DE LA SALLE UNIVERSITY-DASMARIÑAS
COLLEGE OF ENGINEERING, ARCHITECTURE, AND TECHNOLOGY
ENGINEERING DEPARTMENT
MECHANICAL ENGINEERING PROGRAM
FLUID MACHINERY
MODULE 7
BY:
ENGR. EDARDO L. ALASCO
TABLE OF CONTENTS
VII. MODULE 6 – HEAD LOSS / MOODY’S EQUATION / MOODY’S CHART
7.1 GOSPEL READING
7.2 INTRODUCTION
7.3 TOPIC LEARNING OUTCOMES
7.4 HEAD LOSS / / MOODY’S EQUATION / MOODY’S CHART
7.1 GOSPEL READING
Let us put ourselves in the presence of God..
In the name of the Father, the Son, and the Holy Spirit..
Proverbs 2:6 and 10, “For the Lord gives wisdom, and from His mouth come
knowledge and understanding. For wisdom will enter your heart, and knowledge
will be pleasant to your soul”
St. John Baptist de la Salle..pray for us.. Live Jesus in our hearts..Forever
7.2 INTRODUCTION
The head loss (or the pressure loss) represents the reduction in the total head or
pressure (sum of elevation head, velocity head and pressure head) of the fluid as it
flows through a hydraulic system. The head loss also represents the energy used in
overcoming friction caused by the walls of the pipe and other technological equipment.
The head loss is unavoidable in real moving fluids. It is present because of the friction
between adjacent fluid particles as they move relative to one another (especially in
turbulent flow).
The head loss that occurs in pipes is dependent on the flow velocity, pipe
diameter and length, and a friction factor based on the roughness of the pipe and
the Reynolds number of the flow. Although the head loss represents a loss of energy,
it does not represent a loss of total energy of the fluid. The total energy of the fluid
conserves as a consequence of the law of conservation of energy. In reality, the head
loss due to friction results in an equivalent increase in the internal energy (increase in
temperature) of the fluid.
Classification of Head Loss:
The head loss of a pipe, tube or duct system, is the same as that produced in a
straight pipe or duct whose length is equal to the pipes of the original systems plus the
sum of the equivalent lengths of all the components in the system.
As can be seen, the head loss of piping system is divided into two main categories,
“major losses” associated with energy loss per length of pipe, and “minor losses”
associated with bends, fittings, valves, etc.
Major Head Loss – due to friction in pipes and ducts.
Minor Head Loss – due to components as valves, fittings, bends and tees.
The head loss can be then expressed as:
hloss = Σ hmajor losses + Σ hminor losses
Major losses, which are associated with frictional energy loss per length of pipe
depends on the flow velocity, pipe length, pipe diameter, and a friction factor based on
the roughness of the pipe, and whether the flow is laminar or turbulent (i.e.
the Reynolds number of the flow).
Although the head loss represents a loss of energy, it does not represent a loss of
total energy of the fluid. The total energy of the fluid conserves as a consequence of
the law of conservation of energy. In reality, the head loss due to friction results in an
equivalent increase in the internal energy (increase in temperature) of the fluid.
By observation, the major head loss is roughly proportional to the square of the flow
rate in most engineering flows (fully developed, turbulent pipe flow).
The most common equation used to calculate major head losses in a tube or duct is
the Darcy–Weisbach equation (head loss form).
Minor Losses. In industry any pipe system contains different technological
elements as bends, fittings, valves or heated channels. These additional components
add to the overall head loss of the system. Such losses are generally termed minor
losses, although they often account for a major portion of the head loss. For relatively
short pipe systems, with a relatively large number of bends and fittings, minor losses
can easily exceed major losses (especially with a partially closed valve that can cause
a greater pressure loss than a long pipe, in fact when a valve is closed or nearly closed,
the minor loss is infinite).
The minor losses are commonly measured experimentally. The data, especially for
valves, are somewhat dependent upon the particular manufacturer’s design.
