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

h

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

_{2}= Height of higher specific gravity liquid above the datum line
h

_{2}= L x Sin θ
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}
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 + 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}
As,
value of Δh will be quite small and therefore we may neglect the term Δh (ρ

_{2}g - ρ_{1}g) and we will have following equation.###
**P = L x Sin θ x ρ**_{2}g_{ }-
ρ_{1}g h_{1}

_{2}g

_{ }- 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

Great job. You need to mention that with this analysis you’re measuring the gauge pressure since the open end is in contact with the atmosphere.

ReplyDeleteIn the last line written in bold, as the overall formula, it should be P = L x Sin thetha x(ρ2gh2- ρ1gh1)

ReplyDelete