this is a slid presentation about flow through simple pipe and flow through compound pipe.

1. WELCOME TO OUR
PRESENTATION
1

2. GROUP MEMBERS
2
No Name Id Programe
01 Lipon islam 14207082 BSME
02 Jilani al mamun 14207099 BSME
03 Rashedujjaman 14207069 BSME
04 Nazmul Hossain 14207088 BSME
05 ALamin 13107061 BSME

3. TOPIC
3
Flow Through Simple pipes
And
Flow through compound pipes
3

4. CONTENTS:
4
Flow through simple pipes
 Loss of head in pipes
 Darcy’s formula for loss of head in pipes
 Chezy’s formula for loss of head in pipes
 Transmission of power through pipe
 Time of emptying a Tank through a long pipe
Flow through compound pipes
 Discharge from one reservoir to another through a pipe line
 Discharge through a compound pipe
 Discharge through pipes in parallel

5. LOSS OF HEAD IN PIPES
5
Whenever the water is flowing in apipe ,it experiences some resistance to its
motion,whos efect is to reduse the velocity and ultimately the head of water
available .though there are many types of losses ,yet the major loss is due to
frictional resistance of the pipe only.the frictional resistance of the pipe depends
upon the roughness of the inside surface of the pipe,grseater will be the
resistance .this friction is known as fluid friction and the resistance is known as
frictional resistance . the earlier experiment on the fluid friction were conducted
by froude we concluded on that
1.The frictional resistance varies approximately with the square of thee velocity
of the liquid.
2.The frictional resistance varies with the name of the surface lare on ,some
empirical formula were derived for the loss of head due to the friction out of
which following two are important from the subject point of view.
3.dercy’s formula for loss of head in pipes and
4.chezy’s formula for loss of head in pipes

6. DARCY’S FORMULA FOR LOSS OF HEAD IN PIPE
Fig: uniform long pipe
Consider a uniform long pipe through which is
flowing at a uniform rate.
6
1
1
2
2

7. Let,
l = length of pipe
d = diameter of a pipe
v = velocity of water in the pipe
f' =frictional resistance per unit area per unit velocity
ℎ 𝑓=𝑙𝑜𝑠𝑠 𝑜𝑓 ℎ𝑒𝑎𝑑 𝑑𝑢𝑒 𝑡𝑜 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛
let us consider sections (1-1) and (2-2) of the pipe .
Now let
𝑃1= 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1 − 1,
𝑃2 = 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 2 − 2 ,
Now considering horizontal forces on water between sections (1-1)
and (2-2)and equating the same ,
𝑝1 𝐴 = 𝑝2 𝐴 + 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐
or frictional resistance
=𝑝1 𝐴 − 𝑝2 𝐴
∴
𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑤
=
𝑝1𝐴−𝑝2 𝐴
𝑤
7
7

8. 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝐴𝑤
=
𝑝1
𝑤
−
𝑝2
𝑤
But ,
𝑝1
𝑤
−
𝑃2
𝑤
= hf = loss of pressure head due to friction
ℎ
𝑓=
𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝐴𝑤
=
𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝜋
4
×𝑑2×𝑤
We know that as per froudes experiment frictional resistance
ℎ 𝑓
= 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑎𝑟𝑒𝑎 𝑎𝑡 𝑢𝑛𝑖𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
× 𝑤𝑒𝑡𝑡𝑒𝑑 𝑎𝑟𝑒𝑎 × (𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦)2
f × 𝜋𝑑𝑙 × 𝑣2
substituting the value of friction resistance in the above
equation
ℎ 𝑓 =
𝑓𝜋𝑑𝑙×𝑣2
𝜋
4
×𝑑2 𝑤
=
4𝑓𝑙𝑣2
𝑤𝑑
8

9. let us introduce another coefficient (𝑓′
)𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡
𝑓′
=
𝑓𝑤
2𝑔
ℎ 𝑓 =
4
𝑤𝑑
×
𝑓𝑤
2𝑔
× 𝑙𝑣2
=
4𝑓𝑙𝑣2
2𝑔𝑑
We know that the discharge
Q=
𝜋
4
× 𝑑2
× 𝑣 𝑜𝑟 𝑣 =
4𝑄
𝜋𝑑2
𝑉2
=
16𝑄2
𝜋2 𝑑4
Substituting the value of 𝑉2 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
ℎ 𝑓 =
4, 𝑓𝑙
2𝑔𝑑
×
16𝑄2
𝜋2 𝑑4=
𝑓𝑙𝑄2
3𝑑5
9

