Fluid mechanics problems A fluid with a kinematic viscosity 1500 ft2/sec flows with an average velocity, v = 200 ft/sec through a roundsteel pipe with an inside diameter D=8 in, a relative roughness, 900 and a length of 60 ft. Within the lengthof the pipe there are two 90 deg standard elbow and a fully open globe valve. Determine the total energy losswithin this pipe.Note: consider the velocity is same throughout the pipe.

Answered: A fluid with a kinematic viscosity 1500…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Fluid mechanics

problems

A fluid with a kinematic viscosity 1500 ft2/sec flows with an average velocity, v = 200 ft/sec through a round
steel pipe with an inside diameter D=8 in, a relative roughness, 900 and a length of 60 ft. Within the length
of the pipe there are two 90 deg standard elbow and a fully open globe valve. Determine the total energy loss
within this pipe.
Note: consider the velocity is same throughout the pipe.

Expert Solution

Step 1

Given:

The kinematic viscosity is $ν = 1500 {ft}^{2} / s$ .

The average velocity is $v = 200 ft / \sec$

The inside diameter of the pipe is $8 in$ .

The relative roughness is $e = 900$ .

The length of the pipe is $l = 60 ft$

Step 2

The head loss in elbow is given by,

$H_{e l b o w} = k \frac{v^{2}}{2 g}$ ... (1)

The pipe bend coefficient for the curve bend is,

$k = 0.75$

Substitute the values in equation (1).

$\begin{array}{rcl} H_{e l b o w} & = & 0.75 \frac{200^{2}}{2 (32.17)} \\ = & 466.27 ft \end{array}$

For fully open global valve.

$k = 6$

The head loss for fully open global valve is calculated as,

$\begin{array}{rcl} H_{f u l l y o p e n} & = & 6 \frac{200^{2}}{2 (32.17)} \\ = & 3730.183 ft \end{array}$

Step 3

The Reynolds number is calculated as,

$\begin{array}{rcl} R_{e} & = & \frac{V D}{ν} \\ = & \frac{200 \times (8 \times \frac{1}{12})}{1500} \\ = & 0.0888 \end{array}$

The friction factor is calculated as,

$\begin{array}{rcl} f & = & \frac{1.325}{{[\ln (\frac{e}{3.7 D} + \frac{5.74}{{R_{e}}^{0.9}})]}^{2}} \\ = & \frac{1.325}{{[\ln (\frac{900}{3.7 (0.666)} + \frac{5.74}{{0.0888}^{0.9}})]}^{2}} \\ = & 0.03644 \end{array}$

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