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Worked Out Examples
(Bernoulli Equation)
Example 1. (use of Bernoulli's equation):
Water flows steadily up the vertical 0.1 m diameter pipe and out the nozzle, which is 0.05 m
in diameter, discharging to atmospheric pressure. The stream velocity at the nozzle exit
must be 20 m/s. Calculate the minimum gage pressure required at section .
V2

4m

1. Statement of the Problem
a) Given
 Fluid is water ( = 999 kg/m3 assuming at 15 C).
 The flow goes steadily up a vertical channel.
 Steady state condition & Gravity must be concerned.
 D1  0.1m & D2  0.05m
 p2 = atmospheric pressure
 V2  20m / s
b) Find
 The minimum gage pressure required at section 
2. System Diagram
V2, z2, p2

4m

z1
Streamline
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3. Assumptions
 Steady state condition
 Incompressible fluid flow (working fluid, water, is usually assumed to be incompressible)
 Inviscid fluid flow
 2 - D problem
4. Governing Equations

Bernoulli's Equation:
p


V2
 gz  const.
2
Restrictions:
V1, p1

(1)
(2)
(3)
(4)
Steady flow
Incompressible flow
Frictionless flow
Flow along a streamline
0

t



CV
dV   V  dA … Integral version of mass conservation (to find V1 first.)
CS
Imagine a control volume (one possibility is shown on the diagram by the box sketched
in red) such that control surface intersects with the inlet and the outlet areas. Now, for an



incompressible fluid flow problem, the equation above  0  V  dA
CS
1 inlet () and 1 outlet ()  0  V1 A1  V2 A2
5. Detailed Solution
The Bernoulli's equation can be applied between any two points on a streamline provided that
the other three restrictions are satisfied. The result is
2
p1
2
V
p
V
 1  gz1  2  2  gz 2

2

2
where subscripts 1 and 2 represent any two points on a streamline that correspond to  and
 in this particular problem.
Now, the Bernoulli's equation can be written as:
p1  p 2




1 2
2
V2  V1  g  z 2  z1 
2
Substituting 0  V1 A1  V2 A2  V1 
p1  p 2


1 2 A
 V2   2 V2 
2
 A1 

2
A2
V2 into the above equation,
A1

  g z 2  z1 

Ghosh - 550
p1  p 2

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2
1 2   A2  
 V2 1      g z 2  z1 
2
  A1  
2
p1  p 2 1 2   D2


 V2 1   2

2
  D1



  g  z 2  z1 


2
D2
2
A2
D2
4


 2
A1  2
D1
D1
4
2
2




 p1gage  patm    p2 gage  patm     1 V2 2 1   D22    g z 2  z1 
  D1  
 2



Finally, p1gage  p 2 gage




2
2 2


 1 2   D2  

   V2 1   2   g z 2  z1 
2
  D1  






p2gage is 0 kPa because p2 = atmospheric pressure. Thus,
1

  0.05m 2  2 


2
2




p1gage  0kPa  999kg / m  20m / s  1  
 9.81m / s 4m   0m 
2 
2
  0.1m   






Therefore, the minimum gage pressure required at section , p1gage  226.5kPa

3



6. Critical Assessment
Remember that four restrictions must be satisfied to use the Bernoulli's equation.
Example 2. (Bernoulli's equation and Manometry)
Water flows from a very large tank through a 2 in. diameter tube. The dark liquid in the
manometer is mercury. Estimate the velocity in the pipe and the rate of discharge from the
tank.
12 ft
2 in. i.d.
Flow
2 ft
6 in.
Mercury
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1. Statement of the Problem
a) Given
 All information is described in the figure above.
b) Find
 Velocity in the pipe
 Rate of discharge from the tank
2. System Diagram
, p1, V1, z1
Streamline
12 ft
2 in. i.d.
, p2, V2, z2
A
h1 = 2 ft
Flow
C
B
h2 = 6 in. = 0.5 ft
Mercury
3. Assumptions
 Steady state condition
 Incompressible fluid flow (water, working fluid, is usually considered as incompressible)
 Inviscid fluid flow
 2 - D problem
4. Governing Equations

Bernoulli's Equation:
p


V2
 gz  const.
2
Restrictions:
(5) Steady flow
(6) Incompressible flow
(7) Frictionless flow
(8) Flow along a streamline
5. Detailed Solution
Physical quantities:
S .G.  13.55 (Specific gravity of mercury)
mercury
 water  1.94slug / ft 3 at 59 F
g  32.174 ft / s 2
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Pressure variation in any static fluid is described by the basic pressure-height relation
dp
  g (This equation is from fluid statics.)
dz
Thus,
dp
  g 
dz

p
p0
dp    gdz  p  p0  g z 0  z   p  p0  gh
z
z0
Consider the portion of the figure A, B, and C as a U-tube manometer. Using this statics
relationship (Pascal's Law),
p A  p B    water gh1
p B  pC   mercury gh2
Adding the two equations gives
p A  pC   mercury gh2   water gh1
Since pC  p atm , p A  p atm  S .G.  water gh2   water gh1   water g  S .G. h2  h1 
 mercury

