This question paper consists of 3 printed pages,
each of which is identified by
the Code Number CIVE 140001
Solution
 UNIVERSITY OF LEEDS
May/June 2003
Examination for the degree of
BEng/ MEng
Civil Engineering
FLUID MECHANICS
Time allowed: 2 hours
Attempt 4 questions
L = 0.9m
x = 0.9 cos 60 = 0.45m
The gate opens when the moment at the pivot is clockwise.
That is when the moment due to the water > (0.45/2)w.
Method 1
Force on plane = Area × Pressure at centroid
Useful formulae:
Parallel axis theorem I oo = I GG + Ax 2 ,
2nd moment of area for a rectangle, I GG =
1.
y

Force = (1.0 × 0.9 )×  2.0 −  ρg
2

= 0.9 × (2.0 − 0.779 / 2 )1000 × 9.81
= 14219 N
bd 3
bd 3
, and for a triangle I GG =
12
36
A rectangular sluice gate, 0.9m × 1.0m, is fitted at the base of a reservoir wall with a pivot
in the arrangement shown in Figure 1. The gate will open when the the level of water in
the reservoir reaches a certain depth.
Determine the weight, W, that must be applied at the centre of the gate, if a water level of
h = 2.0m will just cause the gate to open.
(25 marks)
Method 2
Horizontal force = Hf = Area of projection on vertical plane × pressure at centroid
y

H f = 1.0 y 2.0 −  ρg
2

= 12307 N
Vertical force = weight of water above gate
x× y 

V f =  2.0 x −
 ρg1.0
2 

= 7109
Resultant force
2
R = H f +Vf
= 14213 N
Point of action of the force = centre of pressure
Figure 1
1
2
2
2
= 2.309m
sin 60
0.9
x = L2 −
= 1.859m
2
L2 =
2nd moment of area about O I oo
Sc =
=
Ax
1st moment of area about O
2
I oo = I GG + Ax
Sc =
For a rectangle, I GG =
2.
A comparison of the sensitivity of depth change to discharge is to be made between a
rectangular and v-notch a weir.
Initially a rectangular notch weir of width 0.5m and coeeficient of discharge of 0.9 is placed
at the downstream end of a channel carrying 2m3/s of water.
What will be the height above the base of the notch.
[8 marks]
I GG
+x
Ax
The experiment is repeated this time with a 70o V-notch weir with a the coefficient of
discharge of 0.8.
bd 3
12
d2
+x
12 x
0.9 2
=
+ 1.859
12 × 1.859
= 1.895m
Sc =
What will be the height above the base of the notch for the same discharge?
[8 marks]
When the height above the base in either weir is 0.5m what will be the percentage increase
in flow if the level rises by 0.1m?
[7 marks]
Need to find the lever arm, i.e. the distance from the pivot to the centre of pressure specified by Sc.
First find the position of the pivot, x2, from the surface (along the inclined plane)
x2 = L2 – 0.9 = 1.409m
Lever arm, x1 = Sc - x2 = 1.895 – 1.409 = 0.486 m
(Derive all formulae assuming the Bernoulli equation)
Solution
A General Weir Equation
Consider a horizontal strip of width b and depth h below the free surface, as shown in the figure below.
Take moments to fine the weight of the gate, w
Rx1 = 0.225w
w=
14213 × 0.486
= 30700 N
0.225
b
H
h
δh
Elemental strip of flow through a notch
Assuming the velocity is only due to the head.
velocity through the strip, u = 2 gh
discharge through the strip, δQ = Au = bδh 2 gh
Integrating from the free surface,
discharge
h = 0 , to the weir crest, h = H gives the expression for the total theoretical
H
Qtheoretical = 2 g ∫ bh 2 dh
1
0
This will be different for every differently shaped weir or notch. To make further use of this equation we need an
expression relating the width of flow across the weir to the depth below the free surface.
3
4
For a rectangular weir the width does not change with depth so there is no relationship between b and depth h. We
have the equation,
b = constant = B
8
θ 
2 g tan  H 5/ 2
 2
15
Rectangular weir
B
H
Qactual = Cd
Q (m^3/s)
C_d
B (m)
H
check Q
2.000
0.900
0.500
1.313
2.000
V-notch weir
A rectangular weir
Q (m^3/s)
θ
C_d
H
check Q
Substituting this into the general weir equation gives
H
Qtheoretical = B 2 g ∫ h 2 dh
1
0
=
2
B 2 g H 3/ 2
3
To calculate the actual discharge we introduce a coefficient of discharge, Cd , which accounts for losses at the
edges of the weir and contractions in the area of flow, giving
Qactual = Cd
2
B 2 g H 3/ 2
3
For the “V” notch weir the relationship between width and depth is dependent on the angle of the “V”.
b
θ
“V” notch, or triangular, weir geometry.
θ
Vnotch
H (m)
Q (m^3/s)
H (m)
Q (m^3/s)
% increase
0.500
0.234
0.600
0.369
57.744
Rectangular
H (m)
Q (m^3/s)
H (m)
Q (m^3/s)
% increase
0.5
0.470
0.6
0.618
31.453
h
H
If the angle of the “V” is
2.000
70.000 1.221729
0.800
1.180
2.000
then the width, b, a depth h from the free surface is
θ 
b = 2( H − h) tan 
 2
So the discharge is
θ 
Qtheoretical = 2 2 g tan  ∫ ( H − h) h 1/ 2 dh
 2 0
H
2

