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Darcy Weisbach equation is an empirical equation that relates the loss of pressure (head loss) due to friction along the length of a pipe to the average velocity of an incompressible fluid.
- It is used to determine the head loss in the pipe due to friction.
- This equation is named after French engineer Henry Darcy and mathematician Julius Weisbach.
- According to the Darcy-Weisbach equation, the head loss or pressure loss due to friction along the length of a pipe for an incompressible fluid is given by
Hf = \(\frac{Lf_Dv^2}{2gd}\)
- This equation contains a dimensionless friction factor fD, known as Darcy friction factor.
- According to Bernoulli's principle, a drop in pressure results in a rise in fluid velocity.
- Fluid moves from high-pressure to low-pressure levels inside a system.
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Key Terms: Darcy Weisbach Equation, Friction, Fluid, Pressure Loss, Bernoulli’s Principle, Frictional Resistance, acceleration due to gravity
Read More: Mechanical Properties of Fluids MCQ
Darcy Weisbach Equation
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According to Darcy Weisbach equation, for an incompressible fluid flowing through a cylindrical pipe of length L and uniform diameter d, the pressure loss or head loss is given by
Hf = \(\frac{Lf_Dv^2}{2gd}\)
Where
- Hf is the pressure loss or head loss
- fD is known as Darcy friction factor. It is also known as the resistance coefficient, Darcy–Weisbach friction factor, friction factor, or flow coefficient.
- v is the mean flow velocity of incompressible fluid.
- g is the acceleration due to gravity.
Let Q be the volumetric flow rate of the fluid through the pipe, which is given by
Q = Area of the cross-section of the pipe (A) x mean fluid velocity (v)
⇒ v = Q/A = 4Q/πd2
Therefore, head loss in terms of volumetric flow rate, is given by
Hf = \(\frac{16Lf_DQ^2}{2g \pi^2d^5}\)
Let Cf be the coefficient of friction between the fluid and the surface of the pipe, then Darcy friction factor is given by
fD = 4Cf
Therefore pressure loss or head loss can be expressed in terms of coefficient of friction as
Hf = \(\frac{4Lv^2C_f}{2gd}\)
In terms volumetric flow rate, the head loss can be expressed as
Hf = \(\frac{32LQ^2C_f}{g \pi ^2d^5}\)
Read More: NCERT Solution for Class 11: Mechanical Properties of Fluids
Derivation of Darcy Weisbach Equation
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Consider a horizontal pipe having fixed diameter d and uniform cross-sectional area A that allows a flow of incompressible fluid.
Uniform horizontal pipe with a steady flow of fluid
Consider 2 sections of the pipes; S1 and S2 separated by the distance L.
- In section S1, the pressure at all the points is P1 and the velocity is v1.
- In section S2, the pressure at all the points is P2 and the velocity is v2.
- The pressure at section S1 is greater than at section S2 i.e., P1 > P2. This difference in pressure makes the fluid flow through the pipe.
- While the fluid flows through the pipe, there will be a loss of energy due to friction.
Now, we will derive an equation for the head loss of the fluid while flowing through the pipe in 4 steps.
Step 1: Assumptions
The frictional resistance i.e. loss of energy due to friction (frictional head) is:
- Varies with v2. Where v means flow velocity of the fluid.
- Proportional to the density of fluid
- Proportional to the area of the surface in contact
- Independent of pressure
- Depends on the nature of the surface in contact
Step 2: Applying Bernoulli’s Principle
Bernoulli's Principle states that the velocity of the fluid increases, as the potential energy or pressure of the fluid decreases.
Also, it states that the sum of potential energy, velocity, and pressure of an incompressible fluid remains constant.
On applying Bernoulli's principle in sections S1 and S2, we get
P1 + 1/2ρv12 + ρgh1 = P2 + 1/2ρv22 + ρgh2 + HF …(1)
Where
- Ρ is the density of the fluid
- h1 and h2 height of the fluid column in the pipe at sections S1 and S2 respectively.
