Fluids
Spring 2013

Overview
Topics:
• Flow measurements
• Fluid properties
• Fluid statics
• Energy, impulse-momentum equations (fluid dynamics)
• Pipe and other internal flow
• Similitude

Basic Definitions
Density = ρ = mass / unit volume (kg/m 3 , slugs/ft 3 )
ρ water
= 1000 kg/m 3 = 1.94 slugs/ft 3
Specific weight = γ = weight / unit volume (N/m 3 , lb/ft 3 )
γ water
= 9810 N/m 3 = 62.4 lb/ft 3
Specific gravity = SG = ρ fluid
/ ρ water
= γ fluid
/ γ water

Basic Definitions (cont.)
Pressure = force per unit area (N/m 2 , psi) gage pressure = measured pressure with respect to atmospheric pressure absolute pressure = measured pressure with respect to absolute zero pressure (complete vacuum) atmospheric pressure = 101 KN/m 2 = 14.7 psi
So P absolute
= P gage
+ P atmosphere

Basic Definitions (cont.)
Stress = force per unit area (N/m 2 , psi) normal stress = area resisting stress is perpendicular
(normal) to force tangential (shear) stress = area resisting stress is parallel to force

Basic Definitions (cont.)
Fluid viscosity = measure of its resistance to flow when acted upon by an external force such as a pressure gradient or gravity
Newton’s law of viscosity: τ = µ dV/dY

Basic Definitions (cont.)
Absolute viscosity: µ = τ δ / V (N • s/m 2 or lb • s/ft 2 )
Kinematic viscosity: υ = µ / ρ (m 2 /s or ft 2 /s)

Basic Definitions (cont.)
Cohesive force = attraction of liquid molecules to each other
Adhesive force = attraction of liquid molecule to a another surface
Surface tension = force that holds a droplet of some fluid together at the fluid/gas interface (cohesive forces greater than adhesive forces towards gas)
Capillary action = attraction of some fluid to a solid surface
(adhesive forces greater than cohesive forces)

Basic Definitions (cont.)
σ = surface tension, N/m or lb/ft
ß = angle of contact

Example 1

Example 1 - Solution
From Kaplan

Example 2

Example 2 - Solution
From Kaplan

Example 3

Fluid Statics
Hydrostatic pressure p acting at some depth h
• p = ρ g h =
h
• same magnitude in all directions
• pressure on a surface will act normal to the surface

Center of Pressure
Y r
F r
F r
= P c
A = γ h c
A
Y r
= Y c
+ I xc
/ (Y c
A)
Y c
Magnitude of resultant force is equal to pressure of centroid of surface x surface area
Result force acts at the centroid of the pressure distribution,
NOT centroid of surface!

Example 4
What is the resultant force on a the end of a square tank measuring
20 ft x 20 ft and containing 10 ft of water, and where does the resultant force act?
F r
h = 10 ft
w = 20 ft

Example 4
Pressure Prism with height of 10 m and width of 20 m.
Pressure distribution is 0 at top to 624 lb/ft 2 ( γ x h) at bottom
Resultant force = Volume of pressure-area prism
= ½ base x h x w
= (1/2) (624 lb/ft 2 ) (10 ft) (20 ft) = 62,400 lb
Acts at centroid of prism (not surface)
= 1/3 h up from the bottom = 10 ft / 3 = 3.33 ft
= 1/2 w over = 20 ft / 2 = 10 ft

Pressure Measurements
From Ref Hdbk

Example 5
Based on the manometer readings shown in the figure, what is the pressure in water pipe?

Example 5 - Solution p pipe
= γ m h m
- γ w h w
γ m h m
= (2.0 ft) (13.6) (62.4 lb/ft 3 ) = 1297 lb/ft 2
γ w h w
= (1.5 ft) (62.4 lb/ft 3 ) = 94 lb/ft 2 or water pressure in pipe = p pipe
= 1203 lb/ft 2 = 8.4 psi

Buoyancy
F
B
=
f
V
D
F
B
= buoyancy force
f
= specific weight of fluid
V
D
= volume displaced by an object
A body immersed in a fluid is buoyed up by a force equal to the weight of fluid displaced.
A floating body displaces its own weight of the liquid in which it floats.

