head loss calculations - memphis · 2012-10-10 · head loss calculations bernoulli and pipe flow !...

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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

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¢ Consider flow in a constant-diameter pipe

Pipe Flow 3


¢  If we look at the energy at points 1 and 2.

Pipe Flow 4

p1ρ

+v12

2+ z1g =

p2ρ

+v22

2+ z2g

3


¢ Since this system is horizontal, z1 = z2 so

Pipe Flow 5

p1ρ

+v12

2=p2ρ

+v22

2


¢ The pipe has a constant diameter and the flow is constant at both sections so the velocity at each point is the same.

Pipe Flow 6

v1 = v2

p1

ρ+

v12

2=

p2

ρ+

v22

2p1

ρ=

p2

ρ

4


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

Pipe Flow 7

p1ρ

=p2ρ

⇒ p1 = p2


¢  If we ran the system experimentally and measured the two pressures, they would not be the same.

Pipe Flow 8

p1ρ

=p2ρ

⇒ p1 = p2

5


¢ The pressure at point 2 would be lower than the pressure at point 1.

Pipe Flow 9

p1ρ

=p2ρ

⇒ p1 = p2


¢ The pressure is being lost (actually the pressure energy) due to friction as the flow moves along the pipe.

Pipe Flow 10

p1ρ

=p2ρ

⇒ p1 = p2

6

Friction Losses

¢ To be able to quantify what is happening we need to look an element of flow in the pipe.

Pipe Flow 11

Friction Losses

¢ The grey element on the right is an element of the flow, a slice like a quarter.

Pipe Flow 12

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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

Pipe Flow 13

Friction Losses

¢ Then the force from the wall on the section of fluid (opposing the fluid flow) will be equal to

Pipe Flow 14

Fretarding = τwPdz = τwπDdz

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Friction Losses

¢ The pressure on the upstream (left face) of the section will produce a force accelerating the section

Pipe Flow 15


Faccelerating = pA = p πD2

4

Friction Losses

¢ As we move from the left face to the right face (dz) the change in pressure will equal dp

Pipe Flow 16



4

9

Friction Losses

¢ So this will add an additional retarding force

Pipe Flow 17

Fretarding = τwπDdz + p + dp( )πD2

4


4

Friction Losses

¢ Since both the velocity and mass flow rates as the same on both sides of the section, there is no net force on the section (conservation of momentum)

Pipe Flow 18


4


4Faccelerating − Fretarding = 0

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Friction Losses

¢ Therefore

Pipe Flow 19


4


4Faccelerating − Fretarding = 0

p πD2

4−τwπDdz − p + dp( )πD

2

4= 0

−τwπDdz −πD 2

4dp = 0

Friction Losses

¢ Rearranging

Pipe Flow 20


4dp = 0

−τwπDdz =πD 2

4dp

−τwπDπD 2

4

= dpdz

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Friction Losses

¢ So we have an expression for the rate at which the pressure changes as the flow moves downstream

Pipe Flow 21


4dp = 0

−τwπDdz =πD 2

4dp

−τwπDπD 2

4

= dpdz

Friction Losses

¢  In the simplest form

Pipe Flow 22

−τwπDπD 2

4

= dpdz

−4τwD

= dpdz

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Friction Losses

¢ Generalizing to any shape pipe we substitute the hydraulic diameter for the pipe diameter

Pipe Flow 23

−4τwDh

= dpdz

Friction Losses

¢ At this point, we can introduce another dimensionless ratio, the friction factor, f

Pipe Flow 24

−4τwDh

= dpdz

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Friction Losses

¢ The friction factor, f, is the ratio of the friction forces to the inertia forces.

Pipe Flow 25

−4τwDh

= dpdz

f =4τw12ρv 2

Friction Losses

¢ Combining the two expressions.

Pipe Flow 26

−f 12ρv 2

Dh= dpdz

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Friction Losses

¢  Isolating the differentials

Pipe Flow 27

− f ρv2

2Dhdz = dp

Friction Losses

¢ Now back to the pipe we originally considered

Pipe Flow 28

− f ρv 2

2Dhdz

0

L

∫ = dp∫

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Friction Losses

¢  Integrating

Pipe Flow 29

− f ρv 2

2Dhdz

0

L

∫ = dp∫

− fLρv2

2Dh= p2 − p1

Friction Losses

¢ Usually we break this expression into two terms

Pipe Flow 30

− fLρv2

2Dh= − ρv 2

2fLDh

= p2 − p1

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Friction Losses

¢ So we have a modified form of the Bernoulli equation that takes into account the friction losses in the system

Pipe Flow 31

p1ρ

+v12

2+ z1g −

ρv 2

2fLDh

ρ

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=p2ρ

+v22

2+ z2g

The pressure change term takes that form because we are using energy terms in this expression.

