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CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
Fluid Properties Examples
Explain why the viscosity of a liquid decreases while that of a gas increases with a temperaturerise.The following is a table of measurement for a fluid at constant temperature.Determine the dynamic viscosity of the fluid.
du/dy (rad s
-
W
(N m
Using Newton's law of viscosity
u y
w w
P
W
where
P
is the viscosity. So viscosity is the gradient of a graph of shear stress against vellocity
gradient of the above data, or
u y
w w
W
P
Plot the data as a graph: Calculate the gradient for each section of the line
du/dy (s
-
W
(N m
Gradient
Thus the mean gradient = viscosity = 4.98 N s / m
2
0
1
2
3
4
0
1
du/dy
Shear stress
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
The density of an oil is 850 kg/m
3
. Find its relative density and Kinematic viscosity if the dynamic
viscosity is 5
u
kg/ms.
U
oil
= 850 kg/m
3
U
water
= 1000 kg/m
3
J
oil
Dynamic viscosity =
P
u
kg/ms
Kinematic viscosity =
Q
P
U
1
2
6
3
u
u
s
m
P U
Q
The velocity distribution of a viscous liquid (dynamic viscosity
P
= 0.9 Ns/m
2
) flowing over a
fixed plate is given by u = 0.68y - y
2
(u is velocity in m/s and y is the distance from the plate in
m).What are the shear stresses at the plate surface and at y=0.34m?
y
u y
y
y
u
2
w w
At the plate face y = 0m,
w w
u y
Calculate the shear stress at the plate face
2
m
N
u y
u
w w
P
W
At y = 0.34m,
u
w w
y u
As the velocity gradient is zero at y=0.34 then the shear stress must also be zero.
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
5.6m
3
of oil weighs 46 800 N. Find its mass density,
U
and relative density,
J
Weight
46 800 = mg
Mass
m = 46 800 / 9.81 = 4770.6 kg
Mass density
U
= Mass / volume = 4770.6 / 5.6 = 852 kg/m
3
Relative density
water
U
U
J
From table of fluid properties the viscosity of water is given as 0.01008 poises.What is this value in Ns/m
2
and Pa s units?
P
= 0.01008 poise
1 poise = 0.1 Pa s = 0.1 Ns/m
2
P
= 0.001008 Pa s = 0.001008 Ns/m
2
In a fluid the velocity measured at a distance of 75mm from the boundary is 1.125m/s. The fluidhas absolute viscosity 0.048 Pa s and relative density 0.913. What is the velocity gradient andshear stress at the boundary assuming a linear velocity distribution.
P
= 0.048 Pa s
J
s
Pa
u y
s
u y
1
u
w w
w w
P
W
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
Pressure and Manometers 1.1What will be the (a) the gauge pressure and (b) the absolute pressure of water at depth 12m below thesurface?
U
water
= 1000 kg/m
3
, and p
atmosphere
= 101kN/m
2
[117.72 kN/m
2
, 218.72 kN/m
2
]
a)
p
gh
N
m
Pa
kN
m
kPa
gauge
u
u
U
2
2
b)
p
p
p
N
m
Pa
kN
m
kPa
absolute
gauge
atmospheric
2
2
1.2At what depth below the surface of oil, relative density 0.8, will produce a pressure of 120 kN/m
2
? What
depth of water is this equivalent to?[15.3m, 12.2m]a)
U
J
U
U
U
u
u u
water
kg
m
p
gh
h
p
g
m
3
3
of oil
b)
U
u
u
3
3
kg
m
h
m
of water
1.3What would the pressure in kN/m
2
be if the equivalent head is measured as 400mm of (a) mercury
J
(b) water ( c) oil specific weight 7.9 kN/m
3
(d) a liquid of density 520 kg/m
3
[53.4 kN/m
2
, 3.92 kN/m
2
, 3.16 kN/m
2
, 2.04 kN/m
2
]
a)
U
JU
U
u
u
u
u
water
kg
m
p
gh
N
m
3
3
2
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
1.7Determine the resultant force due to the water acting on the 1m by 2m rectangular area AB shown in thediagram below.[43 560 N, 2.37m from O]
.0m
1.22m
2.0 m
A B
C
D
O
P
2.0 m
The magnitude of the resultant force on a submerged plane is:
R = pressure at centroid
u
area of surface
R
g
z A
N
m
u
u
u
u
U
2
This acts at right angle to the surface through the centre of pressure.
