CIVIL , ENVIRONMENTAL & GEOMATIC ENGINEERING DEPARTMENT
1st YEAR FLUID MECHANICS
CEGE1009 MECHANISMS
SOLUTIONS for TUTORIAL SHEET 5
Water  =1000 kg/m3; Air R = 287 J/(KgK),  = 1.24 kg/m3 at STP; mercury  = 13560 kg/m3
1.
A horizontal pipe 350 mm in diameter carries a fluid for which  = 0.95 A tapered section is to be
incorporated in the pipe so as to provide a throat of reduced diameter at one point and the difference
in pressure developed between static pressure holes at the throat and at a section upstream is to be
used to operate a control system. To activate the control, a pressure difference greater than 24 kPa is
required when the volume flow rate through the main pipe exceeds 0.15 m3 /s. Neglecting friction,
calculate the throat diameter required.
[Ans: 162 mm]
0.15 m3/s
d1=350 mm
2
1
3
Rephrasing the question, for a flow rate of 0.15 m3 /s, what value of d2 produces a pressure difference (P1 P2) of 24000 Pa?
As friction can be neglected, apply Bernoulli along the centre streamline:
P1  1 2 u1  P2  1 2 u 2 …. assuming a horizontal pipe
2
2
 Q 2 Q 2  8Q 2   1
1 
P1  P2  1 2   2  2  



 2  d 2 4 d1 4 
A1 
 A2
 1   1  2  2 x 24 x10 3
Re-arranging:  4    4  
 66.6  1386
2
 d 2   d1  16 x0.15 x950
d  0.000688 m
4
4
2
And d 2  0.162 m = 162 mm
2.
A venturi-meter has its axis vertical, the inlet and throat diameters being 150 mm and 75 mm
respectively. The throat is 225 mm above the inlet, the coefficient of discharge is 0.96, and a liquid
with  = 0.78 flows through at a rate of 40 litres/s. Calculate (a) the pressure difference between
inlet and throat; (b) the difference of level which would be registered by a vertical U-tube
manometer containing mercury, the tubes above the mercury being full of the liquid flowing
through the venturi.
[Ans: 34.24 kPa, 259.4
mm]
P2 , z 2
u
2
Let ideal flow rate be
Q0 = u1 A1 = u2 A2 ;
z
Actual flow rate Q = Cd Q0
P1 , z 1
u1
l
C
C’
Applying Bernoulli to the streamline at the centre of the Venturi:
P1  1 2 u1   g z1  P2  1 2 u 2   g z 2
2
2
Q 2 Q 2 
2
2
 P1  P2  1 2  (u 2  u1 )   g ( z 2  z1 )  1 2   0 2  02    g ( z 2  z1 )
A1 
 A2
 P1  P2  1 2 
Q2  1
1 
2

  g ( z 2  z1 ) : but A1  1 4 d1

2
2
2
C d  A2
A1 
 P1  P2  1 2 x780 x
40 2 x10 6
0.96 2
and
A2  1 4 d 2


16
16
 2
 780 x9.81x0.225  32500  1743  34240 Pa
 2
4
4 
 ( x0.075 ) ( x0.15 ) 
Mercury U-tube manometer: take pressures at level C-C’ in both arms and equate them:
P1   g z1  P2   g[( z1  0.225)  l ]   hg gl
Hence: l 
2
( P1  P2 )  gz1  gz1  g x0.225 34240  780 x9.81x0.225

 0.259 m
 hg g  g
9.81x(13560  780 )
3.
Show that the volume enclosed by a paraboloid of revolution height h, radius a, is half the volume of
its circumscribing cylinder.
A cylindrical tank 0.6 m in diameter and 1.2 m high is half filled with water. What speed of rotation
about its vertical axis will cause water to reach the top? At what speed will the water depth at the
wall be 0.9 m and how deep is it then at the centre?
[Ans: 154 rev/min, 109 rev/min, 0.3 m]
z
Paraboloid of revolution has the equation:
hr 2
z
or
a2
z r2

h a2
Volume of elemental disc =  r 2 z
δz
Therefore, total volume of paroboloid is:

r
a
h
 r
0
2
dz  1 2  a 2 h
Therefore paraboloid volume = half volume of
cylinder of same height and diameter.
z
ω
For equilibrium of a small radial cylinder of area δA and length
δr,
q2
P

