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UNIT – IV
BOUNDARY LAYER AND FLOW THROUGH PIPES
Definition of boundary layer – Thickness and classification – Displacement and
momentum thickness – Development of laminar and turbulent flows in circular pipes –
Major and minor losses of flow in pipes – Pipes in series and in parallel – Pipe network
Hydraulic gradient line.
It is defined as the line which gives the sum of pressure head ( P/ρg) and datum
head (z) of a flowing fluid in a pipe with respect to some reference line or is the line
which is obtained by joining the top of all vertical ordinates, showing the pressure head
( P/ρg) of a pipe from the center of the pipe. It is briefly written as H.G.L
Major energy loss and minor energy loss in pipe
The loss of head or energy due to friction in pipe is known as major loss while the
loss of energy due to change of velocity of the flowing fluid in magnitude or direction is
called minor loss of energy.
Total Energy line
It is defined as the line, which gives sum of pressure head, datum head and kinetic
head of a flowing fluid in a pipe with respect to some reference line.
Equivalent pipeline
An Equivalent pipe is defined as the pipe of uniform diameter having loss of head
and discharge of a compound pipe consisting of several pipes of different lengths and
diameters.
Water Hammer in pipes.
In a long pipe, when the flowing water is suddenly brought to rest by closing the
valve or by any similar cause, there will be a sudden rise in pressure due to the
momentum of water being destroyed. A pressure wave is transmitted along the pipe. A
sudden rise in pressure has the effect of hammering action on the walls of the pipe. This
phenomenon of rise in pressure is known as water hammer or hammer blow.
2
Pipes in series:
Pipes in series or compound pipes is defined as the pipes of different lengths and
different diameters connected end to end (in series) to form a pipe line.
Pipes in parallel:
The pipes are said to be parallel, when a main pipe divides into two or more
parallel pipes, which again join together downstream and continues as a mainline. The
pipes are connected in parallel in order to increase the discharge passing through the
main.
Boundary layer.
When a solid body is immersed in a flowing fluid, there is a narrow region of the
fluid in neighbourhood of the solid body, where the velocity of fluid varies from zero to
free stream velocity. This narrow region of fluid is called boundary layer.
laminar sub layer
In turbulent boundary layer region, adjacent to the solid boundary velocity for a
small thickness variation in influenced by various effect. This layer is called as laminar
sub layer.
Boundary layer thickness.
It is defined as the distance from the boundary of the solid body measured in the y
– direction to the point where the velocity of the fluid is approximately equal to 0.99
times the free stream (v) velocity of the fluid.
momentum thickness.
It is defined as the distance, measured perpendicular to the boundary of the solid
body, by which the boundary should be displaced to compensate for the reduction in
momentum of the flowing fluid of boundary
δ
∫
θ = u / v(1 − u / v)dy
0
3
Incompressible flow.
It is define as the type of flow in which the density is constant for the fluid flow.
Mathematically ρ = constant. Examples: Subsonic, aerodynamics.
Different methods of preventing the separation of boundary layers
1.
Suction of slow moving fluid by suction slot
2. Supplying additional energy from a blower
3. Providing a bypass in the slotted wring
4. Rotating boundary in the direction of flow.
5. Providing small divergence in diffuser
6. Providing guide – blades in a bend.
Examples laminar flow / viscous flow
(i)
Flow past tiny bodies,
(ii)
Underground flow
(iii)
Movement of blood in the arteries of human body,
(iv)
Flow of oil in measuring instruments,
(v)
Rise of water in plants through their roots etc.,
Characteristics of laminar flow
(i)
No slip at the boundary
(ii)
Due to viscosity, there is a shear between fluid layers, which is given by
τ = µ(du / dy) for flow in x- direction
(iii)
The flow is rotational.
(iv)
Due to viscous shear, there is continuous dissipation of energy and for
maintaining the flow must be supplied externally.
(v)
Loss of energy is proportional to first power of velocity and first power of
viscosity.
(vi)
No mixing between different fluid layers (except by molecular motion,
which is very small)
Differentiate between laminar boundary layer and turbulent boundary layer
4
The boundary layer is called laminar, if the Renolds number of the flow is
defined as Re = U x X / v is less than 3X105
If the Renolds number is more than 5X105, the boundary layer is called
turbulent boundary.
Where, U = Free stream velocity of flow
X = Distance from leading edge
v = Kinematic viscosity of fluid
Chezy’s formula.
Chezy’s formula is generally used for the flow through open channel.