Generally, most of methods that are used in industry, define a coefficient K as a value
for certain technological component.
Reference:
https://www.thermal-engineering.org/what-is-head-loss-pressure-loss-definition/
7.3 COURSE LEARNING OUTCOMES:
By the end of this course, students are expected to:
TLO1. Learn the different head loss in the pipe and fittings.
TLO2. Formulate different equations for the computation of different head loss.
TLO3.Solve problems using different equations for the computations of head loss
for different pipe set up and fittings
7.4 Head Loss / Moody’s Equation / Moody’s Chart
Example 1. A pump is to draw water from the reservoir A to reservoir B, as shown in
the figure. Recommend the nominal diameter of pipes (suction and discharge) and the
size of pump motor to be used for the following specifications:
Suction pipe line:
Length of straight pipe . . . . . . . . . . . . . . . . . . . . 10 m
Pipe material . . . . . . . . . . . . . . . . . . . . . . . . . . . cast iron
Absolute roughness . . . . . . . . . . . . . . . . . . . . . . . 0.50 m
Schedule number . . . . . . . . . . . . . . . . . . . . . . . . 40
Discharge pipe line:
Length of straight pipe . . . . . . . . . . . . . . . . . . . . . 40 m
Pipe material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cast iron
Absolute roughness . . . . . . . . . . . . . . . . . . . . . . . . 0.50 mm
Schedule number . . . . . . . . . . . . . . . . . . . . . . . . . 40
Other data:
Pump capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 lps
Pump efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78%
Water viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.001002 Pa-s
𝑚 𝑚
Allowable velocity . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0 to 2.5
𝑠 𝑠
Pipe sizes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 𝐷𝑆 > 𝐷𝐷
Considering the suction pipe line:
𝑚
Solving for the pipe inside diameter, try 𝑉𝑆 = 1.0
𝑠
From Q = A V
Q = 𝐴𝑆 𝑉𝑆
𝛱
Q= 𝐷𝑆 2 𝑉𝑆
4
𝑚3
4 (0.0.40) 0.03937 𝑖𝑛
𝐷𝑆 = √ 𝛱 ( 1.0 𝑚
𝑠
)
= 0.2257 m = 225.7 mm x
𝑚𝑚
= 8.88 in
𝑠
For schedule 40, with ID = 8.81 in,
use: 10 in nominal diameter
1 𝑚𝑚
ID = 10.020 in x 0.03937 𝑖𝑛 = 254.51 mm = 0.25451 m
To check the actual suction velocity,
𝑚3
𝑄 0.040 𝑚
𝑠
V= = 𝛱 2
= 0.786
𝐴 ( 0.25451) 𝑚2 𝑠
4
𝑚
Since, it is less than 1.0 , use the lower nominal size, 8 in
𝑠
use: 8 in nominal diameter
1 𝑚𝑚
ID = 7.981 in x 0.03937 𝑖𝑛 = 202.72 mm = 0.20272 m
To check the actual suction velocity,
𝑚3
𝑄 0.040 𝑠 𝑚
V= = 𝛱 2
= 1.24 → ok (1.0 < 1.24 > 2.5)
𝐴 ( 0.20272) 𝑚2 𝑠
4
hence, the recommended nominal size diameter is 8 in with
inside diameter (ID) = 202.72 mm, schedule 40, cast iron pipe.