Sc
I
Ax
OO
2nd moment of area about a line through O
1st moment of area about a line through O
By the parallel axis theorem (which will be given in an exam),
I
I
Ax
oo
GG
2
, where
I
GG
is the 2
nd
moment of area about a line through the centroid and can be found in tables.
Sc
I
Ax
x
GG
G
G
d
b
For a rectangle
I
bd
GG
3
As the wall is vertical,
Sc
D
x
z
and
Sc
m
u
u
3 .
from O
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
1.8Determine the resultant force due to the water acting on the 1.25m by 2.0m triangular area CD shown inthe figure above. The apex of the triangle is at C.[43.
u
3
N, 2.821m from P]
G
G
d
b
d/
For a triangle
I
bd
GG
3
Depth to centre of gravity is
z
m
cos
R
gz A
N
m
u
u
u
u
U
2
Distance from P is
x
z
m
cos
Distance from P to centre of pressure is
Sc
I Ax
I
I
Ax
Sc
I
Ax
x
m
oo
oo
GG
GG
u
2
3
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
Forces on submerged surfaces 2.1Obtain an expression for the depth of the centre of pressure of a plane surface wholly submerged in afluid and inclined at an angle to the free surface of the liquid.A horizontal circular pipe, 1.25m diameter, is closed by a butterfly disk which rotates about a horizontalaxis through its centre. Determine the torque which would have to be applied to the disk spindle to keepthe disk closed in a vertical position when there is a 3m head of fresh water above the axis.[1176 Nm]The question asks what is the moment you have to apply to the spindle to keep the disc vertical i.e. tokeep the valve shut?So you need to know the resultant force exerted on the disc by the water and the distance x of this forcefrom the spindle.We know that the water in the pipe is under a pressure of 3m head of water (to the spindle)
F
h
h’
x
Diagram of the forces on the disc valve, based on an imaginary water surface.
h
m
, the depth to the centroid of the disc
h’ =
depth to the centre of pressure (or line of action of the force)
Calculate the force:
F
gh A
kN
u
u
u
U
S
2
Calculate the line of action of the force,
h’
h
I Ah
oo
2nd moment of area about water surface
1st moment of area about water surface
By the parallel axis theorem 2
nd
moment of area about O (in the surface)
I
I
Ah
oo
GG
2
where
I
GG
is the
2nd moment of area about a line through the centroid of the disc and
I
GG
S
r
4
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
h
I
Ah
h
r
r
r
m
GG
S S
4 2
2
So the distance from the spindle to the line of action of the force is
x
h
h
m
And the moment required to keep the gate shut is
moment
Fx
kN m
u
2.2A dock gate is to be reinforced with three horizontal beams. If the water acts on one side only, to a depthof 6m, find the positions of the beams measured from the water surface so that each will carry an equalload. Give the load per meter.[58 860 N/m, 2.31m, 4.22m, 5.47m]First of all draw the pressure diagram, as below:
d
1
d
2
d
3
f f
f
R
2h/
h
The resultant force per unit length of gate is the area of the pressure diagram. So the total resultant forceis
R
gh
N
per m length
u
u
u
2
U
2
Alternatively the resultant force is, R = Pressure at centroid
u
Area , (take width of gate as 1m to give
force per m)
R
g
h
N
per m length
u
u
U
h
This is the resultant force exerted by the gate on the water.The three beams should carry an equal load, so each beam carries the load
f
, where
f
R
N
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
As this force act normal to the surface, it must act through the centre of radius of the dam wall. So thedepth to the point where the force acts is,
y = 30sin 39.
q
=19m
2.4The arch of a bridge over a stream is in the form of a semi-circle of radius 2m. the bridge width is 4m.Due to a flood the water level is now 1.25m above the crest of the arch. Calculate (a) the upward force onthe underside of the arch, (b) the horizontal thrust on one half of the arch.[263.6 kN, 176.6 kN]The bridge and water level can be drawn as:
1.25m 2m
a) The upward force on the arch = weight of (imaginary) water above the arch.
