in the tank by 50 per cent if the discharge is via a \(40 \mathrm{~mm}\) diameter pipe, \(6 \mathrm{~m}\) long, for which a friction factor of \(0.005\) may be assumed and the separation losses may be represented by a \(k\) value of \(0.9\). Assume final discharge \(2 \mathrm{~m}\) below tank base level. [1276s]

For flow through pipes at high Reynolds number, the coefficient of friction is given by the following relation, $$ \frac{1}{\sqrt{f}}-4 \log _{10}\left(\frac{r}{\varepsilon}\right)=3.48 $$ where \(r=\) pipe radius and \(\varepsilon=\) mean height of roughness projections. A pipe of internal diameter \(0.15 \mathrm{~m}\) is formed of a material for which \(\varepsilon\) is \(0.00038 \mathrm{~m}\). The pipe is \(1524 \mathrm{~m}\) long and it connects two water reservoirs whose surface levels are maintained at the same height. Water may be pumped along the pipe and the maximum pumping power available is \(82 \mathrm{~kW}\). Calculate the maximum rate of flow in the pipe. $$ \left[0.06 \mathrm{~m}^{3} \mathrm{~s}^{-1}\right] $$

Two reservoirs whose difference of level is \(15 \mathrm{~m}\) are connected by a pipe \(\mathrm{ABC}\) whose highest point \(\mathrm{B}\) is \(2 \mathrm{~m}\) below the level in the upper reservoir A. The portion \(\mathrm{AB}\) has a diameter of \(200 \mathrm{~mm}\) and the portion BC a diameter of \(150 \mathrm{~mm}\), the friction coefficient being the same for both portions. The total length of the pipe is \(3 \mathrm{~km}\). Find the maximum allowable length of the portion \(\mathrm{AB}\) if the pressure head at \(\mathrm{B}\) is not to be more than \(2 \mathrm{~m}\) below atmospheric pressure. Neglect the secondary losses. \([1815 \mathrm{~m}]\)

There is a pressure loss of \(300 \mathrm{kN} \mathrm{m}^{-2}\) when water is pumped through pipeline \(\mathrm{A}\) at a rate of \(2 \mathrm{~m}^{3} \mathrm{~s}^{-1}\) and there is a pressure loss of \(250 \mathrm{kN} \mathrm{m}^{-2}\) when water is pumped at a rate of \(1.4 \mathrm{~m}^{3} \mathrm{~s}^{-1}\) through pipeline B. Calculate the pressure loss which will occur when \(1.5 \mathrm{~m}^{3} \mathrm{~s}^{-1}\) of water are pumped through pipes \(\mathrm{A}\) and \(\mathrm{B}\) jointly if they are connected \((a)\) in series, \((b)\) in parallel, assuming that junction losses may be neglected. In the latter case calculate the volume rate of flow through each pipe. $$ \begin{array}{r} {\left[(a) 456 \mathrm{kNm}^{-2},(b) 54.1 \mathrm{kNm}^{-2}\right.} \\ \left.0.849 \mathrm{~m}^{3} \mathrm{~s}^{-1}, 0.651 \mathrm{~m}^{3} \mathrm{~s}^{-1}\right] \end{array} $$

A smooth walled tube is used in a \(3000 \mathrm{~m}\) long pipeline carrying water at \(15^{\circ} \mathrm{C}\) between two reservoirs whose surface elevations are \(6 \mathrm{~m}\) apart. Entry is sharp edged and the outlet is also abrupt to the downstream reservoir. The pipeline contains six \(45^{\circ}\) bends and two globe valves. Determine the necessary pipe diameter so that the discharge should be 28 litre \(\mathrm{s}^{-1}\) to the lower reservoir. Take the equivalent length of each bend as \(26.5\) diameters, the valves as 75 diameters and the entry as 30 diameters. \([222 \mathrm{~mm}]\)
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Dynamics

Read Explanation

Space Physics

Read Explanation

Scientific Method Physics

Read Explanation

Medical Physics

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.