Divide equation (1) by ρg, we get
P1/ρg + v12/2g + h1 = P2/ρg + v22/2g + h2 + HF …(2)
For horizontal pipe with uniform diameter, the area of the cross-section at sections 1 and 2 are equal i.e.
A1 = A2 = A
From the equation of continuity, we have
A1v1 = A2v2
⇒ v1 = v2
Since the pipe is horizontal, therefore
h1 = h2
- v1 = v2
Hence, equation (2), becomes
P1/ρg = P2/ρg + HF
Rearranging above for pressure loss, we get
HF = (P1 - P2)/ρg …(3)
Step 3: To Find Frictional Resistance
The frictional resistance can be expressed using Froude’s formula. Let f’ is the frictional resistance per unit area per unit velocity, then
Total frictional resistance (F) = f’ x (surface area of cylindrical pipe between sections 1 and 2) x (v)2
⇒ F = f’ x 2πrL x v2
We have, 2r = d (diameter of the pipe)
Where r is the radius of the pipe
⇒ F = f’ x πdL x v2 …(4)
Step 4: Calculation of the Total Force Acting at Sections S1 and S2
The total force is the sum of fluid friction and forces acting on S1 and S2 due to pressure. If the flow of fluid is steady and in +x direction, then various forces acting on the fluid between sections S1 and S2 are given by
- Force at Section 1, F1 = P1A
Where P1 is the pressure at S1
The direction of this force is toward +ve x-axis.
- Force at Section 2, F2 = P2A
Where P2 is the pressure at S2
The direction of this force is toward the -ve x-axis.
- Fluid frictional force (F): It is a resistive force thus its direction is -ve.
Thus, the net force can be calculated as
P1A - P2A - F = 0
⇒ P1A - P2A = F
⇒ (P1 - P2) = F/A …(5)
Substituting the value of F from equation (4) in the above equation, we get
(P1 - P2) = (f’ x πdL x v2)/A
But area of cross-section of the pipe, A = πd2/4
⇒ (P1 - P2) = (f’ x πdL x v2)/(πd2/4)
⇒ (P1 - P2) = (4f’Lv2)/d
Dividing both sides by ρg, we get
(P1 - P2)/ρg = (4f’Lv2)/dρg
Comparing the above equation with equation (3), we get
HF = (4f’Lv2)/dρg
HF = (f’/ρg) x (4Lv2/d) …(6)
Substitute: f’/ρ = Cf /2
Where Cf be the coefficient of friction between the fluid and the surface of the pipe
Therefore,
Hf = \(\frac{4Lv^2C_f}{2gd}\)
Also, we have
4Cf = fD, Darrcy’s friction factor
Therefore, the head loss can be given as
Hf = \(\frac{Lf_Dv^2}{2gd}\)
Also Read:
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Applications of Darcy Weisbach Equation
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Following are the applications of Darcy Weisbach equation:
- The head loss per unit length is considered to be an important factor in Hydraulic engineering.
- The practical consequence is that the head loss Hf decreases with the inverse fifth power of the pipe diameter (d), provided that the volumetric flow rate is to be constant.
- On doubling the diameter of a pipe, the amount of material required per unit length doubles, and thus its installed cost increases. Meanwhile, the head loss is decreased by a factor of 32 i.e. reduction of about 97%.
- Thus the energy consumed in moving a given fluid decreases dramatically.
Advantages of Darcy Weisbach Equation
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Following are the advantages of the Darcy Weishbach equation
- There is dimensional consistency in it.
- This equation is useful for any fluid, including gas, oil, sludges, and brine.
- It is useful in the transition zone between laminar flow and turbulent flow.
- It is based on the fundamentals.
Read Also: Mechanical Properties of Solids
Solved Examples
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Ques. Water flows with a velocity of 5 m/s along a 60 mm diameter horizontal pipe having a 120 m length and a friction factor equal to 0.02. Find head loss due to friction. Take g = 10 m/s2.
