We
were discussing the basic definition and significance of Pascal’s
Law along with its derivation , Vapour
pressure and cavitation, Absolute
pressure, Gauge pressure, Atmospheric pressure and Vacuum pressure, pressure
measurement, Piezometer and
also the basic concept of U-tube
manometer in our previous posts.
Today
we will understand here the basic concept of inclined single column manometer
to measure the pressure at a point in fluid, in the subject of fluid mechanics,
with the help of this post.
Single column manometer
Single column manometer is one modified form of U-Tube manometer. There will be one
reservoir with large cross-sectional area about 100 times as compared to the
area of glass tube. One limb (let us say left) of the glass tube will be
connected with the reservoir and another limb (right) of glass tube will be
open to atmosphere as displayed here in following figure.
This
complete set-up will be termed as single column manometer. Pressure will be
measured at a point in the fluid by connecting the single column manometer with
the container filled with liquid whose pressure needs to be measured. Rise of
liquid in right limb of glass tube will provide the pressure head.
There
are basically two types of single column manometers, on the basis of right limb
of manometer, as mentioned here.
Inclined
single column manometer
Inclined single column manometer
In
case of inclined single column manometer, other limb of glass tube will be inclined
as displayed here in following figure.
Let
us consider we have one container filled with a liquid and we need to measure
the pressure of liquid at point A in the container. Let us consider that we are
using the pressure measuring device “Inclined single column manometer” here to
measure the pressure of liquid at point A as displayed here in following
figure.
Let
us consider the following terms from above figure.
XX
is the datum line between in the reservoir and in the inclined limb of
manometer
P
= Pressure at point A and we need to measure this pressure or we need to find
the expression for pressure at this point.
Let
us consider that container, filled with a liquid whose pressure is to be
measured, is connected now with inclined single column manometer.
Once inclined single column manometer will be connected with container, heavy liquid
in reservoir will move downward due to the high pressure of liquid at point A
in the container. Therefore heavy liquid will rise in inclined limb of the
manometer.
Inclined
single column manometer will be quite sensitive manometer as distance moved by
heavy liquid in the right limb will be more due to inclination of right limb of
the manometer.
L
= Length of heavy liquid moved in the inclined limb from XX
h1 =
Height of lower specific gravity liquid above the datum line
h2 =
Height of higher specific gravity liquid above the datum line
h2 =
L x Sin θ
Δh
= Fall of heavy liquid in the reservoir
S1 =
Specific gravity of the light liquid i.e. specific gravity of liquid in
container
S2 =
Specific gravity of the heavy liquid i.e. specific gravity of liquid in
reservoir and in right limb of the manometer
ρ1 =
Density of the light liquid = 1000 x S1
ρ2 =
Density of the heavy liquid = 1000 x S2
YY
= Datum line after connecting the manometer with container
Level
of heavy liquid in the reservoir will be dropped and therefore there will be
respective rise in the level of heavy liquid in the right limb.
Pressure
in the left limb above the datum line YY = P + ρ1g (Δh + h1)
Pressure
in the right column above the datum line YY = ρ2g (Δh + h2)
As pressure is same for the horizontal surface and therefore we will have
following equation as mentioned here
P
+ ρ1g (Δh + h1) = ρ2g (Δh + h2)
P
= ρ2g (Δh + h2) - ρ1g (Δh + h1)
P
= Δh (ρ2g - ρ1g) + ρ2g h2 - ρ1g
h1
As,
value of Δh will be quite small and therefore we may neglect the term Δh (ρ2g
- ρ1g) and we will have following equation.
P = L x Sin θ x ρ2g - ρ1g h1
Do
you have any suggestions? Please write in comment box.
Reference:
Fluid mechanics, By R. K. Bansal
Image
Courtesy: Google
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