R
g
m
R
kN
v v
u
u
·¸ u¹
u
u
U
S
volume of water
volume
2
3
b) The horizontal force on half of the arch, is equal to the force on the projection of the curved surface ontoa vertical plane.
F
g
kN
h
u
u
u
pressure at centroid
area
U
2.5The face of a dam is vertical to a depth of 7.5m below the water surface then slopes at 30
q
to the vertical.
If the depth of water is 17m what is the resultant force per metre acting on the whole face?[1563.29 kN]
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
q
h
1
h
2
x
h
2
= 17.0 m, so h
1
x = 9.5/tan 60 = 5.485 m.
Vertical force = weight of water above the surface,
F
g h
x
h
x
kN
m
v
u
u
u
u
u
u
U
2
1
The horizontal force = force on the projection of the surface on to a vertical plane.
F
gh
kN
m
h
u
u
u
2
2
U
The resultant force is
F
F
F
kN
m
R
v
h
2
2
2
2
And acts at the angle
tan
T T
F
F
v h
D
2.6A tank with vertical sides is square in plan with 3m long sides. The tank contains oil of relative density0.9 to a depth of 2.0m which is floating on water a depth of 1.5m. Calculate the force on the walls and theheight of the centre of pressure from the bottom of the tank.[165.54 kN, 1.15m]Consider one wall of the tank. Draw the pressure diagram:
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
d
1
d
2
d
3
F
f
1
f
2
f
3
density of oil
U
oil
U
water
kg/m
3
Force per unit length, F = area under the graph = sum of the three areas =
f
1
+ f
2
+ f
3
f
N
f
N
f
N
F
f
f
f
N
1 2 3
1
2
3
u
u
u
u
u
u
u
u
u
u
u
u
To find the position of the resultant force F, we take moments from any point. We will take momentsabout the surface.
DF
f d
f d
f d
D D
m
m
u
u
u
1
1
2
2
3
3
from surface
from base of wall)
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
Application of the Bernoulli Equation 3.1In a vertical pipe carrying water, pressure gauges are inserted at points A and B where the pipe diametersare 0.15m and 0.075m respectively. The point B is 2.5m below A and when the flow rate down the pipe is0.02 cumecs, the pressure at B is 14715 N/m
2
greater than that at A.
Assuming the losses in the pipe between A and B can be expressed as
k
v
2 g
where
v
is the velocity at A,
find the value of
k
If the gauges at A and B are replaced by tubes filled with water and connected to a U-tube containingmercury of relative density 13.6, give a sketch showing how the levels in the two limbs of the U-tubediffer and calculate the value of this difference in metres.[
k
= 0.319, 0.0794m]
Rp
A
B
d
A
= 0.2m
d
B
= 0.2m
Part i)
d
m
d
m
Q
m
s
p
p
N
m
h
kv
g
A
B
B
A
f
3
2
2
Taking the datum at B, the Bernoulli equation becomes:
p
g
u
g
z
p
g
u
g
z
k
u
g
z
z
A
A
A
B
B
B
A
A
B
U
U
2
2
2
By continuity: Q = u
A
A
A
= u
B
A
B
u
m s
u
m s
A B
2
2
S S
giving
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
3.3A Venturimeter is used for measuring flow of water along a pipe. The diameter of the Venturi throat istwo fifths the diameter of the pipe. The inlet and throat are connected by water filled tubes to a mercuryU-tube manometer. The velocity of flow along the pipe is found to be
H
m/s, where
H
is the
manometer reading in metres of mercury. Determine the loss of head between inlet and throat of theVenturi when
H
is 0.49m. (Relative density of mercury is 13.6).
