01 Pressure Measurement - yamnuska.ca

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


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







pressure



pabs = patm +pgauge







pressure



pabs = patm +pgauge







pressure



pabs = patm +pgauge



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






atmosphere






101.3 kPa










atmosphere






101.3 kPa






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.



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.



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.



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.



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)



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)


∆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



h

p1

p1

A

Derivation:












uniform...

...and Fup = p2A



h

A

p1

A

Derivation:












uniform...

...and Fup = p2A



h

p2

p2

Ap

1 A

Derivation:












uniform...

...and Fup = p2A



h

A

p1

A

p2

A

Derivation:












uniform...

...and Fup = p2A



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



V h

A

p1

A

p2

A

Derivation:



Express W as W = γ ·V

The cylinder is in equilibrium so


V = Ah so

p2A− γ ·Ah−p1A = 0

p2− γh−p1 = 0

p2−p1 = γh

∆p = γh



V h

A

p1

A

p2

A

Derivation:





V = Ah so

p2A− γ ·Ah−p1A = 0

p2− γh−p1 = 0

p2−p1 = γh

∆p = γh



V h

A

p1

A

p2

A

Derivation:





V = Ah so

p2A− γ ·Ah−p1A = 0

p2− γh−p1 = 0

p2−p1 = γh

∆p = γh


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.


Water Tower


Water Tower


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


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



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



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



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



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


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)


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


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


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.


Manometers

h

AB

The di�erential manometer illustrated is used to measure

the di�erence in pressure between A and B


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


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


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