10. CHEZY'S FORMULA FOR LOSS OF HEAD IN
PIPES:
Consider a uniform long pipe through which water is following at
a uniform rates shown in fig:
Let
I = Length of the pipe and
D = Diameter of the pipe
Area of pipe 𝐴 =
𝜋
4
×𝑑2
And perimeter of pipe P = 𝐴 = 𝜋d
V = velocity of water in pipe
f = Frictional resistance, per unit area of v'etted surface per unit
velocity and
ℎ1, Loss of head due to frication
Now let us consider section (1-1) and (2-2) of the pipe Let,
𝑝1= intensity of pressure at section 1 –land P2= Intensity of
pressure at section 2,2

11. A LITTLE CONSIDERATION WILL SHOW THAT P1 AND P2 WOULD HAVE BEEN
EQUAL , IF THERE WOULD HAVE BEEN NO FRICTIONAL RESISTANCE . NOW
CONSIDERING HORIZONTAL FORCES ON WATER BETWEEN SECTIONS 1-1 AND 2-
2 AND EQUATING THE SAME ,
𝑝1𝐴=𝑝2𝐴+𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑐𝑒
OR, FRICTIONAL RESISTANCE=𝑝1 𝐴 − 𝑝2A
OR,
𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑤
=
𝑝1𝐴−𝑝2𝐴
𝑤
OR,
𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑎𝑡𝑎𝑛𝑐𝑒
𝐴𝑤
=
𝑝1
𝑤
-
𝑝2
𝑤
11

12. But ,
𝑝1
𝑤
−
𝑝2
𝑤
= ℎ 𝑓= loss of pressure head due to friction
ℎ𝑓=
(𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒)
𝐴𝑤
We know that’s experiment , frictional resistance
=frictional resistance per unit area at velocity × wetted
area×(𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦)2
=f´×𝜋𝑑𝑙 × 𝑣2
Substituting the value of frictional resistance in the above
equation
ℎ 𝑓=
f´× 𝜋𝑑𝑙×𝑣2
𝐴𝑤
=
f𝑙𝑣2´
𝐴𝑤
(∴ 𝜋𝑑 = 𝑝 𝑖. 𝑒 𝑝𝑟𝑒𝑚𝑖𝑡𝑒𝑟)
=
f𝑙𝑣2´
𝐴𝑤
×
𝑝
𝐴
12

13. Now substituting another term of hydraulic mean depth in the above
equation ,such that hydraulic mean depth
m=
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑓𝑙𝑜𝑤
𝑤𝑒𝑦𝑦𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟
=
𝐴
𝑝
∴ ℎ 𝑓=
f´𝑙𝑣2
𝐴𝑤
×
1
𝑚
𝑣
2=
ℎ𝑓.𝑤.𝑚
𝑓´𝑙 =
𝑤
𝑓´
×m×
ℎ𝑓
𝑙
V=
𝑤
𝑓´
×m×
ℎ𝑓
𝑙
……(1)
Now substituting two more terms in the above equation such that
C=
𝑤
𝑓´
=c
ℎ 𝑓=𝑖
Now substituting the above two values in equation (1)
V=c 𝑚𝑖 13

14. Transmission of
Power through a Pipe
1
4

15. Whenever water is allowed to fall from higher level to lower level ,
we can always generate some power as a matter of fact ,whenever
we come across a waterfall, we do not allow the water simply to
fall. But it is made to follow through a pipe , so that the direction of
the water may be set in some convenient way from which we may
produce some power .
A little consideration will show that some hade of water will be
lost due to friction in the pipe through which the water is flowing
Fig: transmission of pipes
15

16. Consider a high level strong tank .let a pipe AB lead
water from this tank from A to a power house at B
as shown in figure:
𝐻 =Hade of water at a power house AB in meters
𝑙 = length of the pipe AB in metre
𝑣 = velocity of water in the pipe in m/s
ℎ1=loss of heat in the pipe AB due to friction in
meters
𝑓 = coefficient of friction, and
d=diameter of the pipe AB in meters
16

