An Accurate and Computationally Efficient Explicit Friction Factor ...
FRICTION FACTOR
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FRICTION FACTOR
• A common parameter used in LAMINAR and especially in TURBULENT flow is the Fanning friction factor, f
• ‘f ’is as the ratio defined of wall shear stress to the product of the density and the velocity head
22vf
We know that, for flow of fluid in a circular tube,
Sub. the above eqn in the previous one…
For laminar flow,
by comparing above two equations,….
2
4 DPDL
LvPDf 22
2
32
D
vLP
Dvf
16
Re
16
Nf
For LAMINAR flow
General Eqn for “f”
F.L in B.Eqn• In a straight pipe…..the friction losses to be used
in B.Eqn is…..• We know,
• But all the terms in B.Eqn are having the units of J/kg…..so friction losses is written by modifying the above eqn as…
LvPDf 22
2v
DL
4fh2
f
P
lossesfriction
2v
DL
4f2
22
22b
bb
Pa
aa v
gZp
Wv
gZp
• In turbulent flow, it is not possible to predict the value of ‘f ’ theoretically
• It should be determined empirically (experimentally)
• It also depends on surface roughness of the pipe.
• To predict the value ‘f ’ …. Friction factor chart (MOODY chart) is available
FRICTION FACTOR---TURBULENT FLOW
‘f’ vs NRe
Friction factor chart (MOODY chart)
• Relative roughness factors ε/D where ‘ε’ is the roughness parameter, represents the average ht in ‘m’ for projections from the wall
• For commercial steel pipe, ε = 4.6 x 10-5m
Prob 1
• Water flows in a smooth plastic pipe of 200mm dia @ a rate of 0.1m3/s. Det the friction factor for this flow:
• Re = 6.36x105------Turbulent flow
• For smooth pipe, ε/D = 0
• From Moody’s chart…..
• f = 0.003
Prob 2
• Calculate the press. drop along 170m of 5cm dia, horizontal steel pipe thro which olive oil at 20ºC is flowing at a rate of 0.1m3/min. density 910kg/m3 & viscosity 84x10-3 Ns/m2
• We know,
• NRe = 460
• For Laminar flow, f = (16/NRe )= 0.0347
• ∆P=154.709kPa
LvPDf 22
Prob 3• Water is to flow thro 300m of horizontal pipe at a
rate of 0.06 m3/s. A head of 6m is available. What must be the pipe dia? Take f = 0.0056
• We know, • v =(Q/A)= 0.0764/D2
• and ∆P = hg• D = 0.201cm
ghD
vLfP
fLv
PD
2
2
2
2