1 dimensions · bee 5330 –fluids fe review, feb 24, 2010 5 2.5 surface tension the water molecule...

BEE 5330 –Fluids FE Review, Feb 24, 2010 1

A fluid is a substance that can not support a shear stress.

Liquids differ from gasses in that liquids that do not completely fill a container will form

a free surface in a gravitational field (and mix minimally with any atmosphere) while a

gas will form an atmosphere (and eventually mix with an existing atmosphere).

1 Dimensions

Dimensions represent classes of units we use to describe a physical quantity. Most fluid

problems involve four primary dimensions

• Mass [M]

• Length [L]

• Time [T]

• Temperature [Θ]

For example velocity has the dimensions of LT−1.

1.1 System of Units

Units are the bane of the United States! Remember the NASA Jet Propulsion Lab (JPL)

satellite disaster?! In September 1999 we lost a $125,000,000 Mars Orbiter because a

subcontractor to NASA was working in English units while NASA had converted to

metric units in 1990.

A system of units is a particular method of attaching a number to a dimension. A major

source of calculation error is units errors ⇒ check your units! Use your engineering

common sense, you should always have a rough estimate of the answer you expect, at


least to an order of magnitude. If the answer is outside this range there is a good chance

you have made a units error.

1.1.1 British Gravitational (BG)

• Length [L] ∼ foot

• Mass [M] ∼ slug; F = ma ⇒ 1 lbs = 1 slug · 1 ft/s2

• Time [T] ∼ second

• Temperature [Θ] ∼◦R (degrees Rankine — absolute temperature scale)=◦F+459.67

1.1.2 International System (SI)

• Length [L] ∼ meter

• Mass [M] ∼ kilogram; F = ma ⇒ 1 Newton = 1 kg · 1 m/s2

• Time [T] ∼ second

• Temperature [Θ] K (Kelvin — absolute temperature scale)=◦C+273.15

2 Thermodynamic Properties

• Temperature Measure of internal energy level.

• Pressure Measure of compressive (normal) stress at a point.

P =F

A

• Density ρ = MassVolume


2.1 Specific Weight

γ = ρg =weight

volume

γwater=62.4 lbs/ft3

2.2 Specific Gravity

The specific gravity is the density of a substance normalized by the density of water

at a certain temperature, often 4◦C, the temperature of maximum density at normal

pressures. Hence we write

S.G. =ρ

ρwater @ 4◦C

=ρ

1000 kg/m3in S.I. units

S.G. of sands and gravels is about 2.6 - 2.7.

2.3 Viscosity

τ = µdu

dz

Therefore

µ =τ

du

dz

=[MLT−2L−2]

[LT−1]

[L]

=[M]

[LT]

2.3.1 Kinematic Viscosity

If we normalize the viscosity by the density we have the kinematic viscosity.

ν =µ

ρ=

[L2]

[T]

At 20◦C water has an absolute or dynamic viscosity of 1.0 × 10−3 N s m−2 (or Pa·s) and

a kinematic viscosity of 1.0 × 10−6 m2s−1.


2.4 Example - A block sliding down an inclined plane

If the block has mass 1 kg:

1. Determine the viscosity, µ, of the lubricant fluid in the gap.

2. What speed will the block travel if the angle, θ, is adjusted to 10◦ and the gap, δ,

is decreased to 0.5 mm

1) µ = 0.049kg

m · s (=N · sm2

= Pa · s); 2) V = 0.070m

s


2.5 Surface Tension

The water molecule is polar. The O− attracts the H+. Within the fluid this attraction

is in balance, i.e., the net force due to all of the polar pairs is zero. However, at the

surface half of this force is missing and the surface is pulled toward the fluid interior with

a certain energy.

surface energy =J

m2=

N m

m2=

N

m=

force

length= tension

hence we refer to this energy as the surface tension.

2.6 Example – Rise/drop in a capillary tube

3 Hydrostatics

In many fluid problems the velocity is zero or the velocity is constant ⇒ τ = 0.

