Energy in Steady Flow

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Energy in Steady Flow

Motion of a real fluid, Review

A

P

Dec. 18, 2013 Dr. Celik

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General Energy Equation, steady flow, incompressible fluid

Review

Change in the internal energy per unit

weight of a fluid (e.g. due to friction,

hL=hf) coincides with change in

temperature:

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Power considerations in fluid flow: Review

P = γhQ

Horsepower = P = γhQ/550

( [Q] = cfs, [h] = ft, [γ] = pcf )

Power in BG units

Kilowatts = P = γhQ/1000

( [Q] = m3/s, [h] = m, [γ] = N/m3 )

Power in SI units

Pump efficiency,

ηpump = (γhpumpQ ) / (power input)

P : power put into flow by a

pump, then h = hpump

P : power extracted from

flow by a turbine, then

h = hturbine

Turbine efficiency,

ηturbine = (power output) / (γhturbineQ )

P : power lost because of friction,

then h = hL

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Energy and Power considerations in fluid flow, Stories

Why are electricity rates

cheaper at night in some

places?

http://www.planetware.com/ http://www.energymanagertraining.com

How can I transfer fermented

beer (homebrew) over to

another vessel in a way that

does not expose it to air or

other contaminants?

Because at night (low

demand) the generated energy

will be otherwise wasted.

You can use a simple siphon. http://www.planetware.com/

How should I eliminate

cavitation in a pipe flow?

You can use a control valve.

http://www.valvedirectory.com

or reduce the operating

temperature.

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Power considerations in fluid flow Story

There are

approximately 95,000 miles

nationwide of refined

products pipelines

It has been estimated that

industrial pumping systems

account for nearly 25% of

industrial

electrical energy demand in

the United States.

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Definition of Hydraulic Grade Line and Energy Line

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Definition of Hydraulic Grade Line and Energy Line

Real fluid

EL (also called

Energy Grade Line,

EGL) is positioned

above the HGL by

an amount equal to

velocity head (V2/2g)

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Tips for Drawing HGL & EL

1) As the velocity goes to zero, the HGL and the EL approach each other.

Thus, in a reservoir, they are identical and lie on the surface.

2) The EL and HGL slope downward in the direction of the flow due to the head loss in the pipe

(exceptions: pump, diffuser). For steady flow in a pipe, the slope will be constant (ΔhL/ΔL =

const.)

3) A jump occurs in the HGL and the EL

whenever energy is added to the fluid as

occurs with a pump, and a drop occurs if

energy is extracted from the flow, as in the

presence of a turbine.

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5) At points where the HGL passes through the centerline of the pipe, the pressure is zero. If

the pipe lies above the HGL, there is a vacuum in the pipe, a potential location for cavitation,

also a condition that is often avoided, if possible, in the design of piping systems; an

exception would be in the design of a siphon.

4) When a flow passage changes diameter, the distance between the EL and HGL will

change, because velocity changes. In addition, for real fluids, the slope of the EL will

change because the head loss per length will be larger in a conduit with larger flow

velocity.



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6) When a fluid discharges with velocity V from the end of the pipe into a tank reservoir

that is so large that the velocity within it is negligible, the entire kinetic energy of the flow

dissipates. This situation is called the loss of head at submerged discharge.



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Problem 2

In the pipe: hL = 1.6V2/2g

patm = 90 kPa

For the liquid in the suction

pipe, V = 1.8 m/s

What is the maximum allowable

value of z if the liquid were

a) Water at 20° C

b) Gasoline with pv = 49 kPa

abs with a specific weight of

8 kN/m3

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Problem 2 , Solution

2) For water at 20° C, γ = 9.789

kN/m3 and pv = 2.34 kN/m2 abs

1) Apply energy equation between

A and B. When maximum z is

considered, to prevent cavitation,

pb abs = p min = pv

zmax = (90-pv)/γ – 0.264

zmax = 8.69 m

3) For gasoline with pv = 49 kPa

abs and γ =8 kN/m3

zmax = 4.86 m

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Problem 1

1

2

pump 10 m

L = 5000 m

Q

D = 0.5 m

40 m

hL = 0.01(L/D)V2/2g

Q = 3 m3/s

What is the power

(hp) supplied to the

flow?

Water

5 m

Draw the HGL and

EL for the system.

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Problem 3

A piping system with a black box shows a large EGL change at the box

(steady flow-uniform diameter).

What is the flow direction?

What could be in the “black box”? A pump, or a turbine?

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Problem 4

The EGL and HGL are as shown for a certain system.

Find:

(a) Direction of flow.

(b) Whether there is a reservoir.

(c) Whether the diameter at E is uniform or variable.

(d) Whether there is a pump.

(e) Sketch a physical set up that could exist between C and D.

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Problem 5

Two tanks are connected by a uniformly tapered pipe. Draw the HGL and EGL.

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Momentum and Forces in a Fluid

dt

VmdF S)(

Newton’s 2nd law:

The sum of external forces on a body of fluid or system is equal to the

rate of change of linear momentum of that body (of the fluid or system)

Change in momentum must be in the same direction as the force (acting

on the fluid)

Momentum principle is important in flow problems where we

need to determine the forces.

