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. hf = 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+hf ————(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 hf = (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 = F= f1xPxLxV2

(P1-P2) = (f1xPxLxV2 / A ).

But from (1) we get

P1 -P2 = ?ghhf

Equating the values of (P1 -P2) we get     

?ghf   = h(f1xPxLxV2 / A ).

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

(P/A) = (?d / (?d2/4)) = (4/d)

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

hf = 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.


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.