Pressure as a scalar quantity as it is a quantity with no dependence on direction (as

opposed to velocity, which is a vector).

−∇P − γ~k = ρ~a


Now, if the fluid is at rest (or at least moving at a constant velocity):

~a = 0 ⇒ ∇P = −γ~k

Hence we can write:∂P

∂x= 0,

∂P

∂y= 0

∂P

∂z= −γ

We see that P = P (z) only, hence the pressure at a given elevation (z position) is

constant. In the vertical we have

dP

dz= −γ where we have replaced ∂ with d now.

Hence as z ↑ P ↓

Incompressible Fluids

It is usually reasonable to assume that all liquids essentially have a constant γ, certainly

true for all fluids at constant temperature and pressure. Under this condition we have

P1 = γh + P2 or h =P1 − P2

γ

where we refer to h as the pressure head as it is a pressure measured in units of length.

3.1 Measurement of Pressure

Absolute pressure is the pressure relative to a perfect vacuum, hence it is always a positive

(or zero) value.

Gage pressure is the pressure relative to the local atmospheric value

P > Pa ⇒ Pgage = P − Pa > 0

P < Pa ⇒ Pgage = P − Pa < 0 Pvacuum = Psuction = Pa − P > 0


3.2 Manometry

A manometer is a vertical or inclined tube used to measure pressure. There are three

fundamental types of manometers: Piezometer tube, U-tube, and inclined tube.

Manometer analysis is straight forward hydrostatics - the key is to keep track of the signs

of the pressure terms! Consider the following U-tube manometer:

P2 −P1 = −γ(z2 − z1) Now we track the pressure from A to 1 and from 1 to 2. A simple

rule, based on our understanding that as we move down in a static fluid the pressure

increases, is:

Pdown = Pup + γ|∆z|

This obviates the need to keep track of signs of directions in the vertical. Hence for our

U-tube problem we have

PA + γ1|zA − z1| − γ2|z2 − z1| = Patm

and we solve for PA


Example - Manometer

3.3 Hydrostatic Force on a Plane Surface

Note: The approach presented here is different than the approach presented

in the text but I feel it is more powerful and makes more sense!

Consider

FR = PCA

Therefore the magnitude of the force depends on

• PC – the pressure acting at the centroid of the surface S.

• A – the area of the surface S.


Note the above is a bit different than most books present this material. If we make two

more assumptions

4. The column of fluid above the surface S is exposed to atmospheric pressure.

5. The density of the entire fluid, from the deepest part of the surface S all the way

to the free surface of the fluid, is constant.

Then we arrive at the form our book (and most books) present, namely

FR = γ sin θ

∫

A

y dA

where θ is the angle between vertical and our surface S (e.g., θ = 0 for a vertical surface).

But from mechanics we recognize 1A

∫

Ay dA as the first moment of the area A with respect

to the x axis which we will denote yc since this is the position along the y axis of the

centroid of the area. Hence we have

yc =1

A

∫

A

y dA ⇒ and hence

FR = γ sin θ ycA ⇒ but we can write hc = sin θ yc hence

FR = γhcA

where hc is the depth of the centroid (e.g., now in a direction parallel to gravity).

But this assumes assumptions 4 and 5 are true! Examples of problems that

violate these assumptions will appear in the problem sets and Lab #2 so

proceed with caution if you like the books approach.

3.3.1 Where is the force located on the surface?

yR =Ixcγ cos θ

PCA(1)

where

Ixc =

∫

y2 dA (2)


3.4 Pressure Prism

Consider the following example:

We can solve this directly by applying the analysis presented above (left as an exercise

for the student), however, for many situations (or just for many people who prefer to

think in a different manner!) a decomposition of the pressures into a series of pressure

prisms is often easier. Consider the decomposition such that

FyR = F1y1 + F2y2

= (γbh2h1)

(

h1 +h2

2

)

+

(

γbh2

2

2

) (

h1

2h2

3

)

Therefore yR =

h1

(

h1 +h2

2

)

+h2

2

(

h1 +2h2

3

)

h1 +h2

2

= h1 +h2

2

h1 +2h2

3

h1 +h2

2

= 4 + 3

(

4 + 4

4 + 3

)

= 7.43′

3.5 Buoyancy - Archimedes’ Principal

FB = Weight of fluid displaced by a body or a floating body displaces its own weight of

the fluid on which it floats.


