Describes the theory behind pressure

1. 1
PRESSURE & HYDROSTATICS
Pressure = F / A
• Pressure is typically measured with respect to a reference
level
• Reference could be atmosphere or vaccum.
• The reference level = atmospheric pressure (95-105kPa)
• Pressure measured with respect to atmosphere = Gage
pressure
• Pressure measured with respect to perfect vaccum =
Absolute pressure
The three related by the equation –
Pabs = Pgage + Patm
See Figure 3.1

2. 2
Points –
• Perfect vacuum = lowest possible pressure
• Absolute pressure will always be positive
• Gage pressure above atmospheric is +ve
• Gage pressure below atmospheric will be –ve
• Units = Pa (abs) or Pa (gage) – always mention what the
reference is!
• Range of atmospheric pressure = 95 kPa to 105 kPa

3. 3
Assumed as 101 kPa in the text.
Or 14.7 psi(a)
Problem 3.1
Given Pressure (gage) = 155 kPa
Atmospheric Pressure = 98 kPa
What is the Absolute Pressure?????
Absolute pressure =
Pressure and Elevation:
Pressure is directly related to elevation and changes with
elevation
- example – swimming pool or diving deep into a water body
The greater the depth/elevation of fluid – the greater the pressure
Elevation in water – measured with respect to a reference;
positive upwards; negative downwards
See Figure 3.2

4. 4
It is advisable to choose the lowest point for a reference level so
that you don’t have to deal with negative elevation values!
****Change in pressure =
Δp = γ h
Where γ is the specific weight of the fluid and h is the
elevation

5. 5
******Key Points:
• Equation is only valid for homogenous liquids at rest
• Points on the same horizontal level have the same
pressure
• Pressure varies linearly with change in elevation/depth
• Change in pressure is proportional to the specific weight
of the fluid
Above equation does not apply to gases because the specific
weight of gas changes with elevation –
However the change in gas pressure with elevation is small!
An increase in elevation of 300 m changes gas pressure by only
3.4kPa
Problem 3.5
Determine the pressure at a depth of 5m for water.
Δp = γ h
Specific wt of water = 9.81 kN/m3
Therefore,
Δp = 9.81 x 5 = 49.05 kN/m2
= 49.05 k Pa

6. 6
Problem 3.7 (SI units)
Compute the gage pressures at pts. A, B, C, D, E and F.
Specific gravity of Oil = 0.9
Always start with the point of know pressure – Pt A in this case.

7. 7
Answers:
Pa = 0 (gage) – atmospheric pressure
γ oil = 0.9 x 9.81 = 8.83 kN/m2
Pb = 3.0 x 8.83
Pb = 26.5 kN/m2
= 26.5 kPa
Pc = 6.0 x 8.83
Pc = 53 kN/m2
Pd = Pb (same level)
Pd = 26.5 kN/m2
Pe = Pa – same level as A
Pe = 0 kN/m2
Pf = 0 – 1.5*8.83
Pf = -13.2 kN/m2

8. 8
Key observations from example above –
• Pressure increases with depth
• Pressure changes linearly
• Points at the same elevation have the same pressure
• Pressure decreases at higher elevations

9. 9
3.6 Pascal’s paradox
Pressure depends only on the elevation and the type of the
fluid; NOT on the size of the fluid container
See Figure 3.7
All containers have the same pressure at the bottom!!
-referred to as the Pascal’s Paradox

10. 10
The role of elevation also comes into play with respect to
pressure in a water distribution system – See Figure 3.8
The Supply point should be higher than the receiving points to
allow pressure to be maintained by gravity.

11. 11
3.7 Manometers
Instrument to measure pressure.
• Simplest kind – U tube manometer
• One end – open to the atmosphere
• Other end – connected to the fluid whose pressure is to be
measured
• Contains liquid (gage fluid)– whose deflection indicates the
pressure
• Gage liquid – should not mix with the other liquid

12. 12
Procedure for measurement:
• Start from the point which is exposed to atmosphere, and
move towards the point at which the pressure is desired.
• If you move down in the fluid, pressure increases; and vice-
versa

13. 13
Problem 3.8 (SI units)
Compute the pressure at A?

