PRESSURE
Mass and Weight:
Mass (m) is the amount of matter in an object (SI unit = kg, British unit = slug)
As long as the amount of matter remains the same, the mass is constant.
Weight () is the force of gravity on an object (SI unit = N, British unit = lb)
An astronaut’s body weight on the moon is only about 1/6 of that on earth. His mass is
unchanged but the force of gravity is different and so his weight is not constant.
m = 1 slug
m = 1 kg
a = 1 m/s2
F = 1N
a = 1 ft/s2
F = 1 lb(f)
g = 9.81 N/Kg
w = 9.81 N
g = 32.2 lb/slug
w = 32.2 lb
a = 9.81 m/s2 = 32.2 ft/s2
a = 9.81 m/s2 = 32.2 ft/s2
Units of Force:
SI:
A Newton (N) is a force which will cause a 1 kg mass to accelerate at 1 m/s2.
The force of gravity on earth (g) is 9.8 N/kg.
Gravity causes a 1 kg mass to accelerate (fall) at a rate of 9.81 m/s2.
British:
A pound of force (lbf) will cause a mass of 1 slug to accelerate at 1 ft/s2.
The force of gravity on earth (g) is 32.2 lb/slug.
Gravity causes a 1 slug mass to accelerate (fall) at a rate of 32.2 ft/s2.
cgs:
A dyne of force (dyn) will cause a mass of 1 g to accelerate at 1 cm/s2.
The force of gravity on earth (g) is 981 dyn/g.
Gravity causes a 1 g mass to accelerate (fall) at a rate of 981 cm/s2.
Conversions:
1 N = 1 kgm/s2
1 lb = 1 slugft/s2
1 dyn = 1 gcm/s2
force-force: 1 lb = 4.45 N
1 N = 105 dynes
mass-mass: 1 slug = 14.6 kg
1 kg = 103 g
mass-weight: 1 kg = 9.807 N
1 kg = 2.205 lb
Pressure
1 slug = 32.17 lb
454 g = 1 lb
1
There are 2 definitions for density:
a) Mass Density and
b) Weight Density
a) Mass Density ( - pronounced “rho”) is mass per unit volume
 =
m
v
kg
m3
units:
or
g
cm3
slugs
ft 3
or
b) Weight Density (D) is weight (gravitational force) per unit volume.
D

v
units :
N
m
or
3
lb
ft
3
or
dyn
cm
3
Note: D 
Complete the following table:
*
g/cm3
H2 O
1.00 (= s.g.)
Hg
13.596 (= s.g.)
air
0.0013
kg/m3

v

m g
 g
v
1 m = 3.2808 ft
g/L
dyn/cm3
N/m3
lb/ft3
slug/ft3
*H2O at 3.98C, Hg at 20C, air at STP

Convert the mass density of water in kg/m3 to various units of weight density
Pressure
2
Gravitational Acceleration
The “constant” acceleration due to gravity varies due to several factors. One of these factors is the latitude of
the observer. Another is the altitude.
Latitude
Acceleration due to gravity decreases as the inverse of the square of the distance from the center of mass of the
body imparting gravitational acceleration; g = k/x2, where x = centre to centre distance.
Because the earth is not perfectly spherical, the distance from the center of the earth for any person standing on
the surface depends upon the latitude, L.
The center-to-surface distance is less at the poles than at the equator. As a result, gravitational acceleration
varies from 9.8322 m/s2 at the poles to 9.7803 m/s2 at the equator.
The acceleration due to gravity at any latitude, L, (at sea level) is given by the following equation:
g = 9.780327  [1 + 0.0053024  sin2(L) – 0.0000058  sin2(2L)]
For example, g at sea level at 42° north latitude (Windsor and Rome, Italy) is 9.8035 m/s2.
N
Earth is an ellipsoid. It is flattened at its poles.
r = 6,378,140 m
g = 9.7803 m/s2
r = 6,356,755 m
g = 9.8322 m/s2
Altitude
The equatorial radius of the earth is approximately 6,378,140 meters.
The polar radius is approximately 6,356,755 meters
Modeling the earth as an ellipsoid, the radius, R, of the earth at any latitude, L, is given by the following
equation:
R = 6,356,755  sqrt [1 + 0.0067396  cos2(L)]
At the 42nd parallel R is calculated to be 6,368,574 meters at sea level.
A person at an altitude of H meters above sea level experiences an acceleration due to gravity of:
a = g  R2 / (R + H)2
For example, Windsor, Ontario is ca. 581 ft (177 m) above sea level. This gives an acceleration due to gravity
of 9.8029 m/s2.
Not withstanding the foregoing calculation, we will use a standard value for g of 9.80665 m/s2, rounding to 4
sig figs = 9.807 m/s2, (32.17 ft/s2) and 9.807 N/kg (32.17 lb/slug).
Pressure
3
There are 2 definitions for density:
a) Mass Density and
b) Weight Density
c) Mass Density ( - pronounced “rho”) is mass per unit volume
 =
m
v
kg
m3
units:
or
g
cm3
slugs
ft 3
or
d) Weight Density (D) is weight (gravitational force) per unit volume.
D

