CIVE 1400: FLUID MECHANICS
Examination May/June 1996
Model answers.
1(a)
State Buckingham’s  Theorems and explain the uses of dimensional analysis.
(8 marks)
1(b)
An apparatus is used to measure the pressure drop in a pipe of 3cm diameter in which water is
flowing at 1.1 m/s. Use Buckingham’s  Theorems to calculate the velocity of air in a 2 cm
diameter pipe which will give kinematically similar conditions.
If the pressure drop over a certain length of pipe bearing water is 1 kN/m2, what is the equivalent
pressure drop in the pipe bearing air?
For water kinematic viscosity was 1.31  10-6 m2/s and the density 1000 kg/m3. For air those
quantities were 15.1  10-6 m2/s and 1.19 kg/m3.
(12 marks)
1(a):
There are two theorems accredited to Buckingham, and know as his  theorems.
1st  theorem:
A relationship between m variables (physical properties such as velocity, density etc.) can be expressed as
a relationship between m-n non-dimensional groups of variables (called  groups), where n is the number
of fundamental dimensions (such as mass, length and time) required to express the variables.
2nd  theorem
Each  group is a function of n governing or repeating variables plus one of the remaining variables.
In engineering the application of fluid mechanics in designs make much of the use of empirical results
from a lot of experiments. This data is often difficult to present in a readable form. Even from graphs it
may be difficult to interpret. Dimensional analysis, for which the Buckingham  theorems give a good
strategy to perform, provides a method for choosing relevant data and how it should be presented.
If it is possible to identify the factors involved in a physical situation, dimensional analysis can form a
relationship between them.
Often hydraulic structures are too complex for simple mathematical analysis and a hydraulic model is
build. Usually the model is less than full size but it may be greater. The real structure is known as the
prototype. Measurements taken from the model require a suitable scaling law to predict the values in the
prototype. Dimensional analysis can help derive this.
1
1(b):
If p is the pressure drop over the length of pipe.
The variables which govern laminar flow in a pipe are:
Name
Symbol
Dimension
pressure drop
p
ML-1T-2
length
L
L
density

ML-3
diameter
D
L
velocity
u
LT-1
coeff. Dynamic viscosity

ML-1T-1
roughness height
k
L
So the defining function can be written:
 ( p, L, , u, D, , k ) = 0
There are 7 variables so m = 7
There are 3 dimensions so n = 3
Number of  groups = m -n = 7 - 3 = 4 groups.
i.e.
 ( 1, 2, 3, 4 ) = 0
Choose  , u, D as the governing (or repeating variables).
Group 1:
1 = a ub Dc p
In terms of dimensions:
M0 L0 T0 = Ma L-3a Lb T-b Lc M L-1 T-2
M:
L:
T:
0 = a +1
0 = -3a + b + c -1
0 = -b -2
a = -1 , b = -2 , c = 0
1 =  u-2 p =
p
u 2
2
Group2
2 = a ub Dc L
In terms of dimensions:
M0 L0 T0 = Ma L-3a Lb T-b Lc L
M:
L:
T:
0=a
0 = -3a + b + c +1
0 = -b
a = 0 , b = 0 , c = -1
2 = D-1 L =
L
D
Group3
3 = a ub Dc 
In terms of dimensions:
M0 L0 T0 = Ma L-3a Lb T-b Lc M L-1 T-1
M:
L:
T:
0=a+1
0 = -3a + b + c -1
0 = -b - 1
a = -1 , b = -1 , c = -1
3 =  u-1 D-1  =

uD
Group4
4 = a ub Dc k
In terms of dimensions:
M0 L0 T0 = Ma L-3a Lb T-b Lc M L-1 L
M:
L:
T:
0=a
0 = -3a + b + c +1
0 = -b
a = 0 , b = 0 , c = -1
4 = D-1 k =
k
D
3
Note that this is the same as 2
So
 p L  k 
, ,
, 0
 u 2 D uD D 
2 
writing 1a
L
= 2 / 1 =
D
p
L u 2
, And inverting 3 which gives Re.

u 2 p D
 L u 2 L  k 
2 
, ,
,  0
 p D D uD D 
p u 2  L k


3  , , Re
D D

L
D
For kinematically similar conditions the Reynolds number is the same for both air and water:
uD uD
Re 



Re air  Re water
u  0.02
11
.  0.03
6 
151
.  10
1.31  10  6
u air  19.02 m / s
For pressure drop:
 L u 2 
 L u 2 

