Prepared by: Melvin R.

Esguerra Review Notes in Hydraulics Page 1

PIPES



A pipe is a closed conduit through which fluids flow. In hydraulics, pipes are generally understood to be
conduits of circular cross section which flow full. Conduits flowing partially full are considered open channels.

Head Losses in Pipes
There are generally two types of head losses in pipes:
1. Major loss, which is commonly referred to as the loss of head due to pipe friction
Darcy - Weisbach Formula
2 2
5
0.0826
2
f
L v fLQ
h f
D g
D
= =
Manning Formula
2 2 2 2
4 16
3 3
6.35 10.29
f
n Lv n LQ
h
D D
= =
Where
f
h = loss of head due to pipe friction
f = friction factor
L = length of pipe
D = diameter of pipe
v = velocity of the stream
Q = discharge
n = roughness of coefficient

The friction factor f and the roughness coefficient n may be related by the equation.
1
6
8 1 g
R
f n
=

2. Minor losses which consist of the following:
Loss, of head due Lo contraction of the cross section of the stream
2
2
c c
v
h k
g
| |
=
|
\ .

Loss of head due to enlargement of the cross section of the stream
2
2
e e
v
h k
g
| |
=
|
\ .

Loss of head due to obstructions such as gates or valves
2
2
o o
v
h k
g
| |
=
|
\ .

Loss of head due to bends in the pipes
Prepared by: Melvin R. Esguerra Review Notes in Hydraulics Page 2

2
2
b b
v
h k
g
| |
=
|
\ .


Hydraulic and Energy Gradients
Hydraulic gradient is a graphical representation with respect to any selected datum, of the potential
(pressure + elevation) head or energy which the liquid possesses at ails sections of the pipe.
Energy gradient is a graphical representation with respect to any selected datum, of the total head or energy
possessed by the liquid. It is then above the hydraulic gradient a distance equal to the velocity head at each
section.


SAMPLE PROBLEMS:
1. A pump at elevation 380 m is pumping 70 L/s through 2300 m of 200-mm pipe to a reservoir whose level is
at elevation 450 m. Find the pressure at a point 1280 m from the pump (measured along the pipe) where the
elevation is 400 m above the datum. Assume f = 0.0225.



2. Two water reservoirs A and B are 8 km apart and are connected by a750-mm diameter steel pipe (f = 0.02).
Reservoir A is 10 m higher than reservoir B. If the pressure at B is 4 kg/cm
2
greater than that at A,
determine the amount and direction of the flow of water inside the pipe. Apply Darcy Weisbach formula for
major head loss and disregard other minor losses.
Prepared by: Melvin R. Esguerra Review Notes in Hydraulics Page 3



3. From a reservoir whose surface elevation is 30 m, water is pumped to another reservoir whose surface
elevation is 95 m. The length of a 600 mm diameter suction pipe is 1500 m and that of a 500 mm diameter
discharge pipe is 1000 m. Find the required horsepower output the pump if the discharge is to be maintained
at 0.48 m
3
/s. Use f = 0.02 and assume minor losses to be 10% of the major losses.

Pipes in Series
Discharge and Frictional Loss Equations
1.
1 2 3
Q Q Q = =

2.
1 2 3
HL hf hf hf = + +

SAMPLE PROBLEMS:
1. Three cast-iron pipes are connected in series as shown in the figure.
a. Determine the total lost head when the discharge is 0.1 m
3
/s.
b. What must be the elevation of the outlet end of the pipe with respect to the water surface elevation in the
reservoir?
c. Draw the hydraulic and energy gradients.
Assume n = 0.011, a coefficient of contraction in the reservoir of
c
K = 0.5 and in the pipes of
c
K = 0.1.



2. Three pipes of different diameters and lengths discharge 160 liters per second. If the coefficient of friction is
0.012 and disregarding other minor losses
a. Determine the head loss in each pipe
b. Determine the diameter of an equivalent

Prepared by: Melvin R. Esguerra Review Notes in Hydraulics Page 4




Pipes in Parallel
Discharge and frictional loss equations:
1.
1 5
Q Q =

2.
1 2 3 4
Q Q Q Q = + +
3.
2 3 4
hf hf hf = =
4.
1 3 5
HL hf hf hf = + +

SAMPLE PROBLEMS:
1. A 900-mm main carrying 1.20 m
3
/s branches into two pipelines between points x and y, one 1200 m long
with a diameter of 450 mm and the other 750 m long with a diameter 600 mm. Both pipes come together at
point y and continue as a single 900 mm pipe. Assuming values of f as 0.022, 0.033 and 0.024 for 900 mm,
600 mm, and 450 mm pipes, respectively, determine the rate of flow in the branch pipes.