 θ  2
= 2 2 g tan   Hh 3/ 2 − h 5/ 2 
 2 5
5
0
H
=
8
θ 
2 g tan  H 5/ 2
 2
15
The actual discharge is obtained by introducing a coefficient of discharge
5
6
φ ( u, d , ρ , ν , p ) = 0
3.
Water flows through a 2cm diameter pipe at 1.6m/s. Calculate the Reynolds number and
find also the velocity required to give the same Reynolds number when the pipe is
transporting air.
[5 marks]
Assuming the pressure loss along a pipe, p, can be expressed in terms of the following
fluid density
ρ
kinematic viscosity ν
diameter
d
velocity
u
show that the pressure loss can be expressed as:
p = ρu 2φ (Re )
φ (π 1 , π 2 ) = 0
π1 = ua d b ρ c ν
1
Water
1000 kg/m3
1.31×10−6m2/s
air
1.19kg/m3
15.1×10−6m2/s
M 0 L0 T 0 = ( LT −1 )
M]
L]
Draw up the table of values you have for each variable:
variable
water
air
u
1.6m/s
uair
p
pwater
pair
1000 kg/m3
1.19kg/m3
ρ
−6 2
ν
1.31×10 m /s
15.1×10−6m2/s
d
0.02m
0.02m
Kinematic viscosity is dynamic viscosity over density = ν = µ/ρ.
The Reynolds number =
ρud ud
Re =
=
µ
ν
ud
ν
=
( L)b ( ML−3 ) L2 T −1
1
c1
T]
π1 = u −1d −1 ρ 0ν
ν
=
ud
And the second group π2 :
(note p is a pressure (force/area) with dimensions ML-1T-2)
M 0 L0 T 0 = ( LT −1 )
a1
( L) b ( ML−3 ) MT −2 L−1
1
c1
0 = c2 + 1
0 = a2 + b2 - 3c2 - 1
-2 = a2 + b2
0 = -a2 - 2
a2 = - 2
b2 = 0
T]
π 2 = u −2 ρ −1 p
=
p
ρu 2
So the physical situation is described by this function of nondimensional numbers,
 ν
p 
,
 =0
 ud ρu 2 
φ (π1 , π 2 ) = φ 

