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.

h

_{1}= Height of lower specific gravity liquid above the datum line
h

_{2}= Height of higher specific gravity liquid above the datum line
Δh = Fall of heavy liquid in the
reservoir

S

_{1}= Specific gravity of the light liquid i.e. specific gravity of liquid in container
S

_{2}= Specific gravity of the heavy liquid i.e. specific gravity of liquid in reservoir and in right limb of the manometer
ρ

_{1}= Density of the light liquid = 1000 x S_{1}
ρ

_{2}= Density of the heavy liquid = 1000 x S_{2}
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 h

_{2}
Δh = a /A x h

_{2}
Pressure in the left limb above the
datum line YY = P + ρ

_{1}g (Δh + h_{1})
Pressure in the right column above the
datum line YY = ρ

_{2}g (Δh + h_{2})
As pressure is same for the horizontal
surface and therefore we will have following equation as mentioned here

P + ρ

_{1}g (Δh + h_{1}) = ρ_{2}g (Δh + h_{2})
P = ρ

_{2}g (Δh + h_{2}) - ρ_{1}g (Δh + h_{1})
P = Δh (ρ

_{2}g - ρ_{1}g) + ρ_{2}g h_{2 }- ρ_{1}g h_{1}
P = (a /A x h

_{2}) (ρ_{2}g - ρ_{1}g) + ρ_{2}g h_{2 }- ρ_{1}g h_{1 }
Do
you have any suggestions? Please write in comment box.

###
**Reference:**

Fluid mechanics, By R. K. Bansal

Image
Courtesy: Google

if the ratio of a and A would be soo small than what will be the answer...

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