College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
February 5 Homework Solutions
2.5
Bourdon gages (see Video V2.2 and Fig. 2.13) are
commonly used to measure pressure. When
such a gage is attached to the closed water tank
of Fig. P2.5 (copied from the text at the right),
the gage reads 5 psi. What is the absolute
pressure in the tank? Assume standard
atmospheric pressure of 14.7 psia.
The absolute pressure at the gage is is 5 psi + 14.7
psi = 19.7 psia. The fact that the gage is 6 in above
the measuring line is irrelevant. The only difference
in depth that contributes the pressure at the gage is
the 12 in difference between the gage level and the
air. Thus we can write that pair = 19.7 psia +
water(12 in). Using water = 62.4 lbf/ft3 from Table 1-5
on the inside front cover gives the following result
for the air pressure.
pair 
19.7 lb f
in 2

62.4 lb f
ft 3
12 in 
ft 3
1728 in 3
pair = 20.1 psia
2.11
In a certain liquid at rest measurements of the specific weight at various depths show the
following variation:
h(ft)
0
10
20
30
40
5
60
70
80
90
100
lb f
70
76
84
91
97
102
107
110
112
114
115

ft
3
The depth, h = 0, corresponds to a free surface at atmospheric pressure. Determine,
through numerical integration of Eq. 2.4, the corresponding variation in pressure and
show the results on a plot of pressure (in psf) versus depth (in feet).
z
We have to numerically integrate dp/dz = - or p = p0 –
 dz , where p
0
= 0 (given). We can use
z0
b
a simple trapezoid rule

f ( x)dx 
a
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
N 1

x 


f
a

2
f a  kx   f b  where N = (b – a)/x.

2 
k 1


Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
February 5 homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 2
In this problem the variable of integration is z and the step size, z = –10 ft. The values of f(z)
are the specific weight taken from the table. Some example integrations give the following
results:
p10
ft
 0
lb f 
lb f
 10 ft  lb f
70 3  76 3   730 2
2  ft
ft 
ft
p 20
ft
 0
lb f 
lb f
 lb f 
 10 ft  lb f
70 3  2 76 3   84 3   1540 2
2  ft
ft 
ft 
ft

p30
ft
 0
lb f 
lb f 
lb f
 lb f
 10 ft  lb f
70 3  2 76 3  84 3   91 3   2405 2
2  ft
ft
ft 
ft 
ft

The complete results and graph, done in Excel, are shown below.
h(ft)
0
10
20
30
40
50
60
70
80
90
100
 (lbf/ft3)
70
76
84
91
97
102
107
110
112
114
115
p (psf)
0
730
1530
2405
3345
4340
5385
6470
7580
8710
9855
February 5 homework solutions
2.26
ME 390, L. S. Caretto, Spring 2008
Page 3
A U-tube manometer contains oil, mercury, and water
as shown in Fig. P2.26 (copied at the right). For the
column heights indicated what is the pressure
differential between pipes A and B?
We see that there is a line of equal pressure near the
bottom of the U-tube at the lower arrow for the dimension of
the 12 in. Designating the two equal pressures on either
side of the manometer at this point as pleft and pright allows
us to write the following equations.
pleft  p A   oil 7 in    Hg 12 in 
 pright  p B   H 2O 3 in  12 in 
Take the data for specific weights from table 1-5 on the
inside front cover (assuming the oil is SAE 30 weight oil and ignoring differences in temperature)
are water = 62.4 lbf/ft3, Hg = 847 lbf/ft3, and oil = 57.0 lbf/ft3. Rearranging the equation to solve for
pB – pA and substituting the data gives.
p B  p A   oil 7 in    Hg 12 in    H 2O 3 in  12 in 
847 lb f
62.4 lb f
 57.0 lb f
 ft 3
psia  in 2







