fc-civ hidrcana: channel hydraulics flow mechanics...

Eusebio Ingol Blanco, Ph.D.

Civil Engineering Program, San Ignacio de Loyola University

FC-CIV HIDRCANA: Channel Hydraulics Flow Mechanics Review

Pressure

Objective

• Determine the variation of pressure in a

fluid at rest.

• Compute pressure using manometers

Universidad San Ignacio de Loyola Eusebio Ingol Blanco, Ph.D.

Pressure


• Pressure is defined as a normal force exerted by a

fluid per unit area.

1Pa = 1 N/m2

1kgf/cm2 = 14.223 psi

Pressure


Pgage = Pabs - Patm

Pvac = Patm - Pabs

Absolute pressure: The actual pressure at a given position. It is measured

relative to absolute vacuum (i.e., absolute zero pressure).

Gage pressure: The difference between the absolute pressure and the local

atmospheric pressure. Most pressure-measuring devices are calibrated to

read zero in the atmosphere, and so they indicate gage pressure.

Vacuum pressures: Pressures below atmospheric pressure.

Example


Pressure at a Point


• Pressure is the compressive force per unit area but it is not a vector.

Pressure at any point in a fluid is the same in all directions.

Forces acting on a wedge-shaped fluid element

in equilibrium.

PPPP

zgPP

PP

yxandzybyDividing

lzlx

zyxgmgW

zyxgylPxyPmaF

ylPzyPmaF

zz

xx

321

32

31

32

31

02

1

0

sincos

2/

02

1cos:0

0sin:0

The last term of the last equation drops out as ∆z →0 and the

wedge becomes infinitesimal

Pressure Variation with Depth


• In static fluid, pressure varies only with elevation in the

fluid.

zzgPPP

yxbyDividing

zzz

zyxgmgW

zyxgyxPyxPmaF zz

12

12

21 0:0

Free-body diagram of a rectangular

fluid element in equilibrium.

z2

z1

gdz

dP

2

1

12 gdzPPP

If density changes significantly with elevation, dividing previous

equation by ∆z, and taking the limit as ∆z→0. This yields:

When the variation of density with elevation is known



zPzgPP aboveabovebolow

ghPorghPP gageatm

The pressure of a fluid at rest increases with

depth (as a result of added weight). Pressure in a liquid at rest increases linearly

with distance from the free surface.

The vertical distance is sometimes used

as a measure of pressure, and it called

the Pressure head



• If is a constant dz

dp

zp

zzpp

dzdpz

z

p

p

)( 1212

2

1

2

1

head) ic(piezometrconstant zp

h

z = 0

1

2

z = z1, p = p1

z = z2, p = p2

zp

22

11 z

pz

p

Elevation head

Pressure head

Piezometric head

http://www.ce.utexas.edu/prof/mckinney/ce319f/ce319f.html


Pressure Head


z = 0

1

2

1z

1p /2p

2z

constant zp

Open Tank



Pressure Head


z = 0

3

1z

1p

2z

constant zp

3z

2p

3p

2

1

Pressurized Tank



Under Hydrostatic Conditions


The pressure is the same at all points on a horizontal plane in a

given fluid regardless of geometry, provided that the points are

interconnected by the same fluid.

Pressure using manometers


The manometer is commonly

used to measure small and

moderate pressure differences.

ghPP

PP

atm

2

21

The pressure anywhere in the tank and

at position 1 has the same value due to

the gravitational effects are negligible.

Why P1 =P2?

Example 3-5


A manometer is used to measure the pressure of a gas in a tank. The fluid

has SG = 0.85, and the manometer column height is 55 cm. Local

atmospheric pressure is 96 kPa. Find the absolute pressure within the tank.

Solution

Assumption:

The density of the gas in the tank is much lower

than the density of the manometer fluid

Properties:

SG = 0.85, ρwater = 1000 kg/m3

Analysis:

ρ = SG (ρHO2 ) = (0.85)(1000kg/m3)= 850 kg/m3

P = Patm + ρgh

kPa

mN

kPa

smkg

NmsmmkgkPa

ghPP atm

6.100

/1000

1

/.1

1)55.0)(/81.9)(/850(96

22

3

Pressure using manometers


For immiscible fluids:

1. Pressure change across a fluid column of height h is ΔP = ρgh

2. Pressure increases downward in a given fluid and decrease upward

3. Two points at the same elevation in a continues fluid at rest are at the same pressure

Example 3-6


Solution

Assumption: The air pressure in the tank is uniform

Properties: The densities water, oil, and mercury are given

Analysis: Starting with the pressure at point 1 at the air-

water interface.

P1 + ρwater g h1 + ρoil g h2 - ρmercury g h3 = P2 = Patm

P1 = Patm - ρwater g h1 - ρoil g h2 + ρmercury g h3

P1 = 129.65 kPa

1N = kg.m/s2, 1kPa = 1000 N/m2

fc-civ hidrcana: channel hydraulics flow mechanics...

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