v
N
units :
m
3
or
lb
ft
3
or
dyn
cm
Note: D 
3

v

m g
 g
v
*
g/cm3
kg/m3
g/L
dyn/cm3
N/m3
lb/ft3
slug/ft3
H2 O
1.00 (= s.g.)
103
103
980.7
9807
62.43
1.9404
Hg
13.596 (= s.g.)
13,600
13,600
13,300
133,000
849
26.3
air
0.00129
1.29
1.29
1.27
12.7
0.0805
0.00250
*H2O at 3.98C, Hg at 20C, air at STP
The values in the table are accurate to 3 sig figs using the factors listed below
* USING THE FOLLOWING CONSTANTS/FACTORS
2
2
2
 g = 9.80665 m/s or N/kg = 980.7 cm/s or dyn/g = 32.174 ft/s or lb/slug
 1 lb = 4.4482 N = 453.592 g
 1 kg = 2.20462 lb
 1 slug = 14.594 kg = 32.174 lb

1 m = 3.28083 ft

1 m3 = 35.3143 ft3
H2O @ 4º = 1000 kg/m3 = 62.429 lb/ft3 = 1.94035 slug/ft3

Although the gravitational constant in Windsor, Ontario is accurately calculated as 9.803 N/kg,
the standard conversions between are based on a value of 9.80665 N/kg. The values in the table
use the standard value of g rather than the local value.
Pressure
4
Specific Gravity (sg) =
sg =
weight of a given volume of substance
weight of an equal volume of water
Ds
s
=
Dw
w
The standard density for water is 1.00 g/cm3 which occurs at 3.98 ºC. Water is densest at
this temperature and is less dense at all other temperatures, above and below this
temperature.
D4º = 62.4 lb/ft3 = 9810 N/m3
and
4º = 1.00 g/cm3 = 1000 kg/m3 = 1.94 slug/ft3
Note that for sg all units cancel so sg has no units.
13.6 g / cm3 13600 kg / m3


  e.g. Hg s.g. =
SI and cgs: s.g. =
= 13.6
1.0 g / cm3
1000 kg / m3
10
. g / cm3
British:
s.g. =
Ds
62.4 lb / ft 3
e.g. Hg s.g. =
849 lb/ft 3
62.4 lb/ft 3
 13.6
Note that the density of a substance in the cgs system is numerically the same as its s.g.,
however the density of a substance in the British or SI system is different than s.g.
What about gases?
The discussion of density and specific gravity above applies to liquids and solids.
Since gases are much less dense than liquids or solids, the density of gases is usually
reported in units of grams per litre (gpL) rather than g/mL or g/cm3
For example, Cl2 gas has a density of 3.17 gpL, while air is 1.29 gpL.
The s.g. of a gas is the ratio of the density of the gas to the density of air at standard
conditions called "STP", i.e., at 0 ºC and 1.00 atmosphere pressure.
s.g.=
density of gas
density of air @ STP
=
density of gas (g / L)
1.29 g / L
s.g. Cl 2 =
3.17 gpL
1.29 gpL
= 2.46

Carbon dioxide has a specific gravity (relative vapor density) of 1.53 at STP.
Calculate its density. (ans = 1.97 g/L). Is it heavier or lighter than air?

Hydrogen, the lightest gas, has a density of 0.089 g/L. What is its relative vapor
density? ( ans = 0.069)
Pressure
5
Pressure Measurement:
A fluid at rest exerts a force and pressure on the walls of its container. The force is
perpendicular to the container walls.
Pressure (p) = force per unit area: p 
F
A
units :
N
lb
 Paschal (Pa) or 2 (psf)
2
m
ft
A = r2
or
lb
(psi)
in 2
h1
Within a fluid, pressure increases with the
depth where there is greater force per unit
area.
h2
h3
V = Axh = r2 x h = d2/4 x h
To describe the pressure at various depths,
consider the height (h) of a column of the
liquid above an area (A).
The mass of fluid in a column is obtained
from:
 = m/V
p =
m = V = (Ah)
(   A  h)  g
F

m g
=
=
=
=  gh = Dh
A
A
A
A
This is called the pressure-depth equation:


p = gh = Dh
Using the pressure depth equation, calculate the pressure exerted by a column of
H2O 251 cm high in both Pa and psf, then interconvert the two values as a check.
(ans. = 2.46  104 Pa, 514 psf)
Pressure
6
Pressure Exerted by a Fluid:
low air pressure
air pressure is exerted equally in all directions
high air
pressure
1. Pressure is proportional to
height of a fluid, e.g., air
pressure is less in Mexico
City or in Denver Colorado
compared to sea level.
Mexico City and Denver
are both about 1 mile
above sea level.
2. Pressure is equal in all directions.
Paschal’s Principle states that external
pressure applied to an enclosed fluid is
transmitted undiminished throughout
the fluid. This is true for static fluids
(fluids at rest). We will later see that
when fluids move (flow) they lose
pressure due to friction.
3. The slope of a container has no influence on pressure at a given depth in a fluid.
h
P1
P2
P3
P4
P1 = P2 = P3 = P4
Pressure
7
4. The total pressure in a fluid is the sum of the pressure of the liquid plus
atmospheric pressure above the fluid. Total pressure is called “absolute
pressure”.
atmospheric pressure
atmospheric
pressure +
fluid pressure =
fluid pressure
total pressure or
‘absolute pressure’
absolute (total) pressure = fluid pressure + atmospheric pressure
pabs = gh + patm
or
pabs = Dh + patm

Without the aid of scuba tanks, pearl divers are know to dive to a depth of 50 ft.
Calculate the fluid pressure and absolute pressure in psi on a diver at a 50.0 ft depth of
sea water.
Assume the density of sea water is 1.027 g/cm3.
Atmospheric pressure at sea level is 14.7 psi. (ans. 22.3 and 37.0 psi)
Pressure
8
Atmospheric Pressure results from the weight of air above us. Air, like water, is a fluid
and has mass. We live at the bottom of a sea of air.
Standard atmospheric pressure is
1.0 atm. (14.7 psi)
Atmospheric pressure is the
same pressure as exerted by 33.9
ft. of fresh water (33 ft. sea
water). Water is much denser
than air so the depth of water is
much less than the depth of air
for a given pressure.
Air pressure at the earth’s
surface is due to the
weight of all the air above
that point.
ca. 80 km to
the top of the
stratosphere
A column of Hg 76cm (29.92 in)
exerts the same pressure as
atmospheric pressure. Mercury
is 13.6 times denser than water.
Mercury exerts a pressure 13.6
times greater than water for the
same height of fluid…
0.76 m Hg  13.6 = 33.9 ft H2O
What height of water in
meters equals atmospheric press?

Standard Atmospheric Pressure at sea level is:
2
5
2
6
2
14.696 lb/in2 = 2116 lb/ft = 1.01325  10 N/m (Pa) = 1.01325  10 dyn/cm
= 101.325 kPa = 1 atm = 760 mmHg. (Torr) = 76 cm Hg = 29.9213 inHg
= 33.90 ft fresh water (~33 ft. sea water) = 1.01325 bar = 1013.25 mb
3
Note:
5
2
6
2
1 dyn/cm = 1 bar = 1 barye
Also: P 


2
10 mb = 1 bar = 10 N/m = 10 dyn/cm
1 bar = 105 Pa
F F x
work
energy
(pressure can be calculated as energy/volume)



A A  x volume volume
Convert 14.696 psi to kg/cm2. Convert 101325 Pa to kg/cm2
Do problems 1a) through e)
Pressure
9

A student measures a pressure of 61.5” water on the flow table in the lab. Convert
this to units of Paschals. (ans = 1.53  104 Pa)
Atmospheric Pressure and Weather Forecasting:
‘Standard’ atmospheric pressure is a physical constant defined as 1.0 atm (760 mmHg,
29.92 in Hg, 101.325 kPa). However, the actual atmospheric pressure is not constant. It
varies with geographic location, height above sea level, temperature, weather patterns,
seasons, and time.
glass column,
sealed at the
top and open
at the bottom
Atmospheric Pressure
vacuum
height of
Hg column
h = 76 cm
Atmospheric pressure is
measured by means of a Cistern
barometer. A glass tube, open at
one end is completely filled with
Hg and then inverted into a pool
of Hg so that the opening is
below the surface of the Hg and
no air is allowed in.
The column of Hg will fall,
leaving a vacuum above it. The
column will stop falling when its
height exerts the same pressure
as atmospheric pressure.
This is found to be 76 cm Hg.

Hg
How many meters high
would a column of water
be to exert the same
pressure?
This mercury barometer measures
atmospheric pressure.
Atmospheric pressure is meausured by meteorologists and reported on local radio/TV
broadcasts because it is a good indicator for weather forecasts. ‘High’ atmospheric
pressures (> 760 mm Hg), e.g., 765 mm Hg, usually indicate clear skies, whereas ‘low’
atmospheric pressures (< 760 mm Hg), e.g., 745 mm Hg, usually indicate overcast
(cloudy/rainy) weather.
Pressure
10
Pressure Measurement
Absolute Pressure refers to the total pressure.
Absolute Zero pressure is total vacuum (no pressure).
Gauge pressure is the pressure read by a gauge that has been set to read zero at normal
atmospheric pressure. Gage pressure = (absolute pressure – atmospheric pressure).
For example, scuba divers can determine their depth by wearing a depth gauge on their
wrist. This depth gauge is really a pressure gauge that is set to read zero pressure at the
surface. Below the surface, the depth gage will show only the pressure due to the depth
of water above them, since the pressure due to the atmosphere has already been
subtracted.
A pressure reading that excludes (ignores) the normal atmospheric pressure is called
‘gage pressure’. An example follows:
For every 33 feet that a scuba diver descends in sea water, pressure increases 1 atm.
For a diver at a depth of 99 feet, the gauge pressure would read 3 atm but the total or
absolute pressure is really 4 atm (including air pressure above the sea).
Gage pressure in such situations can be calculated by the ‘pressure/depth equation’.
pgauge = gh = Dh
pabs = gh + patm =
Thus:

pgauge + patm
An ocean diver’s depth gage indicates a depth of 43 ft. Calculate the gage
pressure and absolute pressure at her depth in units of kPa.
(ans. = 130 kPa g and 230 kPa abs.)
Note that if a diver’s pressure gauge were placed in a container and the container were
partially evacuated, the gauge pressure would read negative even though the absolute
pressure is still positive.
e.g. A gauge pressure of -0.3 atm means that absolute pressure remaining is +0.7 atm.
i.e.,
pabs (0.7 atm) = pgauge (-0.3 atm) + patm (1 atm)
It is impossible to have a negative absolute pressure and the largest possible negative
gage pressure is –1.0 atm.

Do problem 1f) through l).
Pressure
11
Diver’s depth gages are not the only pressure gages that are set to read zero pressure at
normal atmospheric pressure. Gage pressure gages are common in industrial and
domestic applications. Tire pressure gages read gage pressure, not absolute (the gage
reads zero when not connected to a tire).

A safe pressure for most standard automobile tires is 35 psi, however, some
service stations have their air pumps calibrated in N/cm2. Calculate the equivalent
pressure in N/cm2 and state whether this is absolute or gage pressure.
(ans. = 24 N/cm2).
On many occassions we wish to express pressure (or vacuum) not as the total but as a
pressure (or vacuum) compared to atmospheric pressure. In general, pressure (not
including atmospheric pressure) is called ‘gage pressure’; even if no gage is being used to
measure it.
pgage =
pabsolute - patm
For example, bottles of carbonated beverages (soda
pops) are generally pressurized to ca. 4 atm absolute
(a), which is the same as saying they are pressurized
to 3 atm gage (g).

A balloon is pressurized to 0.4 atm greater
than atmospheric pressure. State the
absolute and gage pressure in the balloon.
30
pabs = ? atm
pg = ? atm
40 50 60
70
80
20
10
10 20 30
0
40
-10
50
-20
-30
0
inpsi
Hg
aa
‘absolute pressure’ gage in
inches Hg reading
atmospheric pressure
inpsi
Hg
ag
‘gage pressure’ gage in inches
Hg reading atmospheric
pressure
Gages such as these, which can read both pressure and vacuum, are sometimes referred to
as ‘pressure/vacuum gages’.
Pressure
12

A student vacuum filters her precipitate in a
Buchner funnel. The vacuum in the lab is set at
25 in. Hg. State the gage pressure and absolute
pressure in inches Hg.
25 in
Hg
vacuum
Pressure in Closed Containers:
Note that when a fluid is contained in a closed, rigid-walled container, the pressure within
the vessel is independent of atmospheric pressure. A strong walled vessel, when tightly
sealed, insulates its contents from external pressures.
COMPRESSED AIR
Metal pipes, pop bottles, aerosol cans, fire extinguishers, gas cylinders (welder’s fuel,
scuba diving air, etc.), industrial process tanks and vessels, among others, are designed to
withstand elevated pressures. In such cases, the absolute pressure (total pressure) in the
container does not include atmospheric pressure. Absolute pressure includes atmospheric
pressure only when a system is open to the atmosphere.
Pressure
13
Complete all the cells in the following table
* pabs = pgauge + patm
**pgage = pabsolute - patm
Absolute Pressure*
Gauge **
Pressure
2.5
atm
Torr
mm Hg
in Hg
ft H2O
psia
psfa
atm
Torr
mm Hg
in Hg
ft H2O
psig
psfg
1.0
atm
Torr
mm Hg
in Hg
ft H2O
psia
psfa
atm
Torr
mm Hg
in Hg
ft H2O
psig
psfg
atm
Torr
mm Hg
in Hg
ft H2O
psiv
psfv
0.6
atm
Torr
mm Hg
in Hg
ft H2O
psia
psfa
atm
Torr
mm Hg
in Hg
ft H2O
psig
psfg
atm
Torr
mm Hg
in Hg
ft H2O
psiv
psfv
atm
Torr
mm Hg
in Hg
ft H2O
atm
Torr
mm Hg
in Hg
ft H2O
atm
Torr
mm Hg
in Hgv
ft H2Ov
2.5 atm (A)
e.g., aerosol
can
1.0 atm (A)
standard
atmospheric
pressure
0.6 atm (A)
e.g., vacuum
filtration
Absolute Zero
Pressure
total
vacuum
0.0

It is not possible for an absolute pressure of less than zero.

