Darcy Weisbach Equation Derivation - Explanation and Applications

Darcy Weisbach Equation statement

It is an empirical equation in fluid mechanics named after Henry Darcy and Julius Weisbach. The Darcy Weisbach Equation relates the loss of pressure or head loss due to friction along the given length of pipe to the average velocity of the fluid flow for an incompressible fluid.

Darcy Weisbach Equation

\(H_{F}=\frac{4fLv^{2}}{2gd}\)

Where,

HF is the head loss or pressure loss.

f is the coefficient of friction or friction factor.

v is the velocity of incompressible fluid.

L is the length of the pipe.

d is the diameter.

g is the acceleration due to gravity. (g = 9.8m/s2)

Derivation of Darcy Weisbach Equation

Derivation of Darcy Weisbach Equation

Figure(1)Uniform horizontal pipe with a steady flow of fluid.

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.

For simplicity consider two sections; S1 and S2 of the pipe separated by the distance L.

At all the point of S1, The pressure is P1, velocity is V1.

At all the point of S2, the pressure is P2 and velocity is V2.

Consider the fluid flow as shown in the figure(1) Thus, the pressure at S1is greater than the pressure at S2 i.e.,(P1>P2) This pressure difference makes the fluid flow along the pipe.

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

Bernoulli’s principle

which states that a decrease in the pressure or potential energy of the fluid increases the velocity/speed of the fluid flow or in other words, “For incompressible fluid, the sum of its potential energy, pressure, and velocity remains constant.”

Step 2: Applying Bernoulli’s principle

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

\(P_{1}+\frac{1}{2}\rho v_{1}^{2}+\rho gh_{1}=P_{2}+\frac{1}{2}\rho v_{2}^{2}+\rho gh_{2}+H_{F}\) —–(1)

Where,

HF is the head loss due to friction

On dividing above equation (1) by 𝜌g we get-

\(\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2g}+h_{1}=\frac{P_{2}}{\rho g}+\frac{v_{2}^{2}}{2g}+h_{2}+H_{F}\) ——-(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)

h1=h2.

Here, the diameter is uniform, for uniform diameter-

v1=v2

On substituting them, the equation(2) becomes-

\(\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2g}+h_{1}=\frac{P_{2}}{\rho g} +\frac{v_{1}^{2}} {2g}+h_{1}+H_{F}\) \(\Rightarrow \frac{P_{1}}{\rho g}=\frac{P_{2}}{\rho g}+H_{F}\) —–(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)

Step 3: Find frictional resistance

Due to the combined effect of wet surface and surface roughness, the resistance is offered to the flow of fluid due to friction. As a result, speed is reduced. The Froude was the first person to observe the dependency of frictional resistance with surface roughness.

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’ × 2𝜋rL × v2

= f’ × 𝜋dL × v2

F = f’ × PL × v2 ——-(5)

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

Step 4: Net force acting on the fluid at section S1 and S2

The net force is the sum of Force due to pressure at S1, S2, and Fluid friction.

  • At S1, 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 S2, 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-

    P1A – P2A – F = 0

    P1A – P2A = F

    (P1– P2)A = F

    \(\Rightarrow P_{1}-P_{2}=\frac{F}{A}\) ——(6)

    Substitute the values for F and (P1– P2) from equation (5) and (4) respectively.

    \(\Rightarrow \rho gH_{F}=\frac{f’PLv^{2}}{A}\)

    On rearranging the terms we get-

    \(H_{F}=\frac{f’PLv^{2}}{A \rho g}\) \(H_{F}=\frac{f’}{\rho g}\times Lv^{2}\times \frac{P}{A}\) —–(7)

    But, \(\frac{P}{A}=\frac{Wetted\;perimeter}{Area}=\frac{\pi d}{\frac{\pi }{4}d^{2}}=\frac{4}{d}\)

    Substituting \(\frac{P}{A}=\frac{4}{d}\) in equation(7),

    \(\left ( 7 \right )\Rightarrow H_{F}=\frac{f’}{\rho g}\times \frac{4Lv^{2}}{d}\) ——-(8)

    Now substitute \(\frac{f’}{\rho}=\frac{f}{2}\)

    Where,

    f’ is a frictional resistance

    ⍴ is the density of the fluid.

    f is the coefficient of friction

    Because the fluid is incompressible, which means that with the application of the external force there will not be any change in the density.

    \(\left ( 8 \right )\Rightarrow H_{F}=\frac{f}{2g}\times \frac{4Lv^{2}}{d}\)

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

    \(H_{F}=\frac{4fLv^{2}}{2gd}\)

Application of Darcy Weisbach Equation:

Is used to calculate the loss of head due to friction in the pipe.

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Practise This Question

A cricket ball of mass 150 gm is moving with a velocity of 12 m/s and is hit by a bat so that the ball is turned back with a velocity of 20 m/s. The force of blow acts for 0.01s on the ball. The average force exerted by the bat on the ball is

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