Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

The intake in Figure has cross-sectional area of $0.74 m^{2}$ and water flow at $0.40 m / s$ . At the outlet, distance $D = 180 m$ below the intake, the cross-sectional area is smaller than at the intake, and the water flows out at $9.5 m / s$ into the equipment. What is the pressure difference between inlet and outlet?

Answer

The pressure difference between inlet and outlet of the pipe is $1.7 \times 10^{6} P a$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The cross-sectional area of the inlet, $A_{t} = 0.74 m^{2}$ .
The speed of the water at the inlet, $v_{t} = 0.40 m / s$ .
The depth of the outlet, $D = 180 m$ .
The speed of the water at the outlet, $v_{0} = 9.5 m / s$ .

Step 2: Determining the concept

By applying Bernoulli’s principle, determine the pressure difference between inlet and outlet. According to Bernoulli’s equation, as the speed of a moving fluid increases, the pressure within the fluid decreases.

The equation is as follows:

$p v + \frac{1}{2} p g^{2} h + c o n s t a n t$

Where, p is pressure, v is velocity, h is height, g is the acceleration due to gravity, h is height and $p$ is density.

Step 3: Determining the pressure difference between the inlet and outlet of the pipe

The water flow should obey Bernoulli’s principle,

$p_{t} v + \frac{1}{2} p g^{2} h + p {1 = {p_{0}}_{}}_{} v + \frac{1}{2} p {g^{2}}_{0} h + 2_{}$

Simplifying,

$p_{l} v + \frac{1}{2} p g^{2} h + p_{l} = p_{0} v + \frac{1}{2} p {g^{2}}_{0} h +_{0} p_{l} v p_{0} v \frac{1}{2} (p g^{2} - h l^{2}) h (o - l) Δ p v \frac{1}{2} v ({^{}}^{2}_{0} - p g^{2}) h h (_{0} - l)$

Where, $h_{0} - h_{l} = D = 180 m$ and density of water $p = 1000 k g / m^{3}$

Thus, putting the values,

$Δ p = \frac{1}{2} 1000 k g / m^{3} \times ({(0.4 m / s)}^{2} - {(9.5 m / s)}^{2} + 1000 k g / m^{3} \times (9.8 m / s^{2}) \times (180 m)) = 1.7 \times 10^{6} p a$

Hence, the pressure difference between inlet and outlet of the pipe is $1.7 \times 10^{6} p a$ .

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Determining the concept

Step 3: Determining the pressure difference between the inlet and outlet of the pipe

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Fundamentals Of Physics

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Short Answer

The intake in Figure has cross-sectional area of 0.74m2 and water flow at 0.40m/s. At the outlet, distance D=180m below the intake, the cross-sectional area is smaller than at the intake, and the water flows out at 9.5m/s into the equipment. What is the pressure difference between inlet and outlet?

Step by Step Solution

Step 1: Given data

Step 2: Determining the concept

Step 3: Determining the pressure difference between the inlet and outlet of the pipe

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