Pharos University

ME 259 Fluid Mechanics II

Review of Previous Fluid Mechanics

Dr. Shibl

Fluid Properties:Liquid or Gas

• Liquids are:– Incompressible, V ≠ f(P)– Viscous (high viscosity)– Viscosity decreases with temperature

• Gases are:– Compressible, V = f(P)– Low viscosity– Viscosity increases with temperature

Equations for Fluid Property

• Circular Area: • Weight: w = m*g Newton• Density: = m/V Kg/m3

• Specific Weight: = w/V N/m3

• Specific gravity: SG=/water

Area = /4*D2

Viscosity• Dynamic Viscosity

= Shear Stress/Slope of velocity profile

• Kinematic Viscosity cS (centistokes) or m2/Sec.

yv

AF

/

/

Slope = v/y

v

y

F

cP (centipoise) or Pa-sec

Pressure and Elevation

• Change in pressure in homogeneous liquid at rest due to a change in elevationP = h

Where,

P = change in pressure, kPa = specific weight, N/m3

h = change in elevation, m

Pressure-Elevation Relationship

• Valid for homogeneous fluids at rest (static)

Free Surface Free Surface

P1

P2

P1 > P2

P2 = Patm + gh

Example: Manometer

• Calculate pressure (psig) or kPa (gage) at Point A. Open end is at atmospheric pressure.

A

Hg: SG = 13.54

Water

0.4 m

0.15 m

Forces due to Static Fluids

• Pressure =Force/Area (definition)

• Force = Pressure*Area

• Example:– If a cylinder has an internal diameter of 50

mm and operates at a pressure of 20 bar, calculate the force on the ends of the cylinder.

Force-Pressure: Rectangular Walls

P = *h

Patm

d

Vertical wall

Flow Classification • Classification of Fluid Dynamics

Apr 19, 2023 10/27

Inviscidµ = 0

Viscous

Laminar

Turbulent

Compressible Incompressibleϱ = constant

Internal External

Definitions

• Volume (Volumetric) Flow Rate– Q = Cross Sectional Area*Average Velocity

of the fluid– Q = A*v

• Weight Flow Rate– W = *Q

• Mass Flow Rate– M = *Q

Volumev

Q = Volume/Unit timeQ = Area*Distance/Unit Time

Key Principles in Fluid Flow

• Continuity for any fluid (gas or liquid)– Mass flow rate In = Mass Flow Rate out

– M1 = M2

– 1*A1*v1 = 2*A2*v2

• Continuity for liquids– Q1 = Q2

– A1*v1 = A2*v2

M1 M2

Newton’s Laws

• Newton’s laws are relations between motions of bodies and the forces acting on them.– First law: a body at rest remains at rest, and a body in motion

remains in motion at the same velocity in a straight path when the net force acting on it is zero.

– Second law: the acceleration of a body is proportional to the net force acting on it and is inversely proportional to its mass.

– Third law: when a body exerts a force on a second body, the second body exerts an equal and opposite force on the first.

Momentum Equation

• Steady Flow

• Average velocities

• Approximate momentum flow rate

Total Energy and Conservation of Energy Principle

• E = FE + PE + KE

• Two points along the same pipe: E1 = E2

• Bernoulli’s Equation:

Pz

v

g

Pz

v

g1

112

22

22

2 2

g

vwzw

PwE

2

2

g

wvwz

wp

g

wvwz

wp

22

22

22

21

11

Conservation of Energy

Pz

v

gh h h

Pz

v

gA R L1

112

22

22

2 2

17

1

211

2

222 z

2g

VPz

2g

VP

gg

Bernoulli’s Equation

18

zρg

P

2g

V2

piezometric head

kinetic head

1

211

2

222 z

2g

VPz

2g

VP

gg

19

Flow through a contraction

constant2g

Vz

ρg

P 2

position

head Total Energy

Piezometric Head

Kinetic Head

1 2

20

position

Energy Grade Line

Piezometric Head

Kinetic Head

head

Hydraulic Grade Line

21

Bernoulli’s Equation• No shaft work• Steady state• Constant temperature• Incompressible

• No heat transfer• No shear work (frictionless)• Single uniform inlet and

single uniform outlet

22

Frictional Effects

• Pipes are NOT frictionless• Add a loss due to friction to Bernoulli’s eq.