7
in

12
in

15
in

3
3
1 lb f
ft 3
ft 3
 ft
 1728 in
pB – pA = 5.57 psi
2.40
The differential mercury manometer of Fig.
P2.40 (copied at the right) is connected to
pipe A containing gasoline (SG = 0.65), and
to pipe B containing water. Determine the
differential reading, h, corresponding to a
pressure in A of 20 kPa and a vacuum of
150 torr [use torr instead of mm Hg] in B.
Call the pressure at the gasoline-mercury
interface p1 and the pressure at the watermercury interface p2. The difference between
these two pressures is due to the column of
mercury with height h.
p2
p1
p1  p 2   Hg h
We can relate p1 and p2 to the pressures in A and B as follows.
p1  p A   gasol h  0.3 m 
p B  p 2   water h  0.3 m 
Combining the two equations above gives the following expression for p A – pB:
p A  p B  p1  p 2   water h  0.3 m    gasol h  0.3 m 
February 5 homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 4
From the first equation we have p1 – p2 = Hgh. Substituting this result into the equation above for
pA – pB gives.
p A  p B   Hg h   water h  0.3 m   gasol h  0.3 m 
 Hg   water   gasol h  0.3 m  water   gasol




Solving this equation for h gives
h

p A  p B  0.3 m   water   gasol

 Hg   water   gasol
The data for specific weights of water and mercury are taken from table 1-6 on the inside front
cover (ignoring differences in temperature): water = 9.80 kN/m3 and Hg =133 kN/m3. The
gasoline has a specific gravity of 0.65 and, assuming that its reference specific weight is that of
water just found, we find the gasoline specific weight as (SG)(water) = (0.65)(9.80 kN/m3) = 6.37
kN/m3.
The pressure in A is a gage pressure. The pressure in B is given as a vacuum of 150 mm Hg. A
vacuum is the difference between the atmospheric pressure and the absolute pressure. By
definition, pvacuum = patmospheric – pabsolute and pgage = –pvacuum. Thus the 150 torr vacuum is a gage
pressure of –(150 torr)(101.325 kPa)/(760 torr) = –20.00 kPa. Substituting this pressure for pB,
the given value of pA = 20 kPa, and the specific weight data from the previous paragraph into the
equation for h gives
h

p A  p B  0.3 m   water   gasol
 Hg   water   gasol


 9.81 kN 6.37 kN 



0
.
3
m



kPa  m 2
m3 
 m3
133 kN 9.81 kN 6.37 kN


m3
m3
m3
20 kPa   20 kPa
kN
h = 0.384 m
February 5 homework solutions
2.42
ME 390, L. S. Caretto, Spring 2008
Page 5
The manometer fluid in the manometer of Fig. P2.42
(copied at the right) has a specific gravity of 3.46.
Pipes A and B both contain water. If the pressure in
pipe A is decreased by 1.3 psi and the pressure in
pipe B increases by 0.9 psi, determine the new
differential reading of the manometer.
There is a common pressure on both sides of the
manometer at the point of the lower water-gage-fluid
interface. Writing the two equal pressures at this point in
terms of the pressures in A and B gives the following
equations.
pleft  p A   water hwater ,left   gage hgage
 pright  p B   water hwater ,rigfht
For the configuration shown above, hwater,left = 2 ft, hgage = 2 ft, and hwater,right = 1 ft. We can solve
this equation for pB – pA to obtain the following result.
p B  p A   water hwater ,left   gage hgage   water hwater ,right
Use the specific weight of water as 62.4 lbf/ft3 from Table 1-5 on the inside front cover for both the
value of water and the reference specific weight for the gage fluid. This gives the specific weight
of the gage fluid as (3.46)(62.4 lbf/ft3) = 215.9 lbf/ft3. Substituting these values and the original
gage readings into the equation for pB – pA gives
pB  p A 
62.4 lb f
ft 3
2 ft  
215.9 lb f
ft 3
2 ft  
62.4 lb f
ft 3
1 ft  
494.2 lb f
ft 2
The new value of pB – pA is found as follows.
 p B   p A,old  p A    p B  p A old  p B  p A
494.2 lb f 0.9 lb f 144 in 2  1.3 lb f 144 in 2 811.0 lb f




ft 2
in 2
ft 2
in 2
ft 2
ft 2
 p B  p A new  p B,old
With the increased value of pB, the water-gage-fluid interface on the left will become lower by a
value of h so that the new value of hwater,left will become 1 ft + h. This change will be reflected
throughout the manometer system. The water-gage-fluid interface on the right will rise by the
same value h so that hwater,righy becomes 2 ft – h. Both of these differences will change the
height of the gage fluid by 2h to a new value of hgage = 2 ft + 2h. Substituting these new values
of the gage heights and the new value of pB – pA into the original equation for pB – pA gives
 p B  p A new 
811.0 lb f
ft
2