  2 r
r
r
zi
dP 
nit
r
d
also,
P
  g
z
P
P
r 
z   2 r dr   g d z
r
z
Integrating:
P  1 2  2 r 2   g z  const k
Taking the z origin at the base of the cylinder, the liquid surface on the axis (r=0) is at z=z0
For a liquid, the free surface is at atmospheric pressure, which we take as constant.
Therefore, P  Patm at z  z 0 and r  0
Substituting, Patm   gz 0  k and hence: k  Patm  gz 0
Giving: P  Patm  1 2  2 r 2   g ( z 0  z ) anywhere in the fluid. But at the surface, P  Patm
Hence, the equation describing the free surface is: 0  1 2  2 rs   g ( z 0  z s )
2
or ( z s  z 0 ) 
 2 rs 2
2g
, which is the equation of a paraboloid!
The volume of a paraboloid is half the volume of its enclosing cylinder. So, as the cylindrical tank was halffilled initially, the fluid surface will eventually reach the outer rim at the same time it reaches the
bottom at the centre as it is spun around. In other words, at that speed of rotation,  0 ,
z s  1.2m at r  0.3m when z s  0 at r  0
Hence: (1.2  0) 
 2 (0.3) 2
2 x9.81
, and   16.2 Hz
For Part 2, when the surface is at 0.9m above the base at the rim, the volume outside the paraboloid
must equal the initial volume above the new level at the centre, that is:
1
 a 2 ( z r  z 0 )  a 2 (h0  z 0 ) and z 0  h0  z r
2
Given h0  0.6m and z r  0.9m, z 0  0.3m
Then, ( z s  z 0 ) 
 2 rs 2
2g
; 
2g(zr  z0 )
a
2

2 x9.81(0.9  0.3)
 11.43 Hz.
0.3 2
4.
In order to act as dust collector, a column of air is made to rotate as a forced vortex in a pipe 1 m in
diameter with an angular velocity 10 s-1 . If the pressure at the wall is atmospheric, find the gauge
pressure at the centre.
[Ans: - 15.5 Pa]
As the fluid is a gas, the density is relatively small and the term  g z can be neglected; hence the pressure
variation is only significant in the radial direction:
q2
P

  2 r or dP   2 r dr for equilibrium.
r
r
Integrating:
P  1 2  2 r 2  const k
Given that the pressure at the wall is atmospheric, P  P at r  a ,
Then Patm  1 2  2 a 2  const k allows us to calculate k  Patm  1 2  2 a 2
P  Patm  1 2  2 (r 2  a 2 )
Hence, pressure at the centre is: P  Patm  1 2  2 a 2
And gauge pressure is: P  Patm   1 2  2 a 2   1 2 x1.23 x10 2 x0.5 2
kg m 2
 15.4 Pa
m3 s 2
A ventilating duct of square section with 200 mm sides, includes a 90 0 bend in which the centre-line
of the duct follows a circular arc of radius 300 mm. Assuming that frictional effects are negligible
and that upstream of the bend the air flow is completely uniform, determine the way in which the
velocity varies with radius in the bend. Find the mass flow rate when a water-filled U-tube
manometer connected between the mid-points of the outer and inner walls reads 11.5 mm.
[qr = constant; 0.538 kg/s]
5.
As the fluid is a gas, the density is relatively small and the
term  g z can be neglected;
r2 =0.4m
Plan
1. Upstream flow is uniform, hence: P  1 2  q 2  P0  const
r1 =0.2m
.. and the same constant for all streamlines.
2. Negligible friction, so P0  const along a streamline.
r
q
Hence, P0 remains independent of position racross the bend.
So,
( P 
1
2
q2 )
r
 0, or
Now, considering dynamic equilibrium of a small element in the bend:
Or
q2
r
 q
q
 0 or
r
P q 2

r
r
dq dr

 0 which integrates to qr  const k
q
r
For this velocity distribution (as a free vortex),
Hence, P   1 2 
q
P
 q
0
r
r
P q 2 k 2

