2. Moody Chart
• Figure 8.11 – Moody chart for pipe friction
factor (Stanton diagram - The plot of the airflow
friction coefficient against the Reynolds number)
• Four zone of pipe flows:
(i) Laminar flow
(ii) Critical zone
(iii) Transition flow
(iv) Complete turbulence (fully rough pipe flow)

3. Moody Chart
• Thomas Edward Stanton (1865–1931)
• The value of f independent of Reynolds
number and relative roughness,
• Moody chart
D
e

5. Categories of most single pipe flow problems
are:-
Type Find Given Remark
Head loss
problem
hf D, Q or V, L, e, v
The simplest, can be solved
directly.
Discharge
problem
Q or V D, hf, L, e, v
Quite difficult, direct solution is not
possible, require trial & error
method, assume the initial f.
Sizing
problem
D Q, hf, L, e, v
Quite difficult, direct solution is not
possible, require trial & error
method, assume the initial f.

6. Type 1 Problem
Exercises (pg 284)
• 8.13.2 Oil (s = 0.94) with viscosity 0.0096
N.s/m2 flows in a 90 mm diameter welded
steel pipe (see Table 8.1) at 7.2 L/s. What
is the friction head loss per meter pipe?

7. • Solution
 
 
sm
D
Q
A
Q
V /13177.1
1090
102.744
23
3
2







    40001097.9
0096.0
090.013177.1100094.0
Re 3




VD
The flow is turbulent!
From Table 8.1: for welded steel pipe, e = 0.046 mm

8. 000511.0
90
046.0

D
e
From Moody chart, f  0.032
 
 
mm
gD
fV
L
hf
/02321.0
090.062.19
13177.1032.0
2
22


9. Type 2 Problem
• Example 8.4
Sebatang paip yang bergarispusat dan
panjang masing-masingnya 100 mm dan
120 m mengalirkan minyak pada jumlah
turus 5 m. Tentukan kadaralir minyak
melalui paip besi tuangan itu. Ambil
v = 1 x 10-5 m2/s.

10. • Solution
D = 100 mm = 0.1 m
L = 120 m
hf = 5 m
v = 1 x 10-5 m2/s
From Table 8.1, for cast iron pipe, e = 0.25
mm

11. 0025.0
100
25.0

D
e
Assume the flow is turbulent.
gD
fLV
hf
2
2
   
  ff
V
28592.0
120
1.062.195

  V
V
v
VD
000,10
101
1.0
Re 5


 

12. Use trial and error
Try f V (m/s) Re Obtained f Remark
0.025 1.80831 1.81 x 104 0.0305 Try again
0.0305 1.63717 1.64 x 104 0.0301 Try again
0.0301 1.64801 1.65 x 104 0.0301 Converged!
    smAVQ /01294.064801.1
4
1.0 3
2


Therefore,
This f is obtain from your trial value
This f is obtain from moody graph

13. Type 3 Problem
• Example 8.5
Tentukan garispusat paip besi galvani
yang diperlukan untuk mengalirkan air
sejauh 180 m pada kadaralir 85 L/s
dengan kehilangan 9 m. Ambil
v = 1.14 mm2/s.

14. Solution
L = 180 m
hf = 9 m
Q = 85 L/s = 0.085 m3/s
v = 1.14 mm2/s = 1.14 x 10-6 m2/s
From Table 8.1, for galvanized iron pipe, e = 0.15 mm
DD
e 00015.0

Assume the flow is turbulent.

15.  
222
10823.0085.044
DDD
Q
V 

D
D
Dv
VD 52746.94934
1014.1
10823.0
Re 62








 
52
2
2
22
2
16
22 Dg
fLQ
gDA
fLQ
gD
fLV
hf



16.   
 
f
f
hg
fLQ
D
f
01193958.0
962.19
085.018016
2
16
2
2
2
2
5


51
41248.0 fD 
Start by assuming a mid range value of f
Try f D (m) e/D Re Obtained f Remark
0.030 0.20456 0.00073 4.641 x 105 0.0188 Try again
0.0188 0.18631 0.00081 5.096 x 105 0.0195 Try again
0.0195 0.18767 0.00080 5.058 x 105 0.0195 Converged!
This f is obtain from your trial value
This f is obtain from moody graph

17. Therefore, accept
D = 0.18767 m
= 187.67427 mm