Fluid Mechanics
Principles of Hydrostatics Manometer
Chapter 2
Principles of Hydrostatics - Manometer
Introduction
This module discusses the process of determining pressure or pressure differences using a
manometer
MANOMETER
A manometer is a tube, usually bent in a form of U,
containing a liquid of known specific gravity, the surface
of which moves proportionally to changes of pressure. It is
used to measure pressure or pressure differences.
Types of Manometer
Open Type
It has an atmospheric surface in one leg and is capable of measuring gage pressure.
(ang manometer na ito ay may butas or subjected sa atmosphere sa isang dulo)
It is not advisable to rely on general formulas for the solution of manometer problems. Each
problem should be considered individually and solved in accordance with fundamental
principles of variation of hydrostatic pressure with depth. It is ordinarily easier to work in
units of pressure head rather than pressure.
Suggested Steps in Solving Open Manometer Problems:
1. Decide on the fluid in feet or meter, of which the heads are to be expressed. (water is
most advisable)
2. Starting with the atmospheric surface in the manometer, number in order, the interface
of different fluids.
3. Identify the points of equal pressure (taking into account that for a homogeneous fluid
at rest, the pressure along the same horizontal plane are equal). Label these points
with the same number.
4. Proceed from level to level, adding (if going down) or subtracting (if going up)
pressure heads as the elevation decreases or increases, respectively with due regard
for the specific gravity of the fluids.
Differential Type
It has no atmospheric surface and capable of measuring only pressure differences.
(ang manometer na ito ay walang butas. Ito ay closed vessel)
Suggested Steps in Solving Differential Manometer Problems:
1. Number the “strategic points” indicated by the levels of contact of the fluids.
2. Starting with the unknown pressure head at one of the end points, write a continued
algebraic summation of heads, progressing from point to point, and equating the
continued sum to the unknown head at the other end point.
3. Solve the equation for the pressure-head difference and reduce to pressure difference
if desired.
Sample Problem 1 (Open Type Manometer)
Measuring Pressure with a Manometer
A manometer is used to measure the pressure of a
gas in a tank. The fluid used has a specific gravity of
0.85, and the manometer column height is 55 cm. If
the local atmospheric pressure is 96 KPa, determine:
a) the gage pressure in the tank
b) the absolute pressure in the tank
a) Solving for gage pressure in the tank
Step 1(optional in this case)
Decide on the fluid in feet or meter, of which the heads are to be expressed. (water is most
advisable)
Ang pag-convert ng pressure head ng liquid sa loob ng manometer into pressure head ng
water (most common) ay ginagawa kapag higit sa isang uri ng liquid ang nasa loob ng
manometer. Bukod dito, ginagawa ang pag-convert ng pressure heads upang mapadali ang
pagsosolve dahil lalabas na iisang uri na lang ng liquid ang nasa loob ng manometer.
Pero sa problem na ito, isang uri lang ng liquid ang nasa loob ng manometer. Pwede na
nating gamitin ang pressure head ng original na liquid.
Step 2
Starting with the atmospheric surface in the manometer, number in order, the interface of
different fluids.
1
2
Step 3
Identify the points of equal pressure (taking into account that for a homogeneous fluid at rest,
the pressure along the same horizontal plane are equal). Label these points with the same
number.
The pressure in both points 2 is the same
because they have the same elevation in the same
liquid
Step 4
Proceed from level to level, adding (if going down) or subtracting (if going up) pressure
heads as the elevation decreases or increases, respectively with due regard for the specific
gravity of the fluids.
0 + ℎ𝑓𝑙𝑢𝑖𝑑 =
(0.55𝑚) =
𝛾𝑓𝑙𝑢𝑖𝑑
𝑃2
𝐾𝑁
(0.85)(9.81 3 )
𝑚
(0.85) (9.81
4.586
𝑃2
𝐾𝑁
) (0.55𝑚) = 𝑃2
𝑚3
𝐾𝑁
= 𝑃2
𝑚2
Note: Since the fluid in the tank is a gas, we can say that the pressure in point 2 is the same
to any point within the tank. Therefore, the pressure at 2 is the pressure in the tank.
𝑃2 = 𝑔𝑎𝑔𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑡𝑎𝑛𝑘 = 𝟒. 𝟓𝟖𝟔 𝑲𝑷𝒂
Another Solution:
Proceed from level to level, we can also add (if going down) or subtract (if going up) pressure
as the elevation decreases or increases.