V = C mi
Where , C = chezy’s constant, m = hydraulic mean depth and i = hf/ L.
A crude of oil of kinematic viscosity of 0.4 stoke is flowing through a pipe
of diameter 300mm at the rate of 300 litres/sec. find the head lost due to friction
for a length of 50m of the pipe.
Given :
Kinematic viscosity
v = 0.4 stoke = 0.4 cm2/s = 0.4 X 10-4 m2/s
Dia. Of pipe
d = 300mm = 0.3m
Discharge
Q = 300 Lit/S = 0.3 m3/s
Length of pipe
L = 50m
Velocity
V = Q/ Area = 0.3 / (
Renold number
Re = ( V X d )/ v = (4.24 X 0.30) / 0.4X10-4 = 3.18 X 104
π
( 0.3)2) = 4.24 m/s
4
As Re lies between 4000 and 100,000, the value of “f” is given by
f=
0.079
( Re )
1/ 4
=
0.079
(3.18 X 10 4 )1 / 4
= 0.00591
Head lost due to friction hf = 4 f L V2/ 2 g d
= (4 X 0.00591 X 50X 4.242) /( 0.3 X 2 X 9.81)
= 3.61 m
Find the type of flow of an oil of relative density 0.9 and dynamic viscosity 20
poise, flowing through a pipe of diameter 20 cm and giving a discharge of 10 lps.
Solution :
5
s = relative density = Specific gravity = 0.9
µ = Dynamic viscosity = 20 poise = 2 Ns/m2.
Dia of pipe D = 0.2 m; Discharge Q = 10 lps = (10 / 1000) m3/s; Q = AV.
So V = Q / A = [10 / (1000 X (
π
( 0.2)2) )] = 0.3183 m/s.
4
Kinematic viscosity = v = µ / ρ = [ 2 / (0.9X1000)] = 2.222X10-3 m2 / s.
Reynolds number Re = VD / v
Re = [0.3183 X 0.2 / 2.222X10-3] = 28.647;
Since Re ( 28.647) < 2000,
It is Laminar flow.
Formula for finding the loss of head due to entrance of pipe hi
hi = 0.5 ( V2 / 2g)
Formula to find the Efficiency of power transmission through pipes
n = ( H – hf) / H
where, H = total head at inlet of pipe.
hf = head lost due to friction
Hydro dynamically smooth pipe carries water at the rate of 300 lit/s at 20oC
(ρ = 1000 kg/m3, ν = 10-6 m2/s) with a head loss of 3m in 100m length of pipe.
Determine the pipe diameter. Use f = 0.0032 + (0.221)/ (Re)0.237 equation for f
where hf = ( fXLXV2)/ 2gd and Re = (ρVD/µ)
Given:
Discharge,
Q = 300 lit/sec=
Density
ρ = 1000 kg/m3
0.3m3/s
Kinematic viscosity ν = 10-6m2/s
Head loss
hf = 3m
Length of pipe, L = 100m
Value of friction factor, f = 0.0032 + 0.221 / (Re)0.237
Renolds number Re = (ρVD/µ) = (VXD) / ν
VXD/ 10-6 = VXDX106
Find diameter of pipe.
(µ/ρ = ν)
6
Let D = diameter of pipe
Head loss in terms of friction factor is given as
hf = ( fXLXV2)/ 2gXD
3 = (fX100XV2)/ 2X9.81XD
f = ( 3XDX2X9.81)/ 100V2
f = 0.5886D / V2 ------------------------------( i)
now Q = AXV
0.3 =
π
( D)2 X V or D2 X V = ( 4 X 0.3 / π ) = 0.382
4
V2 = 0.382 / D2 --------------------------- (ii)
f = 0.0032 + (0.221)/ (Re)0.237
0.5886/ D2 = 0.0032 + (0.221)/ (VXDX106)0.237
{ from equation (i), f = 0.5886D/V2 and Re = VXDX106}
0.221
2 2
0.5886D / ( 0.382/D ) = 0.0032 +
 0.382

XDX 10 6 

2
 D

0.237
{ from Equation (ii), V = 0.382 / D2}
0.221
5
2
0.5886 x D / 0.382 = 0.0032 +
 (0.382 X 10 6 ) 0.237


D 0.237





4.0333 D5 = 0.0032 + 0.0015 X D0.237
4.0333 D5 – 0.0105 D 0.237 – 0.0032= 0 ---------------(iii)
the above equation (iii) will be solved hit trial method
(i). Assume D = 1m, then L.H.S of the equation (iii), becomes as
L.H.S = 4.033 X 1 5 – 0.0105 X 10.237 – 0.0032
= 4.033 – 0.0105 – 0.0032 = 4.0193
by increasing the value of D more than 1m , the L.H.S. will go on increasing.
