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01 Pressure Measurement(Water Resources I)
Dave Morgan
Prepared using Lyx, and the Beamer class in LATEX2ε ,
on September 12, 2007
01 Pressure Measurement WRI 1/25
Readings for �Pressure Measurement�
Suggested readings for this topic are from Applied Fluid Mechanics
by Mott:
Read sections:
3.3 - 3.9
Study Example Problems: 3.1 - 3.13
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Absolute and Gauge Pressure
Pressure measurements are made relative to some reference
pressure, usually atmospheric (a.k.a. ambient) pressure
Pressure relative to the atmosphere is called gauge pressure
Pressure relative to a perfect vacuum is called absolute
pressure
Absolute, atmospheric and gauge pressures are related by the
following expression:
pabs = patm +pgauge
01 Pressure Measurement WRI 3/25
Absolute and Gauge Pressure
Pressure measurements are made relative to some reference
pressure, usually atmospheric (a.k.a. ambient) pressure
Pressure relative to the atmosphere is called gauge pressure
Pressure relative to a perfect vacuum is called absolute
pressure
Absolute, atmospheric and gauge pressures are related by the
following expression:
pabs = patm +pgauge
01 Pressure Measurement WRI 3/25
Absolute and Gauge Pressure
Pressure measurements are made relative to some reference
pressure, usually atmospheric (a.k.a. ambient) pressure
Pressure relative to the atmosphere is called gauge pressure
Pressure relative to a perfect vacuum is called absolute
pressure
Absolute, atmospheric and gauge pressures are related by the
following expression:
pabs = patm +pgauge
01 Pressure Measurement WRI 3/25
Absolute and Gauge Pressure
Pressure measurements are made relative to some reference
pressure, usually atmospheric (a.k.a. ambient) pressure
Pressure relative to the atmosphere is called gauge pressure
Pressure relative to a perfect vacuum is called absolute
pressure
Absolute, atmospheric and gauge pressures are related by the
following expression:
pabs = patm +pgauge
01 Pressure Measurement WRI 3/25
Absolute and Gauge Pressure
When using a tyre-gauge to check the pressure in a car or a
bike tyre, the tyre-gauge reports gauge pressure; this is the
amount of pressure in the tyre in excess of the pressure of the
atmosphere
If a tyre-gauge reports a pressure of 275 kPa and the
atmospheric pressure is 101.3 kPa(abs), then the absolute
pressure inside the tyre is 376.3 kPa
Normal pressures near the earth's surface are in the range of
95 kPa(abs) to 105 kPa(abs); the average at sea-level is about
101.3 kPa
Atmospheric pressure changes with the weather; at 10:00 pm
on Tuesday 28th August, 2007, atmospheric pressure at
Calgary International Airport was reported by Environment
Canada to be 102.4 kPa
01 Pressure Measurement WRI 4/25
Absolute and Gauge Pressure
When using a tyre-gauge to check the pressure in a car or a
bike tyre, the tyre-gauge reports gauge pressure; this is the
amount of pressure in the tyre in excess of the pressure of the
atmosphere
If a tyre-gauge reports a pressure of 275 kPa and the
atmospheric pressure is 101.3 kPa(abs), then the absolute
pressure inside the tyre is 376.3 kPa
Normal pressures near the earth's surface are in the range of
95 kPa(abs) to 105 kPa(abs); the average at sea-level is about
101.3 kPa
Atmospheric pressure changes with the weather; at 10:00 pm
on Tuesday 28th August, 2007, atmospheric pressure at
Calgary International Airport was reported by Environment
Canada to be 102.4 kPa
01 Pressure Measurement WRI 4/25
Absolute and Gauge Pressure
When using a tyre-gauge to check the pressure in a car or a
bike tyre, the tyre-gauge reports gauge pressure; this is the
amount of pressure in the tyre in excess of the pressure of the
atmosphere
If a tyre-gauge reports a pressure of 275 kPa and the
atmospheric pressure is 101.3 kPa(abs), then the absolute
pressure inside the tyre is 376.3 kPa
Normal pressures near the earth's surface are in the range of
95 kPa(abs) to 105 kPa(abs); the average at sea-level is about
101.3 kPa
Atmospheric pressure changes with the weather; at 10:00 pm
on Tuesday 28th August, 2007, atmospheric pressure at
Calgary International Airport was reported by Environment
Canada to be 102.4 kPa
01 Pressure Measurement WRI 4/25
Pressure and Elevation
Pressure decreases with increased elevation in the atmosphere;
the atmospheric pressure in Quito (2850 m above sea-level) is
(usually) less than than the atmospheric pressure in Death
Valley (86 m below sea-level). Pressure usually decreases
about 3.4 kPa for every increase in elevation of 300 m.
