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 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.
Vertical
single column manometer
Inclined
single column manometer
Vertical single column manometer
In
case of vertical single column manometer, other limb of glass tube will be
vertical 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 “Vertical 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 right 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 vertical
single column manometer.
Once vertical 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 right limb of the manometer.
Once vertical 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 right limb of the manometer.
h1 = Height of lower specific
gravity liquid above the datum line
h2 = Height of higher
specific gravity liquid above the datum line
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
A = Cross-sectional area of the
reservoir
a = Cross-sectional area of the right
limb
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.
A x Δh = a x h2
Δh = a /A x h2
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
P = (a /A x h2) (ρ2g
- ρ1g) + ρ2g h2 - 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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