 3
r
r
r
k2
B
r2
Let P  P1 at r  r1
P  P2 at r  r2
and
1
1
Then, P2  P1   1 2  k 2  2  2 
r1 
 r2
This relation determines k when pressures are known:
1000 x9.81x11.5
P2  P1   water g h 
 112.82 Pa
1000
Hence k 2 
 2( P2  P1 )
1
1

2
2 
 r2 r1 
 k2 

 2 x112 .82
183 .45
m4
m2

 9.784 2 ; k  3.13
s
 1
s
1  (25  6.25)
1.23 x  2 

0.2 2 
 0.4
*
Taking duct as s high, mass flux across radial element, dm   (s dr) q and the total flux
*
r2

*
r2
r2
r1
r1

m  dm  qs dr   s
r1
k
r2
 r dr   s k ln( r )  1.24 x0.2 x3.13x ln 2  0.538 kg/s
1
6.
A cylindrical air-filled container is fitted with an impeller which, in the region 0 < r < b,
drives the air round as a forced vortex at angular velocity , but allows the air in the region with b <
r < R to move as a free vortex. A small hole in the centre of the upper cover plate communicates
with the ambient air at atmospheric pressure. Derive expressions for the pressure distribution in the
container, the pressure at the rim, and force on the upper plate.
The fluid is air, so the variation of pressure with z is negligible and P varies with r only.
R
For equilibrium,
b
In a free vortex, q  k / r
Free vortex
Forced vortex
r
P q 2

in both parts of the vortex.
r
r
q
In a forced vortex, q   r
At r=b, velocities and pressures must match.
Therefore, q   b  k / b and k  b 2
In the forced vortex core, where 0<r<b, dp   2 r dr
and P  1 2  r 2  2  Patm
Given P  Patm at r  0, C  Patm
In the free outer vortex, where b<r<R,
Integrating, P   1 2 
hence, integrating: P  1 2  r 2  2  C
P q 2
k2
b 4 2

 3  3
r
r
r
r
b 4 2
D
r2
For equality of pressures at r=b,
1
2 2
1
2  b   Patm   2 
And in the free vortex, P   b 2  2  1 2 
b 4 2
 D ; hence, D   b 2  2  Patm
r2
b 4 2
 Patm
r2
Pressure at the rim r=R is: PR   b 2  2  1 2 
2
b 4 2
 Patm  Patm   b 2 2 (1  b
)
2
2R 2
R
Force on elemental ring dF  ( P  Patm ) 2 r dr
b
R


Hence, total force on upper plate F  ( P  Patm ) 2 r dr  ( P  Patm ) 2 r dr
0
b
F 

0
R
1
2
2
2
2
b
2  r  .2 r dr    b  (1 
b
2
b

And F  b 2  2 R 2  3 4 b 2  b 2 ln( R b )


4
2 r 
2
2 2
2
).
2

r
dr


.
=
   b  r  b ln r
2R 2
4
 

7.
Determine the head over a 600 V-notch weir when the discharge is 170 litres/s. [Take the
coefficient of discharge as 0.60]
Applying Bernoulli’s equation to flow over a V-notch weir (assuming negligible velocity upstream and
remembering the width of flow over the weir varies with head), it can be shown that the discharge is given as:
8
 5
Q
2 g tan H 2
15
2
Substituting given values into this formula, we find that H = 0.534m
Background can be found in many textbooks.