14. 14
γm = 9.81 x 13.54 = 132.8 kN/m3
P1 = 0
P2 = 0.25 x 132.8 = 33.2 kN/m2
P3 = P2 = 33.2 kN/m2
P4 = P3 – 0.4 x 9.81 = P3 - 3.92 = 29.28 kN/m2
Answer = Pa = 29.28 kN/m2
= 29.28kPa

15. 15
Problem 3.9 (US Units)
Determine the difference in pressure between points A and B
Specific wt of water = 62.4 lb/ft3
.

16. 16
P1 = Pa + 33.75 x γo
P1 = P2
P3 = P1 – 29.5 x γw
P4 = P3 – 4.25 x γo
Pb = P4 = Pa + 33.75 x γo – 29.5 x γw – 4.25 x γo
Or
Pb – Pa = 33.75 x γo – 29.5 x γw – 4.25 x γo
= 29.5 γo – 29.5 x γw
= 29.5 (γo – γw)
γo = 0.86 x 62.4 = 53.7 lb/ft3
Pb-Pa = 29.5 in x (53.7 – 62.4) lb/ft3
= 29.5 in x (-8.7 lb/ft3
) x (1 ft3
/ 1728 in3
)
Answer : Pb – Pa = -0.15 lb/in2

17. 17
Other types of manometers:
Well-type manometer
Inclined well-type manometer

18. 18
Barometers
• Device for measuring atmospheric pressure.
• Filled with mercury
• Mercury filled tube is inverted in mercury bath. Mercury
column drops a little – filled with mercury vapor at 0.17 Pa.
The height of the mercury provides the atmospheric pressure
0 + γmh = Patm
Patm = γmh

19. 19
• Mercury depth decreases 1.0 inch every 1000 ft of increase
in altitude. (pressure will decrease as you go up in the
atmosphere).
• Specific wt of mercury changes with temp! So adjustments
with temp have to be made!

20. 20
Pressure gages and transducers
Gage - Pressure sensed mechanically.
Pressure transducer – pressure measured at one point, displayed
at another – pressure sensed mechanically and converted into an
electrical signal.

21. 21

22. 22
Assignment # 2
• 3.48M
• 3.54M
• 3.62M
• 3.63E
• 3.65M
• 3.66M
Some other types of problems –

23. 23
Q. A pressure gage at 19.0 ft from bottom of tank reads =
13.19 psi.
Another at 14 ft, reads = 15.12 psi
Compute – specific wt, density, and specific gravity of fluid in
tank
We have two known pressure points and the distance between
them!
Δp = γ Δh
Apply the equation
(15.12 – 13.19) * 144 = γ * (19.0-14.0)
Therefore γ = 55.6 lb/ft3
Remember γ = ρg
Therefore, ρ = 55.6/ 32.2 = 1.73 slug/ft3
SG = γf/ γw = 55.6/62.4 = 0.891.
Q.

24. 24
A reservoir of CCl4 has mass of 500 kg and a volume of 0.315
m3
. Find the weight, density, specific weight and specific
gravity.
m = 500 kg
g = 9.81 m/s2
W = mg = 500*9.81 = 4905 N = 4.905 kN
Density = ρ = m/V = 500/0.315 = 1587 kg/m3
Specific wt = γ = W/V = 4.905/0.315 = 15.57 kN/m3
SG = 15.57/9.81 = 1.59

25. 25
Q.
Fluid is Oil = SG = 0.85
Find the pressure heads at A and B?
γw = 9.81 kN/m3
γo = 0.85*9.81 = 8.33 kN/m3
pA = 0 – 2.8*8.33 = - 23.35 kN/m2
pB = -23.35 + 2.2*8.33 = -5.02 kN/m2
2.2 m
0.6m
A
B