AMEE 202
Introduction to Fluid Mechanics
Instructor: Marios M. Fyrillas
Email: m.fyrillas@frederick.ac.cy
HOMEWORK ASSIGNMENT ON
BERNOULLI’S EQUATION
1. Conventional spray-guns operate by achieving a low pressure through high
speed air passing through a nozzle (see figure) that pumps the liquid (paint)
up into the stream of air. In an experiment, a compressor is used to supply
air at 3 kPa (gage). Determine the velocity of the air through the nozzle if the
water in the container is elevated 10 cm.
2
h= 10 cm
4
1
1
3
Point 1 is the compressor tank where the pressure is 3000 Pa.
Take Bernoulli's equation between point 1 and 2.
p1 
1 2
1
 u1  
 gh1  p2   u22  
 gh2
2
2

0
0
0
Because of the small density of air the gravity terms are
negligible. Also the velocity in the compressor tank is  0.
Hence
1 2
 u2 .
2
Because the streamlines are parallel, only the hydrostatic pressure
p1  p2 
need to be incorporated to relate the pressures between sections 2 and 3.
p2   gh  p3 (neglecting small pressure variation across the air stream)
But p3  patm and combining above equations we obtain
p1  patm   gh 

2  p1  patm   gh 
1 2
 u 2  u2 
2

2  3000  1000  9.81  0.1
 2.82 m/s
1000
2. Water flows up ramp in a constant width rectangular channel at a rate
Q  0.534 m 3/s . The upstream depth is y1  0.7 m and the width of the
channel is b  1 m .
a. Write the Bernoulli’s equation for a streamline between sections 1 and 2.
Show the streamline
b. Write the mass conservation for sections 1 and 2.
c. If the velocity downstream V2  0.79 m/s determine y2 and z2 .
z2
Consider Bernoulli Equation on the streamline
joining the water interface at point 1 and point 2:
 y1  z1  
V12
V2
  y 2  z2   2
2g
2g
Eq. 1
Mass conservation
m 1  m 2  1Q1   2Q2 (incompressible)  Q1  Q2  Q
 V1  y1  b  V2  y2  b  Q
V1 
Q1
0.534

 0.76 m/s
y1  b 0.7  1
Q2  y2  b  V2  y2 
Eq. 2
Eq. 3
Q2
0.534

 0.68 m
b  V2 1  0.79
Assuming that the datum is at the level z1 , i.e. z1  0
and substituting eq. 2 and eq. 3 into eq. 1 we obtain
0.7  0+
0.762
0.79 2
 0.68  z2 
2  9.81
2  9.81
 z2  0.7+0.04-0.68-0.032=0.028
3. The flowrate of water under a sluice gate as shown in the figure is a
function of the water depths upstream and downstream of the gate and the
width of the sluice gate. Use the Bernoulli and continuity equations to
determine the theoretical flowrate.
Consider Bernoulli Equation on the streamline
joining the water interface at point 1 and point 2:

p1 V12
p
V2

 z1  2  2  z2  V12
V22
 z1 
 z2
 g 2g
 g 2g

2g
2g

p1  p2  patm

Eq. 1
Mass conservation
m 1  m 1 (incompressible)  Q1  Q2
V1  z1  b  V2  z2  b (same width b)  V2 
V1  z1
z2
2
V  z  1
V12
Substitude in eq.1 
 z1   1 1 
 z2
2g
 z2  2 g
2
V12   z1  
1       z1  z2   0  V1 

2 g   z2  


Volumetric flowrate Q  A1 V1 

2 g  z2  z1 
1   z / z  
2
2 g  z2  z1 
1   z1 / z2 
2
1
2

 z1.b
4. A vertical nozzle projects a jet of water 50 m into the air. Use Bernoulli’s
equation to calculate the water velocity at the exit from the nozzle.
Take Bernoulli's equation between point 2 and 3.
1
1
p2   u22  
 gh2  
p3   u32   gh3

2
2 0
=0
p
p
atm

atm
1 2
 u2   gh3  u2  2 gh3  31.4 m/s
2
5. In a rectangular channel the velocity is measured using a pitot tube as
shown on Figure 4. Based on the measurements it was concluded that the
height ( H ) is related to the depth ( y ) by the relation H  0.2 y 2 . Determine:
a. The velocity profile and the shear stress.
b. If the cross-sectional area has dimensions 50  50 (cm) determine the
volumetric flow rate.
c. What second order correction should be included to the velocity
equation in order to obtain a no shear stress condition on the surface
yh.
H
h
(1)
y
Flow in a channel
1
1

 u12   gh1  p2   u22   gh2 
2
2

1
2
h1  h2
  p2  p1   u1
2

u2  0


p1 
The hydrostatic pressure at points (1) and (2) can be obtained from statics:
p1   g ( h  y )  patm
p2   g ( H  h  y )  patm
Combine all three equations to obtain:
1
 u12
2
Above equations can be simplified to obtain
 g ( H  h  y )  patm   g ( h  y )  patm 

2
2
  u1  2 g  0.2 y  u1  1.98 y
but H  0.2 y 2 
u12  2 gH
y
0.5
u  1.98 y
0.99
Q  area of triangular prism 
u
0.99  0.5
 0.5  0.124 m3/s
2
or Integrate