Considering the discharge pipe line:
𝑚
Solving for the pipe inside diameter, try 𝑉𝐷 = 2.5 𝑠
From Q = A V
Q = 𝐴𝐷 𝑉𝐷
𝛱
Q= 𝐷𝐷 2 𝑉𝐷
4
𝑚3
4 (0.0.40) 0.03937 𝑖𝑛
𝐷𝑆 = √ 𝛱 ( 2.5 𝑚
𝑠
= 0.143 m = 143 mm x = 5.6299 in
𝑠
) 𝑚𝑚
For schedule 40, with ID = 5.6299 in,
use: 6 in nominal diameter
1 𝑚𝑚
ID = 6.065 in x 0.03937 𝑖𝑛 = 154.05 mm = 0.15405 m
To check the actual suction velocity,
𝑚3
𝑄 0.040 𝑠 𝑚
V= = 𝛱 2
= 2.147 → ok (1.0 < 2.147 > 2.5)
𝐴 ( 0.15405) 𝑚2 𝑠
4
hence, the recommended nominal size diameter is 6 in with
inside diameter (ID) = 154.05 mm, schedule 40, cast iron pipe.
For pump motor size,
𝑃
𝑛𝑃 = 𝑃 x 100%
𝑖𝑛
𝑃
𝑃𝑖𝑛 = 𝑛
𝑃
and P = Q ɤ H
𝐻 = [ 𝑍𝐵 - ( -𝑍𝐴 ) ] + 𝐻𝐿 = [ 𝑍𝐵 + 𝑍𝐴 ) ] + 𝐻𝐴−𝐵
𝐻𝐴−𝐵 = ℎ𝑓𝑠 + ℎ𝑓𝐷
𝐿𝑇𝑠 𝑉𝑠2
ℎ𝑓𝑠 = 𝑓𝑠
2 𝐷𝑠 𝑔
1
Ɛ 106
𝑓𝑠 = 0.0055 [ 1 + (20,000 + )3 ]
𝐷 𝑁𝑅
𝑁𝑠 𝑘𝑔
Water viscosity = 0.001002 Pa-s = 0.001002 = 0.001002
𝑚2 𝑚𝑠
𝑚 𝑘𝑔
𝑉𝐷𝜌 ( 1.24 𝑠 )(0.20272 m )(1,000 )
𝑚3
and 𝑁𝑅 = = 𝑘𝑔 = 0.25087 x 106
µ 0.001002
𝑚𝑠
1
0.50 106
𝑓𝑠 = 0.0055 [ 1 + (20,000 + ) ] = 0.0256
3
202.72 0.25087 x 106
solving for equivalent length of fittings and valves:
for 8 in nominal diameter:
1𝑚
standard elbow = 20 ft x 3.28 𝑓𝑡 = 6.1 m
1𝑚
gate valve (fully open) = 4.6 ft x 3.28 𝑓𝑡 = 1.4024 m
Total = 7.5024 m
𝐿 𝑇𝑠 , total length of suction pipe = 10 + 7.5024 = 17.5024 m
𝑚2
( 17.5024 𝑚) ( 1.24)2
𝑠2
ℎ𝑓𝑠 = (0.0256) 𝑚 = 0.1732 m
2 (0.20272 𝑚)( 9.81) 2
𝑠
for discharge pipe friction head loss, ℎ𝑓𝐷
𝐿𝑇𝐷 𝑉𝐷2
ℎ𝑓𝐷 = 𝑓𝐷
2 𝐷𝐷 𝑔
from, Moody’s equation for 𝑓𝐷 :
1
Ɛ 106
𝑓𝐷 = 0.0055 [ 1 + (20,000 𝐷 + 𝑁 )3 ]
𝑅
𝑚 𝑘𝑔
𝑉𝐷𝜌 ( 2.147 𝑠 )(0.15405 m )(1,000 )
𝑚3
and 𝑁𝑅 = = = 𝑘𝑔 = 0.330 x 106
µ 0.001002
𝑚𝑠
1
0.50 106
𝑓𝐷 = 0.0055 [ 1 + (20,000 154.05 + 0.330 x 106 ) ] = 0.02763
solving for equivalent length of fittings and valves:
for 6 in nominal diameter:
1𝑚
1 check valve (same as close return bend) = 35 ft x 3.28 𝑓𝑡 = 10.67 m
1𝑚
1 gate valve (fully open) = 3.5 ft x 3.28 𝑓𝑡 = 1.067 m
1𝑚
2 standard elbow = 2 x 16 ft = 32 x 3.28 𝑓𝑡 = 9.76 m
Total = 21.497 m
𝐿 𝑇𝐷 , total length of suction pipe = 40 + 21.497 = 61.497 m
𝑚2
( 61.497 𝑚) ( 2.147)2
𝑠2
ℎ𝑓𝐷 = (0.0276) 𝑚 = 2.59 m
