College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
Solutions to Exercise One – Fluid Statics and Manometers
1. The drain plug in a bathtub is designed to seal properly when there is a 0.05 psi pressure
applied to it. How deep must the water be in the tub for the plug to seal? (Problem 2.8 in
text.)
This is a simple application of the basic equation of fluid statics: p1 + z1 = p2 + z2 or p2 = p1 +
(z2 – z1) = p1 + h. In this case we want to find the depth, h = (p2 – p1)/. p1 = 0 psi(gage) at the
free surface of the liquid and we are given that p2 = 0.05 psi(gage). Using  = 62.4 lbf/ft3 for water
from Table 1.5 on the inside front cover gives the following result.
 0.05 lb f
 144 in 2



0
2
ft 2
p 2  p1  in

h

 0.115 ft = 1.38 in
62.4 lb f

ft 3
2. The basic elements of a hydraulic press are shown in Figure
P2.12 copied from the text at the right. The plunger has an
area of 1 in2, and a force, F1, can be applied to the plunger
through a lever mechanism (not shown) having a mechanical
advantage of 8 to 1. If the large piston has an area of 150 in2,
what load, F2 can be raised by a force of 30 lbf applied to the
lever? Neglect the variation of pressure with depth. (Problem
2.12 in text.)
We first have to determine F1 from the information given about the
applied force of 30 lbf and the mechanical advantage of 8 for the lever that is used to apply the
force F1. From the definition of mechanical advantage as the ratio of the resultant force to the
applied force we conclude that F1 = 8(30 lbf) = 240 lbf.
We are told to neglect the variation of pressure with depth. This means that the pressure is
constant throughout the hydraulic fluid. This pressure is found by dividing the value of F 1 = 240
lbf by the 1 in2 area of the piston through which F1 is applied. This gives a pressure of 240 lbf/in2.
Since the same pressure is exerted on the 150 in2 area of the piston that generates the force F2,
we conclude that F2 = PA2 = (240 lbf/in2)(150 in2) so that F2 = 36,000 lbf .
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Solutions to Exercise One
ME 390, L. S. Caretto, Spring 2008
3. A closed cylindrical tank filled with
water has a hemispherical dome and
is connected to an inverted piping
system as shown in figure P2.25,
copied from the text at the right. The
liquid in the top part of the piping
system has a specific gravity of 0.8,
and the remaining parts of the
system are filled with water. If the
pressure gage reading at A is 50 kPa
determine (a) the pressure in pipe B,
and (b) the pressure head in torr
(millimeters of mercury) at the top of
the dome (point C). (Problem 2.25
from text.)
Page 2
pA
(a) From the diagram we see that the fluid with a specific gravity of 0.8 fills the piping system on
both sides at a point 4 m from the top branch. Thus the pressure on both sides of the piping
system is the same at this point. We also see that this the level of the water on the left side of the
piping system is the same as the level of the pressure gage in the tank which reads 50
kPa(gage). From these two observations we conclude that the manometer fluid at a distance 4 m
from the top of the piping system has a pressure of 50 kPa(gage) on both sides of the system,
which is pA.
On the right side of the piping system, the difference in pressure between this level where the
pressure = pA and the level of the pipe B, is given by 3 m of the fluid with the 0.8 specific gravity
and 2 m of water. Thus we can write our basic equation for the variation of pressure with depth
as follows:
pA + top(3 m) + water(2 m) = pB.
The specific weight of the top fluid is found by multiplying its specific gravity of 0.8 times the
specific weight of water water= 9.80 kN/m3 from Table 1. on the inside front cover. This gives top
= 7.84 kN/m3. Substituting the values for the specific weight of the top fluid and water and the
value of pA = 50 kPa(gage) into the equation for pb gives the following result.
 7.84 kN
3 m  9.80 3kN 3 m kPa  m
p B  p A   top 3 m   water 2 m  50 kPa  
3
m
 m
 1 kN
2
PB = 93.1 kPa(gage)
Because pB is measured relative to pA which is a gage pressure, pB is also a gage pressure.
(b) We apply our usual relation between the levels C and A: pA = pC + water(3 m)
Notice that the curvature of the tank has on effect on this relationship. We can solve thie
equation for pC, and substitute the values of 50 kPa for pA and the specific weight of water used
previously to give the desired result.
pC  p A   water 3 m   50 kPa 
9.80 kN
kPa  m 2


3
m
 20.6 kPa
1 kN
m3
Solutions to Exercise One
ME 390, L. S. Caretto, Spring 2008
Page 3
We are asked to express this as a pressure head in mmHg. To do this we simply divide the
pressure by the specific weight of mercury which is found from table 1.5 on the inside front cover
to be 133 kN/m3. Thus the pressure head is
h
pC
 Hg