Hence decrease the value of D.
(ii) Assume D = 0.3 than L.H.S of equation (iii)
becomes as L.H.S = 4.033 X 0.30.237 – 0.0032
= 0.0098 – 0.00789 – 0.0032 = - 0.00129
7
as this value of negative, the values of D will be slightly more than 0.3
(iii) Assume D = 0.306 then L.H.S of equation (iii) becomes as
L.H.S = 4.033 X 0.3060.237 – 0.0105X0.3060.237 – 0.0032
= 0.0108 – 0.00793 – 0.0032 = - 0.00033
This value of L.H.S is approximately equal to equal to zero. Actually
the value of D will be slightly more than 0.306m say 0.308m.
Expression for loss of head due to friction in pipes.
Or
Darcy – Weisbach Equation.
Consider a uniform horizontal pipe, having steady flow as shown figure. Let 1 -1
and 2-2 are two sections of pipe.
Let P1 = pressure intensity at section 1-1.
Let P2 = Velocity of flow at section 1-1.
L = length of the pipe between the section 1-1 and 2-2
d = diameter off pipe.
f1 = Frictional resistance per unit wetted area per unit velocity.
hf = loss of head due to friction.
And P2,V2 = are the values of pressure intensity and velocity at section 2-2.
Applying Bernoulli’s equation between sections 1-1 & 2-2
Total head 1-1 = total head at 2-2 + loss of head due to friction between 1-1&2-2
(P1/ρg) + (V12 / 2g) + Z1 = (P2/ρg) + (V22 / 2g) + Z2+hf ------------(1)
but Z1 = Z1 [ pipe is horizontal ]
V1= V2 [ diameter of pipe is same at 1-1 & 2-2]
(1) becomes,
(P1/ ρg) = (P2/ρg)+hf
hf = (P1/ ρg) - (P2/ρg)
frictional resistance = frictional resistance per unit wetted area per unit velocity X
wetted area X velocity 2.
F = f1 X π d l X V2 [ Wetted area = π d X L, and Velocity V = V 1 = V2]
8
F1 = f1XPXLXV2 ----------- (2). [π d = wetted perimeter = p]
The forces acting on the fluid between section 1-1 and 2-2 are,
1) Pressure force at section 1-1 = P1X A
2) Pressure force at section 2-2 = P2 X A
3). Frictional force F1
Resolving all forces in the horizontal direction.,
P1 A – P2A – F1 = 0
(P1-P2)A = F1 = f1XPXLXV2
(P1-P2) = (f1XPXLXV2 / A ).
But from (1) we get
P1 – P2 = ρg hf
Equating the values of (P1 – P2) we get
ρg hf = (f1XPXLXV2 / A ).
hf = (f1 / ρg) X (P/A) X LX V2
(P/A) = (π d / (π d2/4)) = (4/d)
hence, hf = ( f1 / ρg) X ( 4/d) X LXV2.
Putting (f1 / ρ) = ( f / 2) , where f is the co – efficient of friction
hf =
4 fLV 2
2 gd
This equation is known as Darcy – Weisbach equation. This equation is
commonly used to find loss of head due to friction in pipes
The rate of flow through a horizontal pipe is 0.25 m3/s. the diameter of the
pipe which is 200mm is suddenly enlarged to 400mm. the pressure intensity in
the smaller pipe is 11.772 N/cm2. Determine (i). Loss of head due to s
udden enlargement (ii). Pressure intensity in large pipe. (iii). Power lost
due to enlargement.
Given:
Discharge
Q = 0.25 m3/s.
9
Dia. Of smaller pipe
D1 = 200mm = 0.2m
Area
A1 =
Dia of large pipe
D2 = 400mm = 0.4m
Area
A2 =
Pressure in smaller pipe
p1 = 11.772 N/cm2 = 11.772 X104 N/m2.
Now velocity
V1 = Q / A1 = 0.25 / 0.03414 = 7.96 m/s.
Velocity
V2 = Q / A2 = 0.25 / 0.12566 = 1.99 m/s.
π
(0.2) 2 = 0.03141 m2.
4
π
(0.4) 2 = 0.12566 m2.