p 
=0
ρu 2 

p = ρu 2φ (Re )
. × 0.02
16
= 24427
. × 10 − 6
131
To calculate Reair we know,
Re water = Re air
For dynamic similarity these non-dimensional numbers are the same for the both water and air in the pipe.
π 1air = π 1water
uair 0.02
15 × 10 − 6
= 18.44m / s
24427 =
uair
a1
φ  Re,
Reynolds number when carrying water:
Re water =
c2
0 = c1
0 = a1 + b1 - 3c1 + 2
-2 = a1 + b1
0 = -a1 - 1
a1 = -1
b1 = -1
M]
c2 = -1
L]
Solution
1
π2 = u d ρ p
b2
As each π group is dimensionless then considering the dimensions, for the first group, π1:
Hence find the ratio of pressure drops in the same length of pipe for both cases.
[20 marks]
You will need to use these physical properties:
variable
ρ
ν
1
a2
π 2 air = π 2 water
We are interested in the relationship involving the pressure i.e. π2
To obtain the ratio of pressure drops we must obtain an expression for the pressure drop in terms of governing
variables.
Choose the three recurring (governing) variables; u, d, ρ.
From Buckinghams π theorem we have m-n = 5 - 3 = 2 non-dimensional groups.
 p 
 p 
 2 = 2
 ρu  air  ρu  water
2
pwater ρwater u water
=
2
ρair uair
pair
=
7
. 2
1000 × 16
1
=
= 6.327
. × 18.44 2 0158
.
119
8
4
Describe with the aid of diagrams the following phenomena explaining why and when
they occur.
(Each part requires at least a half page description of the phenomenon plus diagrams.)
(i) The laminar boundary layer
(5 marks)
(ii) The turbulent boundary layer
5. Water flows horizontally along a 200mm pipeline fitted with a 90o bend that moves the
water vertically upwards. The diameter at the outlet of the bend is 100mm and it is 0.5m
above the centreline of the inlet. If the flow through the bend is 150 litres/s, calculate the
magnitude and direction of the resultant force the bend support must withstand. The volume
of the bend is 0.01m3 and the pressure at the outlet is 100 kN/m2.
[25 marks]
(5 marks)
100mm
(iii) The laminar sublayer
(5 marks)
(iv) Boundary layer separation
(5 marks)
(v) Methods to prevent boundary layer separation
(5 marks)
Solution
p = 100 kN/m2
200mm
Answer: As the question says - EACH PART REQUIRES AT LEAST HALF A PAGE DESCRIPTION PLUS
DIAGRAMS - take from lecture notes AND other books.
Figure 2
Solution
Force on a pipe bend question
Enter values in the yellow boxes
Pressure in (kN/m^2)
Inflow (litres/s)
Volume of bend (m^3)
Bend angle (degrees)
Height difference (m)
Head loss in bend (m)
Inlet diameter (m)
outlet diameter (m)
100
150
0.01
90
0.5
0
0.2
0.1
Q (m^3/s)
inlet area (m^2)
outlet area (m^2)
inlet vel (m/s)
outlet vel (m/s)
Angle (rad)
0.15000
0.03142
0.00785
4.77465
19.09859
1.57080
Total force
Ftx (N)
Fty (N)
Pressure force
P outlet (N/m^2)
Fpx
Fpy
Body force
Fbx
Fby
Resultant force
9
-716.19724
2864.78898
-75884.49735
3141.59265
595.99545
0.00000
-98.10000
10
Frx
Fry
Fr
Angle (degrees)
-3857.78990
2366.89353
4526.00573
-31.53062
Force acting on bend (N)
-4526.00573
Starting with the Bernoulli and Continuity equations derive the following expression
that can be used to measure flow rate with a Venturi meter.
6 (a)
Q actual = C d A1 A2

 p − p2
2g  1
+ z1 − z 2 

 ρg
A12 − A22
Also show that when the pressure difference is measured using a manometer the
following expression can be used
ρ

2 gh man − 1
 ρ

A12 − A22
Qactual = Cd A1 A2
[15 marks]
(b)
A venturimeter is used to measure the flow of water in a pipe of diameter 100mm. The
throat diameter of the venturimeter is 60mm and it has a coefficient of discharge of 0.9.
When a flow of 100 litres/s is flowing the attached maonmeter shows a head difference
of 60cm, what is the density of the manometric fluid of the manometer?
Solution
6(a):
Applying Bernoulli along the streamline from point 1 to point 2 in the narrow throat of the Venturi meter we have
p1 u12
p2 u22
+
+ z1 =
+
+z
ρg 2 g
ρg 2 g 2
By the using the continuity equation we can eliminate the velocity u2,
Q = u1 A1 = u2 A2
u2 =
u1 A1
A2
Substituting this into and rearranging the Bernoulli equation we get
2

p1 − p2
u 2  A 
+ z1 − z2 = 1  1  − 1
ρg
2 g  A2 

u1 =

 p − p2
+ z1 − z2 
2g 1

 ρg
2
 A1 
  −1
 A2 
= A2

 p − p2
+ z1 − z2 
2g 1

 ρg
A12 + A22
To get the theoretical discharge this is multiplied by the area. To get the actual discharge taking in to account the
losses due to friction, we include a coefficient of discharge
11
12
Qideal = u1 A1
Qactual = Cd Qideal = Cd u1 A1
Qactual = Cd A1 A2
6 (b)

Q 

  C d A1 A2 

ρ man = ρ  
 p1 − p2

2g
+ z1 − z2 
ρ
g


A12 − A22
p1 + ρgz1 = p2 + ρman gh + ρg ( z2 − h)
ρ

p1 − p2
+ z1 − z2 = h man − 1
ρg
 ρ

a1 (m^2)
a2 (m^2)
Thus the discharge can be expressed in terms of the manometer reading::
Qactual = Cd A1 A2
out
1000
100
60
Q (m^3/s)
0.1
h (m)
0.6
rho man (kg/m^3) 115182.5
1
z2
in
z1
datum
13
0.9
0.007854
0.002827
2
h
 A12 − A22  
 + 1

 2 gh  
cd
rho man (kg/m^3) ?
rho (kg/m^3)
d1 (mm)
d2 (mm)
This can also be expressed in terms of the manometer readings
ρ

2 gh man − 1
 ρ

A12 − A22
2
14