This preview shows page 1. Sign up to view the full content.
Unformatted text preview: aminar flow,
n =1 but for turbulent flow n is between 1 and 2 and its precise value depends upon the
roughness of the pipe surface. Surface roughness promotes turbulence and the effect is shown in
the following work.
Relative surface roughness is defined as ε = k/D where k is the mean surface roughness and D
the bore diameter.
An American Engineer called Moody conducted exhaustive experiments and came up with the
Moody Chart. The chart is a plot of Cf vertically against Re horizontally for various values of ε.
In order to use this chart you must know two of the three coordinates in order to pick out the
point on the chart and hence pick out the unknown third coordinate. For smooth pipes, (the
bottom curve on the diagram), various formulae have been derived such as those by Blasius and
Lee.
BLASIUS Cf = 0.0791 Re0.25
LEE Cf = 0.0018 + 0.152 Re0.35. The Moody diagram shows that the friction coefficient reduces with Reynolds number but at a
certain point, it becomes constant. When this point is reached, the flow is said to be fully
developed turbulent flow. This point occurs at lower Reynolds numbers for rough pipes.
A formula that gives an approximate answer for any surface roughness is that given by Haaland. ⎧ 6.9 ⎛ ε ⎞1.11 ⎫
⎪
⎪
= −3.6 log10 ⎨
+⎜
⎟⎬
Cf
⎪ R e ⎝ 3.71 ⎠ ⎪
⎭
⎩
1 © D.J.DUNN 25 © D.J.DUNN Fig. 3.2 CHART
26 WORKED EXAMPLE 3.1
Determine the friction coefficient for a pipe 100 mm bore with a mean surface roughness of
0.06 mm when a fluid flows through it with a Reynolds number of 20 000.
SOLUTION
The mean surface roughness ε = k/d = 0.06/100 = 0.0006
Locate the line for ε = k/d = 0.0006.
Trace the line until it meets the vertical line at Re = 20 000. Read of the value of Cf
horizontally on the left. Answer Cf = 0.0067.Check using the formula from Haaland. ⎧ 6.9 ⎛ ε ⎞1.11 ⎫
⎪
⎪
= −3.6 log10 ⎨
+⎜
⎟⎬
Cf
⎪ R e ⎝ 3.71 ⎠ ⎪
⎩
⎭
1 1.11
⎧ 6.9
⎪
⎪
⎛ 0.0006 ⎞ ⎫
= −3.6 log10 ⎨
+⎜
⎟⎬
Cf
⎪ 20000 ⎝ 3.71 ⎠ ⎪
⎩
⎭ 1 1.11
⎧
⎫
⎪ 6.9
⎛ 0.0006 ⎞ ⎪
= −3.6 log10 ⎨
+⎜
⎟⎬
Cf
⎪ 20000 ⎝ 3.71 ⎠ ⎪
⎩
⎭
1
= 12.206
Cf 1 C f = 0.0067 WORKED EXAMPLE 3.2
Oil flows in a pipe 80 mm bore with a mean velocity of 4 m/s. The mean surface roughness
is 0.02 mm and the length is 60 m. The dynamic viscosity is 0.005 N s/m2 and the density
is 900 kg/m3. Determine the pressure loss.
SOLUTION
Re = ρud/µ = (900 x 4 x 0.08)/0.005 = 57600
ε= k/d = 0.02/80 = 0.00025
From the chart Cf = 0.0052
hf = 4CfLu2/2dg = (4 x 0.0052 x 60 x 42)/(2 x 9.81 x 0.08) = 12.72 m
∆p = ρghf = 900 x 9.81 x 12.72 = 112.32 kPa. © D.J.DUNN 27 ASSIGNMENT 3 1. A pipe is 25 km long and 80 mm bore diameter. The mean surface roughness is
0.03 mm. It carries oil of density 825 kg/m3 at a rate of 10 kg/s. The dynamic
viscosity is 0.025 N s/m2.
Determine the friction coefficient using the Moody Chart and calculate the
friction head. (Ans. 3075 m.) 2. Water flows in a pipe at 0.015 m3/s. The pi...
View
Full
Document
This document was uploaded on 02/07/2014.
 Spring '14

Click to edit the document details