[0.23m of water]
h
Z
Z
H
For the manometer:
p
gz
p
g z
H
gH
p
p
gz
gH
gH
gz
w
w
m
w
w
m
w
1
1
2
2
1
2
2
1
U
U
U
U
U
U
U
For the Venturimeter
p
g
u
g
z
p
g
u
g
z
Losses
p
p
u
gz
u
gz
L
g
w
w w
w
w
w
w
1
21
1
2
22
2
1
2
22
2
1
2
1
U
U U
U
U
U
U
Combining (1) and (2)
p
g
u
g
z
p
g
u
g
z
Losses
L
g
Hg
u
u
w
w
w
m
w
w
1
1
2
1
2
22
2
22
1
2
U
U
U
U
U
U
but at 1. From the question
u
H
m
s
u A
u A
d
u
d
u
m
s
1
1
1
2
2
2
2
2
2
u
S
S
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
Substitute in (3)
Losses
L
m
u
u
2
2
3.4Water is discharging from a tank through a convergent-divergent mouthpiece. The exit from the tank isrounded so that losses there may be neglected and the minimum diameter is 0.05m.If the head in the tank above the centre-line of the mouthpiece is 1.83m. a) What is the discharge?b) What must be the diameter at the exit if the absolute pressure at the minimum area is to be 2.44m ofwater? c) What would the discharge be if the divergent part of the mouth piece were removed. (Assumeatmospheric pressure is 10m of water).[0.0752m, 0.0266m
3
/s, 0.0118m
3
/s]
h
From the question:
d
m
p
g
m
p
g
m
p
g
2
2
1
3
minimum pressure
U
U
U
Apply Bernoulli:
p
g
u
g
z
p
g
u
g
z
p
g
u
g
z
1
1
2
1
2
2
2
2
3
23
3
U
U
U
If we take the datum through the orifice:
z
m
z
z
u
1
2
3
1
negligible
Between 1 and 2
22
2
2
2
2
3
u
u
g
u
m
s
Q
u A
m
s
S
Between 1 and 3
p
p
1
3
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
3
2
3
3
3
23
3
u
u
g
u
m
s
Q
u A
d
d
m S
If the mouth piece has been removed,
p
p
1
2
p
g
z
p
g
u
g
u
gz
m
s
Q
m
s
1
1
2
2
2
2
1
2
3
U
U
S
3.5A closed tank has an orifice 0.025m diameter in one of its vertical sides. The tank contains oil to a depthof 0.61m above the centre of the orifice and the pressure in the air space above the oil is maintained at13780 N/m
2
above atmospheric. Determine the discharge from the orifice.
(Coefficient of discharge of the orifice is 0.61, relative density of oil is 0.9).[0.00195 m
3
/s]
.66m
oil
d
o
= 0.025m
P = 13780 kN/m
From the question
V
U U
U
o w
o d
C
Apply Bernoulli,
p
g
u
g
z
p
g
u
g
z
1
1
2
1
2
2
2
2
U
U
Take atmospheric pressure as 0,
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
2
2
2
2
3
U
S
o
g
u
g
u
m
s
Q
m
s
u
u
3.6The discharge coefficient of a Venturimeter was found to be constant for rates of flow exceeding a certainvalue. Show that for this condition the loss of head due to friction in the convergent parts of the meter canbe expressed as
KQ
2
m
where
K
is a constant and
Q
is the rate of flow in cumecs.
Obtain the value of
K
if the inlet and throat diameter of the Venturimeter are 0.102m and 0.05m
respectively and the discharge coefficient is 0.96.[
K
=1060]
3.7A Venturimeter is to fitted in a horizontal pipe of 0.15m diameter to measure a flow of water which maybe anything up to 240m
3
/hour. The pressure head at the inlet for this flow is 18m above atmospheric and
the pressure head at the throat must not be lower than 7m below atmospheric. Between the inlet and thethroat there is an estimated frictional loss of 10% of the difference in pressure head between these points.Calculate the minimum allowable diameter for the throat.[0.063m]
d
= 0.15m
d
From the question:
d
m
Q
m
hr
m
s
u
Q
A
m
s
p
g
m
p
g
m
1
3
3
1
1
2
U
U
Friction loss, from the question:
h
p
p
g
f
1
2
U
Apply Bernoulli:
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
Tank emptying 4.1A reservoir is circular in plan and the sides slope at an angle of tan
(1/5) to the horizontal. When the
reservoir is full the diameter of the water surface is 50m. Discharge from the reservoir takes placethrough a pipe of diameter 0.65m, the outlet being 4m below top water level. Determine the time for thewater level to fall 2m assuming the discharge to be
a
gH
cumecs where
a
is the cross sectional
area of the pipe in m
2
and
H
is the head of water above the outlet in m.