It is not possible for a gage pressure to be less than –1 atm.

It is not possible for a vacuum to be greater than 1 atm.
Pressure
Vacuum
---------------
14
Mechanical Pressure Measuring Devices
Liquid filled devices use H2O, Hg, brominated organics (s.g. > 1), and hydrocarbon
liquids (s.g. <1).
Cistern Barometer (Hg Barometer):
As already discussed, the cistern barometer measures
atmospheric pressure because its well is open to the
atmosphere and there is zero pressure at the top of the
closed tube. The height of the Hg column is a direct
measure of pressure (cm or in Hg). The pressure/depth
equation can be used to calculate atmospheric pressure in
any units. pgauge = gh = Dh

Qualitatively describe the height of the Hg column
during a tornado?

during clear, cold weather?
If, instead of being open to atmosphere at one end, the
vented connection on this enclosed barometer were
connected to enclosed vessels, this barometer would read
the absolute pressure of the vessel attached.

Qualitatively describe the height of the Hg column
when connected to a vacuum cleaner?

when connected to an aerosol can?

Do problem 2
Cistern Barometer
vacuum
glass column,
sealed at the
top and open
at the bottom
h
Hg well
J-tube (closed tube) manometer
Closed Tube or J-Tube Manometer:
The cistern barometer is an accurate instrument, however,
its large well of Hg requires it remain wall mounted or
otherwise secured. A more versatile, portable version of
this barometer is called the J-tube or ‘closed tube,
manometer. The well of Hg is replaced with a simple loop
at the bottom, giving the manometer a J-shape.
h
The difference in height of the mercury columns gives the
absolute (total) pressure of any vessel to which it is
connected (or atmospheric pressure, if its connector is left
open to atmosphere). ‘h’ in., cm, or mm Hg is the
pressure.

Do problem 3
Pressure
15
Using a non evacuated, closed tube manometer to measure absolute pressure.
Alternately, the closed leg does not have to be evacuated but may contain an air-filled
space above the liquid Hg.
p2
p1
L2
L1
h2
zero line
1. both at atm. press.
vol. of trapped air = L1  area
2. p2 is applied
vol. of trapped air = L2  area
The pressure is equal at equal depths in the same fluid
Thus p2 = ptrapped air in L2 + gh2
p on trapped air = p1 = patm
p on trapped air = p2 - gh2
From Boyle’s Law: pv = constant
patm  L1  area = (p2 -gh2)  L2  area
L1 , L2 , and h2 are easily measured; and if patm is known, then p2 can be calculated.
Sample Problem:
A closed tube manometer is zeroed by opening both legs to atmosphere. One leg is
closed by means of a stop cock and the other leg is connected to a tank used to store
methane gas. Calculate the absolute pressure (in atm.) in the methane storage tank given
the following manometer readings: L1 = 30.0 cm, L2 = 20.0 cm, and h2 = 15.0 cm. Unless
otherwise stated, assume atmospheric pressure is at standard pressure.
(ans. = 1.70 atm).

Do problem 4
Pressure
16
Also, patm can be determined by adding more Hg after initially sealing of one leg at
atmospheric pressure.
Patm
Patm
10.8”
14.5”
3.30”
zero line
Using a non evacuated closed tube manometer
to measure
pressure.
=
patmospheric
2  v2
p1 v1
Let the unknown pressure, p1, be ‘h’ inches Hg and ‘a’ is the internal area of uniform
manometer tube. Its dimensions aren’t needed because the terms cancel.
p1v1 = p2v2
(h in Hg)  (10.8  a) = [h + (14.5 -3.3)]  [(10.8 - 3.3)  a]
10.8 h = (h + 11.2) 7.5
1.44 h = h + 11.2
0.44 h = 11.2  h = 25.5 in Hg.
‘h’ is not the height of the Hg column. It is simply pressure (p1), which is unknown, in
units of inches Hg.

Do problem 5
Pressure
17
U-tube Manometer (Open tube Manometer)
Patm
zero line
h
It is a U-shaped tube usually made of glass or clear rubber tubing and may be filled with Hg
or sometimes with water. Both “legs” are initially open to the atmosphere and there is no
difference in the height of the liquid columns since there is no pressure differential.
When one leg of the manometer is connected to a container of gas under pressure (>1atm),
the liquid column is forced down on the high pressure side and raised up on the side open
to atmosphere (the lower pressure side). The difference in height between the liquid levels
in the two legs of the manometer is a direct measure of gauge pressure, i.e., the difference
from atmospheric pressure.
pgauge = gh
Similarly, if one side of the manometer is connected to a partially evacuated bottle (<1atm),
the liquid column rises on the low pressure side and is forced downwards on the
atmospheric side (now the higher pressure side).
Manometers are frequently used in industry to measure pressure differentials, i.e., the
gauge pressure in some part of a process (storage tank, distillation column, etc.).
Caution: Volatile liquids may evaporate under vacuum. Recall that the height of the fluid
is independent of the diameter and shape of the legs, thus tubing irregularities give no error.