 

 p D  air  p D  water
p air
Dw u a2 a
 p w
Da u w2 w
p air  533.7 N / m2
4
CIVE 1400: FLUID MECHANICS
Examination May/June 1996
Model answers.
2(a)
Obtain the expression for the centre of pressure of an irregular plane surface wholly submerged in
a fluid.
(8 marks)
2(b)
A gate which is a quarter of a circle or radius holds back 2.0 m of water as shown in the diagram.
Figure 1
Calculate the magnitude of the resultant hydrostatic force on a unit length of the gate.
(12 marks)
2(a):
This plane surface is totally submerged in a liquid of density  and inclined at an angle of  to the
horizontal. Taking pressure as zero at the surface and measuring down from the surface, the pressure on
an element A , submerged a distance z, is given by
p  gz
and therefore the force on the element is
F  pA  gzA
The resultant force can be found by summing all of these forces i.e.
R  g  zA
(assuming  and g as constant).
5
The term
Az i.e.
 zA is known as the 1st Moment of Area of the plane PQ about the free surface. It is equal to
 zA  Az
= 1st moment of area about the line of the free surface
where A is the area of the plane and z is the depth (distance from the free surface) to the centroid, G. This
can also be written in terms of distance from point O ( as z  x sin )
 zA  Ax sin
 1st Moment of area about a line through O  sin
Thus:
The resultant force on a plane
R  gAz
 gAx sin
This resultant force acts at right angles to the plane through the centre of pressure, C, at a depth D. The
moment of R about any point will be equal to the sum of the moments of the forces on all the elements
A of the plane about the same point. We use this to find the position of the centre of pressure.
It is convenient to take moments about the point where a projection of the plane passes through the
surface, point O in the figure.
Moment of R about O = Sum of moments of force
on all elements of A about O
We can calculate the force on each elemental area:
Force on A  gzA
 g s sin  A
And the moment of this force is:
Moment of Force on A about O  g s sin  A  s
 g sin  As 2
 , g and  are the same for each element, so the total moment is
Sum of moments of forces on all elemets of A about O  g sin  s2A
We know the resultant force from above R  gAx sin , which acts through the centre of pressure at C,
so
Moment of R about O = gAx sin  S c
Equating gives,
gAx sin Sc  g sin  s 2A
Thus the position of the centre of pressure along the plane measure from the point O is:
 s A

2
Sc
Ax
6
It look a rather difficult formula to calculate - particularly the summation term. Fortunately this term is
known as the 2nd Moment of Area , I o , of the plane about the axis through O and it can be easily
calculated for many common shapes. So, we know:
2nd moment of area about O  I o   s2A
And as we have also seen that Ax  1st Moment of area about a line through O,
Thus the position of the centre of pressure along the plane measure from the point O is:
Sc 
2 nd Moment of area about a line through O
1st Moment of area about a line through O
and
depth to the centre of pressure is
D  S c sin 
To calculate the 2nd moment of area of a plane about an axis through O, we use the parallel axis theorem
together with values of the 2nd moment of area about an axis though the centroid of the shape obtained
from tables of geometric properties.
The parallel axis theorem can be written
I o  I GG  Ax 2
where I GG is the 2nd moment of area about an axis though the centroid G of the plane.
Using this we get the following expressions for the position of the centre of pressure
I GG
x
Ax
I

D  sin   GG  x 
 Ax

Sc 
7
2(b):
Horizontal force:
Rh 
gh 2
2
 9810  2 2  0.5  19628.4 N
Vertical force:
Sector from centre of gate to where water surface touches is angle
cos  = 2/4,
 = 60 which is 60/360 =1/6 of a circle
Rv = weight of imaginary water
Rv = g ( 1/6 of the circle - the triangle )
42 = 22 + x2
x = 3.46 m
  4 2 3.46  2 
Rv  9810  

  48260 N
2 
 6
Total thrust:
R  Rh2  Rv2  52099 N
This acts a the angle:
 Rv 
  67.87 
 Rh 
  tan 1 
8
CIVE 1400: FLUID MECHANICS
Examination May/June 1996
Model answers.
3(a)
Where does most of the energy loss occur in a Venturi meter and why is this the case?
(8 marks)
3(b)
A Venturi meter is being calibrated in a laboratory. The meter is lying horizontally and has a
diameter of 75 mm at the entrance and 50 mm at the throat. The flow rate is obtained by measuring
the time required to collect a certain quantity of water. The average number of such measurements
gives 0.614 m3 of water flowing in 55.82 seconds. If the pressure gauge at the throat reads 20
kN/m2 less than that at the entrance, calculate the head loss due to friction using the Bernoulli
equation.
(12 marks)
3(a):
3(b):
d1 = 75 mm = 0.075 m
d2 = 50 mm = 0.05 m
p2 - p1 =20 kN / m2 = 20 000 N / m2
Apply Bernoulli from 1 to 2
p1 u12
p 2 u 22

z 

 z  hf
g 2g 1 g 2g 2
As horizontal then z1 = z2, rearranging gives:
p1  p2 u12  u22

 z1  hf
g
2g
p1 - p2 = 20 000 N/m2
By continuity
Q = au = a1u1 = a2u2
so
d12 u1 = d22 u2
4Q
 2.49
d 12
4Q
u2 
 5.602
d 22
u1 
Substituting in the equation for hf gives;