[1325 seconds]
x
r
0m
H
From the question:
H = 4m
a =
S
2
= 0.33m
2
Q
a
gh h
In time
G
t
the level in the reservoir falls
G
h
, so
Q t
A h
t
A
Q
h
G
G
G
G
Integrating give the total time for levels to fall from
h
1
to
h
2
T
A
Q
dh
h
h
³
1
2
As the surface area changes with height, we must express
A
in terms of
h.
A =
S
r
2
But
r
varies with
h
It varies linearly from the surface at
H = 4m, r = 25m
, at a gradient of tan
r = x + 5h25 = x + 5(4)x = 5
so
A =
S
( 5 + 5h )
2
S
S
h
2
S
h )
Substituting in the integral equation gives
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
T
h
h
h
dh
h
h
h
dh
h
h
h
h h
dh
h
h
h
dh
h
h
h
h
h
h
h
h
h h
h
h h
³
³
³ ³
2 2
2
1/
3 2
1/
1/
53 2
3 2
1
2
1
2
1
2 1
2
2 1
S
S
S
S
/
/
/
From the question, h
1
= 4m
h
2
= 2m, so
@
@
T
u
u
u
u
u
u
1/
53 2
3 2
1/
53 2
3 2
/
/
/
/
sec
4.2A rectangular swimming pool is 1m deep at one end and increases uniformly in depth to 2.6m at the otherend. The pool is 8m wide and 32m long and is emptied through an orifice of area 0.224m
2
, at the lowest
point in the side of the deep end. Taking
C
d
for the orifice as 0.6, find, from first principles,
a) the time for the depth to fall by 1m b) the time to empty the pool completely.[299 second, 662 seconds]
.6m
1.0m
32.0m
L
The question tell us
a
o
= 0.224m
2
, C
d
Apply Bernoulli from the tank surface to the vena contracta at the orifice:
p
g
u
g
z
p
g
u
g
z
1
1
2
1
2
2
2
2
U
U
p
1
= p
2
and
u
1
u
gh
2
We need
Q
in terms of the height
h
measured above the orifice.
Q
C a u
C a
gh
h
h
d
o
d
o
u
u
u
2
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
And we can write an equation for the discharge in terms of the surface height change:
Q t
A h
t
A
Q
h
G
G
G
G
Integrating give the total time for levels to fall from
h
1
to
h
2
T
A
Q
dh
A
h
dh
h
h
h
h
1
2
1
2
a) For the first 1m depth, A = 8 x 32 = 256, whatever the
h.
So, for the first period of time:
T
h
dh
h
h
h
h
1
2
1
2
sec
b) now we need to find out how long it will take to empty the rest.We need the area
A
, in terms of
h
A
L
L h A
h
So
T
h
h
dh
h
h
h
h
1
2
1
3 2
2
3 2
3 2
3 2
sec
/
/
/
/
Total time for emptying is,
T = 363 + 299 = 662 sec
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
4.3A vertical cylindrical tank 2m diameter has, at the bottom, a 0.05m diameter sharp edged orifice forwhich the discharge coefficient is 0.6.a) If water enters the tank at a constant rate of 0.0095 cumecs find the depth of water above the orificewhen the level in the tank becomes stable.b) Find the time for the level to fall from 3m to 1m above the orifice when the inflow is turned off.c) If water now runs into the tank at 0.02 cumecs, the orifice remaining open, find the rate of rise in waterlevel when the level has reached a depth of 1.7m above the orifice.[a) 3.314m, b) 881 seconds, c) 0.252m/min]
h
d
o
= 0.005m
Q = 0.0095 m
/s
From the question:
Q
in
= 0.0095 m
3
/s, d
o
=0.05m, C
d
Apply Bernoulli from the water surface (1) to the orifice (2),
p
g
u
g
z
p
g
u
g
z
1
1
2
1
2
2
2
2
U
U
p
1
= p
2
and
u
1
u
gh
2
With the datum the bottom of the cylinder,
z
1
= h, z
2
We need
Q
in terms of the height
h
measured above the orifice.
