Other Friction Losses

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Other Friction Losses

• Valves and Fittings

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Goals

• Calculate frictional losses in a system
  containing valves, fittings, and sudden
  expansions and contractions
• Express frictional losses in terms of velocity
  head
• Assess relative contributions of different
  sources to total viscous dissipation

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Sudden Expansion
Frictional losses occur as result of turbulence
generated immediately downstream of the expansion
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Sudden Expansion
Assume
Ke is the expansion loss coefficient which we
will attempt to describe in terms of flow
properties.
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Sudden ExpansionMass Balance
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Sudden ExpansionMomentum Balance
Assume turbulent b1 b2 1
0
0
Replaced Sa with Sb because pa is at the point
of expansion.
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Momentum Balance
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Mechanical Energy Balance
Assume turbulent a1 a2 1
0
0
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Combining
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Final Result
Recall Mass Balance Result

• Notes
• Velocity head is based on smaller cross section
• What if flow becomes laminar in large pipe?

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For Tank Filling
Sb
Sa
Va
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Sudden Contractions
At sudden contractions, flow streamlines converge
causing the downstream developed flow to have an
area smaller than the downstream pipe diameter.
This flow constriction is called the vena
contracta. Viscous dissipation occurs in the
vortices developed in this area.
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Sudden Contraction
Development of an expression for sudden
contraction proceeds in much the same way as that
for sudden expansion with the definition of a
contraction coefficient.
For laminar flow experimentally, Kc lt 0.1 and hfc
is usually neglected
Turbulent (empirical)
Note Calculations again based on small cross
section.
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Tank Emptying
Sa
Sb
Vb
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Velocity Heads
The above expression shows that friction loss in
a complicated flow system can be expressed as a
number of velocity heads. It is a measure of
momentum loss resulting from flow through the
system. For instance in making a 90 turn all
x-momentum is turned into y-momentum.
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Alternate Method
The previous equation can be manipulated to
change the Kf values into equivalent lengths of
pipe (see attached table) of diameter D. When
this method is used the equivalent lengths are
add to the length of the actual pipe sections and
the equation becomes.
Note The values in the table are L/D and must
be multiplied by D to get equivalent lengths.
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Example
Water is pumped at 250 gpm from tank 1 to tank 2
as shown. Calculate the required power input to
the pump assuming a pump efficiency of 70.
Pa 0 psig
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