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Chapter 7: Problem 25

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.

Short Answer

Expert verified
The diameter of the pipe is required to be calculated using principles from fluid dynamics, specifically continuity and Bernoulli's equations, while incorporating head loss due to friction.

Step by step solution

01

- Calculate the Required Volume of Water

The depth of water in the tank needs to increase from 0.1 m to 2.7 m. With a plan area of 6 m², the volume of water required is:\[ V = \text{Area} \times \text{Height Increase} = 6 \times (2.7 - 0.1) = 6 \times 2.6 = 15.6 \text{ m}^3 \]
02

- Calculate the Required Flow Rate

The time given for filling the tank is 15 minutes (900 seconds). Therefore, the flow rate is:\[ Q = \frac{V}{\text{time}} = \frac{15.6}{900} = 0.01733 \text{ m}^3/\text{s} \]
03

- Apply Continuity Equation

For flow rate Q, using the continuity equation: \[ Q = A \times v \] where A is the cross-sectional area of the pipe and v is the velocity of water. Therefore,\[ A = \frac{Q}{v} \]
04

- Calculate the Pressure Difference

The pressure difference due to the height difference is:\[ \text{Pressure Difference} = \text{Gauge Pressure} - \text{Hydrostatic Pressure} \]\[ \text{Hydrostatic Pressure} = \rho g h = 1000 \times 9.81 \times 2.9 = 28449 \text{ Pa} \]\[ \text{Pressure Difference} = 300,000 - 28449 = 271551 \text{ Pa} \]
05

- Calculate Velocity Using Bernoulli's Equation

Using Bernoulli's and Darcy-Weisbach equations:\[ \frac{1}{2} \rho v^2 = \text{Pressure Difference} - f \frac{L}{d} \frac{ \rho v^2 }{2} \]Rearranging to solve for velocity:\[ v = \frac{\text{Pressure Difference}}{\rho \times (1 + f \frac{L}{d})} \]Inserting knowns:\[ v = \frac{271551}{1000 (1 + 0.007 \frac{35}{d})} \]
06

- Calculate Diameter of the Pipe

Using the continuity equation: A = π(d/2)^2 and Q = A × v. Solving these for d using the values calculated for v and Q:\[ 0.01733 = \frac{π (d/2)^2 271551}{1000(1 + 0.007 35 / d)} \]Solving for diameter d.

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Key Concepts

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

fluid dynamics
Fluid dynamics involves the study of fluids (liquids and gases) in motion. The underlying principles help us understand how fluids flow and how forces and energy interact with them. In our exercise, we deal with water flowing through pipes. Key parameters include:
  • Pressure (force per unit area)
  • Velocity (speed of the fluid)
  • Density (mass per unit volume)
The relationship between these parameters helps solve complex flow problems. Understanding fluid dynamics is crucial for assessing water supply systems, like the one in our problem, as it ensures efficient and effective distribution of water.
Bernoulli's equation
Bernoulli's equation is fundamental to fluid mechanics. It states that in a flowing fluid, the total mechanical energy remains constant if there's no energy added or removed. The equation is written as:
Bernoulli's equation helps us link different parts of the fluid flow, making it easier to determine unknown variables like velocity or pressure loss. In our exercise, Bernoulli's equation is used to calculate the velocity of water entering the tank. It brings together the pressure difference and the velocity, incorporating the height difference between the water main and the tank.
continuity equation
The continuity equation ensures that the fluid mass is conserved as it flows. This concept is mathematically illustrated as:

Continuity Equation


The exercise uses this equation to connect the flow rate (Q) and the cross-sectional area (A) of the pipe. It helps determine how changes in the pipe diameter affect the velocity and flow rate, ensuring steady and continuous supply of water.
Darcy-Weisbach equation
The Darcy-Weisbach equation calculates the pressure losses due to friction in a pipe. This frictional loss is critical as it impacts the efficiency of fluid transport. The equation is given by:
    Darcy-Weisbach Equation:

    Where:


    • f: Friction factor (given as 0.007 in the problem)
    • L: Length of the pipe
    • d: Diameter of the pipe
    • v: Velocity of the fluid
    • g: Acceleration due to gravity
    • h: Height difference

    The problem demonstrates using the Darcy-Weisbach equation to adjust for flow resistance throughout the pipe length. It highlights how pipe diameter directly influences pressure loss and flow rate, ensuring optimal water supply to the tank.

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Most popular questions from this chapter

A large tank with vertical sides is divided by a vertical partition into two sections \(A\) and \(B\), with plan areas of \(1.5 \mathrm{~m}^{2}\) and \(7.5 \mathrm{~m}^{2}\) respectively. The partition contains a \(25 \mathrm{~mm}\) diameter orifice \(\left(C_{d}=0.6\right)\) at a height of \(300 \mathrm{~mm}\) above the base. Initially section \(A\) contains water to a depth of \(2.15 \mathrm{~m}\) and section \(B\) contains water to a depth of \(950 \mathrm{~mm}\). Calculate the time required for the water levels to equalize after the orifice is opened.

A tank \(1.5 \mathrm{~m}\) high is \(1.2 \mathrm{~m}\) in diameter at its upper end and tapers uniformly to \(900 \mathrm{~mm}\) diameter at its base (its axis being vertical). It is to be emptied through a pipe \(36 \mathrm{~m}\) long connected to its base and the outlet of the pipe is to be \(1.5 \mathrm{~m}\) below the bottom of the tank. Determine a suitable diameter for the pipe if the depth of water in the tank is to be reduced from \(1.3 \mathrm{~m}\) to \(200 \mathrm{~mm}\) in not more than 10 minutes. Losses at entry and exit may be neglected and \(f\) assumed constant at \(0.008\).

Two vertical cylindrical tanks, of diameters \(2.5 \mathrm{~m}\) and \(1.5 \mathrm{~m}\) respectively, are connected by a \(50 \mathrm{~mm}\) diameter pipe, \(75 \mathrm{~m}\) long for which \(f\) may be assumed constant at \(0.01\). Both tanks contain water and are open to atmosphere. Initially the level of water in the larger tank is \(1 \mathrm{~m}\) above that in the smaller tank. Assuming that entry and exit losses for the pipe together amount to \(1.5\) times the velocity head, calculate the fall in level in the larger tank during 20 minutes. (The pipe is so placed that it is always full of water.)

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.

Two water reservoirs, the surface levels of which differ by \(1.5 \mathrm{~m}\), are connected by a pipe system consisting of a sloping pipe at each end, \(7.5 \mathrm{~m}\) long and \(75 \mathrm{~mm}\) diameter, joined by a horizontal pipe \(300 \mathrm{~mm}\) diameter and \(60 \mathrm{~m}\) long. Taking entry head losses as \(0.5 u^{2} / 2 g\) and \(f=0.005(1+25 / d)\) where \(d \mathrm{~mm}\) is the pipe diameter, calculate the steady rate of flow through the pipe.
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