4
(i). Loss of head due to sudden enlargement,
he = ( V1 – V2)2/ 2g = (7.96 – 1.99) 2 / 2X 9.81 = 1.816 m.
(ii). Let the pressure intensity in large pipe = p2.
Then applying Bernoulli’s equation before and after the sudden enlargement,
(P1/ρg) + (V12 / 2g) + Z1 = (P2/ρg) + (V22 / 2g) + Z2+he
But Z1 = Z2
(P1/ρg) + (V12 / 2g) = (P2/ρg) + (V22 / 2g) +he
Or (P1/ρg) + (V12 / 2g) = (P2/ρg) + (V22 / 2g) + Z2+hf
(P2/ρg) = (P1/ρg) + (V12 / 2g) - (V22 / 2g) - he
=
11.772 X 10 4
7.96 2
1.99 2
+
−
− 1.816
1000 X 9.81 2 X 9.81 2 X 9.81
= 12.0+3.229-0.2018-1.8160
= 15.229 – 20.178 = 13.21m of water
p2
= 13.21 X ρg = 13.21 X 100X 9.81 N/m2
= 13.21X1000X9.81X10-4 N/cm2. = 12.96N/cm2.
(iii). Power lost due to sudden enlargement,
P = (ρg Q he) / 1000 = (1000X9.81X0.25X1.816)/1000 = 4.453kW.
10
A horizontal pipeline 40m long is connected to a water tank at one end and
discharges freely into the atmosphere at the other end. For the first 25m of
its length from the tank, the pipe is 150mm diameter is suddenly enlarged to
300mm. the height of water level in the tank is 8m above the centre of the
pipe. Considering all losses of head, which occur. Determine the rate of flow.
Take f = 0.01 for both sections of the pipe.
Given:
Total length of pipe, L = 40m
Length of 1 st pipe,
L1 = 25m
Dia of 1st pipe
d1 = 150mm = 0.15m
Length of 2nd pipe
L2 = 40 – 25 = 15m
Dia of 2nd pipe
d2 = 300mm = 0.3m
Height of water
H = 8m
Co-effi. Of friction
f = 0.01
Applying the Bernoulli’s theorem to the surface of water in the tank and
outlet of pipe as shown in fig. and taking reference line passing through
the center of the pipe.
0+0+8 = (P2/ρg) + (V22 / 2g) +0+all losses
8.0 = 0+(V22 / 2g)+hi+ hf1+ he+ hf2
11
Where, hi = loss of head at entrance = 0.5 V12/ 2g
hf1 = head lost due to friction in pipe 1 =
4 XfXL1 XV12
d1 X 2 g
he = loss of head due to sudden enlargement = (V1- V2)2/2g
hf2 = head lost due to friction in pipe 2
4 XfXL2 XV22
=
d 2 X 2g
But from continuity equation, we have
A1V1 = A2V2
π 2
d 2 XV 2 
d
4
=  2
V1 = (A2V2/A1) = π
 d1
d 12
4
2
2

0.3 
 XV 2 = 
 XV 2 = 4V 2
 0.15 

Substituting the value of V1 in different head losses, we have
hi = 0.5 Vi2/ 2g = (0.5 X (4V2)2 )/ 2g = 8V22/2g
hf1 =
4 X 0.01X 25 X (4V22 ) 4 X 0.01X 25 X 16 V22
V2
=
X
= 106.67 2
0.15 X 2 g
0.15
2g
2g
he = (V1 – V2)2/ 2g
= (4V2 – V2)2/2g = 9V22/2g
4 X 0.01X 15 X (V22 ) 4 X 0.01X 15 V22
V22
hf2 =
=
X
= 2.0
0.3 X 2 g
0.3
2g
2g
Substituting the values of these losses in equation (i), we get
8.0 =
=
V2 =
V 22 8V 22
V 2 9V 2
V2
+
+ 106.67 2 + 2 + 2 X 2
2g
2g
2g
2g
2g
V22
2g
[ 1+8+106.67+9+2] = 126.67
V22
2g
8.0 x 2 xg
8.0 X 2 X 9.81
=
= 1.113 m/s
126.67
126.67
12
Rate of flow Q = A2XV2 =
π
( 0.3)2 X 1.113 = 0.07867 m3/s = 78.67 litres/sec.
4
A pipe line, 300mm in diameter amd 3200m long is used to pump up 50kg
per second of an oil whose density is 950n kg/m3.and whose Kinematic
viscosity is 2.1 stokes. The center of the pipe at upper end is 40m above than
at the lower end. The discharge at the upper end is atmospheric. Find the
pressure at the lower end and draw the hydraulic gradient and the total
energy line.