Example - Buoyancy

4 Conservation of Mass

If we have one-dimensional inlets and outlets

∑

(ρiAiVi)in =∑

(ρiAiVi)out

or∑

(mi)in =∑

(mi)out

Incompressible one-dimensional flow then we can write

∑

(ViAi)out =∑

(ViAi)in or∑

(Q)out =∑

(Q)in

Example - Pipe Entrance Flow

u = Umax

(

1 − r2

R2

)

If the inlet flow is uniform and denoted U0, what is Umax


Umax = 2U0

5 Conservation of Linear Momentum for a constant

Control Volume 1-D system

∑

~FC.V. =∑

(mi~vi)out −∑

(mi~vi)in

Example


6 Conservation of Energy and Bernoulli Equations

Incompressible 1-D Flow With No Shaft Work in head form:(

P

γ+

v2

2g+ z

)

out

=

(

P

γ+

v2

2g+ z

)

in

− hf

where we can think of hf as the friction losses and we see that hf > 0

Example – Gas Pipeline

Consider the following pipe flow:

If Q = 75 m3/s, the pipe radius is r = 6 cm, the inlet pressure is maintained at 24 atm

by a pump, the outlet vents to the atmosphere, the pipe rises 150 m from inlet to outlet

and the pipe length is 10 km, what is hf? What is the velocity head?

hf=198 m Therefore the friction loss is greater than the ∆z and the pump must drive

against both!

The velocity head is only 0.17 m!

Note that the length did not come into our solution. hf includes the total losses along


the pipe due to friction effects and hence includes the effect of length implicitly.

6.1 Bernoulli Equation

(

P

γ+

v2

2g+ z

)

out

=

(

P

γ+

v2

2g+ z

)

in

Clearly anywhere along a streamline, as long as no work is done between analysis points

and the assumption of frictionless flow is good, we can write

P

γ+

v2

2g+ z = h0

where the constant h0 is referred to as the Bernoulli constant and varies across stream-

lines.

Bernoulli Equation Assumptions

• Flow along single streamline ⇒ different streamlines, different h0.

• Steady flow (can be generalized to unsteady flow).

• Incompressible flow.

• Inviscid or frictionless flow, very restrictive!

• No ws between analysis points on streamline.

• No q between points on streamline.

6.2 Pressure form of Bernoulli Equation

If we multiply our head form of the Bernoulli equation by the specific weight we arrive

at the pressure form of the Bernoulli Equation:

P + ρv2

2+ γz = Pt


where we call the first term the static pressure, the second the dynamic pressure, the third

the hydrostatic pressure, and the right-hand-side the total pressure. Hence the Bernoulli

Equation says that in inviscid flows the total pressure along a streamline is constant.

If we remain at a constant elevation the above equation reduces to

P + ρv2

2= Ps

where we refer to Ps as the stagnation pressure. Thus by definition the stagnation pressure

is the pressure along horizontal streamlines when the velocity is zero.

6.3 Pitot-Static Tube

The static and stagnation pressures can be measured simultaneously using a Pitot-static

tube. Consider the following geometry:

The equation for the Pitot tube is known as the Pitot formula

V1 =

√

2P2 − P1

ρ

or in terms of heads

V =√

2g(H − h)


6.4 Example – Flow accelerating out of a reservoir

V2 =

2gh

1 −(

A2

A1

)2

1

2

and if A1 ≫ A2 1 −(

A2

A1

)2

≈ 1 ⇒ V2 ≈√

2gh, again!

This was first noted by Torricelli and is known as Torricelli’s equation.

