 Objectives of Class:
 Define and explain Absolute pressure,
Atmospheric pressure and Gage pressure.
 Define pressure at a point.
 Derive pressure variation relation discuss pressure measurement methods and techniques

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2

 Since the unit Pa is too small for pressures encountered in practice, kilopascal (1 kPa =
10 3 Pa) and megapascal (1 MPa = 10 6 Pa) are commonly used.
 1 atm = 101,325 Pa = 101.325 kPa =
1.01325 bar
 Other units include kgf/cm 2 , lbf/in 2 =psi .
 1 kgf/cm 2 = 9.807 N/cm 2 = 14.223 psi

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atm
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ABSOLUTE, GAGE, AND VACUUM PRESSURES
P abs
= P atm
-
P vac
P abs
= P gage
+
P atm
P atm
P gage (+ve)
P abs
P vacuum =
P gage (-ve)
P abs
Perfect Vacuum, P abs
= 0 atm

 Atmospheric pressure is the pressure at any point in the Earth's atmosphere .
In most circumstances atmospheric pressure is closely approximated by the hydrostatic pressure caused by the weight of air above the measurement point.
Source: en. Wikipedia.com.

How does it change?
Changes not only with elevation but also with weather condition)
Also known as “ Barometric Pressure ”
P abs
= P atm
- P vac
P abs
= P gage
+ P atm
P at m
P gage
(+ve)
P abs
P vacuum =
P gage (-ve)
P abs
Perfect Vacuum, P abs
= 0 atm

ABSOLUTE PRESSURE
 Pressure measured relative to perfect vacuum is called ABSOLUTE PRESSURE , Pa( abs ) or psia .
A perfect vacuum is the lowest possible pressure.
Absolute pressure ( always +ve .)
P abs
= P atm
- P vac
P at m
P gage
(+ve)
P abs
P vacuum =
P gage (-ve)
P abs
P abs
= P gage
+ P atm
Perfect Vacuum, P abs
= 0 atm

Pressure measured relative to atmospheric pressure, Pa (gage) or psig.
Most pressure-measuring devices are calibrated to read zero in the atmosphere, and gage pressure , P gage
=P abs
- P atm
.
P abs
= P atm
- P vac
P at m
P gage
(+ve)
P abs
P vacuum =
P gage (-ve)
P abs
P abs
= P gage
+ P atm
Perfect Vacuum, P abs
= 0 atm

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
A gage pressure above atmospheric pressure
(relative to atmospheric) is +ve .
A gage pressure below atmospheric pressure
(relative to atmospheric) is –ve . It is also called vacuum pressure
P abs
= P atm
- P vac
P abs
= P gage
+ P atm
P gage
(+ve)
P at m
P abs
P vacuum =
P gage (-ve)
P abs
Perfect Vacuum, P abs
= 0 atm

Express a pressure of 155 kPa (gage) as an absolute pressure. The local atmospheric pressure is 98 kPa (abs).

Express a pressure of 225 kPa(abs) as a gage pressure.
The local atmospheric pressure is 101 kPa(abs).

Solving algebraically for Pgage gives

Express a pressure of 75.2 kPa (abs) as a gage pressure. The local atmospheric pressure is 103.4 kPa.

Notice that the result is negative.
This can also be read “28.2 kPa below atmospheric pressure” or “28.2 kPa vacuum.”

Express a pressure of –42.7 kPa as an absolute pressure. Assume, P atm=101 kPa:

Pressure: A scalar? A vector?
Pressure at any point in a fluid is the same in all directions .
Magnitude, but not a specific direction, Scalar quantity.
Other words, Pressure at any point in fluid has the same magnitude in all directions.

z
P
3 l
P
1
Δz
Ө
P
2
Δx x
Small wedge shape fluid element of unit length in equilibrium
( Δy = 1)
The force acting on a surface = product of mean pressure and surface area, respectively.
Σ F x
= ma x
= 0: P
1
Δ z
– P
3 lsin Ө = 0 - - - (1)
Σ F z
= ma z
= 0 : P
2
Δ x
– P
3 lcos Ө – ½ ρg Δ x
Δ z
- - - (2)
Δ x = lcos Ө, Δ z = lsin Ө ;
Weight of the fluid element

Therefore, (1) become P
1
– P
3
= 0 ; P
1
= P
3,
(2) become P
2
– P
3
– ½ ρg Δ z
= 0;
The fluid element shrinks to a point, when Δ z
0
Then, P
1
= P
2 =
P
3
= 0.

