Physics Assignment-2 | Task Report | Doc

Task 1
a) Pressure in the pipe = gh = 1000 X 9.81 X 100 = 981000N/⍴ m2 = 981Kn/m2
The pressure in the open channel will equal to atmospheric pressure because velocity
will increase which will compensate with the pressure up to atmospheric pressure.
b) Forces of resistance in the pipe flow
i) Frictional resistance: Due to viscosity of water velocity near the surface of
pipe is zero it gradually increases up to max value at some height above the
surface of pipe. Due to there is loss of head as the fluid flows
ii) Boundary layer resistance: At boundary of pipe local velocity is flow is less at
some distances above the boundary layer the local velocity attains maximum
value. Due to the lesser velocity at boundaries of pipe there is loss head in pipe
flow.
c) Effect of temperature on losses:
Viscosity decreases as the temperature of the fluid increases. Frictional losses and
thickness of boundary layer is directly proportional to viscosity therefore as
temperature increases the amount of these loses in the pipe flow decreases.
d) Difference between laminar flow and turbulent flow(Douglas et al. 2011)
Laminar flow Turbulent flow
In laminar flow subsequent layers do not
interfere with each other as the fluid
flows. Flow is smooth
In turbulent flow subsequent layer
crosses each other’s path resulting in
formation of eddy
Velocity of flow is lower in laminar flow
as compare to turbulent flow
Velocity of flow is higher in turbulent
flow as compare to laminar flow
Reynolds number is lesser than 2000 for
pipe flow and lower than 500 for open
channel flow
Reynolds number is higher than 2000 for
pipe flow and greater than 500 for open
channel flow.
e) Reynolds number:
Reynolds number is defined as ratio of inertia force to the viscous force. Reynolds
number is the criteria for differentiating between laminar, turbulent and transition
flow

f) Boundary layers: Due viscosity of fluid velocity of flow at surface is zero and it
gradually attains maximum velocity at some distance above the surface. Distance at
which fluid attains 99 percent of maximum velocity is called boundary layer. When
roughness of more boundary layer thickness increases. (Douglas et al. 2011)
g) Resistance of the flowing fluid can be reduced by reducing viscosity of fluid.
Viscosity of fluid can be reduced by
i) Increasing temperature: By increasing temperature inter particle forces
resulting in reduced viscosity.
ii) By reducing hardness of water: hardness can be reduced by lime soda process,
ziolite process etc.
h) A parallel pipe can be laid adjacent to exiting pipe which will increase discharge.
Task 2
a) Discharge is 30m3/s, width of flow channel is 2m and n is .02 we need to depth of
flow.
By Manning’s equation
Q = 1
n R
2
3 S
1
2 A
R = A
P =
π
4 D2
πD
= D
4
Slope S is not given assuming it to be 0.004
30 = 1
0.02 ( 0.25 D)
2
3 0.004
1
2 X 30 X 2 X D
D = 0.695m

b) Discharge is 10m3/s, length of pipe is 2KM, friction factor is 0.006 and diameter of
pipe is equal to 1.5m. We need to find head loss in pipe.
Sol) Applying Darcey Weisbach equation.
Hf = 8 f
π2
L
g
Q2
d5
= 8 X 0.006
π2
X 2000 X
g
102
1.55 = 13.05m
c) Discharge is 10m3/s, length of pipe is 2000Km, head available is 50m and friction
factor is 0.006 We need to find the diameter of pipe
Applying Darcy wiesbach equation.
Hf = 8 f
π2
L
g
Q2
d5
50 = 8 X 0.006 X
π2
2000 X
g
102
d5
d = 1.1467m
To get discharge of 10m3/s diameter of pipe required will be 1.1467
d) Difference between pipe flow and an open channel
Flow in pipe flow is due
For a flow of 10m3/s and depth of 2m, the dimension of the open channel flow
required is.
Assuming rectangular channel
The critical section will be required to make channel economical.
Critical depth of flow = 3
√ q2
g = 3
√ Q2
gB2