College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
Solutions to Exercise Four – Bernoulli’s Equation Part II
1
Water flows through the pipe contraction shown in the
figure at the right. For the given 0.2-m difference in
manometer level, determine the flow rate as a function
of the diameter of the small pipe, D. (Problem and
figure 3.30 from Munson et al., Fluid Mechanics text.)
Apply the Bernoulli equation for incompressible, inviscid
flows, shown below, between two points (1) and (2) along a
streamline in the center of the pipe.
z 2  z1  

(1)
(2)

p 2  p1 V22  V12

0
g
2g
The measurement tube at point 1 is facing the flow so it will measure the stagnation pressure.
This means that the elevation difference shown in the diagram (h = 0.2 m) times the specific
weight of the fluid is the difference between the stagnation pressure at point (1) (p 1 +V12/2) and
the static pressure, p2 at point (2). We thus have the following interpretation of the height
difference.
h  gh  p1 
V12
2
 p2

p 2  p1 
V12
2
 gh
We can substitute this expression for p2 into our Bernoulli equation along with the fact that at the
center of the pipe z2 = z1. This gives the Bernoulli equation as


2
2
z 2  z1   p 2  p1  V2  V1  0 
g
2g
p1 
V12
2
 gh  p1
g

V
2
2

 V12
0
2g
We see that the terms in p1 and V12 cancel leaving the following result.
V22
h
0
2g

V2  2 gh
The flow rate is simply V2A2 = V2D22/4, with D2 = D.
Q  A2V2  A2 2 gh 

4
D 2 2 gh 

4
20.2 m
9.81 m 1.556 m 2

D
s
s2
The flow rate will have units of m 3/s when D is in meters.
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Exercise four solutions
2
ME 390, L. S. Caretto, Spring 2008
Page 2
A Pitot-static tube is used to measure the velocity of helium in a pipe. The temperature
and pressure are 40oF and 25 psia. A water manometer connected to the Pitot-static tube
indicates a reading of 2.3 in. Determine the helium velocity. Is it reasonable to consider
the flow as incompressible? Explain. (Problem 3.34 from Munson et al., Fluid Mechanics
text. Figure below from solutions manual.)
The flow can be assumed incompressible if the Mach number, Ma = V/c, where c = (kRT)1/2 is the
sound speed, is less than 0.3. We can assume that the flow is incompressible, then compute the
velocity and Mach number, Ma. If Ma < 0.3 our answer will be correct.
A Pitot-static tube measures the difference between the static pressure, ps, where the fluid has a
velocity, V, and the stagnation pressure, p0 = ps + V2/2, which occurs when the original velocitt,
V, is reduced to zero.. (See the discussion of the Pitotstatic tube in the text and lecture presentations for more
information.) We can obtain the velocity, V, from the
difference in static and stagnation pressures by solving
the equation p0 = ps + V2/2 for V. When the pressure
difference is measured by a manometer, as shown at the
right (the top part of the diagram, linking each side with
the flowing fluid was truncated in copying), the measured value of p0 – ps = (mano – fluid)h. In this
case where the manometer fluid is water and the flowing fluid is helium we can neglect the
specific weight of heilum and write p0 – ps = manoh so the equation for velocity becomes.
V 2
p0  p s
 fluid

2 mano h
 fluid
We can compute the density of the fluid, Heliuim, from the ideal gas law using the value of R =
2.446x104 ft•lbf/slug•R for helium found in Table 1.7 in the inside front conver. We can use the
pressure of 25 psia directly since it is already an absolute pressure, but we have to convert the
temparature of 40oF to a Rankine temperaure by the equation R = oF + 459.67, so we use a
temperature of 499.67 R in both the density calculation and the Mach number calculation below.
25 lb f 144 in 2
P
5.80 x10  4 slugs
in 2
ft 2



RT 2.446 x10 4 ft  lb f
ft 3
499.67 R 
slug  R
We can check the assumption that the specific weight of the helium was neglegible compared to
the manometer fluid, water, as follows:
 mano   He   mano   He g 
62.4 lb f
ft 3
2
62.38 lb f
5.80 x10 4 slugs 32.174 ft lb f  s


3
2
1 slug  ft
ft
s
ft 3
This gives an error of only 0.03%. Other gases, with higher molecular weights, whould give a
larger, but still neglegible, error.
Subsituting the density found above and the specific weight of water, the manometer fluid, as
62.4 lbf/ft3 in the equation for velocity gives.
Exercise four solutions
V
ME 390, L. S. Caretto, Spring 2008
2 mano h
 fluid
2
62.4 lb f