Q
C a u
C a
gh
h
h
out
d
o
d
o
u
2
2
S
For the level in the tank to remain constant:
inflow = out flow Q
in
= Q
out
h
h
m
(b) Write the equation for the discharge in terms of the surface height change:
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
@
@
T
h
h
h
dh
h
h
h
dh
h
h
h
dh
h dh
h
h
h
h
h
h
h
h
h
h h
³
³ ³ ³
2
2 2
3 2
1
2
1
2
1
2
1
2
2 1
. sec
/
4.5Two cylinders standing upright contain liquid and are connected by a submerged orifice. The diametersof the cylinders are 1.75m and 1.0m and of the orifice, 0.08m. The difference in levels of the liquid isinitially 1.35m. Find how long it will take for this difference to be reduced to 0.66m if the coefficient ofdischarge for the orifice is 0.605. (Work from first principles.)[30.7 seconds]
h
= 1.35m
d
= 1.0m
d
= 1.75m
d
o
= 0.108m
A
m
A
m
d
m
a
m
C
o
o
d
1
2
2
2
2
2
2
2
S
S
S
by continuity,
A
h
A
h
Q t
1
1
2
2
G
G
G
defining,
h = h
1
- h
2
G
G
G
h
h
h
1
2
Substituting this in (1) to eliminate
G
h
2
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
A
h
A
h
h
A
h
A
h
h
A
h
A
A
A
A
h
A
A
Q t
1
1
2
1
2
1
2
1
2
1
2
1
2
1
2
G
G
G
G
G
G
G
G
G
From the Bernoulli equation we can derive this expression for discharge through the submerged orifice:
Q
C a
gh
d
o
So
A
A
h
A
A
C a
gh
t
d
o
1
2
1
2
G
G
G
G
t
A A
A
A
C a
g
h
h
d
o
1
2
1
2
Integrating
T
A A
A
A
C a
g
h
dh
A A
A
A
C a
g
h
h
d
o
h
h
d
o
u
u
u
u
u
³
1
2
1
2
1
2
1
2
2
1
1
2
. sec
4.6A rectangular reservoir with vertical walls has a plan area of 60000m
2
. Discharge from the reservoir take
place over a rectangular weir. The flow characteristics of the weir is
Q = 0.678 H
3/
cumecs where
H
is
the depth of water above the weir crest. The sill of the weir is 3.4m above the bottom of the reservoir.Starting with a depth of water of 4m in the reservoir and no inflow, what will be the depth of water afterone hour?[3.98m] From the question
A = 60 000 m
2
, Q = 0.678 h
3/
Write the equation for the discharge in terms of the surface height change:
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
Q t
A h
t
A
Q
h
G
G
G
G
Integrating between
h
1
and
h
2,
to give the time to change surface level
@
T
A
Q
dh
h
dh
h
h
h
h
h
h h
u
³
³
1
2
1
2
2 1
3 2
1 2
/
/
From the question
T = 3600 sec
and
h
1
= 0.6m
@
2
1 2
1 2
2
/
/
h
h
m
Total depth = 3.4 + 0.58 = 3.98m
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
Notches and weirs 5.1Deduce an expression for the discharge of water over a right-angled sharp edged V-notch, given that thecoefficient of discharge is 0.61.A rectangular tank 16m by 6m has the same notch in one of its short vertical sides. Determine the timetaken for the head, measured from the bottom of the notch, to fall from 15cm to 7.5cm.[1399 seconds] From your notes you can derive:
Q
C
g H
d
5 2
tan
/
T
For this weir the equation simplifies to
Q
H
5 2
/
Write the equation for the discharge in terms of the surface height change:
Q t
A h
t
A
Q
h
G
G
G
G
Integrating between
h
1
and
h
2,
to give the time to change surface level
@
T
A
Q
dh
h
dh
h
h
h
h
h
h h
u
³ u
³
1
2
1
2
2 1
5 2
3 2
/
/
h
1
= 0.15m, h
2
= 0.075m
@
T
3 2
3 2
sec
/
/
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
b) Write the equation for the discharge in terms of the surface height change:
Q t
A h
t
A
Q
h
G
G
G
G
Integrating between
h
1
and
h
2,
to give the time to change surface level
@
@
T
A
Q
dh
B
h
dh
B
h
T
h
h
h
h
u
³
³
1
2
1
2
3 2
1 2
0 30 6
1 2
1 2
sec
/
/
..