Q  u dA 
y2
1.98 y  0.5 dy  1.98  0.5
0
2
 u b dy  
0.5
u  1.98 y  ay 2
du
( y  0.5)  0  1.98  2ay  1.98  2a  0.5  a  1.98
dy
0.5
 1.98  0.54  0.124 m3 /s
0
6. Figure shows a pump that draws a liquid of density 920 kg/m3 from a
sealed underground tank. In order to prime the pump with liquid prior to a
pumping operation, the delivery valve is closed and a compressed air
supply at a stagnation pressure of 5 bars is allowed to flow through the
reducer shown in the figure. The tank and the pump body are connected as
indicated on the figure. For the conditions shown, assume incompressible
flow through the reducer and calculate the maximum values of h for which
priming can be achieved.
p =4.98
A2  0.2 A1
A1
Air in
Air
h
Figure 5: Reducer-nozzle
Take Bernoulli's equation between points 1 and 2.
1 2
1 2
p


u


gh

p

 u2  
 gh2  500000
1
1
1
2


2
2
 498000
0
0
Mass conservation between points 1 and 2
1u1 A1   2u2 A2 assume incompressible  u1 A1  u2 A2
 u1 A1  0.2u2 A1  u1  0.2u2
Hydrostatic pressure between points 1 and 2
p2    gh  p1
From Bernoulli's at point 1 and mass conservation
1
1
500000  498000
2
p1    0.2u2   500000   u22 
 50000

2
2
0.22
 498000
From above and Bernoulli's equation at point 2
1
p2  500000   u22  500000  50000  450000
2
From above and hydrostatic pressure equation
p  p2 498000  450000
h 1

 5.32 m
920  9.81
 g
7. Air is drawn in the wind tunnel used for testing automobiles. Determine the
manometer reading, h , when the velocity in the test section is 60 km/hr .
Determine the difference between the stagnation pressure on the front of
the automobile and the pressure in the test section.
patm
Km/hr
  900 kg/m3
Figure 5: Wind tunnel (Question 1)
2
1
 60 
 air 

2
 3.6 
psec tion   water gh  patm  oil g 0.0254
patm  psec tion 
8. A hovercraft (air cushion vehicle) is supported by forcing air into the
chamber created by a skirt around the periphery of the vehicle as shown in
Figure 7. The air escapes through the 10 cm clearance between the lower end
of the skirt and the ground (or water). Assume the vehicle weighs 5000 kg
and is essentially rectangular in shape, 10 by 20 m . The volume of the
chamber is large enough so that the kinetic energy of the air within the
chamber is negligible. Determine the flowrate, Q , needed to support the
vehicle. If the ground clearance were reduced to 4 cm , what flowrate would
be needed? If the vehicle weight were reduced to 2500 kg and the ground
clearance maintained at 10 cm , what flowrate would be needed?
Figure : Hovercraft
The pressure inside the chamber must be greater than atmospheric, and provides
a lift force
mg 5000  9.81

F  pA  mg  p( gage) 
10  20
A
Take a streamline from inside the chamber to the lower end of the skirt:
patm 
1 2
u  p  u 
2
2 p( gage)

Q  uA  u   (10  10  20  20)  10 
9. Water flows through the pipe contraction shown in the figure. For the given
0.2-m difference in the manometer level, determine the flowrate as a
function of the diameter of the small pipe, D.
10. Water flows through the pipe contraction shown in the figure. For the given
0.2-m difference in the manometer level, determine the flowrate as a
function of the diameter of the small pipe, D.
11.
12.
Water in a rectangular channel flows into a gradual contraction section
as is indicated in the figure. If the flowrate is Q  0.708 m3 / s and the
upstream depth is y1  0.61 m , determine the downstream depth, y2 .
Consider Bernoulli Equation on the streamline
joining the water interface at point 1 and point 2:
y1 
V12
V2
 z1  y2  2  z2
2g
2g
Eq. 1
Mass conservation
m 1  m 1 (incompressible)  Q1  Q2 =0.708 m3/s (note that the width is not the same)
Q1
0.708
=
y2  b2 y2  0.914
 Q1  V1  y1  b1  V2  y2  b2  V2 
V1 
Q1
0.708

 0.952 m/s
y1  b1 0.61  1.22
Eq. 2
Eq. 3
Assuming that the datum is at the level z1 , i.e. z1  0
and substituting eq. 2 and eq. 3 into eq. 1 we obtain
2
 0.708 
0.952 2
1
0.61+
 0  y2  
0

2  9.81
 y2  0.914  2  9.81
0.031
0.656=y2 
y2 2
Multiplying by y2 2 and simplifying we obtain
y 32  0.656  y 22  0.031=0
The solutions of the above equation are
y2  -0.191, y2  0.292, y3  0.556
To determine which of the positive roots to choose
we need to compare the values with the critical value.
2
The specific energy head E  y 
 Q  1
V2
 y

2g
 y  b  2g
2
The minimum occurs when
dE
Q 1
 1  
0
3
dy
 b  g  yc
2
Q  1
 yc  3  
b g
Note that because b is not the same the critical value is different
at each section.
2
 0.708  1
At section 1, the critical value is yc1  
 0.325m.

 1.22  g
Since y1  yc1 , it means that the flow upstream is subcritical.
3
The flow will remain subcritical since the reduction in width is
gradual, i.e. we will choose the biggest value, i.e. y2  0.556m.