Given:
Dia of pipe
d = 300mm = 0.3m
Length of pipe
L = 3200m
Mass
M = 50kg/s = ρ. Q
Discharge
Q = 50/ ρ = 50/950 = 0.0526 m3/s
Density
ρ = 950 kg/m3
Kinematic viscosity ν = 2.1 stokes = 2.1 cm2/s = 2.1 X10-4 m2/s
Height of upper end = 40m
Pressure at upper end = atmospheric = 0
Renolds number, Re = VXD/ν, where V = Discharge / Area
= 0.0526/ (
π
(0.3) 2) = 0.744 m/s
4
Re = (0.744X0.30) / (2.1X10-4) = 1062.8
Co – efficient of friction, f = 16/ Re = 16 / 1062.8 = 0.015
Head lost due to friction,hf
=
4 XfXL XV 2
d X 2g
4 X 0.015 X 3200 X (0.744) 2
=
= 18.05m of oil.
0.3 X 2 X 9.81
13
Applying the Bernoulli’s equation at the lower and upper end of the pipe and
taking datum line passing through the lower end, we have
(P1/ρg) + (V12 / 2g) + Z1 = (P2/ρg) + (V22 / 2g) + Z2+hf
but Z1 = 0, Z2 = 40m., V1 = V2 as diameter is same.
P2 = 0, hf = 18.05m
Substituting these values, we have
= 5400997 N/m2 = 54.099 N/cm2.
H.G.L. AND T.E.L.
V2/2g = (0.744)2/2X9.81 = 0.0282 m
p1/ ρg = 58.05 m of oil
p2 / ρg = 0
Draw a horizontal line AX as shown in fig. From A draw the centerline of the
pipe in such way that point C is a distance of 40m above the horizontal line. Draw a
vertical line AB through A such that AB = 58.05m. Join B with C. then BC is the
hydraulic gradient line.
Draw a line DE parallel to BC at a height of 0.0282m above the hydraulic
gradient line. Then DE is the total energy line.
A main pipe divides into two parallel pipes, which again forms one pipe as shown.
The length and diameter for the first parallel pipe are 2000m and 1.0m respectively,
while the length and diameter of 2nd parallel pipe are 2000m and 0.8m. Find the rate
of flow in each parallel pipe, if total flow in main is 3.0 m3/s. the co-efficient of
friction for each parallel pipe is same and equal to 0.005.
Given:
14
Length of Pipe 1
L1= 2000m
Dia of pipe1
d1 = 1.0m
Length of pipe 2
L2 = 2000m
Dia of pipe 2
d2 = 0.8m
Total flow
Q = 3.0m3/s
f1 = f2 = f = 0.005
let Q1 = discharge in pipe 1
let Q2 = discharge in pipe 2
from equation,
Q = Q1+ Q2 = 3.0 ---------------(i)
using the equation we have
4 Xf 1 XL1 XV12
d1 X 2 g
=
4 Xf 2 XL 2 XV 22
d 2 X 2g
4 X 0.005 X 2000 XV12
1.0 X 2 X 9.81
=
4 X 0.005 X 2000 XV 22
0.8 X 2 X 9.81
V12 V 22
V2
=
orV12 = 2
1.0 0.8
0.8
V2
V2
V1 =
=
0.8 0.894
Now, Q1 =
π
π
d12XV1 = ( 1)2X(V2 / 0.894)
4
4
And
π
π
π
d22XV2 = (0.8) 2X(V2) =
(0.64) X (V2)
4
4
4
Q2 =
Substituting the value of Q1 and Q2 in equation (i) we get
π
π
(1)2X(V2 / 0.894)+
( 0.64)X(V2) = 3.0 or 0.8785 V2 + 0.5026 V2 = 3.0
4
4
V 2[0.8785+0.5026] = 3.0 or V = 3.0 / 1.3811 = 2.17 m/s.
Substituting this value in equation (ii),
15
V1 = V2 / 0.894 = 2.17 / 0.894 m/s
Hence Q1 =
π
π 2
d12XV1 =
1 X2.427 = 1.096 m3/s
4
4
Q2 = Q – Q1 = 3.0 – 1.906 = 1.904 m3/s.
Three reservoirs A, B, C are connected by a pipe system shown in fig. Find
the discharge into or from the reservoirs B and C if the rate of flow from
reservoirs A is 60 litres / s. find the height of water level in the reservoir C.
take f = 0.006 for all pipes.