2. The discharge in the pipe system shown is 0. 85 m
3
/s. Determine the diameter of pipe 4 when the total head
loss is 10 m. The coefficient of friction f in each pipe is 0.02.

Pipes in series and Parallel
Discharge and frictional loss equations

Case I
1 5
Q Q =

1 2 3
Q Q Q = +
3 4
Q Q =
2 3 4
hf hf hf = +
1 2 5
HL hf hf hf = + +




Prepared by: Melvin R. Esguerra Review Notes in Hydraulics Page 5

Case 2
1 6
Q Q =

1 2 4 5
Q Q Q Q = + +
3 2 5
Q Q Q = +
6 3 4
Q Q Q = +
2 5
hf hf =
4 2 3
hf hf hf = +
1 4 6
HL hf hf hf = + +

SAMPLE PROBLEMS:
1. If the total loss of head due to friction from 1 to 5 is 4 m, what is the discharge and head loss in each
section? Use f = 0.02.

2. The flow from a to e through the pipe system shown is 300 L/s. If n = 0.011 for all the pipes, determine the
loss of head between a to e.



Pipes Connecting Three Reservoirs
Discharge and Frictional Loss Equations
Case I

1 2 3
Q Q Q = +

1 2
. . hf hf EL A EL B + =
1 3
. . C or hf hf EL A EL + =
3 2
. . C hf hf EL B EL =


Prepared by: Melvin R. Esguerra Review Notes in Hydraulics Page 6





Case 2
1 2 3
Q Q Q + =

1 3
. . C hf hf EL A EL + =
2 3
. B . C or hf hf EL EL + =
1 2
. A . B hf hf EL EL =

SAMPLE PROBLEMS:
1. In the figure shown, the discharge in branch 1 is 0.20 m
3
/s. Using f = 0.02, find the diameter of branch 3.


2. A 1200 mm concrete pipe 1830 m long caries 1.40 m
3
/s from reservoir A, discharging into two concrete
pipes each 1370 m long and 750 mm in diameter. One of the 750 mm pipes discharges into reservoir B, in
which the water surface is 6 m lower than that in A. Determine the elevation of the water surface in
reservoir C, into which the other 750 mm discharges. Use n = 0.011.



Prepared by: Melvin R. Esguerra Review Notes in Hydraulics Page 7

Pipe Network
A pipe network is a pipe line arrangement consisting of many loops and branches.
Problems involving networks are solved by trial and error. A systematic approach developed by Hardy
Cross helps to solve these problems with fewer trials but with reasonable accuracy. It is a method of successive
approximations by which the distribution of flow can be determined.

Brief summary of the method
1. Assume the flow in all pipes, remembering that total inflow equals total outflow at each junction.
2. Compute the frictional losses hf in all the pipes in each loop
3. Compute the flow correction Q A in each loop using the formula
chf cchf
Q
hf hf
n c cc
Q Q

A =
| |
+
|
\ .



Where: , c cc

= summation of quantities in the clockwise and counterclockwise directions, respectively
hf = frictional loss in the pipe
Q = amount of flow in the pipe
n = the exponent of Q in the hf formula
= 2 if Darcy-Weisbach or Manning Formula is used
The sign of Q A indicates direction. It is counterclockwise when positive and clockwise when negative .
4. If in any loop the clockwise losses exceed the counterclockwise losses, the algebraic sign of their
difference is positive, and the clockwise flow must be reduced by an amount Q A and the
counterclockwise flow increased by the same amount. Pipes common to two loops will have two
corrections.
5. A second computation using the corrected flow is then made and then the process is repeated until the
corrections become negligible.

SAMPLE PROBLEMS:
1. Using the Manning Formula, assuming n = 0.011, compute the flow in each pipe of the network shown.



Prepared by: Melvin R. Esguerra Review Notes in Hydraulics Page 8

Water Hammer
Any change in velocity of flow in a pipe fine causes change in momentum of the fluid flowing and
consequently produces pressure variation. This pressure variation does not occur throughout the pipe line
instantaneously due to the elasticity of fluid and the pipe walls; but propagates in the form of pressure waves.
Such pressure surges with alternating pressures are called water hammer. This phenomenon could occur in all
fluids, the magnitude of pressure change of which would depend upon whether the velocity change is sudden or
gradual.

