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INTRODUCTION TO FLUID MECHANICS – CVG 2116

FLUID STATICS – Part 1

(HYDROSTATICS)

Taught by: Mahmoud AL-RIFFAI

Notes by: Ioan NISTOR

CVG 2116

3.1 Definition of Pressure

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CVG 2116

3.1 Pressure (cont.)

Note: As the element shrinks to an infinitesimal point ∆x→0, ∆y→0, ∆z→0 and ∆l→0, which leads to Pn = Pz from ΣFz=0

CVG 2116

Pressure transmission

Blaise Pascal1623-1662

Example: The hydraulic lift

F=100 N; F2=?

The force exerted on the small piston (A1):(AC)F - (BC)F1= 0

F1=1100 N

p1=F1/A1=6.22 106 N/m2

p1=p2 F2=p1A2 F2=12.22 kN

A

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CVG 2116

Absolute Pressure, Gage Pressure and Vacuum

The pressure in the extraterrestrial space is ZERO-, absolute zero

All measured pressures are reported to this pressure absolutes pressures, PA

Example: the atmospheric pressure patm=101 kPa = pa

In the engineering practice many instruments which measure the pressure are doing so

with respect to the atmospheric pressure gage pressure, pg

Pg > 0 & PA > Pa

PA = Pa = atmospheric pressure = 101.325kPa

PA > 101.325kPa

PA = 0kPa = absolute zero

Pg < 0 & PA < Pa

CVG 2116

Absolute Pressure, Gage Pressure and Vacuum (example)

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CVG 2116

3.2 Pressure variation with elevation

ΣFl=0; pΔA - (p+ Δp)ΔA – ΔAΔl sinα = 0,

Δp/Δl = - γsinά dp/dl = - γ dz/dl

dp/dz = - γ

CVG 2116

3.2 Pressure variation with elevation - Example

Example: Compare the rate of change of pressure for air and for fresh water at sea level

(p=101.3kPa, T=15.5oC) with respect to a 4-m decrease in elevation. (the specific

weights are assumed constant).

SOLUTION

Specific weights of water and air (from the ideal gas law!)

ρair=p/(RT) = 101.3 103(N/m2)/[287(J/kgK) (15.5+273)(K)] ρair=1.22kg/m3

γair=gρair = 11.97 N/m3

γwater= 9799 N/m3

Therefore, (dp/dz)air = - 11.97 N/m3, (dp/dz)water = - 9799 N/m3

The total change in pressure for air Δpair= - 11.97 N/m3 x (-4m)=47.9Pa

The total change in pressure for water Δpwater= - 9799 N/m3 x (-4m)=39.2kPa

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CVG 2116

Pressure variation for a uniform density fluid

CVG 2116

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CVG 2116

Pressure variation with elevation for different density fluids – Example

In an open tank, oil with a specific gravity of 0.80 forms a layer 0.9 m

deep above a total depth of 3.0 m of water. What is the pressure

measured at the bottom of the tank?

Fluid properties: Soil = 0.8, water= 9810 N/m3

Pressure at the bottom of the oil layer:

where z1=3m, z2=2.10m

p2 = 7.06kPa,

Pressure at the bottom of the tank

where z3 = 0m

p3 = 27.7kPa

7.06kPa

27.7kPa

CVG 2116

Pressure variation with the altitude

Fluids with non-uniform density (compressible):

The equation of state: p = RT ou = p/RT ( =pg/RT)

p – the absolute pressure [Pa]

T – the absolute temperature [K]

R – the universal gas constant [J/kgK]

in the troposphere: T = To-(z-zo)

dp/dz = - pg/RT p=po[T/To]g/R

In the stratosphere: T = const.

p = poe-(z-zo)g/RT

U.S.National Weather Service: 45oN latitude in July

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CVG 2111

3.3 Pressure measurement

Pressure transducer

CVG 2116

3.3 Pressure measurement (Manometry)

pv~0 pressure of Hg vaporsat normal temperature

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Piezometer

Example: suppose ∆h=60cm, l=180cm

Calculate the pressure in the pipe center!

p1=0 (open atmosphere), therefore p2=0-m∆h

p2=79.8kPa ( mercury m=133kN/m3)

p2=p3 (same elevation), therefore p4=p3+∆p34; ∆p34=-l ∆p34=-17.66kPa

p4=62.1kPa

General equation of the manometer: p2=p1+Σdownihi-Σupihi CVG 2116

CVG 2116

Differential manometer

Example: Find the piezometric pressure variation and the piezometric head

between the two points if the deflection of mercury in the manometer is

2.54cm.(Twater=10oC)

The equation of the manometer: p2= p1+ w(∆y+∆h) – m∆h - w(∆y+z2-z1)

p2 + wz2 - (p1+ wz1) = ∆h(w - m)

or

pz2-pz1

= ∆h(w - m); pz2-pz1

= ?

and

h2-h1=(pz2-pz1

)/w=∆h(1- m/w); h2-h1=?

3.3 Pressure measurement – Example to be solved in class