Weather stations do not normally display this di�erence due to
elevation; the stations adjust the pressure for elevation. Thus,
101.3 kPa would be the usual reported pressure in both Quito
and Death Valley; this enables us to know something about
the weather conditions in a particular location without
knowing the elevation and making calculations.
01 Pressure Measurement WRI 5/25
Pressure and Elevation
Pressure decreases with increased elevation in the atmosphere;
the atmospheric pressure in Quito (2850 m above sea-level) is
(usually) less than than the atmospheric pressure in Death
Valley (86 m below sea-level). Pressure usually decreases
about 3.4 kPa for every increase in elevation of 300 m.
Weather stations do not normally display this di�erence due to
elevation; the stations adjust the pressure for elevation. Thus,
101.3 kPa would be the usual reported pressure in both Quito
and Death Valley; this enables us to know something about
the weather conditions in a particular location without
knowing the elevation and making calculations.
01 Pressure Measurement WRI 5/25
Pressure and Elevation
Pressure increases with increased depth in a �uid. The
pressure at the bottom of the deep end in a swimming pool is
noticeably greater than just below the surface.
In �uid mechanics, elevation refers to the vertical distance
from some reference point to a point of interest (Death Valley
is 86 m below sea-level). Elevation is usually denoted by z.
A di�erence in elevation between two points is usually denoted
by h.
01 Pressure Measurement WRI 6/25
Pressure and Elevation
Pressure increases with increased depth in a �uid. The
pressure at the bottom of the deep end in a swimming pool is
noticeably greater than just below the surface.
In �uid mechanics, elevation refers to the vertical distance
from some reference point to a point of interest (Death Valley
is 86 m below sea-level). Elevation is usually denoted by z.
A di�erence in elevation between two points is usually denoted
by h.
01 Pressure Measurement WRI 6/25
Pressure and Elevation
The change in pressure in a homogeneous liquid at rest due to
change in elevation is given by:
∆p = γh
where
∆p = change in pressure
γ = speci�c weight of liquid
h = change in elevation
Note:
1 This equation does not apply to gases
2 Points at the same elevation (same horizontal level) have the
same pressure (see Pascal's Paradox)
01 Pressure Measurement WRI 7/25
Pressure and Elevation
The change in pressure in a homogeneous liquid at rest due to
change in elevation is given by:
∆p = γh
where
∆p = change in pressure
γ = speci�c weight of liquid
h = change in elevation
Note:
1 This equation does not apply to gases
2 Points at the same elevation (same horizontal level) have the
same pressure (see Pascal's Paradox)
01 Pressure Measurement WRI 7/25
∆p = γ ·h Derivation
h
A
Derivation:
Consider a vertical cylinder of liquid
within a body of liquid.
Let the cylinder have height h and
cross-sectional area A
The pressure, p1, on the top
surface of the cylinder is uniform
since the surface is horizontal
The force exerted on the top
surface is Fdown = p1A
Similarly the pressure, p2, on the
bottom surface of the cylinder is
uniform...
...and Fup = p2A
01 Pressure Measurement WRI 8/25
∆p = γ ·h Derivation
h
p1
p1
A
Derivation:
Consider a vertical cylinder of liquid
within a body of liquid.
Let the cylinder have height h and
cross-sectional area A
The pressure, p1, on the top
surface of the cylinder is uniform
since the surface is horizontal
The force exerted on the top
surface is Fdown = p1A
Similarly the pressure, p2, on the
bottom surface of the cylinder is
uniform...
...and Fup = p2A
01 Pressure Measurement WRI 8/25
∆p = γ ·h Derivation
h
A
p1
A
Derivation:
Consider a vertical cylinder of liquid
within a body of liquid.