Friction Losses

¢ Reducing

Pipe Flow 32

p1ρ

+v12

2+ z1g − v

2

2fLDh

=p2ρ

+v22

2+ z2g

17

Friction Losses

¢ And rewriting the expression in terms of head

Pipe Flow 33

p1ρg

+v12

2g+ z1 −

v 2

2gfLDh

=p2ρg

+v22

2g+ z2

Friction Losses

¢ So the downstream energy (at point 2) is lower than the energy at point 1

Pipe Flow 34

p1ρg

+v12

2g+ z1 −

v 2

2gfLDh

=p2ρg

+v22

2g+ z2

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Example

¢  A 2-nominal pipe is inclined at an angle of 30° with the horizontal and conveys 0.001 m3/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.

Pipe Flow 35

Example


Pipe Flow 36

The 2-nominal refers to the pipe size. At one time, it would have been the inside diameter of the pipe in inches but that is no longer the case. Like the 2 by 4 in lumber, it is now just a reference. Table C.1 in the back of the book gives the critical dimensions for the pipe. If no schedule (has to do with wall thickness and strength of the pipe) you may assume that it is the most common, Schedule 40.

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Example


Pipe Flow 37

We can draw a sketch of the system and use the modified Bernoulli expression to answer the pressure drop.

Example


Pipe Flow 38

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.

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Example


Pipe Flow 39

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


Pipe Flow 40

Using the loss term we developed we can determine the pressure drop (in units of head) due to the friction in the pipe.

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Example


Pipe Flow 41

The substitutions and computation for the problem ate on pages 224 and 225 of the text.

Example

¢  A rectangular galvanized duct 4 in by 2 ft conveys heated air (T=98°F) to a locker room. The duct is 30 ft long, and the mass flow rate is 0.04 slug/s. Assuming ideal gas conditions and relatively constant density in the system, determine the pressure drop in the duct.

Head Loss 42

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Example


Head Loss 43

Example


Head Loss 44

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Example


Head Loss 45

We have the length of the pipe, the hydraulic diameter, the mass density of the fluid, and the velocity. What we don’t have is the friction factor for this flow. There is no simple way to calculate this. Under some flow regimes, there is a good approximation but to make the calculation we first need to calculate the Reynolds number of the flow and the relative roughness of the pipe.

Example


Head Loss 46

For most common piping materials, we can find what is known as the characteristic dimension of wall roughness which is given the symbol, ε (epsilon). Table 5.2 in the chapter gives some examples of values for ε. In this example, the conduit is galvanized so ε in feet is shown.

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Example


Head Loss 47

We now calculate a new dimensionless ratio known as the relative roughness which is the ratio of the characteristic roughness to the hydraulic diameter.

Example


Head Loss 48

We also need to calculate the Reynolds number of the flow.

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Example


Head Loss 49

If the flow had been laminar, we could have used 64/Re to calculate f. Since this flow is turbulent, we have to use the Moody Diagram.

Figure 5.15 page 229

Moody Diagram

Head Loss 50

On the right side of the diagram we locate a curve representing the relative roughness.

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Moody Diagram

Head Loss 51


On the bottom of the diagram, we locate the Reynolds number.

Moody Diagram

Head Loss 52



Move straight up from the Reynolds Number until you intersect with the curve representing the relative roughness.

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Moody Diagram

Head Loss 53



Move straight up from the Reynolds Number until you intersect with the curve representing the relative roughness.

Move from the intersection point to the left hand axis, that value is the friction factor.

Pipe Flow 54

Homework 18-1

¢ A 4-nominal riveted steel pipe 35 m long is laid horizontally and is to convey castor oil at a rate of 0.1 m3/s. Determine the pressure drop in the pipe.

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Pipe Flow 55

Homework 18-2

¢ A 12-standard, type K copper tube conveys water at a rate of 1 800 gpm. If the tube is 15 ft long, determine the pressure drop.

Pipe Flow 56

Homework 18-3

¢ A 15-ft-long annulus made of 4-nominal, schedule 40 pipe and 2-nominal, schedule 40 pipe conveys carbon disulfide at a volume flow rate of 0.3 ft3/s. Determine the pressure drop over the 15-ft length. Both pipes are made of cast iron.

head loss calculations - memphis · 2012-10-10 · head loss calculations bernoulli and pipe flow !...

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