A lecture of fluid mechanics

1. CEE-235
Credit: 3.0
(Session: 2019-20)
Fluid Mechanics
Dr. Bijit Kumar Banik
Professor
Department of Civil and Environmental Engineering
Shahjalal University of Science and Technology
Contact: bijit_sustbd@yahoo.com; bijit-cee@sust.edu
Lecture-6
Force exerted on vane/blade

2. Impulse-Momentum Principle for steady flow with average velocity
Momentum correction factor (β)
Momentum correction factor (β)
As velocity is not
uniform over a section,
the momentum per unit
time transferred across
that section will be
different (for laminar
flow, it is greater than
that computed by mean
velocity)

3. Thus rate of momentum transferred across an elementary area ,
where local velocity is , is,
Momentum transferred for entire section (mV)actual,
(mV)actual
(mV)actual
Momentum computed by mean velocity (mV)mean,
(mV)mean
(mV)mean
Hence, the correction factor,
(mV)actual
(mV)mean
(mV)actual
(mV)mean
For circular pipe,
Laminar flow =4/3
Turbulent flow = 1.005 to 1.05

4. Problem 6.1 [p-176 Franzini]: In laminar flow through a circular pipe the
velocity profile is a parabola, the equation of which is ,
where u is the velocity at any radius r, is the maximum velocity in the
center of the pipe where r=0 and is the radius to the wall of the pipe.
Average velocity for laminar flow in a circular pipe, V =0.5 . Prove =4/3.
Home Work for next class

5. ∑ Fx = P1A1 − P2A2 − (FR/F)x = Q (V2−V1)
∑ Fx = P1A1 − P2A2 − (FR/F)x = Q (V2−V1)
6.3 Force exerted on pressure conduits [PP-151, Franzini]
Flow in the horizontal plane

6. ∑ Fx = P1A1 − P2A2cos − (FB/F)x = Q (V2cos −V1)
∑ Fx = P1A1 − P2A2cos − (FB/F)x = Q (V2cos −V1)
If the fluid undergoes a change in both direction and velocity:
∑ Fy = - P2A2 + (FB/F)y = Q (V2sin )
∑ Fy = - P2A2 + (FB/F)y = Q (V2sin )
Force exerted by bend on fluid
FB/F = [(FB/F)x
2 + (FB/F)y
2]1/2
FB/F = [(FB/F)x
2 + (FB/F)y
2]1/2

7. Force exerted by nozzle
on liquid in y-direction
Force exerted by nozzle
on liquid in y-direction
Example 6.1: Determine the magnitude and direction of the resultant force
exerted on this double nozzle. Both nozzle jets have a velocity of 12 m/s .
The axes of the pipe and both nozzles lie in a horizontal plane.
kN/m3 (Neglect friction) [page 153, Franzini]

8. Solution:
Continuity Equation, A1V1 = A2V2 + A3V3
Continuity Equation, A1V1 = A2V2 + A3V3
A1= 152; A2 = 102; A3= (7.5)2
A1= 152; A2 = 102; A3= (7.5)2
152 V1 = 102 (12) + (7.5)2 (12)
152 V1 = 102 (12) + (7.5)2 (12)
V1 = 8.33 m/s
V1 = 8.33 m/s
Q1 = 2 * (8.33) = 0.147 m3/s
Q1 = 2 * (8.33) = 0.147 m3/s
Q2 = * (0.10)2 * 12 = 0.094 m3/s
Q2 = * (0.10)2 * 12 = 0.094 m3/s
Q3 = * (0.075)2 * 12 = 0.053 m3/s
Q3 = * (0.075)2 * 12 = 0.053 m3/s
Bernoulli’s Equation, at point (1) and (2):
Bernoulli’s Equation, at point (1) and (2):
(P1/ ) + (V1
2/2g) = (P2/ ) + (V2
2/2g)
(P1/ ) + (V1
2/2g) = (P2/ ) + (V2
2/2g)
(P1 / (9.81 1000)) + (8.332/(2 9.81)) = 0 + 122 / (2 9.81)
(P1 / (9.81 1000)) + (8.332/(2 9.81)) = 0 + 122 / (2 9.81) P2 = Atm. pressure = 0
P2 = Atm. pressure = 0

9. P1/(9.81 1000) = 3.80
P1/(9.81 1000) = 3.80
P1 = 37.3 103 N/m2 = 37.3 kN/m2
P1 = 37.3 103 N/m2 = 37.3 kN/m2
Applying Momentum equation:
Applying Momentum equation:
∑ Fx = P1A1 – (FN/L)x = ρQ2V2 cos150 + ρQ3V3 cos300 – ρQ1V1
∑ Fx = P1A1 – (FN/L)x = ρQ2V2 cos150 + ρQ3V3 cos300 – ρQ1V1
ρ = =
. /
. /
= 1.
. /
/
= 103 kg/m3
(37.3× 103) × (0.15)2– (FN/L)x=103×0.094×12cos150 + 103×0.053×12cos300 – 103×0.147 × 8.33
(37.3× 103) × (0.15)2– (FN/L)x=103×0.094×12cos150 + 103×0.053×12cos300 – 103×0.147 × 8.33

