Darcy-Weibach Equation

Darcy-Weibach Equation

Expression for loss of head due to friction in pipes or Darcy -Weisbach Equation.

Consider a uniform horizontal pipe, having steady flow as shown figure. Let 1 -1 and 2-2 is two sections of pipe.

Let P₁ = pressure intensity at section 1-1. Let P₂ = Velocity of flow at section 1-1.

L = length of the pipe between the section 1-1 and 2-2 d = diameter off pipe.

f¹ = Frictional resistance per unit wetted area per unit velocity. h_f = loss of head due to friction.

And P₂,V₂ = are the values of pressure intensity and velocity at section 2-2.

Applying   Bernoulli's - 1 equation &2-2   between   section 1.1,1.2(previous)

Total head 1-1 = total head at 2-2 + loss of head due to friction between 1-1&2-2 (P₁/?g) ₁ ²+/2g)(V+Z₁ = (P₂/?g)₂ ² / +2g) +(VZ₂+h_f ------------(1)

but Z₁ = Z₁ [ pipe is horizontal ]

V₁= V₂ [ diameter of pipe is same at 1-1 & 2-2]

(1) becomes,

(P₁/ ?g)₂/?g)+h=_f(P h_f = (P₁/ ?g)-(P₂/?g)

frictional resistance = frictional resistance per unit wetted area per unit velocity X wetted area X velocity ².

F = f¹ x ?d _l x V² [ Wetted area = ?d x L, and Velocity V = V₁ = V₂] F₁ = f¹xPxLxV² ----------- (2). [?d = wetted perimeter = p]

The forces acting on the fluid between section 1-1 and 2-2 are,

1) Pressure force at section 1-1 = P₁X A

2) Pressure force at section 2-2 = P₂ X A

3). Frictional force F₁

Resolving all forces in the horizontal direction.,

P₁ A -P₂A -F₁ = 0

(P₁-P₂)A = F₁ = f¹xPxLxV²

(P₁-P₂) = (f¹xPxLxV² / A ).

But from (1) we get

P₁ -P₂ = ?ghh_f

Equating the values of (P1 -P2) we get     

?gh_f   = h(f¹xPxLxV² / A ).

hf = (f¹  /   ?g)   X   (P/A)   X   LX   V²

(P/A) = (?d / (?d²/4)) = (4/d)

Hence, h_f = ( f¹ /   ?g)   x   (.   4/d)   x   LxV²

h_f = 4 fLV ² / 2gd

This equation is known as Darcy -Weisbach equation. This equation is commonly used to find loss of head due to friction in pipes.

Problem

Water flows through a pipe AB 1.2m diameter at 3 m/s and then passes through a pipe BC 1.5 m diameter at C, the pipe branches. Branch CD is 0.8m in diameter and carries one third of the flow in AB. The flow velocity in branch CE is 2.5 m/s. Find the volume rate of flow in AB, the velocity in BC, the velocity in CD and the diameter of CE.