Conclusion
 Pressure at any point in a fluid is the same in all directions.
 Pressure has a magnitude, but not a specific direction (as stated above), and thus it is a scalar quantity.

In the presence of a gravitational field, pressure increases with depth because more fluid rests on deeper layers.

HYDROSTATIC FORCE AND PRESSURE
VARIATION WITH DEPTH
The force balance in hydrostatic case:
[p + (dp/dz)dz]dA
∑F = 0
Z- direction pdA+ (-[p + (dp/dz)dz]dA) + (-
ρgdAdz) = 0 dz dp/dz = ρg z pdA p = po + ∫ρgz
When p=po at z=zo p = po + (zo-z) ρg = po + ρgh

VARIATION OF PRESSURE WITH DEPTH
 Pressure in a fluid at rest is independent of the shape of the container.
 Pressure is the same at all points on a horizontal plane in a given fluid.

Calculate the change in water pressure from the surface (exposed to atmosphere) to a depth of 5 m.

If the surface of the water is exposed to the atmosphere, the pressure there is 0
Pa(gage). Descending in the water
(decreasing elevation) produces an increase in pressure. Therefore, at 5 m the pressure is 49.05 kPa(gage).
3
2

*GAS PRESSURE VARIATION WITH ELEVATION
Air density at sea level, 15 degree Celcius is 1.225 kg/m3.
Pressure difference at 5m height difference;
ρgh
= (1.225 kg/m3) * 9.807 ms-2 * 5m * (1 N/1 kgms-2) * ( 1 kPa/ 1000
N/m2)
= 0.06 kPa
1 atm = 101.325 kPa
For gas, the variation of pressure with height is negligible, because of their low density.
Also, weight is too small.
However is accuracy is desired, it becomes significant.
Gravitational Effect g = 9.807 m/s2 at sea level, at elevation 14,000 m above sea level, g = 9.764 m/s2, which is 0.4% change. Therefore, g variation is so small and g can be considered constant.

Figure below shows a tank of oil with one side open to the atmosphere and the other side sealed with air above the oil. The oil has a specific gravity of 0.90. Calculate the gage pressure at points A, B, C, D, E, and F and the air pressure in the right side of the tank.

Point A At this point, the oil is exposed to the atmosphere, and therefore
Point B The change in elevation between point A and point B is 3.0 m, with B lower than A. The specific weight of the oil:
Then we have

Now, the pressure at B is
Point C The change in elevation from point A to point C is 6.0 m, with C lower than A. Then, the pressure at point
C is
Point D Because point D is at the same level as point B, the pressure is the same. That is, we have

Point E Because point E is at the same level as point A, the pressure is the same. That is, we have
Point F The change in elevation between point A and point F is 1.5 m, with F higher than A. Then, the pressure at F is

Air Pressure in the right side of the tank is exposed to the surface of the oil, where p
F
= -13.2kPa the air pressure is also -13.2kPa or 13.2 kPa below atmospheric pressure.

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“A pressure applied to a confined fluid increases the pressure throughout by the same amount ”.
 In picture, pistons are at same height:
P
1

P
2

F
1

F
2

F
2

A
2
A A
1 2
F A
1 1

 Incompressible fluid static,

P
  g
 z
Fluid column measure pressure difference
1 2
Measure small and moderate pressure difference

Gas: Low density, weight is so small to give significance pressure increase.
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An elevation change of
 z in a fluid at rest corresponds to

P/
 g.
A device based on this is called a manometer.
A manometer consists of a
U-tube containing one or more fluids such as mercury, water, alcohol, or oil.
Heavy fluids such as mercury are used if large pressure differences are anticipated.
P
1