2 (0.15405 𝑚)( 9.81) 2
𝑠
𝐻𝐴−𝐵 = ℎ𝑓𝑠 + ℎ𝑓𝐷 = 0.1732 + 2.59 = 2.7632 m
𝐻 = [ 𝑍𝐵 + 𝑍𝐴 ) ] + 𝐻𝐴−𝐵 = 12 + 3.5 + 2.7632 = 18.26 m
𝑚3 𝐾𝑁 𝐾𝑁 𝑚 𝐾𝐽
P = Q ɤ H = 0.040 (9.81 3 ) (18.26 m) = 7.165 = 7.165 = 7.165 KW
𝑠 𝑚 𝑠 𝑠
7.165
𝑃𝑖𝑛 = = 9.19 KW Ans.
0.78
𝑙𝑖
Example 2. What is the power of the pump required to draw 60 of water from a
𝑠
reservoir A to reservoir B, as shown in the figure? Assume 80% over-all efficiency of the
pump and average absolute roughness of cast iron of 0.50 mm. The absolute viscosity of
water is 0.001002 Pa s.
Suction line Discharge line
Straight pipe 20 m 60 m
Nominal diameter 8 in 6 in
Schedule number 80 40
Pipe material Cast iron Cast iron
Absolute roughness 0.50 mm 0.50 mm
Water viscosity 0.001002 Pa s 0.001002 Pa s
𝑙𝑖
Q = 60
𝑠
Pump efficiency = 80%
Solution:
Considering the suction line:
Pipe Table Mechanical Engineering Tables and Charts, MRII ( Manila Review Institute Inc.) p. 121-122
For 8 in nominal diameter, the inside diameter, ID = 193.7 mm
Solving for
𝑚3
𝑄 0.060 𝑠 𝑚
𝑉𝑠 = = 𝛱 2
= 2.037
𝐴𝑠 ( 0.1937) 𝑚2 𝑠
4
𝑚 𝑘𝑔
𝑉𝐷𝜌 ( 2.037 𝑠 )(0.1937 m )(1,000 )
𝑚3
and 𝑁𝑅 = = = 𝑘𝑔 = 3.94 x 105
µ 0.001002
𝑚𝑠
Ɛ 0.0005
and relative roughness, 𝑒𝑠 = = = 0.00258 or 0.003
𝐷𝑠 0.1937
solving for the value of friction factor (coefficient of friction), 𝑓𝑠 using Moody Chart,
Checking for the value of friction factor, 𝑓𝑠 , using Moody’s equation:
1
0.50 106
𝑓𝑠 = 0.0055 [ 1 + (20,000 193.7 + 3.94 x 105 ) ] = 0.0263
3
Solving for equivalent length of fittings and valves:
for 8 in nominal diameter:
1𝑚
standard elbow = 20 ft x 3.28 𝑓𝑡 = 6.1 m
1𝑚
gate valve (fully open) = 4.6 ft x 3.28 𝑓𝑡 = 1.4024 m
Total = 7.5024 m
𝐿 𝑇𝑠 , total length of suction pipe = 20 + 7.5024 = 27.5024 m
𝑚2
( 27.5024 𝑚) ( 2.037)2
𝑠2
ℎ𝑓𝑠 = (0.0263) 𝑚 = 0.7897 m
2 (0.1937 𝑚)( 9.81) 2
𝑠
Considering the discharge line:
For 6 in nominal diameter, the inside diameter, ID = 154.1 mm
Solving for
𝑚3
𝑄 0.060 𝑠 𝑚
𝑉𝐷 = = 𝛱 2
= 3.22
𝐴𝑠 ( 0.1541) 𝑚2 𝑠
4
𝑚 𝑘𝑔
𝑉𝐷𝜌 ( 3.22 𝑠 )(0.1541 m )(1,000 )
𝑚3
and 𝑁𝑅 = = = 𝑘𝑔 = 4.95 x 105
µ 0.001002
𝑚𝑠
Ɛ 0.0005
and relative roughness, 𝑒𝐷 = = = 0.00324