Pressure Rise
a. Due to Sudden Closure of the valve (when
2L
T
c
< )
p vc A =
b. Due to Gradual Closure of the Valve (when
2L
T
c
>
v
p L
T
A =
Where:
p A = pressure rise (Pa)
= mass density of liquid
= 1000 for water
v = velocity of liquid in the pipe at the time of closure (m/s)
c = velocity of sound waves in the liquid sometimes called celerity (m/s)
=
K

(for rigid pipes such as Steel)

=
1
K
KD
Et

| |
+
|
\ .
(for non-rigid pipes such as cast iron
K = bulk modulus of the liquid (Pa)
E = modulus of elasticity of the pipe (Pa)
D = diameter of pipe (mm)
t = thickness of pipe (mm
L = length of pipe (m)
T = time of closure of valve to reduce the velocity from v to zero (s)

Problems:
1. Compute the rise in pressure in a rigid pipe line of 200 m length when water flowing at a velocity of 3
m/s is closed
a. Suddenly
b. in 2 s
Bulk modulus of water is 220 x 10
6
kg/m
2

2. Water flowing in a long pipe in a long pipe is suddenly stopped by closing a valve at the discharge end.
The diameter of the pipe is 150 mm and its thickness is 6 mm. The quantity of water flowing in the pipe
before the valve is closed is 22 liters per second. If the modulus of elasticity of the pipe is 2.1 x 10
6
kg/cm
2
and the bulk modulus of water is 2.1 x 10
4
kg/cm
2
, determine the rise in pressure in the pipe due
to sudden closure of the valve.
Prepared by: Melvin R. Esguerra Review Notes in Hydraulics Page 9

3. A cast-iron pipe of' 150 mm internal diameter has 15 mm wall thickness. If a sudden stoppage of water
flowing through the pipe is not to increase the stress in the pipe more than 200 kg/cm
2
, calculate the rate
of flow through the pipe. E = 132 x 10
8
kg/m
2
, K = 220 x l0
6
kg/m
2
.

Orifices
An orifice is an opening wi.th a closed perimeter through which a fluid flows. Its purpose is the measurement or
control of the flow.



The energy equation from 1 to 2 is
2 2
1 1 2 2
1 2
2 2
v p v p
z z HL
g g
+ + = + + +
Neglecting HL, the velocity of flow
2
v through the orifice becomes theoretical. Thus,
2
1 2 2
1 2
2
1 2
2
0 0 0
2
2
taking total head producing flow,
2
p v p
h
g
p p
v g h
p p
H h
v gH



+ + = + + +
| |
= +
|
\ .
= + =
=


The discharge through the orifice is
Q va =
Where:
a = area of the vena contracta =
c
C A
A = area of the orifice
c
C = coefficient of contraction
v = actual velocity of flow through the orifice
Prepared by: Melvin R. Esguerra Review Notes in Hydraulics Page 10

=
2 v
C v
v
C = coefficient of velocity
C = coefficient of discharge
=
c v
C C
Hence,
( )
2 2
2
c v
Q aV C A C v CcCvAv CA gH = = = =
Problems:
1. An orifice of 50 mm diameter in tank A discharge into tank B as shown in the figure. If the vacuum
gauge reading is 75 kPa below atmosphere, find the rate of flow. C = 0.60.

2. The closed tank shown in the figure has an orifice in the side 50 mm square with C = 0.60. An open
mercury manometer indicates the pressure in the air at the top of the tank. Compute the discharge
a. The liquid in the tank is water; and
b. The upper 3 m of liquid in the tank is oil (rd = 0.82) and the remainder is water.

Flow under Atmospheric Pressure
2 2
1 1 2 2
1 2
2
2
2
2
2 2
0 0 0 0
2
2
2
v p v p
z z
g g
v
h
g
v CA gh
Q CAv CA gh

+ + = + +
+ + = + +
=
= =


Prepared by: Melvin R. Esguerra Review Notes in Hydraulics Page 11


Problems:
1. A right cylindrical tank 6 m in diameter and 12 m high is filled with. 6 m of oil (rd =0.80) and 6 m of
water. A horizontal circular orifice 75 mm in diameter is at the bottom of the tank. If C
c
= 0.61 and Cv=
0.98, determine
a. The velocity of the jet issuing form the orifice at the instant the surface of the liquid in the tank is 6.5
m above the orifice; and
b. The discharge through the orifice at the instant the surface of the liquid is 5.5 m above the orifice.

