# qs04 ```College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
Quiz Four Solutions – Bernoulli
Equation II
JP4 fuel (SG = 0.77) flows through the Venturi
meter shown in the figure at the right. The
inlet velocity in the pipe is 15 ft/s. If viscous
effects are negligible, determine the elevation,
h, of the fuel in the open tube connected to
the throat of the Venturi area. (Problem and
Figure P3.68 taken from Munson et al, Fluid
Mechanics.)
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  
(2)
(1)


p 2  p1 V22  V12

0

2g
The diagram shows that the elevation difference, z2 – z1 = 8 in = 0.6667 ft. We can apply the
continuity equation between points (1) and (2) where we assume that the JP4 density does not
change. This gives
V1 A1  V2 A2
 V1
D12
 V2
4
D22
4
D
D2
 V1  V2 22  V2  2
D1
 D1
2

 ,

Since the diameter at point 2 is the same as the inlet diameter where V = 15 m/s, the continuity
equation tells us that V2 = 15 ft/s as well. Thus we can find the value of V1 from the diametersquared ratio.
D
V1  V2  2
 D1
2
2

15 ft  6 in 
33.75 ft
 

 
,
s  4 in 
s

The measurement tube at point 2 is measuring the static pressure. With the usual use of gage
pressures so that the pressure at the top of this tube is zero, the value of p2 is found from the
fluid-static equation p2 = h2 = (6 ft), where  is the specific weight of the JP4. We can substitute
this expression for p2 into our Bernoulli equation along with the values found above: z2 – z1 =
0.6667 ft, V1 = 33.75 ft/s, and V2 = 15 ft/s. This gives the Bernoulli equation as
z 2  z1  

p 2  p1 V22  V12


2g

2
2
 15 ft   33.75 ft 

 

 6 ft   p1  s  
s
 0
 0.6667 ft 

32
.
174
ft

2
s
Calculating the velocity term and rearranging the numerical equation gives.
Jacaranda (Engineering) 3333
E-mail: [email protected]
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Quiz four solutions
ME 390, L. S. Caretto, Spring 2008
0.6667 ft  6 ft 
p1
 14.205 ft  0 

Page 2
p1
 7.538 ft

The manometer at the Venturi throat is open to the atmosphere so the pressure on the open arm
is zero. Since the bottom of the displacement h is at the same level as the open arm of the
manometer the pressure at this level on the left side of the manometer is zero. This must be less
than the pressure at point 1, p1, by the specific weight, , times the height, h, that we are trying to
find. This gives the manometer equation as follows.
p1  h  0  h  
p1
  7.538 ft 

We see that the manometer height is p1/ which is the same quantity whose numerical value we
found as –7.538 ft from Bernoulli’s equation. This gives us our answer for the manometer height:
h = 7.538 ft