0 + 𝛾𝑓𝑙𝑢𝑖𝑑 ℎ𝑓𝑙𝑢𝑖𝑑 = 𝑃2
(0.85) (9.81
4.586
𝐾𝑁
) (0.55𝑚) = 𝑃2
𝑚3
𝐾𝑁
= 𝑃2
𝑚2
𝟒. 𝟓𝟖𝟔 𝑲𝑷𝒂 = 𝑷𝟐
b. Solving for the Absolute Pressure in the tank
𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 + 𝑃𝑔𝑎𝑔𝑒
𝑃𝑎𝑏𝑠 = 96 𝐾𝑃𝑎 + 4.586 𝐾𝑃𝑎
𝑷𝒂𝒃𝒔 = 𝟏𝟎𝟎. 𝟓𝟖𝟔 𝑲𝑷𝒂
Sample Problem 2 (Open Type Manometer)
The water in a tank is pressurized by air, and
the pressure is measured by a multi-fluid
manometer. The tank is located on a
mountain at an altitude of 1400 m where the
atmospheric pressure is 85.6 KPa.
Determine the gage air pressure in the tank if
h1 = 0.1 m, h2 = 0.2 m, and h3 = 0.35 m.
Take the specific gravity of water, oil, and
mercury to be 1, 0.85, and 13.6, respectively.
Solution:
Step 1 - Decide on the fluid in feet or meter, of which the heads are to be expressed. (water is
most advisable)
Convert the pressure heads into pressure head of
water.
S
Since hwater = S fluid hfluid and swater is always 1, to
water
convert the pressure head of any liquid in meters or
feet of water, just use
𝐡𝐰𝐚𝐭𝐞𝐫 = 𝐬𝐟𝐥𝐮𝐢𝐝𝐡𝐟𝐥𝐮𝐢𝐝
Pwede nating gawin ang conversion sa computation
part na
Step 2 - Starting with the atmospheric surface in the
manometer, number in order, the interface of
different fluids.
Step 3 - Identify the points of equal pressure (taking into
account that for a homogeneous fluid at rest, the
pressure along the same horizontal plane are equal).
Label these points with the same number.
No need to compute the pressures sa mga part na
may X dahil zero lang din ang sum
nila.
Step 4 - Proceed from level to level, adding (if going down) or subtracting (if going up)
pressure heads as the elevation decreases or increases, respectively with due regard for the
specific gravity of the fluids.
Take note: Base sa Step 1, nagdecide tayo na ang gagamitin nating pressure heads ay
pressure head in meters of water. Formula: 𝒉𝒘𝒂𝒕𝒆𝒓 = 𝒔𝒇𝒍𝒖𝒊𝒅𝒉𝒇𝒍𝒖𝒊𝒅
0 + 𝑠𝑚𝑒𝑟𝑐𝑢𝑟𝑦 ℎ3 − 𝑠𝑜𝑖𝑙 ℎ2 − ℎ1 =
𝑃4
𝛾𝑤𝑎𝑡𝑒𝑟
(13.6)(0.35𝑚) − (0.85)(0.2𝑚) − 0.1𝑚 =
4.76𝑚 − 0.17𝑚 − 0.1𝑚 =
4.49 𝑚 =
𝑃4
𝛾𝑤𝑎𝑡𝑒𝑟
𝑃4
𝛾𝑤𝑎𝑡𝑒𝑟
𝑃4
𝛾𝑤𝑎𝑡𝑒𝑟
𝑃4 = (4.49 𝑚)(𝛾𝑤𝑎𝑡𝑒𝑟 )
𝑃4 = (4.49 𝑚) (9.81
𝑃4 = 44.0469
𝐾𝑁
)
𝑚3
𝐾𝑁
𝑚2
𝑷𝟒 = 𝟒𝟒. 𝟎𝟒𝟔𝟗 𝑲𝑷𝒂
Another Solution:
Proceed from level to level, we can also add (if going down) or subtract (if going up) pressure
as the elevation decreases or increases.
0 + 𝛾𝑚𝑒𝑟𝑐𝑢𝑟𝑦 ℎ3 − 𝛾𝑜𝑖𝑙 ℎ2 − 𝛾𝑤𝑎𝑡𝑒𝑟 ℎ1 = 𝑃4
(13.6) (9.81
46.6956
𝐾𝑁
𝑚3
𝐾𝑁
𝐾𝑁
𝐾𝑁
𝐾𝑁
−
1.6677
−
0.981
= 𝑃4
𝑚2
𝑚2
𝑚2
𝑃4 = 44.0469
𝐾𝑁
) (0.35𝑚) − (0.85) (9.81 𝑚3 ) (0.2𝑚) − (9.81 𝑚3 ) (0.1𝑚) = 𝑃4
𝐾𝑁
𝑚2
𝑷𝟒 = 𝟒𝟒. 𝟎𝟒𝟔𝟗 𝑲𝑷𝒂
Note that jumping horizontally from one tube to the
next and realizing that pressure remains the same in
the same fluid simplifies the analysis considerably.
