PDF processed with CutePDF evaluation edition www.CutePDF.com
Draw the schematic diagram of the tank.
Step 2/3
Obtain the specific weight of the water at room temperature from table B.1 “Physical
Properties of Water”.
Calculate the pressure exerted by the water on the gage using the following relation:
Here, the specific weight of the water is
Substitute
for
and height of the water column is h.
and 12 in. for h.
Step 3/3
Calculate the pressure exerted by air as follows:
Here, the gage pressure is
Substitute
for
.
and 5 psi for
.
Calculate the absolute air pressure in the tank.
Here, the atmosphere pressure is
Substitute
for
.
and 14.7 psi for
Therefore, the absolute air pressure in the tank is
.
Draw the differential manometer as follows:
Step 2/3
Here, pressure in the pipe A is expressed in mm of HG. But all other parameters are
in
. Therefore, convert the pressure in the pipe A from mm of HG to
the following conversion formula:
Here,
is specific weight of Mercury,
mercury, and
Substitute
is head developed,
by using
is specific gravity of
is specific weight of water.
for
and
for
.
Step 3/3
Since the specific gravity of the oil is not given, let us assume specific gravity of oil is
0.9.
Write the pressure equation in the left limb and right limb of the manometer and equate
the values (equilibrium condition) to get the pressure in pipe B.
Substitute
negative),
for
for
(Since pressure in pipe A is vacuum, take the value is
, 13.6 for
, and 0.9 for
Since the value obtained is positive value, the pressure in pipe B is gauge pressure.
Therefore, the pressure in pipe B is
.
Draw the schematic diagram of the manometer and indicate the datum line as
.
Step 2/3
Calculate the specific weight of the gage fluid.
Here, the specific weight of the fluid is
Substitute
for
and the specific weight of water is
and
Convert the units of pressure from
.
to
.
Step 3/3
Apply the principal of manometry to the system.
.
Here, the pressures at A and B are
and
, the length of the fluid column in ,
inclination of the tube is , the heights of the water columns as shown in the figure
are
and
.
Substitute
,
,
, and
for
Therefore, the pressure at B is
.
,
for
for
,
for
for
,
for
.
Figure representing the closed cylindrical tank with a hemispherical dome:
Step 2/7
(a)
Calculate the pressure at any point by using the equation,
Calculate the specific weight of water using the relation:
Substitute
Here,
The density of water is
for
and
for
The specific weight of water is
The pressure at a given point is
The acceleration due to gravity is
The height of the fluid in the column is
The specific weight of the given fluid is
Step 3/7
The pressure between the points
and
can be found by starting at one end of the
system and working around to the other end. From the above figure, starting from the
point , the pressure at the point
is
…… (1)
Here,
The pressure at the point
is
The pressure at the point
is
The specific weight of water is
The specific weight of other liquid is
The height of water in the column is
The height of liquid in the column is
Step 4/7
In equation (1) the specific weight of other liquid with specific gravity
determined using the relation:
is
Now substitute
for
and
for
Step 5/7
Substitute
,
for
for
and
,
for
for
,
for
in equation (1)
Therefore, the pressure in the pipe
is
Step 6/7
(b)
Calculate the pressure between the points
…… (2)
Here,
The height of dome is
The pressure at the point
is
and
:
The pressure at the point
is
The specific weight of water is
Substitute
for
,
for
and
for
in equation (2)
Step 7/7
Find the pressure head in millimeters of mercury, at the top of the dome:
Substitute
for
and
for
Therefore, the pressure head in millimeters of mercury at top of the dome
is
Draw the diagram as shown below:
Step 2/3
Pressure at any point in the fluid is as follows:
Here,
is the specific weight of the liquid and h is the height from free surface.
Calculate pressure developed from one end to other end by adding hydrostatic pressure
changes when going down and subtracting it when moving upwards.
At free surface (1) pressure is atmospheric pressure. While moving from point (1) to
point (2) pressure reduced by an amount
pressure increase by an amount
pressure increase to
. When moving to point (2) to point (3)
. When moving to point (4) from point (3)
. Pressure at point (4) is atmospheric pressure.
Step 3/3
Mathematically this can be written as follows:
Here, pa is atmospheric pressure, γw is specific weight of water, h is distance from free
surface to point (2) and Δh is vertical distance between free surfaces in both tanks.
Substitute 9.8 kN/m3 for γw, (0.9×9.8 kN/m3) for γo and 0 for pa:
Therefore, vertical distance between free surfaces of both tanks,
is
.