Examples: forces on vanes, pipe bends, the thrust produced by

a rocket or turbojet, torque produced by a hydraulic turbine

http://www.flowtite.com/

By the law of action and reaction, the fluid exerts an equal and opposite

on the body (boundary, wall) that is producing the change (in velocity)

Such forces occur whenever the velocity of a stream of fluid

changes in direction and/or magnitude

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Momentum Principle

dt

VmdF

)(

Newton’s 2nd law:

The sum of external forces on a body of fluid or system is equal to the

rate of change of linear momentum of that body (of the fluid or system).

Change in momentum must be in the same direction as the force

dt

mVd

dt

mVd

dt

mVd

dt

mVdF

in

CV

out

CVCVS )()()()(

dt

mVd

dt

mVd

dt

mVdF

in

CV

out

CVS )()()(

0 for steady flow

For steady flow, the net force on the fluid mass

is equal to the net rate of outflow of momentum

across the control surface

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Momentum Principle

dt

mVd

dt

mVdF

in

CV

out

CV )()(

Control surface is normal to the

velocity at sections 1 and 2

Velocity is constant across

the control surface

1111111 )()(

VQVmVdt

md

dt

mVd 222

2)(VQ

dt

mVd

111222 VQVQF

QQQm 1122 for steady flow

)()( 1212 VQVVQQVQVF

1

2

VV

VV

in

out

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Momentum Principle , Force and Velocity Considerations

)()( 1212 VQVVQQVQVF

Same direction!

Summation of all forces

acting on the fluid mass

including:

gravity forces (!),

shear forces,

pressure forces exerted by

surrounding fluid and by

solid boundary

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Momentum Principle, Scalar Components

)()( 12 VVQVQF Vectorial

)()( 12 xxxx VVQVQF

ΔV

ΔV ΔVy

ΔV

ΔVx

)()( 12 yyyy VVQVQF

Scalar:

Force components applied on the

fluid. The force of same magnitude but

opposite direction is the one applied

by the fluid on its surroundings

x

y

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Momentum Principle, Comments

inout QVQVF

1)

)( 3355 VQVQF

1

3

)( 442211 VQVQVQ

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Momentum Principle, Comments

2) Draw a suitable system of axes. It is important to establish positive x & y

directions for vectorial relationships.

3) Draw a suitable control volume(CV). The CV should include all the changes

that the velocity experiences (i.e. magnitude and direction)

4) For the summation of forces, all the external forces applied on the CV should

be considered. If we have Fx, then the x component of all the forces should be

included. Gravity is the force applied in the vertical direction

5) The result we find from the momentum principle is a force applied upon the CV.

Most often we are asked to determine the forces applied by the fluid. This is equal

in magnitude but opposite in direction compared to the former force.

6) We don’t need to know the energy losses or we don’t care how complicated the

flow field is when dealing with the momentum. We only need to know the

conditions at the end sections of our CV (and gravity forces).

7) The direction of the forces in x and y axes applied by the fluid are not known

before hand.

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Pressure Conduits

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Momentum Principle, Problem

The flow turns by 90° in a horizontal plane before it discharges into the open.

Determine the forces, Fx and Fy applied by the flow upon the pipe. ρ = 1000 kg/m3

1

2 D2 = 0.5 m Q = 1.5 m3/s

p1 = 30 kPa

D1 = 1 m

F ’x= 49,987.3 N

F’y = 11,459 N

Forces applied

upon the pipe:

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Momentum and Forces in Fluid Flow

Example: Force applied on a spillway

2.1 m

0.6 m

Flow

Spillway

Find the force applied by the flow on the spillway

(assume that the width normal to the page is 1 m,

ρ = 1000 kg/m3)

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Example: Force applied on a spillway

1 2

2.1 m

0.6 m

CV

FH1

FH2

Ff

x

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Free Jet

Gage pressures in the jet are “0”

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Choice of alternative control volumes (Example: Vertical reducing

section)

V1

V2

CV 1

p2

p1

W

Neglect losses but include gravity, determine the force (F) on the

contraction.

z

F

)( 122211 VVQFApApWFz

hg

VVpp

hg

Vp

g

Vp

2

22

2

2

2

112

2

22

2

11

h

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Choice of alternative control volumes (Vertical reducing section)

V1

V2

CV 2

p’2

p1

W’

z

h

h’ F

)'(2

' 2

2

2

112 hhg

VVpp

'' 22 h

pp

''

)(''

2

122211

hAWW

VVQFApApWFz

)(

)()'()'(

122211

1222112

VVQFApApWF

VVQFAhpAphAWF

z

z

Same as

CV 1

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Example: Water jet

(Horizontal plane) A water jet is aimed at the hole in the wall; 25% of the water escapes through the hole, and the remaining water is turned 90° to the

water jet. Find the force on the wall by the jet. ρ = 1000 kg/m3

0.4 m

(diameter)

V1=20 m/s V2=20 m/s

V3

V4

x

Discussion: What If this was a vertical plane? (Vyin =0, Vyout=V3 and V4)

Energy in Steady Flow

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