Minor Losses - University of Memphis · 3 Minor Losses ! Rather than develop a detailed energy loss...

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Civil Engineering Hydraulics

Minor Losses

Minor Losses

¢ Up to this point, all we have considered is pipes conveying flows

¢ The pipes have been a constant cross section and have not had to go around anything, turn any corners, or really change flow direction

Minor Losses 2 Wednesday, October 17, 2012

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Minor Losses

¢  In practice, that usually isn’t so therefore we have to consider the energy losses that these changes in direction cost

¢  In addition to changes in direction, the volumetric flow rate is often controlled by adjusting some sort of flow control device, most commonly a value.


Minor Losses

¢ Any sort of change in the pattern of the flow (the velocity profile) will result in a change (decrease) of the energy of the flow.

¢ Think of it as taking a pack of cards and shuffling it and then having to rearrange the deck.


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Minor Losses

¢ Rather than develop a detailed energy loss description for each type of valve and fitting, there is a characteristic value that we use.


Minor Losses

¢ The pressure drop across or through a fitting or valve is given by

Minor Losses 6

Δp = −K ρv2

2

Wednesday, October 17, 2012

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Minor Losses

¢ The v is the average velocity at the entrance to the fitting or valve

¢ K is the loss coefficient for the fitting or valve

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Δp = −K ρv2

2


Minor Losses

¢ K values for common piping elements are available from most manufacturers as well as in general tables

¢ You may also see the symbol ξ used in place of K

Minor Losses 8

Δp = −K ρv2

2


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Minor Losses

¢ The text has K values starting with Table 5.4

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Δp = −K ρv2

2


Minor Losses

¢ Probably the best way to look at the use of these is to work through some example problems.

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Δp = −K ρv2

2


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Minor Losses

Minor Losses 11

Δp = −K ρv2

2


Minor Losses

Minor Losses 12

Δp = −K ρv2

2


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Minor Losses

Minor Losses 13

Δp = −K ρv2

2


Minor Losses

Minor Losses 14

Δp = −K ρv2

2


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Minor Losses

Minor Losses 15

Δp = −K ρv2

2


Minor Losses

¢ Now we can write the Bernoulli equation and include the loss terms between points 1 and 2

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p1

ρg+

v12

2g+ z1+

fitting lossesρg

+ friction losses

ρg =

p2

ρg+

v22

2g+ z2

Remember that the losses have a negative sign included in their calculationThe losses are divided by rho g because they are calculated in termsof pressure.


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Minor Losses

¢ And rearranging to isolate the pressure change

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p1 − p2= +ρv2

2

2+ ρgz2 −

ρv12

2− ρgz1

− fitting losses( ) − friction losses( )


Minor Losses

¢ Substituting what we know already

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p1 − p2= 870

kgm3

⎛⎝⎜

⎞⎠⎟

1.697ms

⎛⎝⎜

⎞⎠⎟

2

2+ 870

kgm3

⎛⎝⎜

⎞⎠⎟

9.81ms2

⎛⎝⎜

⎞⎠⎟

2.7m( )

−870

kgm3

⎛⎝⎜

⎞⎠⎟

0.763ms

⎛⎝⎜

⎞⎠⎟

2

2− 870

kgm3

⎛⎝⎜

⎞⎠⎟

9.81ms2

⎛⎝⎜

⎞⎠⎟

3.0m( ) - fitting losses( ) - friction losses( )


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Minor Losses

¢ Substituting what we know already

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p1 − p2= 24.29kPa − 25.85kPa

- − K ρv 2

2∑⎛⎝⎜

⎞⎠⎟

- −f12L12

Dh12

v122

2g−

f8L8

Dh8

v82

2g

⎛

⎝⎜

⎞

⎠⎟


Minor Losses

¢ Moving from left to right we have a gate valve, a 90 degree elbow, a 90 degree elbow, a 90 degree elbow, a 90 degree elbow, a reducing connection, a 45 degree elbow, and a 45 degree elbow

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p1 − p2= 24.29kPa − 25.85kPa

- − K ρv 2

2∑⎛⎝⎜

⎞⎠⎟

- −f12L12

Dh12

v122

2g−

f8L8

Dh8

v82

2g

⎛

⎝⎜

⎞

⎠⎟


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Minor Losses

¢ The problem states that all the elbows are standard and flanged so we can find the K values from Table 5-4

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K90 = 0.31K45 = 0.35Kgate = 0.15 (assume that the gate valve is open)Kcontraction = 0.3


Minor Losses

¢ Take care to use the velocity in the section of the pipe that the fitting is in

¢ You may combine the K values for fittings in the same pipe (if the velocity is constant)

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K90 = 0.31K45 = 0.35Kgate = 0.15 (assume that the gate valve is open)Kcontraction = 0.3


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Minor Losses

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p1 − p2= 24.29kPa − 25.85kPa

- − K ρv 2

2∑⎛⎝⎜

⎞⎠⎟

- −f12L12

Dh12

v122

2g−

f8L8

Dh8

v82

2g

⎛

⎝⎜

⎞

⎠⎟


Minor Losses

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p1 − p2= 24.29kPa − 25.85kPa

- −1.15kPa( )

- −f12L12

Dh12

v122

2g−

f8L8

Dh8

v82

2g

⎛

⎝⎜

⎞

⎠⎟


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Minor Losses


Minor Losses

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p1 − p2= 24.29kPa − 25.85kPa

- −1.15kPa( ) - −3.39kPa( )


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Minor Losses


Example 5.9

¢  A water tank is fitted with a drain and outlet pipe as sketched in Figure 5.24. The system has 82 ft of cast iron pipe of 1 1/2-nominal diameter. Determine the flow rate through the pipe. All fittings are threaded and regular.

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Homework 19-1 Water at 15°C is drained from a large reservoir using two horizontal plastic pipes connected in series. The first pipe is 20 m long and has a 10-cm diameter, while the second pipe is 35 m long and has a 4-cm diameter. The water level in the reservoir is 18 m above the centerline of the pipe. The pipe entrance is sharp-edged, and the contraction between the two pipes is sudden. Determine the discharge rate of water from the reservoir.


Homework 19-2 ¢  Water at 70°F flows by gravity from a large

reservoir at a high elevation to a smaller one through a 120-ft-long, 2-in-diameter cast iron piping system that includes four standard flanged elbows, a well-rounded entrance, a sharp-edged exit, and a fully open gate valve. Taking the free surface of the lower reservoir as the reference level, determine the elevation z1 of the higher reservoir for a flow rate of 10 ft3/min.


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Homework 19-3


Minor Losses - University of Memphis · 3 Minor Losses ! Rather than develop a detailed energy loss...

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