Do problems 6 and 7.
Pressure
18
Inclined Manometer
These are similar to U-tube manometers
when one side is left open to the
atmosphere. There are two differences.
The relatively large resevoir of fluid
ensures that the level in the resevoir
remains constant. By pulling one leg
over at an angle, one can measure the
diagonal length rather than the height
(with increased accuracy).
patm
length
angle ()
The height difference (h) is the pressure difference. The height difference is calculated
using a suitable trig function for the angle…
sin  = h/length  h = length  sin 
By choosing a small enough angle, the pressure readings may be magnified by as much as
20:1. The incline manometer is therefore used to measure “draft”, i.e., low pressure air
flows in fumehoods, ducts, baghouses, process stacks, etc.

Do problems 8 and 9
Pressure
19
BiFluid Manometers:
Bifluid manometers contain 2 fluids rather than a single fluid. Bifluid manometers give
larger height differences than single fluid manometers for the same pressure differential
and so, like incline manometers, they are useful for measuring low pressure differentials.
Consider the U-tube manometer shown below. Intially (diagram on the left), a U-tube
mamometer contains a fluid with a density of 2.0 g/mL. Since both arms are open to
atmosphere, the fluid levels must be equal. Pressure is equal at any horizontal depth in the
same fluid.
Patm
These diagonal lines
join points at the same
pressure.
Pressure is
equal at any
horizontal
depth of the
same fluid.
level A
Pressure is equal at
any horizontal depth
of the same fluid.
If 4.0 mL of a second fluid (density = 1.0 g/mL) is poured into the right arm the levels will
be reestablished as shown in the diagram on the right. With fluids of different density, the
pressure is not equal along horizontal lines. The diagonal lines join points of equal
pressure. Consider the density and heights of the fluids and convince yourself of this.
Level ‘A’ is at the same pressure in both arms of the manometer. It is a horizontal level
with the same fluid in both sides of the manometer. There are 2 mL of dense fluid (sg = 2)
above this level in the left arm and 4 mL of light fluid (sg = 1) in the right arm, i.e., the
same weight (pressure) of fluid lies above level A in both arms of the manometer.
Note that the height of liquid in each arm is different even though the same pressure
(atmospheric) is applied to each arm. This behavior is typical of bifluid manometers and is
quite different than that of single fluid manometers.

