College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
Solutions to Exercise Three – Bernoulli’s Equation
1. A fire hose nozzle has a diameter of 1.125 in. According to some fire codes, the nozzle
must be capable of delivering at
least 250 gal/min. If the nozzle is
attached to a 3-in diameter hose,
what pressure must be
maintained just upstream of the
nozzle to deliver this flowrate?
Use the continuity (mass
conservation) equation that 1Q1
= 1V1A1 =2V2A2 = 2Q2to solve
this problem. (Problem 3.18 from Munson et al., Fluid Mechanics text; figure from
solutions manual.)
We start with the usual from of the Bernoulli equation for incompressible flows.


p 2  p1 V22  V12
g z 2  z1  

0

2
At the open outlet the gage pressure is zero. The outlet is at the same elevation as the point (1)
so z2 – z1 = 0. Since the fluid is incompressible the continuity equation for this problem becomes
V1A1 = Q1 = Q2 = V2A2, where A1 = D12/4 =  (3 in)2(1 ft2/144 in2)/4 = 0.049087 ft2and A2 = D22/4
=  (1.125 in)2(1 ft2/144 in2)/4 = 0.006903 ft2. We thus find that V2 = Q2/A2 = (250 gal/min)
(1 ft3/7.4805 gal) / (0.006903 ft2) = 4.841 ft/min = 80.69 ft/s. From the continuity equation we
have V1 = Q1/A1 = Q2/A1 = (250 gal/min)(1 ft3/7.4805 gal) / (0.006903 ft2) = 680.83 ft/min =
11.347 ft/s. We can apply the above results to Bernoulli’s equation with a water density of 1.94
slugs/ft3 from Table 1.5 on the inside front cover.
g z 2  z1  
p 2  p1

V

2
2

2
2
 V12
0  p1
1  80.69 ft   11.347 ft  
 g 0 
 
 
 
1.94 slugs 2 
2
s
s
 
 

3
ft
Solving this equation for p1 gives.
2
2
2
lb f
1.94 slugs 1  80.69 ft   11.347 ft   1 lb f  s
ft 2
p1 

 43.0 2






3
2
2 
s
s
ft
in
 
  slug  ft 144 in
2. Water flowing from the 0.75-in-daimeter outlet shown in
the figure rises 2.8 in above the outlet. Find the
flowrate. (Problem and figure 3.19 from Munson et al.,
Fluid Mechanics text.)
We start with the usual from of Bernoulli’s equation for
incompressible flows.
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Exercise three solutions
ME 390, L. S. Caretto, Spring 2008
g z 2  z1  

Page 2

p 2  p1 V22  V12

0

2
Here we apply the equation between the exit of the pipe (1) and the top of the water (2). Both
locations are open to the atmosphere so we have p1 = p2 = 0. The elevation difference, z2 – z1 =
2.8 in = 0.2333 ft. At the top the velocity, V2, is zero. Setting these values into the Bernoulli
equation gives.
g z 2  z1  
V

p 2  p1
2
2



 V12
V12 
32..174 ft

0.2333 ft    0    0

2
2 
s2

Solving for V1 gives the velocity at point 1 as follows.
V1  2
32..174 ft
0.2333 ft   3.875 ft
2
s
s
We can find the flow rate at point 1 from the equation that Q1 = A1V1..
Q1  V1 A1  V1
3
3.875 ft  0.75 in 
ft 2
0.119 ft 3


4
s
4
s
144 in 2
D12
2
Several holes are punched into a tin can
as shown in the figure on the right. Which
of the figures represents the variation of
water velocity as it leaves the holes?
Justify your choice. (Problem and figure
3.17 from Munson et al., Fluid Mechanics
text.)
The usual from of the Bernoulli equation for
incompressible flows is shown below.


p 2  p1 V22  V12
g z 2  z1  

0

2
For this problem, both the upper surface of the liquid and the fluid flowing out of the can are
atmospheric pressure so p1 = p2. The resulting equation for the exit velocity V2 as a function of
elevation, z2 is shown below.


V22  V12  2 gz1  2 gz 2
 V2 
V
2
1
  2g z
 2 gz1 
1/ 2
2
 a  bz12/ 2
Since b = (2g)1/2 is positive, this equation tells us that V2 increases as z2 decreases, varying as
the square root of z2,. (Recall that z2 is the elevation so as the depth from the top of the liquid
increases, the value of z2 decreases.) All curves show an increase in V2 as z2 increases, but we
can immediately rule out the linear (straight line) curve in (b), because the equation that V2 a z21/2
is non-linear. This leaves the choice between (a) and (c). We see that the slope, dV/dz, in (a)
increases in magnitude as z decreases while the slope in (c) decreases in magnitude as z
decreases. The slope of our equation for V2 is dV2/dz2 = –b/(2z21/2). This slope is negative for
any value of z2; the negative slope tells us that dV2/dz2 increases as z2 decreases. This result is
consistent with the curve in figure (a).