𝐷𝐷 0.1541
solving for the value of friction factor (coefficient of friction), 𝑓𝑠 using Moody Chart
Checking for the value of friction factor, 𝑓𝐷 , using Moody’s equation:
1
0.50 106
𝑓𝐷 = 0.0055 [ 1 + (20,000 154.1 + 4.95 x 105 )3 ] = 0.0278
solving for equivalent length of fittings and valves:
for 6 in nominal diameter:
1𝑚
1 check valve (same as close return bend) = 35 ft x 3.28 𝑓𝑡 = 10.67 m
1𝑚
1 gate valve (fully open) = 3.5 ft x 3.28 𝑓𝑡 = 1.067 m
1𝑚
2 standard elbow = 2 x 16 ft = 32 x 3.28 𝑓𝑡 = 9.76 m
Total = 21.497 m
𝐿 𝑇𝐷 , total length of suction pipe = 60 + 21.497 = 81.497 m
𝑚2
( 81.497 𝑚) ( 3.22)2
𝑠2
ℎ𝑓𝐷 = (0.0278) 𝑚 = 7.77 m
2 (0.1541 𝑚)( 9.81) 2
𝑠
𝐻𝐴−𝐵 = ℎ𝑓𝑠 + ℎ𝑓𝐷 =0.7897 + 7.77 = 8.56 m
𝐻 = [ 𝑍𝐵 + 𝑍𝐴 ) ] + 𝐻𝐴−𝐵 = 20 + 4.5 + 8.56 = 33.06 m
𝑚3 𝐾𝑁 𝐾𝑁 𝑚 𝐾𝐽
P = Q ɤ H = 0.060 (9.81 ) (33.06 m) = 19.46 = 19.46 = 19.46 KW
𝑠 𝑚3 𝑠 𝑠
19.46
𝑃𝑖𝑛 = = 24.23 KW Ans.
0.80
REFERENCES
Online References
Call Reference Material
number or
e-provider
You tube
https://www.youtube.com/watch?v=Pfk-6fgx-vM
Channel
You tube
https://www.youtube.com/watch?v=Okl1GDAuWgs
Channel
You tube
https://www.youtube.com/watch?v=XGnGBo-FrlA
Channel
You tube
https://www.youtube.com/watch?v=clVwKynHpB0
Channel
On-Site References
Call Reference Material
number or
e-provider
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357.F673 Fluid Mechanics And Hydraulic Machines, Authored and edited by 3G E-learning LLC, USA,
2017 [Place of publication not indicated] : 3G E-learning LLC, c2017.
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357.So51 S. K. Som, Gautam Biswas, Suman Chakraborty, Tata McGraw Hill, New Delhi, India 2012
2012
Call # Fluid Power Dynamics,R. Keith Mobley.Boston : Newnes, ©2000
620.1/06
Call # Fluid mechanics and machineries, Samuel R. Limonera, Archie B. Maglaya. De La Salle
TA 357 L629 University Press, c1985
1985
Call # Pumps: Principles and Practice, Jaico Pub. House, 2004.
TJ 900.P969
2003
Call # Schaum's Outline of Theory and Problems of Fluid Mechanics and Hydraulics. Ranald V. Giles
TA 357.
G393 1994 New York: McGraw-Hill, c1994.