Do problems 10 – 14.
Pressure
20
Pressure Gauges for Automatic Process Control
All the foregoing gauges are used in labs by chemists and engineers for calibrating and
testing. For process operation, industry uses simple mechanical pressure gauges, precision
pressure recorders and indicators, and pneumatic and electronic pressure transmitters.
Mechanical pressure gauges operate by bending, deforming or deflecting, or moving some
device.
Bourdon Tube Pressure Gage
This is one of the oldest and most reliable
and is still commonly used. For the C-shaped
Bourdon tube, the outer side has greater
surface area than the inner side. Thus an
applied pressure (force per unit area) has
greater outward force than inward force and
the tube straightens and moves a pointer via a
pinion and gears.
The tube can be wound into a spiral (helix) to
give greater motion. This gauge reads
differential pressure in the range of 10-300 psig.
C-shaped tube
or coil
pressure
gage
air inlet
You may recall the Bourdon tubes are also used as temperature sensors; a rising
temperature causes a pressure increase.
These gages are suitable if the pressure of a non corrosive gas (such as air, N2, He, etc.) is
to be measured. Corrosive gases would attack the interior of the Bourdon tube. For
corrosive gases, other types of pressure gages are available.
Bellows-Type Pressure Gage
scale in psi
These are corrugated hollow chambers that can
expand and contract under pressure/vacuum.
Normally pressure is applied to the inside but
this can be reversed by opening the inside to
atmosphere and applying pressure to the outside.
bellows
Bellows tend to be more sensitive than Bourdon
tubes and are used for lower pressures (0-30 psig).
They can also measure vacuum.
inlet
Pressure
21
Diaphragm-type Presssure Gage
This is a thin flexible membrane which extends
or “dimples” under a pressure difference.
scale in psi
The diaphragm is linked to a pointer and some
amplification of motion is thus achieved.
moveable
diaphragm
The diaphragm measures pg (pressure or
vacuum). Diaphragm gauges are also used as
household “aneroid barometers”.
inlet
Pressure
22
Transducers
Pressure gauges alone cannot be used for automatic process control. Physical motion of
gauges must be converted to electrical or pneumatic signals via “transducers” (literally
meaning ‘ to lead across’). A transducers converts energy from the gauge to a signal,
whose strength is proportional to the magnitude of the pressure, and transmits this to a
recorder or controller. Examples of transducers follow.
1. Strain Gauge
One form (a wire) is bonded to the surface of the moving part (bellows, diaphragm,
Bourdon tube, etc.) or to leveling springs. Any change in the shape of the surface
causes the wire to elongate and change in diameter. This change in diameter and length
causes a change in its electrical resistance, which is easily measured by a Wheatstone
bridge. Thus the physical motion of the gauge is converted into a related electrical
signal. More recently, resistors are replaced by semiconductors which are deposited in
thin films.
2. Capacitative Pressure Transducers
The moving metal diaphragm of the diaphragm gauge is attached to plates of a parallel
plate capacitor and its movement changes the capacitance of this variable capacitor (as
the distance between parallel plates changes). Capacitance changes can be amplified
and used to operate/modulate controllers in a process.
3. Pneumatic Pressure Transmitters
The pressure sensing element is connected to a rod (“pilot-rod”) in a pressure
transmitter which acts like a valve, i.e., opens or closes a high pressure valve sending a
3-15 psi signal to a process controller.
4. Piezoelectric Pressure Sensors
Certain crystals (quartz, tourmaline-B & Al silicate, Rochelle salts-NaK tartrate, and
synthetic crystals- Ba titanate, Pb zirconate) when deformed elastically produce an emf
(and on occasion, sparks as in barbecue ignitors) proportional to the applied force
(40 mV/psi or 5.8 mV/kPa). The emf which these produce when deformed can be
amplified or used directly in transistorized circuits. Peizoelectric sensors are small,
light-weight, and rugged. They have rapid, drift-free response over wide pressure
ranges.
5. Magnetic Pressure Transmitters
The magnitude of emf induced in the secondary winding of an inductor (coil) varies
with the penetration of an iron core in the coil. They are small, rugged, and accurate
( 0.25 %) and exhibit minimal temperature sensitivity.
Other transducers use optics and other electromagnetic, mechanical interactions.
 Complete the remainder of the pressure problems.
Pressure
23
Pressure problems:
1. Convert and report with the correct number of sig figs.
a) 506 Mpa to atm
b) 22.6 mmHg to Pa
c) 25.0 psi to ft water
d) 2100 psi to Pa
e) 4.4 m H2O to Pa
f)
10.0 psig to psia
g) 2.00 psig to mmHg absolute
h) 700 mm Hg absolute to ft. water absolute
i)
–1.25 psig to psia
j)
1.00  103 psfg to psia
k) 12 in Hg vacuum to psia
l)
20.0 psig to psfa
2. The height of a mercury barometer is 76.0 cm. Calculate the height in meters, if oil of
density 0.900 g/mL is used instead of mercury.
3. The height difference of a J-tube manometer reading is 26.2 cm of manometer fluid having
a specific gravity of 10.2. Calculate the height of the manometer reading if the manometer
fluid is changed to one with a specific gravity of 2.62.
4. Air is trapped in a uniform J-tube, sealed at one end, by mercury. When the mercury levels
in the two limbs are the same, the length of the air column is 42.0 cm. When more mercury
is poured in so that the difference in levels is 50.0 cm, the length of the air column is 25.0
cm. Calculate the value of atmospheric pressure in cm Hg.
5. A pressure is applied to one limb of a mercury-filled U-tube manometer and the level of the
open limb is 10.0 in. higher than the other limb
a) Calculate the gage pressure in psi.
b) Calculate the absolute pressure.
6. If on leg of a mercury-filled U-tube manometer is subjected to a pressure of 8.99 psi and the
other leg to a pressure of 6.00 psi, calculate the difference in height of the two columns
expressing the answer in inches Hg.
Pressure
24
7. If the barometric pressure is 14.70 psi, determine the absolute pressure in psi indicated by
the inclined manometer.
to vacuum
 = 30
5.00 in.
Hg
8. An inclined manometer is used to measure the difference in air pressure in a pipe between
two points as shown in the diagram. If the manometer fluid is an oil with a specific gravity of
0.827, what pressure difference (psi) is indicated by the manometer?
2.00 in.
oil
 = 20
9. Water flows through a pipe as shown in the diagram below. Determine the height, h, if the
gage pressure at A is 35.0 kPa.
A water
60.0 cm
patm
h
Hg
Pressure
25
10. Find the gage pressures at points A, B and C in the pipe filled with flowing water as shown
below. Assume all measuements are good to 3 sig figs.
A
C
B
H2O
6”
8”
patm
12”
15”
9”
3”
7”
5”
Hg
11. A manometer is connected to a water line as shown in the diagram. Determine the gage
pressure (psi) in the line at point A.
patm
15.0”
30.0”
Hg
water
A
Pressure
26
12. The mercury manometer shown below is connected to the inlet and outlet of a water pump.
The left leg is connected to the inlet and the right leg is connected to the outlet. Assuming
that the inlet and outlet are the same elevation, determine the pressure increase created by
the pump.
outlet
inlet
P
H2O
h3
Hg
h2
h1
13. The pressure at the liquid surface in the tank shown below is 4.00 psi larger than
atmospheric pressure. Determine the height, h, if the liquid in the tank is
a) water (calculate h in ft. of H2O)
patm
b) mercury (calculate h in cm Hg)
h
Pressure
27
14. The bifluid manometer shown in the diagram below is used to determine small pressure
differences with a better accuracy than a single fluid manometer. Find the pressure
difference, PA – PB (psi), for a deflection of 2.0 in. of the boundary between the two fluids.
At the initial position PA = PB. Fluid x has a specific gravity of 0.80 and fluid y has a specific
gravity of 0.85. The tubing of the manometer has an internal diameter of 0.125 in.
A
fluid Y
B
fluid X
Final
2.0”
Initial
Pressure
28
Answers to Pressure problems.
1.
a)
4990 atm
e)
4.3  104 Pa
i)
13.4 psia
b)
3010 Pa
f)
24.7 psia
j)
21.6 psia
c)
57.7 ft H2O
g)
863 mm Hg
k)
8.8 psia
d)
1.4  107 Pa
h)
30 ft H2O
l)
5.00 103 psfa
9.
2.
11.5 m oil
3.
102 cm
5.
73.5 cm Hg
6.
4.91 psig, 19.6 psia
7.
6.09 in Hg
8.
2.50 in Hg vacuum or 27.4 in Hg
Pressure
0.020 psi or 0.57 in H2O
10. 33.1 cm Hg
11. A = -2.59 in Hg, B = -4.44 in Hg,
C = +1.12 in Hg
12. 0.575 atm or 8.45 psig
13. 12.6 h2 units H2O, 0.93 h2 units Hg
14. 9.22 ft H2O or 20.7 cm Hg
15. 0.0036 psi
29
PRESSURE TEST SUMMARY
1. Sketch the following types of barometers showing liquid levels (if they contain liquids)
and explain how they work, a typical application and what they measure, i.e., gage
pressure, absolute pressure, atmospheric pressure, or vacuum.
a) Closed tube manometer (J-tube manometer)
b) cistern barometer
c) U tube manometer (Open tube manometer)
d) Bourdon tube
e) diaphragm gage
f) bellows type gage
2. Explain what a pressure transucer is and explain how 3 of them work
3. State and explain Paschal’s principle
4. Explain what is meant by ‘head loss’ and what causes it.
5. Be able to do all problems in notes
6. Be able to complete a section of a table like that on page 5 of the notes involving
absolute pressure, gage pressure and vacuum in various units.
7. It is essential that you can interconvert between pressure units, e.g., psig to ft H2Oabs
to kPavacuum, etc.
Lab#8: Head Loss and Equivalent Length:
Calculate head loss per unit length of pipe and equivalent length of a valve