Example 6

Example 7

Example 7 - Solution

Fluid Dynamics
Conservation of mass:
mass entering system = mass exiting system
ρ
1
A
1
v
1
= ρ
2
A
2
v
2
; mass flow rate in kg/s or lbm/s
For incompressible fluids: ρ
1
= ρ
2
A
1
v
1
= A
2
v
2
With Q = volumetric flow rate = A v in m 3 /s or ft 3 /s
Q
1
= Q
2

Example 8

Example 8 - Solution

Fluid Dynamics (Cont.)
Fluid energy - expressed in terms of head (m or ft)
velocity head = v 2 / (2g) pressure head = P/ γ elevation head with respect to a reference datum = z
Hydraulic grade line = pressure head + elevation head
Energy grade line = pressure head + elevation head + velocity head
See handout.

Fluid Dynamics (Cont.)
Conservation of energy (Bernoulli equation):
Final = Initial – losses + gains p
2
/ γ + v
2
2 /2g + z
2
= p
1
/ γ + v
1
2 /2g + z
1
– h f
– h f,fittings
– h t
+ h p
h f
= head loss from friction (“major headloss”)
h f,fittings
= friction losses at fittings, bends, contractions and expansions on right (“minor headloss”). See Reference
Handbook. h f,fittings
= Cv 2 /2g; C ~0.1-2.0
h t
= head removed due to a turbine
h p
= head added due to a pump

Example 9

Example 9 - Solution
From Kaplan

BREAK

Friction Losses
Friction loss for low of velocity v in a pipe of length L and diameter D ( Darcy-Weisbach equation) : h f
= f L/D v 2 /2g f = Darcy friction factor
Look up f as a function of R e
and ε/D on Moody diagram.
Use this equation to estimate hf and plug it into Energy Equation.
See handout – 3 rd to last page!

Pump Power and Efficiency
W = γ Q h p
/ η
W = pump power (N-m/s = J/s = watt; ft-lb/sec, 550 ftlbf/sec = 1 horsepower or hp); η = pump efficiency (~0.7); h p
= pump head in m or ft

Moody Diagram Calculation
TAPS Problem
See Handout

Parallel Pipe Flow
Head loss in each branch will be the same: f
A
L
A
/D
A
v
A
2 /2g = f
B
L
B
/D
B
v
B
2 /2g
Total flow in must equal flow in two branches: Q = Q
A
+ Q
B

Reynolds Number
Unitless or non-dimentional numbers
Re = inertial forces / viscous forces = ( ρ v L ) / μ
For pipe flow Re = ( ρ v D ) / μ where D is pipe diameter
Laminar flow if Re < 2100
Turbulent flow if Re > 4000
Transitional flow in between

Impulse-Momentum
Σ F = ρ Q (v out
– v in
)
May need to consider forces in 1, 2 or 3 directions (treat as vectors, or treat as scaler and handle geometry)
Forces will develop in pipe bends, deflectors, blades
-- see Reference Handbook

Example 11
F z
= ?

Example 11 - Solution
(V z-out
– V z-in
)
(50 – 0 m/s)

Example 12

Example 12 - Solution

Fluid Flow Measurements – Pitot Tube
V = (2 g h) 1/2
From Ref Hdbk

Fluid Flow Measurements – Orifices
From Ref Hdbk

Fluid Flow Measurements – Submerged Orifice
From Ref Hdbk

Fluid Flow Measurements – Freely Discharging
Orifice
From Ref Hdbk

Dimensional Analysis
Dimensionally homogeneous equation = equations that do not depend on the units of measurement
Example: Flow in pipes dominated by viscosity and inertia
Re = ( ρ v D ) / μ
To conduct a test on a scaled model for pipe model (m) and prototype (p):
[( ρ v D ) / μ ] m
= [( ρ v D ) / μ ] p

Similitude
A model must be geometrically, kinematically and dynamically similar to the prototype
For a phenomenon involving n variables consisting of r dimensions (mass, length, time, temp, etc.), then ( n – r ) independent dimensionless groups (referred to as π groups) are needed to properly model the behavior of the prototype = similitude requirements

Example 13

Example 14

Example 14 - Solution
From Kaplan