6.5 Energy and the Hydraulic Grade Line

As we have seen we can write the head form of the Energy equation as

P

γ+

v2

2g+ z = H = Energy Grade Line (EGL)

In the case of Bernoulli flows the energy grade line is simply a constant since by assump-

tion energy is conserved (there is no mechanism to gain/lose energy). For other flows

it will drop due to frictional losses or work done on the surroundings (e.g., a turbine)

or increase due to work input (e.g., a pump). Note that this is the head that would be

measured by a Pitot tube.

We can also writeP

γ+ z = Hydraluic Grade Line (HGL)

and we see that the HGL is due to static pressure – the height a column of fluid would

rise due to pressure at a given elevation or in other words the head measured by a static


pressure tap or the piezometric head.

Example – Venturi Flow Meter

Consider

Q = A2V2 = CvA2

2g∆h

1 −(

A2

A1

)2

1

2

where Cv is known as the coefficient of velocity and lies in the range 0.95 and 1.0 typically.

It accounts for the minor energy losses relative to the ideal Bernoulli flow.

6.6 Frictional Effects

If we have abrupt losses, say at a contraction, a simple way of accounting for this is

through a discharge coefficient. We can write a modified form of Torricelli’s formula for

incompressible flow

Q = A = CdA√

2gh

where A is the area of the orifice and Cd is the discharge coefficient and is 1 for frictionless

(inviscid) flow and can range down to about 0.6 for flows strongly effected by friction.

Note we can handle non-uniform (violation of 1-D assumption) flow effects with a Cd as

well. Note Cd = CcCv where Cc is defined below


6.7 Vena Contracta Effect

For a flow to get around a sharp corner there would need to be an infinite pressure

gradient, which of course does not happen. Hence if the boundary changes directions too

rapidly at an exit, the flow separates from the exit and forms what is known as a vena

contracta

Clearly Aj/A ≤ 1. For a round sharp-cornered exit the coefficient is Cc = Aj/A = 0.61

and typical values of the coefficient fall in the range 0.5 ≤ Cc ≤ 1.0.

7 Dimensional Analysis & Similitude

Dimensional analysis is a method for reducing the number of variables describing the

physics ⇒ if k dimensional variables are important we seek to reduce (condense)

them to n dimensionless variables.

7.1 Buckingham Pi Theorem

If k dimensional variables are described by r physical dimensions then k− r independent

dimensionless variables completely describe the physics.

Before resorting to dimensional analysis try by inspection. Almost always will find a

Reynolds number and if there is a free surface or a ship riding on a free surface will likely

find a Froude number.


7.2 Similitude & Experiments

7.3 Types of Similarity

Total Similarity = Geometric Similarity + Kinematic Similarity + Dynamic Similarity

• Geometric Similarity – length scale ratios the same.

• Kinematic Similarity – velocity scale ratios the same.

• Dynamic Similarity – force scale ratios the same.

If total similarity can not be achieved we accept partial similarity but we let some pa-

rameters go (usually the Reynolds number) and make sure they are in a range (i.e., Re

is high enough to be turbulent at model scale if it is turbulent at prototype scale) that

is consistent with the physics at prototype scale.


7.4 Example – Similitude

7.5 Summary of Dimensionless Parameters

Here are some of the most common dimensionless numbers that show up in fluid me-

chanics:

• Re (Reynolds Number)ρV L

µ=

V L

ν– The ratio of intertial forces to frictional

forces.

• Fr (Froude Number)V√gL

– The ratio of intertial forces to gravitational forces

• We (Weber Number)ρV 2L

σ– The ratio of intertial forces to surface tension forces.

• Eu (Euler Number)∆P12ρV 2

– The ratio of pressure forces to intertial forces.

• St (Strouhal Number)fL

V– The ratio of event frequency (often vortex shedding)

and the advective frequency (inverse of the advective, or inertial, time scale).

• Ma (Mach Number)V

c– The square root of the ratio of intertial forces to

compressibility effect forces – can be thought of as a Froude number for compressible

flows.


8 Viscous Flow in Pipes

• Re < 2100 – Laminar

• Re > 4000 – Turbulent

• 2100 < Re < 4000 – Transition

where for circular pipes we define Re as ReD

ReD =ρV D

µ

These values are approximate and depend on the smoothness of the pipe wall and the

“quietness” of the initial flow (e.g., if the pump supplying the pipe is vibrating the pipe

turbulence is apt to set in at lower Re).