20.6 kPa
 0.155 m
133 kN kPa  m 2
1 kN
m3
4. The cylindrical tank with hemispherical ends
shown in figure P2.35, copied from the text at the
right, contains a volatile liquid and its vapor. The
liquid density is 800 kg/m3, and its vapor density
is negligible. The vapor pressure is 120 kPa(abs)
and the atmospheric pressure is 101 kPa(abs).
Determine (a) the gage pressure reading on the
pressure gage and (b) the height, h, of the
mercury manometer. (Problem 2.35 from text.)
The pressure at the gage will be the vapor pressure
at the liquid vapor interface plus the h term for the h
= 1 m distance from the fluid-favor interface to the
gage: pgage = pvapor = liquidh. Since we are given the density of the liquid is 800 kg/m 3 we have to
compute the specific weight as the g product. We can assume a standard gravity giving lquid=
(800 kg/m3)(9.81 m/s2) = 7848 kg▪m/ m3▪s2 = 7848 N/m3. Thus the absolute pressure at the gage
can be found as follows.
p gage  pvapor   liquid 1 m   120 kPa(abs) 
7848 N
kPa  m 2


1
m
 127.848 kPa(abs)
1000 N
m3
To get the gage pressure reading at the gage we have to subtract the atmospheric pressure of
101 kPa from this absolute pressure giving the result that pgage = 26.8 kPa(gage) .
To find the height, h, of the mercury manometer, we note that the mercury-liquid interface on the
left is at the same level as the pressure gage. So the mercury on both sides of the manometer at
the level of the pressure gage has the same pressure as the gage, pgage = 26.8 kPa(gage). Since
the mercury manometer is open to the atmosphere on the right we have the following equation for
the height of the mercury column: pgage = patm + Hgh, where patm = 0 as a gage pressure. This
gives the result that h = pgage/Hg = (26.8 kPa)(1 kN▪m 2/kPa) / (133 kN/m3), where the specific
weight of mercury is taken from Table 1.6 on the inside front cover. This gives the result that
h = 0.202 m .
Solutions to Exercise One
ME 390, L. S. Caretto, Spring 2008
Page 4
5. Pike’s Peak near Denver, Colorado has an elevation of 14,110 ft. (a) Determine the
pressure at this elevation based on equation 2.12. (b) If the air is assumed to have a
constant specific weight of 0.07647 lbf/ft3, what would the pressure be at this altitude? (c)
If the air is assumed to have a constant temperature of 59 oF, what would the pressure be
at this altitude? For all three cases assume standard conditions at sea level as given in
Table 2.1. (Problem 2.20 from text.)
(a) This problem examines equations not covered in the lecture. Equation 2.12 is shown below
followed by an explanation of the terms in this equation.
g
 z  R
p( z )  p a 1  
 Ta 
The equation accounts for the variation of pressure with distance for a gas, whose gas constant is
R, in an atmosphere where the temperature changes linearly with elevation. From Table 2.1 on
page 47 we find pa = 2116.2 lbf/ft2 and Ta = 518.67 R as the sea level pressure and temperature
for the standard atmosphere. Both of these must be absolute values in this application because
we are using the equation of state for an ideal gas. (R is the gas constant for air which is 1716
ft▪lbf/slug▪R. (See page 12 of the text for this value.)
The temperature profile follows the equation T = T a – z, where z is the elevation. The value of 
= 0.00650 K/m = 0.00357 R/ft for the standard atmosphere is given between equations 2.11 and
2.12 on page 48 of the text. Since this value is positive, the temperature will decrease with
altitude. In deriving this equation the value of g is assumed constant and we will use the standard
value of g = 32.174 ft/s2.
With all the given data we can evaluate the dimensionless exponent g/R in the pressure
equation.
2
32 .174 ft 1 lb f  s
g
slug  ft
s2

 5.252
R 1716 ft  lb f 0.00357 R
slug  R
ft
We can now apply equation 2.12 to find the elevation at 14,110 ft.
 z 
p( z )  p a 1  
 Ta 
g
R
0.00357 R



14110 ft  

2116.2 lb f 
ft


1
2


518.67 R
ft




5.252
 1237 psf(absolute)
(b) If we assume a constant specific weight of 0.07647 lbf/ft3, we have our usual simple formula for
constant specific weight.
p  p a  h 
2116.2 lb f
ft 2

0.7647 lb f
ft 3
(14,110 ft )  1037 psf(absolute)
Solutions to Exercise One
ME 390, L. S. Caretto, Spring 2008
Page 5
Here the predicted pressure is less than in the previous case because the constant specific weight
assumption keeps the value of  too high when it is actually decreasing as accounted for in the
previous calculations in case (a).
(c) The case of an isothermal atmosphere is given by equation 2.10:
p( z )  p a e

gh
RTa
. The
dimensionless exponent in this equation is evaluated as follows:
1 lb f  s 2
32.174 ft


14110
ft
gh
slug  ft
s2

 0.51006
1716
ft

lb
RT a
f
518.67 R 
slug  R
We can now apply equation 2.10 to get the pressure as follows.
p( z )  p a e

gh
RTa

2116.2 lb f
ft
2
e 0.51006 = 1271 psf(absolute)
This answer is closer to the result of part (a) because it accounts for the effect of pressure on
specific weight. From the temperature gradient of 0.00357 R/ft, we see that the temperature at
14,110 feet is 468.3 R. So the difference between this answer and part (a) is that we did not
accounted for the decreasing temperature in the atmosphere in the calculation that we just did for
part (c).