Page 192 - Standard Handbook Of Petroleum & Natural Gas Engineering

Fluid Mechanics 177

At 104°F

Because R > 4,000 the flow is turbulent. From Table 2-9, E = 150 x and E/D = 0.0018.
Interpolating from Figure 2-21 yields f = 0.03. Now apply the Colebrook equation:

I
1
- -2lOg," -
=
0.0018 +
fi 3.7 (18,150)(0.03)"
f = 0.02991
2( 32.2)(0.0833) I
(0.02991)(100)(80.21)2 = 610 psi
P, = 2,000+(~)[(l;0-100)-

In pzping system fittings, valves, bends, etc., all cause additional pressure drops.
For such components the pressure drop can be estimated by modifying the frictional
component of Equation 2-54 with a resistance coefficient, K = fL/D, or an equivalent
length, L/D. Typical resistance coefficients are given in Table 2-12 and typical
equivalent lengths are given in Table 2-13. To correctly apply either the resistance
coefficient or the equivalent length, the flow must be turbulent.

Table 2-12
Representative Values of Resistance Coefficient K [l]

- Sudden contraction

Grodual reduction Sudden enlargement

Gradual enlargement
K K'[1 -(d/Dlz]'
(D-dVSLI 0.05l0.10 lO.ZOlOWb.40 lO.5010.80
d B 1 K' lo.irlo.zolo.~rlo.~Io.9sIi os[t.to
Exit loss=(sharpeoged,pojcctinp.Rounded 1, K.1.0
Compiled from datagivcn in"Pip Friction Manunl.'idrd.. Hydriulic instirutc.
1961.