In general we take an energy equation perspective and use the Darcy–Weisbach equation.

∆P = γ∆z + fL

D

V 2

2g

For laminar flows we have

Eu =64

ReD

L

D

and for turbulent flows we use the Moody Chart or an explicit equation

8.1 Moody Chart

• Laminar flow f independent of ǫ/D

• Fully turbulent flow f independent of Re – inertially dominated

• Turbulent flow at moderate Re – f a function of both Re and ǫ/D

There are curve fits to the Moody chart, the most famous of which is the Colebrook

formula:


Haaland has worked out an explicit relationship that is accurate to within 2%:

1√f

= −1.8 log

ǫ

D3.75

1.11

+6.9

ReD


8.2 Three Types of Pipe Flow Problems

Type Given Find

I D, L, V, ρ, µ, g hL ⇒ ∆P

II D, L, hL, ρ, µ, g V (or Q)

III V (or Q), L, hL, ρ, µ, g D

8.3 Explicit Solutions for 3-types of Pipe Flow - Swamee & Jain

Swamee and Jain extended the concepts of Haaland to find explicit forms of the solutions

for the three types of pipe flow, like Haaland, accurate to within 2% of the Moody diagram

determined iterative solution. They are:

hL = 1.07Q2L

gD5

{

log

[

1

3.7

ǫ

D+ 4.62

(

νD

Q

)0.9]}

−2

(3000 < Re < 3×108; 10−6 <ǫ

D< 0.01)

Q = −0.965

(

gD5hL

L

)0.5

log

[

1

3.7

ǫ

D+

(

3.17ν2L

gD3hL

)0.5]

(Re > 2000)

D = 0.66

[

ǫ1.25

(

LQ2

ghL

)4.75

+ νQ9.4

(

L

ghL

)5.2]0.04

(5000 < Re < 3×108; 10−6 <ǫ

D< 0.01)

Example - pipe flow

9 Drag - (Lift is similar)

In general we rely on experiments to determine the total drag on an object (the sum of

the pressure and friction drags). We express the drag (D) nondimensionally as a drag

coefficient, CD

CD =D

12ρV 2A

where A is an appropriately chosen area with respect to the stresses generating the drag

force.


Characteristic Area

There are three standard area types used in drag coefficients:

1. Frontal Area – the area seen by the flow stream. For drag coefficients dominated

by pressure drag (e.g., blunt objects such as cylinders and spheres) this is often the

appropriate choice of area.

2. Planform Area – area as seen from above (e.g., perpendicular to flow stream). For

drag coefficients dominated by friction drag (e.g., wide flat bodies such as wings

and hydrofoils) this is often the appropriate choice of area.

3. Wetted Area – area of wetted surface. Often used for boats and other surface water

vessels where friction drag is dominant.

10 Drag Coefficient Dependencies – Geometry, Re

& Roughness

In general CD = φ(Re, Fr, ǫ/L, Ma, Geometry). There are nomographs and tables for

regions where the CD is about constant (higher Re).

Example – Drag Consider a parachute with diameter 4 m carrying a person and gear such

that the total mass is 80 kg. If the parachuter jumps from high enough that terminal


velocity (e.g., steady state - no acceleration) is achieved, what is this velocity?

U=8.6 m/s

11 Open-Channel Flow

In uniform flow y1 = y2, V1 = V2 = V , therefore the energy equation becomes:

z1 − z2 = S0L = hf

Manning’s Equation

V ≈ α

nR

2/3

h S1/20 and Q ≈ α

nAR

2/3

h S1/20


where

Rh =A

P = hydraulic radius

where (P) is the length of the wetted perimeter, A is the wetted area, and n is Manning’s

n which varies by a factor of 15 and is tabulated.

Example - Open Channel Flow

1 dimensions · bee 5330 –fluids fe review, feb 24, 2010 5 2.5 surface tension the water molecule...

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