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Bernoulli and Pipe Flow, Lecture notes of Civil Engineering

Head Loss Calculations. Bernoulli and Pipe Flow. ○ The Bernoulli equation that we worked with was a bit simplistic in the way it looked at a fluid system.

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2021/2022

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Civil Engineering Hydraulics
Mechanics of Fluids
Head Loss Calculations
Bernoulli and Pipe Flow
¢The Bernoulli equation that we worked with
was a bit simplistic in the way it looked at a
fluid system
¢All real systems that are in motion suffer
from some type of loss due to friction
¢It takes something to move over a rough
surface
Pipe Flow 2
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a
pf1b
pf1c

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Civil Engineering Hydraulics

Mechanics of Fluids

Head Loss Calculations Bernoulli and Pipe Flow ¢ The Bernoulli equation that we worked with was a bit simplistic in the way it looked at a fluid system ¢ All real systems that are in motion suffer from some type of loss due to friction ¢ It takes something to move over a rough surface 2 Pipe Flow

Bernoulli and Pipe Flow ¢ Consider flow in a constant-diameter pipe 3 Pipe Flow Bernoulli and Pipe Flow ¢ If we look at the energy at points 1 and 2. 4 Pipe Flow p 1 ρ

v 1

2

  • z 1 g = p 2 ρ

v 2

2

  • z 2 g

Bernoulli and Pipe Flow ¢ Since we are dealing with an uncompressible fluid, the pressures at points 1 and 2 should be the same. 7 Pipe Flow

p 1

ρ

p 2

ρ ⇒ p

= p

Bernoulli and Pipe Flow ¢ If we ran the system experimentally and measured the two pressures, they would not be the same. 8 Pipe Flow p

ρ

p

ρ ⇒ p

= p

Bernoulli and Pipe Flow ¢ The pressure at point 2 would be lower than the pressure at point 1. 9 Pipe Flow

p 1

ρ

p 2

ρ ⇒ p

= p

Bernoulli and Pipe Flow ¢ The pressure is being lost (actually the pressure energy) due to friction as the flow moves along the pipe. 10 Pipe Flow p

ρ

p

ρ ⇒ p

= p

Friction Losses ¢ The thickness of the quarter is dz and it has a perimeter length of P (πD) ¢ This makes the area in contact with the sides of the pipe equal to Pdz = πDdz ¢ At the wall there is a shearing stress, τw, which is the stress between the wall and the outer layer of the fluid 13 Pipe Flow Friction Losses ¢ Then the force from the wall on the section of fluid (opposing the fluid flow) will be equal to 14 Pipe Flow Fretarding = τ w Pdz = τ w π Ddz

Friction Losses ¢ The pressure on the upstream (left face) of the section will produce a force accelerating the section 15 Pipe Flow Fretarding = τ w Pdz = τ w π Ddz Faccelerating = pA = p π D

4 Friction Losses ¢ As we move from the left face to the right face (dz) the change in pressure will equal dp 16 Pipe Flow Fretarding = τ w Pdz = τ w π Ddz Faccelerating = pA = p π D

4

Friction Losses ¢ Therefore 19 Pipe Flow Fretarding = τ w π Ddz +( p + dp ) π D^2

Faccelerating = pA = p

π D^2

Faccelerating − Fretarding = 0

p π D^2

− τ w π Ddz −( p + dp ) π D^2

− τ w π Ddz −

π D^2

dp = 0

Friction Losses ¢ Rearranging 20 Pipe Flow

− τ w π Ddz −

π D^2

dp = 0

− τ w π Ddz =

π D^2

dp

τ w π D

π D^2

dp

dz

Friction Losses ¢ So we have an expression for the rate at which the pressure changes as the flow moves downstream 21 Pipe Flow

− τ w π Ddz −^ π^ D^

2

dp = 0

− τ w π Ddz =^ π^ D^

2

dp

τ w π D

π D^2

dp

dz

Friction Losses ¢ In the simplest form 22 Pipe Flow − τ w π D π D

4

dp dz − 4 τ w D = dp dz

Friction Losses ¢ The friction factor, f , is the ratio of the friction forces to the inertia forces. 25 Pipe Flow − 4 τ w Dh = dp dz f = 4 τ w 1 2 ρ v

Friction Losses ¢ Combining the two expressions. 26 Pipe Flow − f 1 2 ρ v

D h = dp dz

Friction Losses

¢ Isolating the differentials

27 Pipe Flow

f ρ v

2 D

h

dz = dp

Friction Losses

¢ Now back to the pipe we originally

considered

28 Pipe Flow

f ρ v

2 D

h

dz

L

= dp

Friction Losses ¢ So we have a modified form of the Bernoulli equation that takes into account the friction losses in the system 31 Pipe Flow p 1 ρ

v 12

+ z 1 g −

ρ v^2

fL Dh ρ

p 2 ρ

v 22

+ z 2 g

The pressure change term

takes that form because we

are using energy terms in this

expression.

Friction Losses ¢ Reducing 32 Pipe Flow p 1 ρ

v 12

  • z 1 gv^2

fL Dh

p 2 ρ

v 22

  • z 2 g

Friction Losses ¢ And rewriting the expression in terms of head 33 Pipe Flow p 1 ρ g

v 12 2 g

  • z 1 − v^2 2 g fL Dh

p 2 ρ g

v 22 2 g

  • z 2 Friction Losses ¢ So the downstream energy (at point 2) is lower than the energy at point 1 34 Pipe Flow p 1 ρ g

v 12 2 g

  • z 1 − v^2 2 g fL Dh

p 2 ρ g

v 22 2 g

  • z 2

Example

¢ A 2-nominal pipe is inclined at an angle of 30° with the

horizontal and conveys 0.001 m^3 /s of water uphill.

Determine the pressure drop in the pipe if it is 7 m long.

Take the friction factor f to be 0.03.

37 Pipe Flow

We can draw a sketch of the system and use the modified Bernoulli expression

to answer the pressure drop.

Example

¢ A 2-nominal pipe is inclined at an angle of 30° with the

horizontal and conveys 0.001 m^3 /s of water uphill.

Determine the pressure drop in the pipe if it is 7 m long.

Take the friction factor f to be 0.03.

38 Pipe Flow

We have the volumetric flow rate so we can calculate the average velocity at

points 1 and 2. Since they have the same cross sectional area, their average

velocities will be the same.

Example

¢ A 2-nominal pipe is inclined at an angle of 30° with the

horizontal and conveys 0.001 m^3 /s of water uphill.

Determine the pressure drop in the pipe if it is 7 m long.

Take the friction factor f to be 0.03.

39 Pipe Flow

If we set the reference plane through the left end of the system at point 1, then

we can calculate the elevation of point 2 from the reference plane from the

information given in the problem.

Example

¢ A 2-nominal pipe is inclined at an angle of 30° with the

horizontal and conveys 0.001 m^3 /s of water uphill.

Determine the pressure drop in the pipe if it is 7 m long.

Take the friction factor f to be 0.03.

40 Pipe Flow

Using the loss term we developed we can determine the pressure drop (in units

of head) due to the friction in the pipe.