 1.94slug / ft 32.174 ft / s 13.550.5 ft   2 ft   298.044lbf / ft 2
mercury
p A  p atm
3
2
This pressure pA is same as the pressure p2 because of the absence of pressure drop for
inviscid fluid flow. Thus,
p 2  298.044lbf / ft 2  p atm
The Bernoulli's equation can be applied between any two points on a streamline provided that
the other three restrictions are satisfied. The result is
p1
2
2
V
p
V
 1  gz1  2  2  gz 2

2

2
where subscripts 1 and 2 represent any two points on a streamline that correspond to  and
 in this particular problem.
Now, the Bernoulli's equation can be written as:


p  p2
1 2
2
V2  V1  1
 g  z1  z 2 
2

 p  p2

2
2
V2  2  1
 g z1  z 2   V1
 

Using an approximation V1  0 ft/s because Areservoir >> Apipe , and p1 = patm,
 p  p2

2
2
V2  2 atm
 g z1  z 2   0 ft / s 



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 
 p  298.044lbf / ft 2  p atm

2
V2  2 atm
 32.174 ft / s 2 12 ft   0 ft   464.914 ft 2 / s 2
3
1.94slug / ft





Finally, V2  21.56 ft / s
Rate of discharge from the tank is:
Q2  V2 A2  V2

4
D2
2
2
2

 21.56 ft / s   ft   0.470 ft 3 / s
4  12 
6. Critical Assessment
This problem illustrates how Bernoulli equation may be applied with manometry.
Review fluid statics if you are not confident with how the Pascal's law was utilized to
relate the pressures pA, pB and pC.
Example 3. (Use of Bernoulli and Rotational Flows):
Consider the flow represented by the stream function  = Ax2y, where A is a dimensional
constant equal to 2.5 ft-1s-1. The density is 2.45 slug/ft3. Is the flow rotational? Can the
pressure difference between points (x,y) = (1,4) and (2,1) be evaluated? If so, calculate it,
and if not, explain why.
1. Statement of the Problem
a) Given
 Stream function  = Ax2y, where A = 2.5 ft-1s-1
  = 2.45 slug/ft3
 p1(1,4) and p2(2,1)
b) Find
 Whether the flow is rotational or irrotational.
 Pressure difference between points (x,y) = (1,4) and (2,1) if it can be evaluated. If
not, explain why.
2. System Diagram
It is not necessary for this problem.
3. Assumptions
 Steady state condition
 Incompressible fluid flow.
 Frictionless fluid flow
 2 - D problem
4. Governing Equations
Ghosh - 550



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

& v
x
y




 w v   u w 
 v u 

Vorticity:   2    V  i 
   j  
  k   
 z x 
 y z 
 x y 
p V2
Bernoulli's Equation:

 gz  const.
 2
Stream function (incompressible fluid flow version) definition: u 
Restrictions:
(9) Steady flow
(10)
Incompressible flow
(11)
Frictionless flow
(12)
Flow along a streamline
5. Detailed Solution
Velocity components can be obtained through the given stream function  = Ax2y using the
stream function definition:



Ax 2 y  Ax 2
y y


v
  Ax 2 y   2 Axy
x
x

u

Since this is 2 - D (xy plane) problem, check z for rotationality of the flow field.
1  v
u 
1


 z        2 Axy  Ax 2    2 Ay   0   Ay  0
2  x y  2  x
y
 2
1
 There is rotation in the flow field. You may wonder that there is no friction (no
viscosity) but why the flow is rotational. Imagine that the fluid flow is rotated initially and
keep rotating forever because of the absence of viscosity.
Because the flow is rotational, Bernoulli's equation cannot be applied between any two
points in the flow. The only way to apply Bernoulli's equation in this case is that two
points must be on the same streamline. This can be verified by:
Point 1 at (x,y) = (1,4): 1 = A(1)2(4) = 4A
Point 2 at (x,y) = (2,1): 2 = A(2)2(1) = 4A
1 =2  Two points are on the same streamline.  Bernoulli's equation can be applied
between these two points.
p1


V1
p V
 gz1  2  2  gz 2
2

2
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Ignoring the elevation difference, p2  p1 
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


 
1
1
 V12  V22   u12  v12  u 22  v22
2
2
u1 = Ax12 = A(1)2 = A
v1 = -2Ax1y1 = -2A(1)(4) = -8A
u2 = Ax22 = A(2)2 = 4A
v2 = -2Ax2y2 = -2A(2)(1) = -4A
Substituting these velocities into Bernoulli's equation,





1
1
33
  A2   8 A2  4 A2   4 A2   33 A2  A2
2
2
2
33 2 33
p2  p1 
A  2.452.52  252.66 lbf/ft2
2
2
p2  p1 
6. Critical Assessment
Make sure four restrictions are all satisfied before using Bernoulli's equation.
Continue