2
2
20000 2.49  5.602

 hf
9810
19.62
hf  0.755 m
9
10
CIVE 1400: FLUID MECHANICS
Examination May/June 1996
Model answers.
4
A pipeline of constant 0.6 m diameter with its centre line in the horizontal plane turns through an
angle of 75. The pipeline carries water at the rate of 0.85 m3/s. A pressure gauge at the bend
indicates that the pressure is equivalent to 41.3 m of water. Calculate the force exerted on the bend
by the water and the direction it acts.
(20 marks)
As constant diameter p1 = p2 = p, A1 = A2 = A and u1 = u2 = u
A = d/4 = 0.2827 m2
u = Q/A = 3.006 m/s
p = 41.3 m of water
p = 41.3  1000  9.18 = 405 153 N/m2
 = 75
Calculate the total force
In the x-direction:

FT x  Q u 2 x  u1 x

u1 x  u
u 2 x  u cos 
FT x  Qu cos   u
In the y-direction:

FT y  Q u 2 y  u1 y
u1 y  u sin 0  0
u 2 y  u sin 
FT y  Qu sin 
11

Calculate the pressure force
FP  pressure force at 1 - pressure force at 2
FP x  p1 A1 cos 0  p 2 A2 cos   p1 A1  p 2 A2 cos 
 pA1  cos  
FP y  p1 A1 sin 0  p 2 A2 sin    p 2 A2 sin 
  pA sin 
Calculate the body force
There are no body forces as the pipe is in the horizontal plane.
Calculate the resultant force
FT x  FR x  FP x  FB x
FT y  FR y  FP y  FB y
FR x  FT x  FP x  0
 Q u cos  u  pA1  pA cos 
 86786 N
(i.e. to the left)
FR y  FT y  FP y  0
 Qu sin   pA sin 
 113102 N
And the resultant force on the fluid is given by
FR 
FR2 x  FR2 y  142 563 N
And the direction of application is
 FR y 
  52.5
 FR x 
  tan 1 
This is in the direction , to the left and up.
The force on the bend is the same magnitude but in the opposite direction
R   FR
12
CIVE 1400: FLUID MECHANICS
Examination May/June 1996
Model answers.
5(a)
Using the Bernoulli equation, show that the discharge through an orifice is given by
Q  Cd Ao 2gh where Ao is the area of the orifice and h is the head of water above the orifice.
(5 marks)
5(b)
A tank of water is 5.6 m by 4.3 m in plan with vertical sides. Water from the tank discharges to the
atmosphere through a 200 mm diameter orifice in the base. Over a period of 5 mins 7 secs the
water level drops from 1.9 m to 0.7 m above the orifice. What is the value of the coefficient of
discharge of the orifice? Work from first principles.
(15 marks)
5(a):
The general arrangement and a close up of the hole and streamlines are shown in the figure below
Tank and streamlines of flow out of the sharp edged orifice
The streamlines contract after the orifice to a minimum value when they all become parallel, at this point,
the velocity and pressure are uniform across the jet. This convergence is called the vena contracta
Apply Bernoulli along the streamline joining point 1 on the surface to point 2 at the centre of the orifice.
p1 u12
p2 u22

z 

z
g 2 g 1 g 2 g 2
At the surface:
velocity is negligible (u1 = 0)
pressure atmospheric (p1 = 0).
At the orifice the jet is open to the air so again
pressure is atmospheric (p = 0).
If we take the datum line through the orifice then z1 = h and z2 =0, leaving
u22
h
2g
u2  2 gh
13
This is the theoretical value of velocity.
Friction losses have not been taken into account. To incorporate friction we use the coefficient of velocity
to correct the theoretical velocity,
uactual  Cv utheoretical
The actual area of the jet is the area of the vena contracta not the area of the orifice. We obtain this area
by using a coefficient of contraction for the orifice
Aactual  Cc Aorifice
So the discharge through the orifice is given by
Q  Au
Qactual  Aactual uactual
 Cc Cv Aorifice utheoretical
 Cd Aorifice utheoretical
 Cd Aorifice 2 gh
Where Cd is the coefficient of discharge, and Cd = Cc  Cv
We can integrate this expression to get the time the level in the tank takes to fall a certain amount.
Tank emptying from level h1 to h2.
The tank has a cross sectional area of A.
In a time dt the level falls by dh or the flow out of the tank is
Q  Av
Q  A
h
t
(-ve sign as h is falling)
Rearranging and substituting the expression for Q through the orifice gives
t 
A
h
Cd Ao 2 g h
14
This can be integrated between the initial level, h1, and final level, h2, to give an expression for the time it
takes to fall this distance
t
A
Cd Ao 2 g
h2
h
h1
h