10. 659 – (FN/L)x = 1089.56 + 550.79 – 1224.51
659 – (FN/L)x = 1089.56 + 550.79 – 1224.51
(FN/L)x = 659 – 415.84 = 243.16
(FN/L)x = 659 – 415.84 = 243.16
(FN/L)x = 0.243 kN
(FN/L)x = 0.243 kN
∑ Fy = (FN/L)y = ρQ2V2 sin150 + ρQ3(-V3 sin300)– ρQ1V1
∑ Fy = (FN/L)y = ρQ2V2 sin150 + ρQ3(-V3 sin300)– ρQ1V1
0
VVI: Remember the sign convention
(FN/L)y = 103 0.094 12 sin150– 103 0.053 12 sin300
(FN/L)y = 103 0.094 12 sin150– 103 0.053 12 sin300
(FN/L)y = = - 26.05 N = - 0.026 kN
(FN/L)y = = - 26.05 N = - 0.026 kN i.e. the assumed direction was wrong
So, FN/L = [(0.243)2 + (0.026)2]1/2 = 0.244 kN
So, FN/L = [(0.243)2 + (0.026)2]1/2 = 0.244 kN (Ans.)
tan-1 .
.
= 6.10
tan-1 .
.
= 6.10 (Ans.)
Solve, Problem 6.4, 6.5, 6.6, 6.7, 6.8, 6.9 and 6.10. Book: Franzini, p 176-177
Solve, Problem 6.4, 6.5, 6.6, 6.7, 6.8, 6.9 and 6.10. Book: Franzini, p 176-177

11. 6.4: Force exerted on a stationary vane or blade (p-155 Franzini)
The mechanism of work and energy from fluid jets to moving vanes is studied
as an application of momentum principle.
Example 6.2 : In the fig. suppose that =30 , v1=30 m/s and the stream is a
jet of water with an initial diameter of 5 cm. Assume friction losses such that
v2=28.5 m/s. Find the resultant force on the blade. Assume that flow occurs in
a horizontal plane , =1000 kg/m3.

12. A= (0.05)2 =0.001963 m2
A= (0.05)2 =0.001963 m2
Momentum eqn at points 1 & 2
Momentum eqn at points 1 & 2
x=(-FB/W)x= Q(v2cos30 -v1)
x=(-FB/W)x= Q(v2cos30 -v1)
=1000 (0.001963 30)(28.5 cos30 -30)
=1000 (0.001963 30)(28.5 cos30 -30)
= -313.2 N
= -313.2 N
(FB/W)x =313.2 N
(FB/W)x =313.2 N
𝑦 = (FB/W)y = (v2sin30 -0)
𝑦 = (FB/W)y = (v2sin30 -0)
=1000 (0.001963 30)(28.5 sin30 )
= 839.2 N
B/W= = 895.7 N
B/W= = 895.7 N
=
.
.
=
.
.
313.2 N
839.2
N
B/W

13. 313.2 N 839.2
N
B/W
313.2 N
839.2
N
W/B
if friction was not considered, v1=v2=30 m/s
if friction was not considered, v1=v2=30 m/s
x=(-FB/w)x=(1000) (0.001963 30)(30cos30 -30)
(FB/w)x = 236.7 N
𝑦 =(FB/w)y=(1000) (0.001963 30) (30sin30 -0)
(FB/w)y =883.4 N
FB/w= = 914.6 N
=
.
.
=75
236.7 N
883.4
N
B/W
A= ×(0.05)2 =0.001963 m2
A= ×(0.05)2 =0.001963 m2

14. 313.2 N 839.2
N
B/W
313.2 N
839.2
N
W/B
x=(-FB/w)x=(1000) (0.001963 30)
(30cos30 -30)
x=(-FB/w)x=(1000) (0.001963 30)
(30cos30 -30)
𝑦 =(FB/w)y=(1000) (0.001963 30)
(30sin30 -0)
𝑦 =(FB/w)y=(1000) (0.001963 30)
(30sin30 -0)
236.7 N
883.4
N
B/W
friction was considered
236.7 N
883.4
N
W/B
friction was not considered
x=(-FB/W)x=1000 (0.001963 30)
(28.5cos30 -30)
x=(-FB/W)x=1000 (0.001963 30)
(28.5cos30 -30)
𝑦 = (FB/W)y =1000 (0.001963 30)
(28.5sin30 -0)
𝑦 = (FB/W)y =1000 (0.001963 30)
(28.5sin30 -0)
if if
Friction increase or decrease (FB/W)x ? Friction increase or decrease (FB/W)x ?
Friction increase or decrease (FB/W)y ? Friction increase or decrease (FB/W)y ?

15. if if
Friction increases (FB/W)x Friction decreases (FB/W)x
Friction decreases (FB/W)y Friction decreases (FB/W)y
If
-FR= 2 V1 ]
You need more force to clear the snow !

16. 6.5 : Relation between absolute and relative velocities (p-156, Franzini )

17. Thank you