Ans. Given
- Fluid velocity, v = 5 m/s
- Diameter of the pipe, d = 60 mm = 60 x 10-3 m
- Length of the pipe, L = 120 m
- Darcy friction factor, fD = 0.02
Head loss in a pipe due to friction is
HF = \(\frac{Lf_Dv^2}{2gd}\)
⇒ HF = (120 x 0.02 x 52) / (2 x 10 x 60 x 10-3)
⇒ HF = 50 m
Ques. A 480 m long pipe with a 60 mm diameter can outflow water at a velocity of 3 m/s. Find the pressure loss if the coefficient of friction in a pipe is 0.005.
Ans. Given
- Fluid velocity, v = 3 m/s
- Diameter of the pipe, d = 60 mm = 60 x 10-3 m
- Length of the pipe, L = 480 m
- Coefficient of friction, Cf = 0.005
Pressure loss due to friction is
HF = \(\frac{4LC_f v^2}{2gd}\)
⇒ HF = [4 x 480 x 0.005 x (3)2]/[2 x 9.8 x 60 x 10-3]
⇒ HF = 7.47 m
Things to Remember
- The Head loss during the flow of fluid is given by Darcy Weisbach Equation i.e. HF = 4LfDv2/2gd.
- This equation is applicable for turbulent flow of fluid having Reynolds number Re > 2000.
- The Darcy factor of friction (fD) is a dimensionless quantity depending on the characteristics of a pipe.
- Bernoulli’s principle says a decrease in pressure causes an increase in fluid velocity.
- In a system, fluid flows from high-pressure to low-pressure levels.
- The total frictional resistance is given by: (F) = f’ x (surface area of cylindrical pipe between sections 1 and 2) x (v)2
Also Read:
Sample Questions
Ques. Water flows with a velocity of 4 m/s along a 50 mm diameter horizontal pipe having a 100 m length and a friction factor equal to 0.02. Find head loss due to friction. Take g = 9.8 m/s2. (3 Marks)
Ans. Given
- Fluid velocity, v = 4 m/s
- Diameter of the pipe, d = 50 mm = 50 x 10-3 m
- Length of the pipe, L = 100 m
- Darcy friction factor, fD = 0.02
Head loss in a pipe due to friction is
HF = LfDv2/2gd
⇒ HF = (100 x 0.02 x 42) / (2 x 9.8 x 50 x 10-3)
⇒ HF = 32.65 m
Ques. A 490 m long pipe with a 100 mm diameter can outflow water at a velocity of 2 m/s. Find the pressure loss if the coefficient of friction in a pipe is 0.004. (3 Marks)
Ans. Given
- Fluid velocity, v = 2 m/s
- Diameter of the pipe, d = 100 mm = 100 x 10-3 m
- Length of the pipe, L = 490 m
- Coefficient of friction, Cf = 0.004
Pressure loss due to friction is
HF = 4LCf v2/2gd
⇒ HF = [4 x 490 x 0.004 x (2)2]/[2 x 9.8 x 100 x 10-3]
⇒ HF = 16 m
Ques. Water flows through a 400 m long iron pipe with a diameter of 200 mm. The volumetric flow rate of water is 0.48 m3/s. Find the head loss due to friction. (Take frictional factor = 0.0225) (3 Marks)
Ans. Given,
- Frictional factor, fD = 0.0225
- Length of the pipe, L = 400 m
- Diameter of the pipe, d = 200 mm = 0.2 m
- Volumetric flow rate of water, Q = 0.48 m3/s
Volumetric flow rate, Q = Area of the cross-section of the pipe (A) x mean flow velocity of the fluid (v)
⇒ v = Q/A
⇒ v = Q/A = 4Q/πd2
⇒ v = (4 x 0.48)/(3.14 x 0.22)
⇒ v = (4 x 0.48)/(3.14 x 0.22) = 15.28 m/s
Now, Head loss in a pipe due to friction is
HF = LfDv2/2gd
⇒ HF = [400 x 0.0225 x 15.282]/[2 x 9.8 x 0.2]