A
2 h
Cd Ao 2 g

 2A
Cd Ao 2 g



h2
h1
h2  h1

5(b):
A = 4.3  5.6 = 24.08 m
h1 = 1.9 m
h2 = 0.7 m
do = 0.20 m
Ao = do2/4 = 0.0314 m2
Time for fall in the level = 5  60 + 7 = 307 sec.
Substituting these into the equation gives:
Cd 
 2A
tAo 2 g

h2  h1

 2  24.08
307  0.0314 19.62
Cd  0.611
Cd 
15

0.7  19
.

CIVE 1400: FLUID MECHANICS
Examination May/June 1996
Model answers.
6(a)
Use the Bernoulli equation to show that the relationship between flow and depth over a sharp8

tan
2 g H 5/ 2
edged triangular weir is given by Q  Cd
15
2
(10 marks)
6(b)
A rectangular weir and a V-notch weir are located in parallel channels of the same dimensions.
Both weirs have an opening 0.3 m wide at the top and 0.3 m deep. Both have a Cd of 0.6. What
head would be required over the rectangular weir to pass the same flow as over the V-notch weir
when it has a head of 0.29 m?
(For a rectangular weir Q  Cd
2
b 2 gH 3/ 2 )
3
(10 marks)
6(a):
A General Weir Equation
Consider a horizontal strip of width b and depth h below the free surface, as shown in the figure below.
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, Q  Au  bh 2 gh
Integrating from the free surface, h  0 , to the weir crest, h  H gives the expression for the total
theoretical discharge
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.
16
For the “V” notch weir the relationship between width and depth is dependent on the angle of the “V”.
“V” notch, or triangular, weir geometry.
If the angle of the “V” is  then the width, b, a depth h from the free surface is
 
b  2 H  h tan 
 2
So the discharge is
 
 2 2 g tan    H  h h 1/ 2 dh
 2 0
H
Qtheoretical
2
   2

 2 2 g tan   Hh 3/ 2  h 5/ 2 
 2 5
5
0
H

8
 
2 g tan  H 5/ 2
 2
15
The actual discharge is obtained by introducing a coefficient of discharge
Qactual  Cd
8
 
2 g tan  H 5/ 2
 2
15
6(b):
Equating the two weir equations:
Cd
8
2
 
2 g tan  HV5/ 2  Cd b 2 g H R3/ 2
 2
15
3
Cd is the same for both equations.
HR = 0.29 m
b = 0.3m
Substituting these into the above equation gives.
8
5/ 2
0.5 0.29  0.2 H R3/ 2
15
H R  01539
.
m
17
CIVE 1400: FLUID MECHANICS
Examination May/June 1996
Model answers.
7
A plunger of diameter 0.1 m and length 0.15 m has five small holes of diameter 2 mm drilled
through it in the direction of its length. The plunger fits closely inside a cylinder containing oil,
such that no oil passes between the plunger and the cylinder. Calculate the force which must be
applied to the plunger, in a downward vertical direction, to make the plunger fall with a speed of
0.0005 m/s. Assume that the upwards flow through the small holes is laminar and that the
coefficient of viscosity of the oil is 0.2 kg/ms.
(20 marks)
Velocity, u = 0.0005 m/s
viscosity  = 0.2 kg m-1 s-1
length = 0.15m
hole diameter, d = 2mm = 0.002m
Plunger diameter D = 0.1m
The Hagen-Poiseuille equation for head loss during laminar flow in a pipe is:
hf 
32 Lu
gd 2
Pressure loss is given by
p  gh f 
32 Lu
d2
Pressure difference between top and bottom of each hole is:
32  0.2  015
.  0.0005
0.002 2
 120 N / m2
p 
So we need a pressure of 120 N/m2 at the bottom of the cylinder.
Pressure = Force / area
   D 2  5d 2  

Force  120  
4


 0.94 N
18