/
/
5.5Develop a formula for the discharge over a 90
q
V-notch weir in terms of head above the bottom of the V.
A channel conveys 300 litres/sec of water. At the outlet end there is a 90
q
V-notch weir for which the
coefficient of discharge is 0.58. At what distance above the bottom of the channel should the weir beplaced in order to make the depth in the channel 1.30m? With the weir in this position what is the depthof water in the channel when the flow is 200 litres/sec?[0.755m, 1.218m]Derive this formula from the notes:
Q
C
g H
d
5 2
tan
/
T
From the question:
T
q
C
d
Q = 0.3 m
3
/s,
depth of water,
Z = 0.3m
giving the weir equation:
Q
H
5 2
/
a) As
H
is the height above the bottom of the V, the depth of water
= Z = D + H,
where
D
is the height
of the bottom of the V from the base of the channel. So
Q
Z
D
D
D
m
5/
5/
b) Find
Z
when
Q = 0.2 m
3
/s
5 2
/
Z
Z
m
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
5.6Show that the quantity of water flowing across a triangular V-notch of angle 2
T
is
Q
C
g H
d
5 2
tan
/
T
. Find the flow if the measured head above the bottom of the V is 38cm, when
T
q
and
C
d
=0.6. If the flow is wanted within an accuracy of 2%, what are the limiting values of the
head.[0.126m
3
/s, 0.377m, 0.383m]
Proof of the v-notch weir equation is in the notes.From the question:
H = 0.38m
T
q
C
d
The weir equation becomes:
Q
H
m
s
5 2
5 2
3
/
/
Q+2% = 0.129 m
3
/s
5 2
/
H
H
m
Q-2% = 0.124 m
3
/s
5 2
/
H
H
m
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
Application of the Momentum Equation 6.1The figure below shows a smooth curved vane attached to a rigid foundation. The jet of water,rectangular in section, 75mm wide and 25mm thick, strike the vane with a velocity of 25m/s. Calculatethe vertical and horizontal components of the force exerted on the vane and indicate in which directionthese components act.[Horizontal 233.4 N acting from right to left. Vertical 1324.6 N acting downwards]
q
q
From the question:
a
m
u
m
s
Q
m
s
a
a
so
u
u
1
3
2
1
3
3
1
2
1
2
u
u
u
u
Calculate the total force using the momentum equation:
F
Q u
u
N
T x
u
U
2
1
cos
cos
cos
cos
F
Q u
u
N
T y
u
U
2
1
sin
sin
sin
sin
Body force and pressure force are 0.So force on vane:
R
F
N
R
F
N
x
t x
y
t y
CIVE1400: Fluid Mechanics
Examples: Answers
Examples: Answers
CIVE1400: Fluid Mechanics
6.2A 600mm diameter pipeline carries water under a head of 30m with a velocity of 3m/s. This water main isfitted with a horizontal bend which turns the axis of the pipeline through 75
q
(i.e. the internal angle at the
bend is 105
q
). Calculate the resultant force on the bend and its angle to the horizontal.
[104.044 kN, 52
q
29’]
θ
u
u
y
x
From the question:
a
m
d
m
h
m
u
u
m
s
Q
m
s
S
2
2
1
2
3
Calculate total force.
F
Q u
u
F
F
F
Tx
x
x
Rx
Px
Bx
U
2
1
F
kN
Tx
u
cos
F
Q u
u
F
F
F
Ty
y
y
Ry
Py
By
U
2
1
F
kN
Ty
u
sin
Calculate the pressure force
p
1
= p
2
= p = h
U
g = 30
u
u
9.81 = 294.3 kN/m
2
F
p a
p a
kN
Tx
u
1
1
1
2
2
2
cos
cos
cos
T
T
F
p a
p a
kN
Ty
u
1
1
1
2
2
2
sin
sin
sin
T
T
There is no body force in the x or y directions.
F
F
F
F
kN
Rx
Tx
Px
Bx