⇒ HF = 535 m
Ques. Find the frictional resistance acting on fluid moving with velocity 3 m/s in a 120 m long pipe having a 100 mm diameter. (frictional factor per unit area per unit velocity = 0.045) (3 Marks)
Ans. Given
- Velocity of the fluid, v = 3 m/s
- Length of the pipe, L = 120 m
- Diameter of the pipe, d = 100 mm = 0.1 m
- Frictional factor per unit area per unit velocity, f’ = 0.045
Total Frictional resistance encounter by the fluid is given by
F = f’ x 2πrL x v2
Where 2r = d (diameter of the pipe)
⇒ F = f’ x πdL x v2
On substituting the values, we get
F = (0.045) x (3.14) x (0.1) x (120) x (3)2
F = 15.26 N
Ques. Find the ratio of the frictional factor to the coefficient of friction in the Darcy Weisbach equation. (3 Marks)
Ans. Darcy Weisbach's Equation in terms of friction factor fD is given as
HF = LfDv2/2gd
⇒ fD = 2gdHF / Lv2 ………(1)
Darcy Weisbach's Equation in terms of coefficient of friction factor Cf is given as
HF = 4LCf v2/2gd
⇒ Cf = 2gdHF / 4Lv2 ………(2)
Compare equation (1) and (2), we get
fD/Cf = 4
Ques. Two pipes of equal length are joined in series. The ratio of the diameter of the second pipe to the first pipe is 2. Find the head loss between pipes. (5 Marks)
Ans. Head loss in a pipe due to friction is
HF = LfDv2/2gd
Volumetric flow rate, Q = Av
⇒ v = Q/A
Where
- A = πd2/4, is the cross-sectional area of the pipe
- v is mean flow velocity of the fluid
⇒ HF = LfDQ2/2gdA2 = (16LfDQ2) / (2gdπ2d4)
⇒ HF = (16LfDQ2) / (2gπ2d5)
Thus,
HF ∝ 1/d5
Let d1 and d2 be the diameter of the two pipes respectively. Given,
d2/d1 = 2
Therefore,
(HF1/HF2) = (d2/d1)5
⇒ (HF1/HF2) = (2)5
⇒ (HF1/HF2) = 32:1
Ques. State Bernoulli’s principle. (2 Marks)
Ans. According to Bernoulli’s principle, if the potential energy or pressure of a fluid decreases, the velocity of the fluid increases.
Or,
The sum of potential energy, velocity, and pressure of incompressible fluid remains constant.
Ques. Fluid flows through a pipe of diameter 200 mm (frictional factor = 0.025) with a velocity of 2 m/s. Find pressure loss per unit length due to friction. (3 Marks)
Ans. The pressure loss per unit length is
Δp/ L = fDv2/ 2d
Given,
fD = 0.025, v = 1 m/s, d = 200 mm = 0.2 m
Δp/ L = [(0.025) x (2)2]/ (2 x 0.2)
Δp/ L = 0.25 Nm-3
Ques. The head loss in a pipe of some length carrying fluid at a rate of Q is H. If the diameter is made twice keeping the same length to carry a flow at a rate of 2Q, find head loss. (3 Marks)
Ans. Let the new head loss be H’
H = (128 x μ x Q x L)/ πd4
Thus,
H ∝ Q/d4
H’/H = Q2d14/Q1d24
Given
Q2 = 2Q d2 = 2d1
H’/H = 2Q (d)4/ Q(2d)4
H’ = H/8
Ques. The pressure difference between the two sections of pipe is 150 mm Hg carrying fluid of density 1.86 kg/m3. Find head loss due to friction in the pipe. (2 Marks)
Ans. The head loss in the pipe due to friction is
HF = (P1 - P2)/ρg
Given,
P1 - P2 = 150 mm Hg, ρ = 1.86 kg/m3
HF = (150)/ (1.86 x 9.86)
HF = 8.22 m
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