Given:
Length of pipe AD,
L1 = 1200m
Dia of pipe AD,
d1 = 30cm = 0.3m
Discharge through AD,
Q1 = 60lit/s = 0.06 m3/s
Height of water level in A from reference line , ZA= 40m
For pipe DB, length
L2 = 600mm, Dia., d2 = 20cm = 0.20m, ZB= 38.0
For pipe DC, length
L3 = 800mm, Dia., d3= 30cm = 0.30m,
pD
Applying the Bernoulli’s equations to point E and, ZA = ZD+ ρg +hf
Where hf =
hf=
4 Xf 1 XL1 XV12
d1 X 2 g
, where V1 = Q1 / Area = 0.006 / (
4 X 0.006 X 1200 X 0.848 2
0.3 X 2 X 9.81
π
( 0.3)2) = 0.848 m/s.
4
= 3.518 m.
p1
{ZD+ ρg } = 40.0 – 3.518 = 36.482 m
Hence piezometric head at D = 36.482m. Hence water flows from B to D.
Applying Bernoulli’s equation to point B and D
16
pD
ZB = {ZD+ ρg }+hf2 or 38 = 36.482 + hf2
hf2 = 38 – 36.482 = 1.518m
But hf2 =
4 XfXL2 XV 22 4 X 0.006 X 600 XV 22
=
d 2 X 2g
0.2 X 2 X 9.81
4 X 0.006 X 600 XV22
0.2 X 2 X 9.81
1.518 =
V2 =
1.518 X 0.2 X 2 X 9.81
= 0.643m / s
4 X 0.006 X 600
Discharge Q2 = V2X
π
π
( d2)2 = 0.643 X X(0.2)2 = 0.0202m3/s = 20.2lit/s.
4
2
Applying Bernoulli’s equation to D and C
pD
{ZD+ ρg }= ZC+hf3
36.482 =
4 XfXL3 XV32
Q3
where, V3 =
ZC + d 3 X 2 g
π 2
d3
4
but from continuity Q1 + Q2 = Q3
Q3 = Q1+Q2 = 0.006 + 0.0202 = 0.0802 m3/s
V3 =
Q3
Q3
=
= 1.134m / s
π 2 π
d3
(0.9) 2
4
4
36.482 = ZC +
4 X 0.006 X 800 X 1.134 2
= ZC+
0. X 2 X 9.81
4.194
ZC = 36.482 – 4.194 = 32.288m
A Pipe line of length 2000 m is used for power transmission. If 110.365 kW
power is to be transmitted through the pipe in which water having pressure
of 490.5 N/cm2 at inlet is flowing. Find the diameter of the pipe and efficiency
17
of transmission if the pressure drop over the length of pipe is 98.1 N/cm2.
Take f = 0.0065.
Given:
Length of pipe
L = 2000m.
H.P transmitted
Pressure at inlet,
= 150
p = 490.5 N/cm2 = 490.5 X104 N/m2.
Pressure head at inlet, H = p / ρg
= 98.1 N/cm2 = 98.1X104/m2
Pressure drop
hf = 98.1 X104 / ρg = 98.1X104/ ( 1000 X 9.81) = 100m
Loss of head
Co – efficient of friction f = 0.0065
Head available at the end of the pipe = H – hf = 500 – 100 = 400m
Let the diameter of the pipe = d
Now power transmitted is given by,
P = [ρg X Q X ( H – hf)] / 1000 kW.
110.3625 = [ 1000X9.81XQX400] / 1000
Q = [ 110.3625 X 1000 / ( 1000X9.81X400) ] = 0.02812m3/s
But discharge Q = AXV =
π 2
d XV
4
π 2
d X V = 0.02812
4
V = (0.02812 X 4) / 3.14 X d2 = 0.0358 / d2----------(1)
Total head lost due to friction,
hf =
but,
4 fXLXV 2
dX 2 g
hf = 100m
100 = hf =
4 XfXL XV 2
d X 2g
=
4 X 0.0065 X 2000 XV 2 2.65 XV 2
=
dX 2 X 9.81
d
= (2.65 / d) X (0.358/d2)2 = 0.003396 / d5
from equation (1),
V = 0.0358 / d2
18
100 = 0.003396 / d5
d = (0.00396 / 100)1/5 = 0.1277m = 127.7mm.
Efficiency of power transmission is given by equation
η=
H −hf
H
=
500 − 100
500
= 0.80 = 80%
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