L2

222

1

211 hz

2g

V

ρg

Pz

2g

V

ρg

P

Head loss due to friction

23

position

head EGL (ideal)

EGL (actual)

HGL (actual)

hL

zρg

P

2g

V2

24

Friction Losses

fmL hhh

2g

VKh

2

mm

hL = head losses due to friction

2g

V

D

Lfh

2

f

Fittings (valves, elbows, etc)

Pipe friction

fmL ΔPΔPΔP

2

ρVKΔP

2

mm

2

ρV

D

LfΔP

2

f

25

Minor LossesKm - minor loss coefficient

2

ρVKΔP

2

mm fittings

mm KK

26

Pipe Friction Losses

2

ρV

D

LfΔP

2

fDarcy-Weisbach equation

Kinetic pressure

Length/diameter ratio

Darcy friction factor- pipe roughness (Table 6.1)- Reynolds number

Reynolds Number

• Dimensionless• Ratio of inertial forces to viscous forces• Used to characterize the flow regime

27

ν

VL

μ

ρVLRe

Reynolds Number

• Describes if the flow is:– Laminar - smooth and steady– Turbulent - agitated, irregular

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

28

Osborne Reynolds Tests

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Laminar

Turbulent

29

Friction Factor

dRe

64f

•For circular pipe Re ≈ 2300 for transition•L = Diameter of pipe

•Laminar flow

•Turbulent flow

1/2d

1/2 fRe

2.51

3.7d

ε2.0log

f

1 Colebrook Equation

30

Piping Systems

• Three examples of piping systems1. Pipes in series

321BA hhhh 321 QQQ

32

33

Home work • Two reservoirs are connected by a pipe as

shown. The volume flow rate in pipe A is 2.2 L/s. Find the difference in elevation between the two surfaces.

A B

D = 2 cmL = 5 m

D = 4 cmL = 5 m

z

= 1000 kg/m3

= 0.001 kg/(m·s) = 0.05 mm

Centrifugal PumpsCopyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

PL2

222

1

211 hhz

2g

V

ρg

Pz

2g

V

ρg

P

Head increase over pump

2g

V

D

Lf

2g

VKzz

2g

V

2g

V

ρg

P

ρg

Ph

22

m12

21

2212

p

2

ρV

D

Lf

2

ρVKVV

2

ρzzρgPPΔP

22

m2

1221212P

Fixed system pressure

Variable system pressure

2Vf

2Qf

2BQAP system

Nature of Dimensional Analysis

Example: Drag on a Sphere

Drag depends on FOUR parameters:sphere size (D); speed (V); fluid density (); fluid viscosity ()

Difficult to know how to set up experiments to determine dependencies

Difficult to know how to present results (four graphs?)

Nature of Dimensional Analysis

Example: Drag on a Sphere

Only one dependent and one independent variable Easy to set up experiments to determine dependency Easy to present results (one graph)

Buckingham Pi Theorem

• Step 1:List all the dimensional parameters involved

Let n be the number of parameters

Example: For drag on a sphere, F, V, D, , , and n = 5

Buckingham Pi Theorem• Step 5

Set up dimensional equations, combining the parameters selected in Step 4 with each of the other parameters in turn, to form dimensionless groups

There will be n – m equations

Example: For drag on a sphere

Dimensional Analysis and Similarity

• Geometric Similarity - the model must be the same shape as the prototype. Each dimension must be scaled by the same factor.

• Kinematic Similarity - velocity as any point in the model must be proportional

• Dynamic Similarity - all forces in the model flow scale by a constant factor to corresponding forces in the prototype flow.

• Complete Similarity is achieved only if all 3 conditions are met.

Flow Similarity and Model Studies

• Example: Drag on a Sphere

For dynamic similarity …

… then …

Flow Similarity and Model Studies

• Scaling with Multiple Dependent Parameters

Example: Centrifugal Pump(Negligible Viscous Effects)

If … … then …

systemP

Q

Static head

System Curve

pumpP

Q

Pump Curve

2

ρV

D

Lf

2

ρVKVV

2

ρzzρgPPΔP

22

m2

1221212P

Pump pressureSystem pressure

P

Q

Operating point

P

Q

P

Q

2BQAP system

changing

P

Q

2BQAP system

changing

Pump Power

• Recall that )( 12 PPvdPvvdPw p

ppp PQwmW

p

pp

PQW

Power provided to fluid

Power required by pump

Home Work Water is pumped between two reservoirs at 0.2 ft3/s (5.6 L/s) through 400 ft (124m) of 2-in (50mm) -diameter pipe and several minor losses, as shown. The roughness ratio is /d = 0.001. Compute the pump horsepower required