P
2
P
2

P atm
  gh

Calculate the pressure of the gas in the tank.
P gas -
ρgh = P atm

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Manometers are well suited to measure pressure drops across valves, pipes, heat exchangers, etc.
Relation for pressure drop
P
1
-P
2 is obtained by starting at point 1 and adding or subtracting
 gh terms until we reach point 2.
If fluid in pipe is a gas,

2
>>

1 and P
1
-P
2
=
 gh

Determine pressure difference between 1 and 2.
Fluid Static
Fluid Dynamic

MANOMETER –PRESSURE DIFFERENCE
Assume pressure at point 1, 2, A and B are P
1,
P
2
, P
A and P
B respectively.
For fluid static, pressure for the same fluid does not vary in horizontal direction.
Therefore, P
A
= P
B.
P
A =
P
1
+ ρ
1 g(h+a)
;
P
B =
P
2
+ ρ
1 ga + ρ
2 gh
P1 + ρ
1 g (a + h) – ρ
2 gh – ρ
1 ga = P2
Therefore,
P1 - P2 = ( ρ
2
ρ
1
) * g * h

* For same fluid (constant density) pressure does not vary in the horizontal direction.
Pressure vary in vertical direction.
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P
A
= P
B
P
A =
P atm +
ρ m gh
3
P
B
=P
C
+ ρ oil gh
2
P
C
=P air +
ρ water gh
1
D
C
B A

P
1
+ ρgh
1
+ ρ
1 gh
2
ρ
2 gh
3
= P
2

*MULTI FLUID MANOMETER – QUIZ 1
Calculate the difference in pressure between points A and B in the following figure and express it as p
B
– p
A
.

P
C
  gh

P atm
P atm
  gh
P
C can be taken to be zero since there is only
Hg vapor above point C, and it is very low relative to P atm
.
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Atmospheric pressure is measured by a device called a barometer ; thus, atmospheric pressure is often referred to as the barometric pressure .
P atm
= 760mmHg at 0C, under standard gravitational acceleration, g (9.807m/s2).
Change in atmospheric pressure due to elevation has many effects: Cooking, nose bleeds, engine performance, aircraft performance.
1 atm = 760 Torr; 1 Torr =
133.3 Pa

PRESSURE MEASUREMENT -
MEDICAL APPLICATION
Most important in biological and medical applications
Complicated – physiological control system
Direct and indirect (auscultatory) measurement

PRESSURE MEASUREMENT NON
INVASIVE TECHNIQUE
 Sphygmomanometer
Use in clinical practice – easy and low cost
Cuff is wrapped around patient’s arm and inflated with air.
There are means for pressurizing the cuff, and gauge for monitoring the pressure inside the cuff.

A cuff of appropriate size is fitted and inflated manually by repeatedly squeezing a rubber bulb until the artery is completely occluded.
Listening with the stethoscope to the brachial artery at the elbow , the examiner slowly releases the pressure in the cuff.

PRESSURE MEASUREMENT – NON
INVASIVE
When blood flow through arteries, turbulent flow makes audible sound known as Korotkoff sounds. The pressure at which the sound is first heard is the systolic blood pressure .
The cuff pressure is further released until no sound can be heard (fifth Korotkoff sound), at the diastolic blood pressure .
Advantage – noninvasive, inexpensive and reliable
Disadvantage – relatively inaccuracy (operator error), low resolution, and difficulty of measuring the diastolic pressure (depend on the operator ability to hear very quiet sound).

PRESSURE MEASUREMENT – INVASIVE
TECHNIQUES
More accurate pressure measurement technique
Intravascular (inside blood vessel) –
Strain gauge – tipped pressure tranducer is inserted into artery to determine whether a specific stenosis is the caused of decreased blood flow.

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Connect middle ear to nasopharynx (throat).
Equalize middle air pressure with environmental pressure
[ Pressure difference between middle ear and outside environment (diving, high altitude)]
Risk of bursting or damaging eardrum
Yawning (lowering and protruding the jaw) helps balancing pressure in the middle ears