Sample Problem 3 (Differential Type Manometer)
A differential manometer shown is measuring the difference in pressure of two water pipes A
and B. The indicating liquid is mercury (s = 13.6)
h1 = 225 mm
h2 = 675 mm
h3 = 300 mm
What is the pressure difference between A and B?
Solution:
Start summing up the pressure heads from left to right (from A to B)
Take note that going down is positive and going up is negative.
Note: lahat ng pressure heads (h) ay converted into pressure heads ng water. Formula:
𝐬
𝐡𝐟𝐥𝐮𝐢𝐝
𝐬
𝐡
𝐡𝐰𝐚𝐭𝐞𝐫 = 𝐟𝐥𝐮𝐢𝐝
= 𝐟𝐥𝐮𝐢𝐝𝟏.𝟎𝐟𝐥𝐮𝐢𝐝 = 𝐬𝐟𝐥𝐮𝐢𝐝𝐡𝐟𝐥𝐮𝐢𝐝
𝐬
𝐰𝐚𝐭𝐞𝐫
𝑷𝑨
𝜸𝒘𝒂𝒕𝒆𝒓
𝑃𝐴
𝛾𝑤𝑎𝑡𝑒𝑟
𝑃𝐴
𝛾𝑤𝑎𝑡𝑒𝑟
+ 𝒉𝟏 + 𝒔𝒎𝒆𝒓𝒄𝒖𝒓𝒚𝒉𝟐 − 𝒉𝟑 = 𝜸
𝑷𝑩
𝒘𝒂𝒕𝒆𝒓
+ 0.225𝑚 + 13.6(0.675𝑚) − 0.3𝑚 = 𝛾
𝑃𝐵
𝑤𝑎𝑡𝑒𝑟
+ 9.105𝑚 = 𝛾
9.105𝑚 =
𝑃𝐵
𝑤𝑎𝑡𝑒𝑟
𝑃𝐵
𝛾𝑤𝑎𝑡𝑒𝑟
−
𝑃𝐴
𝛾𝑤𝑎𝑡𝑒𝑟
𝑃𝐵 − 𝑃𝐴
= 9.105𝑚
𝛾𝑤𝑎𝑡𝑒𝑟
𝑃𝐵 − 𝑃𝐴 = (9.105𝑚)(𝛾𝑤𝑎𝑡𝑒𝑟 )
𝑃𝐵 − 𝑃𝐴 = (9.105𝑚) (9.81
𝐾𝑁
)
𝑚3
𝑷𝑩 − 𝑷𝑨 = 𝟖𝟗. 𝟑𝟐 𝑲𝑷𝒂 − − − −𝒂𝒏𝒔𝒘𝒆𝒓 (gage pressure)
Another Solution:
Start summing up the pressure from left to right (from A to B)
Take note that going down is positive and going up is negative.
𝑷𝑨 + 𝜸𝒘𝒂𝒕𝒆𝒓 𝒉𝟏 + 𝜸𝒎𝒆𝒓𝒄𝒖𝒓𝒚𝒉𝟐 − 𝜸𝒘𝒂𝒕𝒆𝒓 𝒉𝟑 = 𝑷𝑩
𝑃𝐴 + (9.81
𝐾𝑁
𝑚3
) (0.225𝑚) + 13.6 (9.81
𝐾𝑁
𝐾𝑁
𝐾𝑁
𝑚3
) (0.675𝑚) − (9.81
𝐾𝑁
𝑚3
) (0.3𝑚) = 𝑃𝐵
𝐾𝑁
𝑃𝐴 + 2.20725 𝑚2 + 90.0558 𝑚2 − 2.943 𝑚2 = 𝑃𝐵
𝑃𝐴 + 89.32
89.32
𝐾𝑁
= 𝑃𝐵
𝑚2
𝐾𝑁
= 𝑃𝐵 − 𝑃𝐴
𝑚2
𝑷𝑩 − 𝑷𝑨 = 𝟖𝟗. 𝟑𝟐 𝑲𝑷𝒂 − − − −𝒂𝒏𝒔𝒘𝒆𝒓 (gage pressure)
Note: Pwede din kayo mag start from B to A.
References/Additional Resources/Readings


Gillesania, D.I.T, “Fluid Mechanics & Hydraulics 4th Edition”, Copyright 2015 by
Cebu DGPrint, Inc.
C.T. Crowe, J. A. Roberson and D.F. Elger, “Engineering Fluid Mechanics 9th
Edition”, Copyright 2009 by John Wiley & Sons, Inc.
Additional Resources:


R.E. Featherstone and C. Nalluri, “Civil Engineering Hydraulics 3rd Edition”,
Copyright 1995 by Blackwell Science Ltd.
Fluid Mechanics 3rd Edition – Yunus A. Cengel and John M. Cimbala