Lab#9 Polarimetry, Inversion of Sucrose

Do calculations involving the equation for specific rotation
Lab # 11: Distillation:

Calculate concentration (mole fraction or mole%) given refractive index data as done
in the lab.

Calculate the reflux ratio, given flow rates of distillate and reflux
The test is closed book but you will be given a page of test aids (attached).
Pressure
30
TEST AIDS FOR PRESSURE TEST
pabs = gh + patm
or
This pressure-depth equation:
pabs = Dh + patm
p = gh = Dh
The standard density for water is 1.00 g/cm3 which occurs at 3.98 ºC
D4º = 62.4 lb/ft3 = 9810 N/m3
and
4º = 1.00 g/cm3 = 1000 kg/m3 = 1.94 slug/ft3
Standard Atmospheric Pressure at sea level is:
14.696 lb/in2 = 2116 lb/ft2 = 1.01325  105 N/m2 (Pa) = 1.01325  106 dyn/cm2
= 101.325 kPa = 1 atm = 760 mmHg. (Torr) = 76 cm Hg = 29.9213 inHg
= 33.90 ft fresh water (~ 33 ft. sea water) = 1.01325 bar = 1013.25 mb
p
F
A
sg =
Ds
s
=
Dw
w
=
m
v
D

v
Recall units of force:
A Newton (N) is a force that will cause a 1 kg mass to accelerate at 1 m/s2.
The force of gravity on earth (g) is 9.807 N/kg.
Gravity causes a 1 kg mass to accelerate (fall) at a rate of 9.807 m/s2.
A pound of force (lbf) will cause a mass of 1 slug to accelerate at 1 ft/s2.
The force of gravity on earth (g) is 32.17 lb/slug.
Gravity causes a 1 slug mass to accelerate (fall) at a rate of 32.17 ft/s2.
Conversions:
force-force:
mass-mass:
mass-force:
1 lb = 4.45 N
1 slug = 14.6 kg
1 kg = 9.807 N
1 N = 105 dynes
1 kg = 2.205 lb
1 slug = 32.17 lb
 D =  rotation
lc
ln (t - ) = - kt + ln(0 - )
# theoretical plates
Overall Efficiency 
100 %
# actual plates
column height (cm)
HETP 
# theoretical plates - 1
Pressure
Rectifying
(Enriching) Line
Slope 
R
R 1
‘q’ Line
(Feed Line)
Slope 
q
(C  T)  Lv
Lv
q
q 1
31