17. Cross-sectional area of the pipe,
a =
𝜋
4
× d2
𝑚2
We know the weight of water flowing per second `
η = 𝑤𝑄 = 𝑤𝑣𝑎 𝐾𝑁
(AS
𝑣
𝑠
= 𝑄 𝑎𝑛𝑑 𝑤 =
𝑊
𝑣
)
And net head of water available at 𝐵 (neglecting minor losses)
Efficiency of transmission,
η =
ℎ
𝐻
=
𝐻−ℎ 𝑓 𝐴
𝐻
we also know that power available,
𝑝 = Weight of water flowing per second × Head of water
= 𝑤𝑄. ℎ = 𝑤𝑎𝑣 𝐻 − ℎ 𝑓 = 𝑤𝑎𝑣 (𝐻 −
4𝑓𝑙𝑣2
2𝑔𝑑
)
. = 𝑤𝑎(𝐻𝑣 −
4𝑓𝑙𝑣2
2𝑔𝑑
)
17

18. A little consideration will shown that, in that above equation, the
power transmitted depends upon the velocity of water (v),as the
other things are constant. Therefore the power transmitted will
be maximum when
𝑑𝑝
𝑑𝑣
= 0
Or when the differential coefficient of the amount inside the
bucket of equation (𝑖𝑖𝑖) is zero . i e
𝑑 ( 𝐻𝑣 −
4𝑓𝑙𝑣2
2𝑔𝑑
)
𝑑𝑣
= 0
Or 𝐻 – 3
4𝑓𝑙𝑣2
2𝑔𝑑
= 0
Or 𝐻 – 3ℎ 𝑓 = 0 as (
4𝑓𝑙𝑣2
2𝑔𝑑
= ℎ 𝑓)
Or ℎ 𝑓 =
𝐻
3
It means that the power transmission through a pipe is
maximum, when the head lost due to friction in the pipe is equal
to 1/3 of the total supply head.
18

19. TIME OF EMPTYING A TANK THROUGH
A LONG PIPE
19

20. Consider a tank, which is to be emptied through a long pipe as shown in fig.
Let,
𝐻1 = Initial head of water, in the tank, before opening the pipe,
𝐻2= Final head of water, in the tank, after opening the pipe in T seconds,
(𝐻1-𝐻2) =Fall of water level in the tank,
A (𝐻1-𝐻2) = Volume of water discharged through the pipe,
l = Length of the pipe,
d= diameter of the pipe
T=time taken in seconds , to fall the water level in the tank from 𝐻1 to 𝐻2
20

21. Consider an instant, when the head of water in the tank
is h. After a small interval of time dt, at the water level
in the tank fall down by an amount equal to dh.
Volume of water that has passed through the pipe in
the time dt
= -Adh ……. (1)
(Minus value of dh is taken, as the value of h
decreases as the as the discharge increases).
Since the water is being discharged in the
atmosphere, therefore (ignoring all other losses except
friction) there will be some loss of head of water at
outlet also.
21

22. h =
4𝑓𝑙𝑣2
2𝑔𝑑
+
𝑣2
2𝑔
=
𝑣2
2𝑔
(1+
4𝑓𝑙
𝑑
)
v=
2𝑔ℎ
(1+4𝑓𝑙
𝑑
)
In time dt, the volume of water that has pass through the pipe
=Area of pipe ×velocity of water ×time
=
𝜋
4
𝑑2
×
2𝑔ℎ
(1+4𝑓𝑙
𝑑
)
×dt …………..(2)
Equating equation (1) and (2)
-Adh=
𝜋
4
𝑑2×
2𝑔ℎ
(1+4𝑓𝑙
𝑑
)
×dt
22

23. dt=
−4𝐴 1+
4𝑓𝑙
𝑑
𝑑ℎ
𝜋𝑑2 2𝑔ℎ
=
−4𝐴 1+
4𝑓𝑙
𝑑
( ℎ−1
2)𝑑ℎ
𝜋𝑑2 2𝑔
Now the total time T, required to bring the water level
from 𝐻1 𝑡𝑜 𝐻2 may be found by integrating the equation
between the limits 𝐻1 𝑎𝑛𝑑 𝐻2 i,e…
T= 𝐻1
𝐻2
−4𝐴 1+
4𝑓𝑙
𝑑
( ℎ−1
2)𝑑ℎ
𝜋𝑑2 2𝑔
We have seen that the velocity of water
v =
2𝑔ℎ
(1+4𝑓𝑙
𝑑
)
∴ 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑤ℎ𝑒𝑛 𝑡ℎ𝑒 ℎ𝑒𝑎𝑑 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑖𝑠 𝐻1
𝑣1=
2𝑔𝐻1
(1+4𝑓𝑙
𝑑
) 23