62.4 lb f
ft
3
2
ft  h  
215.9 lb f
ft
3
2
ft  2h  
62.4 lb f
ft 3
1 ft  h
The products of specific weights and heights on the right-side of the equation are seen to be
simply the original values for pB – pA = 494.2 lbf/ft2. This gives the following steps for obtaining
h.
February 5 homework solutions
811.0 lb f
ft
2

494.2 lb f
ft
2

ME 390, L. S. Caretto, Spring 2008
316.8 lb f
ft
2
 215.9 lb f 62.4 lb f 62.4 lb f
 h 2



ft 3
ft 3
ft 3

Page 6
215.9 lb f

  h

ft 3

Solving this equation gives h = 1.03 ft. Adding two times this value to the original value of hgage
= 2 ft gives the required answer.
hgage,new = 4.06 ft
2.46
Determine the change in
elevation of the mercury in
the left leg of the
manometer of Fig. P2.46
(copied at the right) as a
result of an increase in
pressure of 5 psi in pipe A
while the pressure in pipe B
remains constant.
The mercury pressures are
equal at the mercury-water
interface and the point in the
incline that is at the same level as this interface. These two equal pressures can be written as
follows. (In this equation we use a variable height for the following distances in the diagram:
hwater= 18 in, hoil = 12 in, and ℓHg = 6 in = the difference between the water-mercury and the oilmrecury interfaces.)
pleft  p A   water hwater  p right  p B   oil hoil   Hg  Hg sin 30 o
Solving for the pressure difference gives
p A  p B   oil hoil   Hg  Hg sin 30 o   water hwater
When the pressure in A increases to pA,new, the water-mercury interface will drop by a value h
and the mercury-oil interface will rise a distance, ℓ along the inclined length. These changes will
cause hwater to increase by h and hoil to decrease by ℓ sin30o. The vertical difference in the
mercury column will increase by the sum of h and ℓ sin30o. Even with all these changes we
will still have the same relationship between the pressure difference and manometer
measurements. This will give the equation below.






p A,new  p B   oil hoil   sin 30 o   Hg h   Hg   sin 30 o   water hwater  h 
If we subtract the equation for the original pressure difference from this equation we obtain the
following result for the increase in pressure in A.


p A,new  p A    oil  sin 30 o   Hg h   sin 30 o   water h
We appear to have two unknowns, h and ℓ. However these two are related because the
volume of the mercury is constant. When the mercury drops by a height change of h, a volume
of mercury equal to rwater2 is displaced from the vertical tube to the inclined tube. The volume, V,
of fluid in an cylinder inclined at an angle  with the horizontal plane is given by the equation V =
February 5 homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 7
r2(ℓ + r/tan), where ℓ is the distance from the bottom of the cylinder to the closest location of the
inclined plane.1 If the value of ℓ is changed by ℓ, keeping r and  constant, the difference in
volume is simply r2ℓ, just as it would be for a vertical cylinder. Equating the two displaced
volumes gives the following relationship between h and ℓ.
2
rwater
h

2
rHg

 rHg
 h  
 rwater
2
2

 0.125 in 

  




4
 0.25 in 

Before substituting this result into our equation for the pressure change, we have to get the
necessary data. Table 1-5 in the inside front cover gives the values for water = 62.4 lbf/ft3 and Hg
= 847 lbf/ft3. The oil has a specific gravity of 0.9 so we find its specific weight using the value of
water = 62.4 lbf/ft3 just found: oil = 0.9(62.4 lbf/ft3)r = 56.16 lbf/ft3. We can now substitute the
equation h = ℓ/4 into the pressure increase equation, set sin30o to its value of 0.5, and
substitute the specific weights just obtained to get the solution for h.




p A,new  p A    oil 4h sin 30 o   Hg h  4h sin 30 o   water   2 oil  3 Hg   water h
h 
p A,new  p A
2 Hg  2 oil   water
5 psi 

144 lb f
psi  ft 2
 0.304 ft
847 lb f
56.16 lb f 62.4 lb f
3
2

ft 3
ft 3
ft 3
The mercury level drops 0.304 ft.
1
Taken from http://www.lmnoeng.com/Volume/InclinedCyl.htm web site accessed January 19,
2008.