Let the cylinder have height h and
cross-sectional area A
The pressure, p1, on the top
surface of the cylinder is uniform
since the surface is horizontal
The force exerted on the top
surface is Fdown = p1A
Similarly the pressure, p2, on the
bottom surface of the cylinder is
uniform...
...and Fup = p2A
01 Pressure Measurement WRI 8/25
∆p = γ ·h Derivation
h
p2
p2
Ap
1 A
Derivation:
Consider a vertical cylinder of liquid
within a body of liquid.
Let the cylinder have height h and
cross-sectional area A
The pressure, p1, on the top
surface of the cylinder is uniform
since the surface is horizontal
The force exerted on the top
surface is Fdown = p1A
Similarly the pressure, p2, on the
bottom surface of the cylinder is
uniform...
...and Fup = p2A
01 Pressure Measurement WRI 8/25
∆p = γ ·h Derivation
h
A
p1
A
p2
A
Derivation:
Consider a vertical cylinder of liquid
within a body of liquid.
Let the cylinder have height h and
cross-sectional area A
The pressure, p1, on the top
surface of the cylinder is uniform
since the surface is horizontal
The force exerted on the top
surface is Fdown = p1A
Similarly the pressure, p2, on the
bottom surface of the cylinder is
uniform...
...and Fup = p2A
01 Pressure Measurement WRI 8/25
∆p = γ ·h Derivation
W h
A
p1
A
p2
A
Derivation:
The other force to be considered is
the weight, W , of the cylinder
Express W as W = γ ·VThe cylinder is in equilibrium so
ΣFy = p2A− γV −p1A = 0
V = Ah so
p2A− γ ·Ah−p1A = 0
p2− γh−p1 = 0
p2−p1 = γh
∆p = γh
01 Pressure Measurement WRI 9/25
∆p = γ ·h Derivation
V h
A
p1
A
p2
A
Derivation:
The other force to be considered is
the weight, W , of the cylinder
Express W as W = γ ·V
The cylinder is in equilibrium so
ΣFy = p2A− γV −p1A = 0
V = Ah so
p2A− γ ·Ah−p1A = 0
p2− γh−p1 = 0
p2−p1 = γh
∆p = γh
01 Pressure Measurement WRI 9/25
∆p = γ ·h Derivation
V h
A
p1
A
p2
A
Derivation:
The other force to be considered is
the weight, W , of the cylinder
Express W as W = γ ·VThe cylinder is in equilibrium so
ΣFy = p2A− γV −p1A = 0
V = Ah so
p2A− γ ·Ah−p1A = 0
p2− γh−p1 = 0
p2−p1 = γh
∆p = γh
01 Pressure Measurement WRI 9/25
∆p = γ ·h Derivation
V h
A
p1
A
p2
A
Derivation:
The other force to be considered is
the weight, W , of the cylinder
Express W as W = γ ·VThe cylinder is in equilibrium so
ΣFy = p2A− γV −p1A = 0
V = Ah so
p2A− γ ·Ah−p1A = 0
p2− γh−p1 = 0
p2−p1 = γh
∆p = γh
01 Pressure Measurement WRI 9/25
Pascal's Paradox
Pascal's Paradox
All three vessels contain the same liquid. The pressure at the
bottom of each vessel is the same because pressure is due only to
the depth of liquid.
01 Pressure Measurement WRI 10/25
Pressure Measurement
Example
A tank, open to the atmosphere in the centre, contains medium
fuel oil. Atmospheric pressure is 102.1 kPa. Calculate the gauge
pressure and the absolute pressure for locations A, B, C, D and E.
