Question

Darcy weisbach equation is used to find loss of head due to

Answer 1

Answer:

Explanation:hey mate your answer is

In fluid dynamics, the Darcy–Weisbach equation is an empirical equation, which relates the head loss, due to friction along a given length of pipe to the average velocity of the fluid flow for an incompressible fluid.

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Answer 2

Answer:

due to friction over a given length of pipe to the average fluid flow velocity.$H_{F}=\frac{4 f L v^{2}}{2 g d}$

Explanation:

Henry Darcy and Julius Weisbach are the names of two empirical equations in fluid mechanics. For an incompressible fluid, the Darcy Weisbach Equation connects the loss of pressure or head due to friction over a given length of pipe to the average fluid flow velocity.

Step 1: Terms and Assumptions

Consider a uniform horizontal pipe with fixed diameter d and area A, which allow a steady flow of incompressible fluid.

When fluid flows there will be the loss of energy due to friction. Thus we can apply Bernoulli’s principle.

Step 2: Applying Bernoulli's principle

On applying Bernoulli's equation at section; S1 and S2 we get-

$P_{1}+\frac{1}{2} \rho v_{1}^{2}+\rho g h_{1}=P_{2}+\frac{1}{2} \rho v_{2}^{2}+\rho g h_{2}+H_{F}$

--(1)

Where,

$\mathrm{H}_{\mathrm{F}}$ is the head loss due to friction

On dividing above equation (1) by$\rho \mathrm{g}$ we get-

$\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2 g}+h_{1}=\frac{P_{2}}{\rho g}+\frac{v_{2}^{2}}{2 g}+h_{2}+H_{F}$

--(2)

For horizontal pipe (That is, the inlet of pipe and the outlet of the pipe are at the same level from the reference plane)

$\mathrm{h}_{1}=\mathrm{h}_{2} \text {. }$

Here, the diameter is uniform, for uniform diameter-

$\mathrm{v}_{1}=\mathrm{v}_{2}$

On substituting them, the equation(2) becomes-

$\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2 g}+h_{1}=\frac{P_{2}}{\rho g}+\frac{v_{1}^{2}}{2 g}+h_{1}+H_{F}$

$\Rightarrow \frac{P_{1}}{\rho g}=\frac{P_{2}}{\rho g}+H_{F}$

--(3)

Thus, on rearranging equation (3) for the head loss we get-

$H_{F}=\frac{P_{1}}{\rho g}-\frac{P_{2}}{\rho g}$

$\Rightarrow H_{F}=\frac{P_{1}-P_{2}}{\rho g}$

$\Rightarrow H_{F} \rho g=P_{1}-P_{2}$

--(4)

The frictional resistance is well expressed through Froude’s formula.

Let f’ be the frictional resistance per unit area(wet) per unit velocity.

Frictional resistance F = f’ × wet area × (velocity)2

= f’ × 2rL × v2

= f’ × dL × v2

F = f’ × PL × v2 ——-(5)

[here, diameter; d=2r and perimeter; P=d]

The net force is the sum of Force due to pressure atThe net force is the sum of Force due to pressure at $\mathrm{S} 1, \mathrm{~S} 2$, and Fluid friction.

- At S 1, pressure is given by-

$P_{1}=\frac{F_{1}}{A}$

Which implies, The net force

$F_{1}=P_{1} A$

For our convenience consider the direction of the force due to pressure as +ve.

- At S 2, the pressure is given by-

$P_{2}=\frac{F_{2}}{A}$

Which implies, The net force

$F_{2}=P_{2} A$

here, the direction of the force due to pressure as -ve.

- Fluid Frictional force(F): It is a resistive force, thus the direction as -ve. Thus, resolving all the forces along horizontal direction we get-

$P_{1} A-P_{2} A-F=0$

$P_{1} A-P_{2} A=F$

$\left(P_{1}-P_{2}\right) A=F$

$\Rightarrow P_{1}-P_{2}=\frac{F}{A}$

--(6)

Substitute the values for F and $\left(P_{1}-P_{2}\right)$ from equation (5) and (4) respectively.

$\Rightarrow \rho g H_{F}=\frac{f^{\prime} P L v^{2}}{A}$

, and Fluid friction.

- At S 1, pressure is given by-

$P_{1}=\frac{F_{1}}{A}$

This implies, The net force

$F_{1}=P_{1} A$

For our convenience consider the direction of the force due to pressure as +ve.

- At S 2, the pressure is given by-

$P_{2}=\frac{F_{2}}{A}$

Which implies, The net force

$F_{2}=P_{2} A$

here, the direction of the force due to pressure as -ve.

- Fluid Frictional force(F): It is a resistive force, thus the direction as -ve. Thus, resolving all the forces along horizontal direction we get-

$P_{1} A-P_{2} A-F=0$

$P_{1} A-P_{2} A=F$

$\left(P_{1}-P_{2}\right) A=F$

$\Rightarrow P_{1}-P_{2}=\frac{F}{A}$

--(6)

Substitute the values for F and

$\left(P_{1}-P_{2}\right)$from equation (5) and (4) respectively.

$\Rightarrow \rho g H_{F}=\frac{f^{\prime} P L v^{2}}{A}$

On rearranging the terms we get-

$H_{F}=\frac{f^{\prime} P L v^{2}}{A \rho g}$

$H_{F}=\frac{f^{\prime}}{\rho g} \times L v^{2} \times \frac{p}{A}$

--(7)

But,

$\frac{p}{A}=\frac{\text { Wetted perimeter }}{\text { Area }}=\frac{\pi d}{\frac{\pi}{4} d^{2}}=\frac{4}{d}$

Substituting

$\frac{P}{A}=\frac{4}{d}$

in equation(7),

$\Rightarrow H_{F}=\frac{f^{\prime}}{\rho g} \times \frac{4 L v^{2}}{d}$

--(8)

Now substitute

$\frac{f^{\prime}}{\rho}=\frac{f}{2}$

Where,

$\mathrm{f}^{\prime}$ is a frictional resistance

$\rho$ is the density of the fluid.

$\mathrm{f}$ is the coefficient of friction

$\text { (8) } \Rightarrow H_{F}=\frac{f}{2 g} \times \frac{4 L v^{2}}{d}$

Thus, On rearranging, we finally arrive at Darcy Weisbach Equation

$H_{F}=\frac{4 f L v^{2}}{2 g d}$

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