Orifice at End of Pipe
1 2
1 1 2 2
2 2
1 1 2 2
1 2
2 2
1 1 2
2
1 1
2
2 2
2 2
0 0 0
2 2
2
2
Q Q
Av CA v
v p v p
z z
g g
v p v
g g
v p
v g
g
Q CA v

=
=
+ + = + +
+ + = + +
| |
= +
|
|
\ .
=



Problem:
1. A 250-mm diameter pipe discharges oil (rd = 0.82) through a 100 mm diameter orifice with C
c
= 1.0 and
C
v =
0.96 as shown in the figure. The liquid in the U-tube is mercury. Determine the discharge

Orifice Inside a Pipe
1 2
1 1 2 2
2 2
1 1 2 2
1 2
2 2
1 1 2 2
2
1 1 2
2
2 2
2 2
0 0
2 2
2
2
Q Q
Av CA v
v p v p
z z
g g
v p v p
g g
v p p
v g
g
Q CA v



=
=
+ + = + +
+ + = + +
| |
= +
|
|
\ .
=



Problem:
1. The pipe orifice shown in the figure has a diameter of 100 mm with C = 0.65. The diameter of the pipe
is 250 mm- The liquid flowing in the pipe is oil (rd = 0.90). The differential manometer attached to the
pipe is partly filled with mercury. When the mercury level difference is 50 mm compute the discharge.
Prepared by: Melvin R. Esguerra Review Notes in Hydraulics Page 12


A 250-mm diameter pipe discharges oil (rd = 0.82) through a 100 mm diameter orifice with C
c
= 1.0 and C
v
=
0.96 as shown in the figure. The liquid in the U-tube is mercury. Determine the discharge




Head Loss in an Orifice
Prepared by: Melvin R. Esguerra Review Notes in Hydraulics Page 13

( )
( )
( )
( )
( )
2
2
2
2
2 2
3 2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
0 2 2
1
1
2
1
1
2
v
v
v
v
v
v
v
v
v
v C gh
v
h
C g
v v gy
C gh gy
y C h
HL h y h C h
C h
v
C
C g
v
g C
=
=
=
=
=
= =
=
=
| |
=
|
|
\ .




Discharge under Falling Head
Let: As = area of liquid surface at any distance h above the orifice
t = time to lower the liquid surface from h
1
to h
2



Then,
2
1
2
h
s
h
dVol
dt
Q
A dh
t
CA gh
=
=
}

Express As in terms of h and integrate. For vessels of constant cross section:

Prepared by: Melvin R. Esguerra Review Notes in Hydraulics Page 14

( )
( )
( )
2
1
2
1
2
1
1
2
1
2
2 1
1 2
1 2
1 2
1 2
1 2
2
2

1
2
2

1
2
2

1
2
2

2
2
volume discharge

average of initial and final discharge
h
s
h
h
s
h
h
s
h
s
s
s
dVol
dt
Q
A dh
t
CA gh
A
t h dh
CA g
A h
CA g
A
h h
CA g
A h h
h h
h h
CA g
A h h
h h
CA g

=
=

=
(
(
=
(
(

=

| |
+
=
|
|
+
\ .

=
+
=
}
}



The head in a vessel with vertical sides at the instant of opening an orifice was 2.75 m, and at closing has
decreased to 1.50 m. Determine the constant head under which in the same time the orifice would discharge the
same volume of water.

A vertical circular tank l.25m in diameter is fitted with a sharp-edged orifice at its base. When the flow of water
in the tank was shut off, the time taken to lower the head from 2 m to 0.75 m was 273 s. Determine the rate of
flow through the orifice under a steady head of 1.5 m.


Prepared by: Melvin R. Esguerra Review Notes in Hydraulics Page 15

Discharqe Under l(lqreasi+q' Head
2
1
h
s s
h
i o
A dh A dh
t
Q Q Q
= =

} }




Where:
t = time for the liquid surface to rise from h
1
to h
2

Q
i
= constant inflow
Q
o
= outflow through the orifice when the head is h
= 2 CA gh


A 1.80 m square tank has a sharp-edged orifice 100 mm in diameter which discharges under a constant head. If
the rate of inflow by which the head is kept constant is suddenly changed from 20 L/s to 35 L/s, how long will it
be, after this change occurs, until the head on the orifice becomes 2 m? Use C = 0.62.

Submerged Orifice Under Atmospheric Pressure