A
B
C
DE625 mm
1375 mm
950 mm
300 mm
01 Pressure Measurement WRI 13/25
Pressure Measurement (Example)
A
B
C
DE625 mm
1375 mm
950 mm
300 mm
Solution
Pressure at B:
B is open to the atmosphere so PB = 0
and PB(abs) = P(atm) = 102.1 kPa
01 Pressure Measurement WRI 14/25
Pressure Measurement (Example)
A
B
C
DE625 mm
1375 mm
950 mm
300 mm
Solution
Pressure at A:
PA = PB −∆p
= 0− γ ·h= −(8.89 kN/m3)(0.30 m)
= −2.667 kN/m2
= −2.67 kPa
PA(abs) = Patm +PA(gauge)
= 102.1 kPa−2.667 kPa
= 99.4 kPa
01 Pressure Measurement WRI 15/25
Pressure Measurement (Example)
A
B
C
DE625 mm
1375 mm
950 mm
300 mm
Solution
Pressure at C :
PC = PB −∆p
= 0− γ ·h= −(8.89 kN/m3)(0.950 m)
= −8.4455 kN/m2
= −8.45 kPa
PC(abs) = Patm +PC(gauge)
= 102.1 kPa−8.4455 kPa
= 93.7 kPa
01 Pressure Measurement WRI 16/25
Pressure Measurement (Example)
A
B
C
DE625 mm
1375 mm
950 mm
300 mm
Solution
Pressure at D:
PD = PB +∆p
= 0+ γ ·h= (8.89 kN/m3)(1.375 m)
= 12.224 kN/m2
= 12.2 kPa
PD(abs) = Patm +PD(gauge)
= 102.1 kPa+12.224 kPa
= 114.0 kPa
01 Pressure Measurement WRI 17/25
Pressure Measurement (Example)
A
B
C
DE625 mm
1375 mm
950 mm
300 mm
Solution
Pressure at E :
PE = PB +∆p
= 0+ γ ·h= (8.89 kN/m3)(2.0 m)
= 17.78 kN/m2
= 17.8 kPa
PE(abs) = Patm +PD(gauge)
= 102.1 kPa+17.78 kPa
= 120.0 kPa
01 Pressure Measurement WRI 18/25
Manometers
h
A
A manometer is a pressure-measuring instrument
It uses the height of a liquid column, h, to measure the
pressure di�erence between two locations
This manometer is open to the atmosphere at one end; it is
used to measure the di�erence in pressure between A and the
atmosphere (that is, it measures gauge pressure)
01 Pressure Measurement WRI 19/25
Manometer Example
Example
Determine the pressure at A given that the temperature of the
water is 25◦C.
105 mm
A
Water
Kerosene, sg = 0.823
1 2
3
47 mm
01 Pressure Measurement WRI 20/25
Manometer Example
105 mm
A
Water
Kerosene, sg = 0.823
1 2
3
47 mm
Solution
P3 = 0
P2 = P3+ γ ·h= 0+(0.823)(9.81 kN/m3)(0.105 m)
= 0.84773 kPa
P1 = 0.84773 kPa
PA = P1− γ ·h= 0.84773 kPa− (9.78)(0.152) kPa
= −0.63883 kPa
PA = −0.639 kPa
01 Pressure Measurement WRI 21/25
Manometer Example
Note
There is not much di�erence in pressure for a di�erence in levels of
0.105 m. For this reason, a gauge �uid with a higher speci�c
gravity, such as mercury, is usually used to measure larger pressure
di�erences.
01 Pressure Measurement WRI 22/25
Manometers
h
AB
The di�erential manometer illustrated is used to measure
the di�erence in pressure between A and B
01 Pressure Measurement WRI 23/25
Di�erential Manometer Example
Example
Determine the pressure di�erence between A and B
105 mm
150 mm
90 mm
AB
Oil, sg = 0.90
Water
Mercury, sg = 13.54
1
2 3
01 Pressure Measurement WRI 24/25
Di�erential Manometer Example
105 mm
150 mm
90 mm
AB
Oil, sg = 0.90
Water
Mercury, sg = 13.54
1
2 3
Solution
P1 = PA+ γ ·h= PA+(9.81 kN/m3)(0.195 m)
= PA+1.913 kPa)
P2 = P1+(13.54)(9.81 kN/m3)(0.15 m)
= PA+(1.913+19.924) kPa
= PA+21.837 kPa
P3 = PA+21.837 kPa
PB = P3−(0.90)(9.81 kN/m3)(0.240 m)
= PA+21.837 kPa−2.119 kPa
∆p = 19.718 kPa
The di�erence in pressure between A and B
is 19.7 kPa
01 Pressure Measurement WRI 25/25