The flow carried by pipe below discharges into the open atmosphere. If the pipe length is 300 m, diameter is 150 mm and roughness of 0.0025 mm, write the equation(s) required to determine the maximum discharge. The flow carried by the pipe below discharges into theopenatmosphere. If the pipe length is 300 m, diameter is 150 mm and roughness of0.0025 mm, write the equation(s) required to determine the maximum discharge.16 mGate valveKv=0.15

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Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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The flow carried by pipe below discharges into the open atmosphere. If the pipe length is 300 m, diameter is 150 mm and roughness of 0.0025 mm, write the equation(s) required to determine the maximum discharge.

The flow carried by the pipe below discharges into the
open
atmosphere. If the pipe length is 300 m, diameter is 150 mm and roughness of
0.0025 mm, write the equation(s) required to determine the maximum discharge.
16 m
Gate valve
Kv=0.15

Expert Solution

Step 1

To find-:

The equation required to determine the maximum discharge.

Given:

The length of the pipe is $L = 300 m m$ .

The diameter of the pipe is $D = 150 m m$ .

The roughness of the pipe is $ε = 0.0025 m m$ .

The elevation difference is $∆ Z = Z_{1} - Z_{2} = 16 m$ .

Formulas used-:

The expression for relative roughness is-:

$k_{R} = \frac{ε}{D}$ (1)

The bernoulli's equation between free stream in the tank and exit of pipe is-:

$\frac{P_{1}}{ρ g} + \frac{{V_{1}}^{2}}{2 g} + Z_{1} = \frac{P_{2}}{ρ g} + \frac{{V_{2}}^{2}}{2 g} + Z_{2}$ (2)

The expression for maximum discharge is-:

$Q = A V$ (3)

Here , $A = πDL$ is the area of the pipe.

Here, $ν$ is the kinematic viscosity.

Calculations-:

Substitute the value of $ε = 0.0025 m m$ and $D = 150 m m$ in equation (1).

$k_{R} = \frac{0.0025}{150} k_{R} = 1.67 \times 10^{- 5}$

Substitute the value of $P_{1} = P_{2} = 0$ , $V_{1} = 0$ $∆ Z = Z_{1} - Z_{2} = 16 m$ in equation (2).

$\begin{array}{rcl} 0 + 0 + Z_{1} & = & 0 + \frac{{V_{2}}^{2}}{2 g} + Z_{2} \\ Z_{1} - Z_{2} & = & \frac{{V_{2}}^{2}}{2 g} \\ {V_{2}}^{2} & = & 16 \times 2 \times 9.81 \\ V_{2} & = & 17.72 \frac{m}{s} \end{array}$

Thus, the value of velocity at exit is $V_{2} = 17.72 \frac{m}{s}$ .

Substitute the value of $D = 150 m m (\frac{1 m}{1000 m m}) = 0.15 m$ , $L = 300 m m (\frac{1 m}{1000 m m}) = 0.3 m$ and $V = 17.72 \frac{m}{s}$ in equation (3).

$Q = π \times 0.15 \times 0.3 \times 17.72 Q = 2.505 \frac{m^{3}}{s}$

Thus, the value of maximum discharge is $Q = 2.505 \frac{m^{3}}{s}$ .

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