24. Velocity of water when the head of water 𝐻2
𝑣2=
2𝑔𝐻2
(1+4𝑓𝑙
𝑑
)
∴average velocity =
𝑣1+𝑣2
2
=
2𝑔𝐻1
(1+4𝑓𝑙
𝑑
)
+
2𝑔𝐻2
(1+4𝑓𝑙
𝑑
)
2
=
2𝑔
(1+4𝑓𝑙
𝑑
)
×( 𝐻1 + 𝐻2)
T=
−4𝐴 1+
4𝑓𝑙
𝑑
𝜋𝑑2 2𝑔 𝐻1
𝐻2
ℎ
1
2 dh
=
−4𝐴 1+
4𝑓𝑙
𝑑
𝜋𝑑2 2𝑔
=
−8𝐴 1+
4𝑓𝑙
𝑑
𝜋𝑑2 2𝑔
×( 𝐻1 + 𝐻2) 24

25. Taking minus out of the bracket,
T=
8𝐴 1+
4𝑓𝑙
𝑑
𝜋𝑑2 2𝑔
×( 𝐻1 + 𝐻2)
If the tank is to be completely emptied ,then
substituting 𝐻2=0,in the above equation
T=
8𝐴 1+
4𝑓𝑙
𝑑
𝜋𝑑2 2𝑔
× 𝐻1
25

26. DISCHARGE FROM ONE RESERVOIR TO
ANOTHER THROUGH A PIPE LINE
sometimes ,discharge from one reservoir to another takes place
through a pipe line.in such a case ,It is assumed that the
difference between the water levels of the two reservoir is lost
due to friction in the pipe in the line . we can discuss the
discharge from one reservoir to another through-
1.Compound pipes
2.Pipes in parallel
3.Branched pipes and
4. Siphon pipes
25

27. DISCHARGE THROUGH A COMPOUND PIPE(I.E
PIPES IN SERIES)
Sometimes while laying a pipeline ,we have to connect pipes of
different lengths and different diameters with one another to from
a pipe line . such a pipe line is called a compound pipe or pipes
in series .a little consideration will show that as the pipes are in
series ,therefore the discharge will be continuous.
Now consider a compound pipe discharge water from one tank
with a higher water level to another with a lower water level as
shown fig.
26

28. Let,
Q=discharge through the pie
H=total loss of head
𝑑1=diameter of pipe 1
𝑙1=length of pipe 1
𝑓1=coefficient of friction of pipe 1
𝑑2 𝑙2 𝑓2=corresponding value of pipe 2
𝑑3 𝑙3 𝑓3 =corresponding value of pipe 3 and so on
Neglecting minor losses except friction , we know that the total
loss of head,
H=loss of head in pipe 1+loss of head in pipe 2+loss of head in
pipe 3
=
4𝑙1 𝑓1 𝑣1
2
2𝑔𝑑1
+
4𝑙2 𝑓2 𝑣2
2
2𝑔𝑑2
+
4𝑙3 𝑓3 𝑣3
2
2𝑔𝑑3
+….
=
4
2𝑔
(
𝑙1 𝑓1 𝑣1
2
𝑑1
+
𝑙2 𝑓2 𝑣2
2
𝑑2
+
𝑙3 𝑓3 𝑣3
2
𝑑3
+…..)
27

29. If the coefficient of friction is the sum for all the
pipes,then
H=
4𝑓
2𝑔
(
4𝑙1 𝑣1
2
𝑑1
+
4𝑙2 𝑣2
2
𝑑2
+
4𝑙3 𝑣3
2
𝑑3
+……)
If the discharge through the pipe line is given then the
total loss of head
H=
𝑓1 𝑙1 𝑄2
3𝑑1
5 +
𝑓2 𝑙2 𝑄2
𝑑2
5 +
𝑓3 𝑙3 𝑄2
𝑑3
5 +………
=
𝑄2
3
(
𝑓1 𝑙1
𝑑1
5+
𝑓2 𝑙2
𝑑2
5+
𝑓3 𝑙3
𝑑3
5+………)
If the coefficient of friction is the same for all pipes,then
H==
𝑓𝑄2
3
(
𝑙1
𝑑1
5+
𝑙2
𝑑2
5+
𝑙3
𝑑3
5+………) 28

30. DISCHARGE THROUGH PIPES IN PARALLEL
Sometimes in order to increase the discharge from one tank into another , anew pipe
has to be laid along with the existing one . such an arrangement is known as pipes in
parallel as shown in fig.
fig: pipes in parallel
A little consideration will show that as the pipes are parallel, therefore the loss
of head for all the pipes will be the same and all the pipes will discharge water
idependently .the total discharge through all the pipes will be the same of
discharge in the various pipes.
29

31. 31