3
Page 3
2.3 in  1 ft
12 in
ft
 203 ft/s
4
5.80 x10 slugs
ft 3
We can now check the compressibliity assumption by computing the Mach number. The value of
k for Helium is also found in Table 1.7 on the inside front cover.
V
Ma  
c
V
kRT

1.66
203 ft
s
4
2.446 x10 ft  lb f
499.67 R 1 slug  2ft
slug  R
 0.063
lb f  s
This value is well below the upper limt of Ma = 0.3 for which we said we could consider a gas flow
incompressible so we conclude that the helium flow can be considered incompressible.
3
Water flows from a large tank as shown
in the figure at the right. Atmospheric
pressure is 14.5 psia and the vapor
pressure is 1.60 psia. If viscous effects
are neglected, at what height, h, will
cavitation begin? To avoid cavitation,
should the value of D1 be increased or
decreased? To avoid cavitation, should
the value of D2 be increased or
decreased?. Explain. (Problem and
figure 3.60 from Munson et al., Fluid
Mechanics text.)
Cavitation is likely to occur at the thin diameter point where D1 = 1 in. (Point 2 is open to the
atmosphere so cavitation will not occur here. To determine if cavitation will occur we have to find
the pressure at the point where D1 = 1 in. To do this we have to know the overall flow rate. We
can find this by applying Bernoulli’s equation between the top of the large tank (point 0) and the
exit (point 2). This gives


p 0  p2 V02  V22
z 0  z 2  

0
g
2g
From the diagram we have z0 – z2 = h, p0 = p2 = 0 since both are open to the atmosphere, and
V02 =( A2V2/A0)2 << V22 and can be neglected because the area ratio, A2/A0 is so small. Inserting
these values into the Bernoulli equation gives.
h


0  0 0  V22

 0  V2  2 gh
g
2g
The flow rate, Q, can then be written as
Q  A2V2  A2 2 gh . Now that we know the flow rate,
we can find the pressure at point (1) where the diameter is 1 in. Applying Bernoulli’s equation
between this point and the exit (point 2) gives.
Exercise four solutions
ME 390, L. S. Caretto, Spring 2008

Page 4

p1  p2 V12  V22
z 1  z 2  

0
g
2g
Here z1 = z2 and p2 = 0. Also, by continuity, V1A1 = V2A2 so that we can rewrite this equation as
follows.
z 1  z 2  


p1  p2 V12  V22
p 0 V 2 A 2 / A 2  V22

 0 1  2 2 1
0
g
2g
g
2g
We can use the previous result that V22 = 2gh and the definition of =g to rewrite the previous
result as follows.
p1 V22 A22 / A12  V22 p1 A22 / A12  1 2 p1 A22 / A12  1
p



V2 

2 gh  1  A22 / A12  1 h  0
g
2g

2g

2g



Solving this equation for the gage pressure, p1, and noting that the area ratio is the same as the
diameter ratio squared gives.


p1   1 D24 / D14 h
If we add the atmospheric pressure to both sides of the equation we will obtain the absolute
pressure at point 1, which must be geater than the vapor pressure to avoid cavitation..


p1,abs  p1  patm   1  D24 / D 4 h  patm  pv
Substituting the given data that D1 = 1 in, D2 = 2 in, patm = 14.3 psia, pv = 1.60 psia, and using a
specific weight of 62.4 lbf/ft3 for water gives the following numerical result.
4
ft 2 1 psia  in 2 62.4 lb f   2 in  
  h  14.3 psia  1.60 psia
1  
lb f
144 in 2
ft 3   1 in  


 6,5 psia
h  12.7 psia  h  1.96 ft
ft
Note than when an inequality is multiplied or divided by a negative number the direction of the
inequality is reversed.
To determine the effects of D1 and D2 we can solve
 1  D24 / D 4 h  p atm  pv
for h. Note
that since D2 > D1 we will be dividing by a negative number so we have to change the direction of
the inequality.
h
pv  patm
patm  pv

4
4
 1  D2 / D1
 D24 / D14  1




This shows that increasing D2 or decreasing D1 will increase h. The physical reasoning for this is
as follows. Neglecting visclous forces, increasing D2 will decrease the overall flow rate and less
material will have to flow through D1 for a given h. Thus we can increase h. However, with D2
fixed, the flow rate is fixed, and decreasing D2 increases the flow rate through point 1; thus we
would have to reduce h to avoid cavitation.