01 Pressure Measurement
(Water Resources I)
Dave Morgan
Prepared using Lyx, and the Beamer class in LATEX 2ε ,
on September 12, 2007
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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
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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
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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
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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:
abs = patm + pgauge
p
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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
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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
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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
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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.
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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 dierence 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.
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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.
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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 dierence in elevation between two points is usually denoted
by h.
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Pressure and Elevation
The change in pressure in a homogeneous liquid at rest due to
change in elevation is given by:
∆p = γ h
where
∆p = change in pressure
γ = specic weight of liquid
h
= change in elevation
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Pressure and Elevation
The change in pressure in a homogeneous liquid at rest due to
change in elevation is given by:
∆p = γ h
where
∆p = change in pressure
γ = specic weight of liquid
h
= change in elevation
Note:
1
2
This equation does not apply to gases
Points at the same elevation (same horizontal level) have the
same pressure (see Pascal's Paradox)
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∆p = γ · h
Derivation
Derivation:
Consider a vertical cylinder of liquid
within a body of liquid.
Let the cylinder have height h and
cross-sectional area A
A
h


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∆p = γ · h
p1
Derivation
p1
A
h
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


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∆p = γ · h
Derivation
p1 A
A
h

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 = p1 A

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∆p = γ · h
Derivation
p1 A
A
h

p2
p2
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 = p1 A
Similarly the pressure, p2 , on the
bottom surface of the cylinder is
uniform...
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∆p = γ · h
Derivation
p1 A
A
h


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 = p1 A
Similarly the pressure, p2 , on the
bottom surface of the cylinder is
uniform...
...and Fup = p2 A
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∆p = γ · h
Derivation
p1 A
W
Derivation:
The other force to be considered is
the weight, W , of the cylinder
A
h


p2 A
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∆p = γ · h
Derivation
p1 A
V
Derivation:
The other force to be considered is
the weight, W , of the cylinder
Express W as W = γ · V
A
h


p2 A
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∆p = γ · h
Derivation
p1 A
V
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
A
h
ΣFy = p2 A − γ V − p1 A = 0


p2 A
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∆p = γ · h
Derivation
p1 A
V
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
A
ΣFy = p2 A − γ V − p1 A = 0
h
V

= Ah so
p2 A

p2 A
− γ · Ah − p1 A = 0
p2
− γ h − p1 = 0
p2
− p1 = γ h
∆p = γ h
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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.
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Water Tower
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Water Tower
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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
950 mm
C
300 mm
B
1375 mm
D
625 mm
E
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Pressure Measurement (Example)
A
950 mm
C
Solution
B:
is open to the atmosphere so PB = 0
and PB (abs ) = P(atm) = 102.1 kPa
300 mm
Pressure at
B
B
1375 mm
D
625 mm
E
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Pressure Measurement (Example)
Solution
Pressure at
P
A
950 mm
C
300 mm
B
1375 mm
D
625 mm
A =
=
=
=
=
A:
B − ∆p
0−γ ·h
−(8.89 kN/m3 )(0.30 m)
−2.667 kN/m2
−2.67 kPa
P
E
P
A(abs ) = Patm + PA(gauge )
= 102.1 kPa − 2.667 kPa
= 99.4 kPa
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Pressure Measurement (Example)
Solution
Pressure at
C
P
A
950 mm
=
=
C
=
300 mm
B
=
1375 mm
D
625 mm
=
C:
B − ∆p
0−γ ·h
−(8.89 kN/m3 )(0.950 m)
−8.4455 kN/m2
−8.45 kPa
P
E
P
C (abs ) = Patm + PC (gauge )
= 102.1 kPa − 8.4455 kPa
= 93.7 kPa
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Pressure Measurement (Example)
Solution
Pressure at
P
A
950 mm
C
300 mm
B
1375 mm
D
625 mm
D =
=
=
=
=
D:
B + ∆p
0+γ ·h
(8.89 kN/m3 )(1.375 m)
12.224 kN/m2
12.2 kPa
P
E
P
D (abs ) = Patm + PD (gauge )
= 102.1 kPa + 12.224 kPa
= 114.0 kPa
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Pressure Measurement (Example)
Solution
Pressure at
P
A
950 mm
E
E:
=
=
C
=
300 mm
B
=
1375 mm
D
625 mm
=
B + ∆p
0+γ ·h
(8.89 kN/m3 )(2.0 m)
17.78 kN/m2
17.8 kPa
P
E
P
E (abs ) = Patm + PD (gauge )
= 102.1 kPa + 17.78 kPa
= 120.0 kPa
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Manometers
A
h
A manometer is a pressure-measuring instrument
It uses the height of a liquid column, h, to measure the
pressure dierence between two locations
This manometer is open to the atmosphere at one end; it is
used to measure the dierence in pressure between A and the
atmosphere (that is, it measures gauge pressure)
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Manometer Example
Example
Determine the pressure at A given that the temperature of the
water is 25◦ C.
A
47 mm
3
Water
105 mm
1
2
Kerosene,
sg = 0.823
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Manometer Example
Solution
A
P3
= 0
P2
=
47 mm
3
Water
105 mm
1
Kerosene,
sg = 0.823
+γ ·h
= 0 + (0.823)(9.81 kN/m3 )(0.105 m)
= 0.84773 kPa
P1
2
P3
= 0.84773 kPa
A = P1 − γ · h
= 0.84773 kPa − (9.78)(0.152) kPa
= −0.63883 kPa
PA
= −0.639 kPa
P
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Manometer Example
Note
There is not much dierence in pressure for a dierence in levels of
0.105 m. For this reason, a gauge uid with a higher specic
gravity, such as mercury, is usually used to measure larger pressure
dierences.
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Manometers
A
B
h
The dierential manometer illustrated is used to measure
the dierence in pressure between A and B
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Dierential Manometer Example
Example
Determine the pressure dierence between A and B
A
105 mm
Water
1
90 mm
150 mm
2
B
Oil,
sg
= 0.90
3
Mercury,
sg = 13.54
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Dierential Manometer Example
Solution
P1
=
=
=
A
105 mm
Water
1
150 mm
2
Mercury,
sg = 13.54
90 mm
P2
B
Oil,
3 sg = 0.90
=
=
=
P3
=
B =
=
∆p =
P
A +γ ·h
3
PA + (9.81 kN/m )(0.195 m)
PA +1.913 kPa)
3
P1 +(13.54)(9.81 kN/m )(0.15 m)
PA +(1.913 + 19.924) kPa
PA +21.837 kPa
PA +21.837 kPa
3
P3 −(0.90)(9.81 kN/m )(0.240 m)
PA +21.837 kPa − 2.119 kPa
19.718 kPa
P
The dierence in pressure between A and B
is 19.7 kPa
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