integral relation for a control volume (part 1)

8/11/2019 Integral Relation for a Control Volume (Part 1)

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Lecture Summary

Mechanics of Fluids 1: Lecture 5: Integral Relations for CV (Part 1)

Department of Mechanical Engineering MEHB223

FlowRatesandDischarge

ControlVolumeAnalysis:GeneralEquation

MassConservationEquation(MCE)

5.1

Introduction



u mo onsaregovern y conserva onlawsmass,momentumandenergy

conservation

Basicconceptsforanalyzingfluidmotionhavebeendescribedinlastlecture

Generalcontrolvolumeanalysis

Applicationtoconservationofmass


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55..22.. Flow Rates and DischargeFlow Rates and DischargeVolume Flow Rate : s/3m



udA

Af

Velocit ro ile

Volume flow rate across dA :

Total Volume flow rate :

Note - Variation of u with A (equation of velocity profile) needed to

evaluate the volume low rate.

Au

t

Atu

t

=

=

=

==

ff AA

udAAu

Volume Flow Rate : Example 1 : -Air low between 2 arallel lates 80 mm a art. The ollowin velocities

55..22.. Flow Rates and DischargeFlow Rates and Discharge



were determined by direct measurement : -

Distance from one plate (mm) 0 10 20 30 40 50 60 70 80

Velocity (m/s) 0 23 28 31 32 29 22 14 0

Plot the velocity distribution and calculates the discharge


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Volume Flow Rate : (Practical Application)55..22.. Flow Rates and DischargeFlow Rates and Discharge



HPP OBALT3 Kaplan turbines,rated power: 3 x 20,4 MW,maximum head 17,3 m,maximum discharge 3 x 137 m3/s

Volume Flow Rate : (Practical Application)





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Volume Flow Rate : (Practical Application)55..22.. Flow Rates and DischargeFlow Rates and Discharge



Average velocity for everyj-th sector

U

R

v r r dr dR

=

12

00

2

( , )

[ ]Qn

R Up j s jj

n* ( )=

= 2

1

m / s3

Flow Rate in eight sectors of measurementcross-section

Volume Flow Rate :




-

Mean velocit : -

f

_

A

_

A

_

AudAudAu

ff

===

=

=

fAff

_

udAAA

u 1

The uniform value of velocity which

corresponds to the same volume flow rate

as in real low


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Mass Flow Rate : s/km




Definition : -

Incompressible flow : -

Att

mm ==

=

=

u

==

ff AA

udAAum

Incompressible and uniform flow : -

===ff A

A

uum

f

___

Auuum ===

ff AA

dAA

Flow Rate of any extensive property : [ ]s/XX




Recall : Extensive property is property that varies directly

proportional to the mass of the system e.g. momentum, kinetic

energy, enthalpy etc.

Define x = X/m : the intensive / specific property of X.

mxX =

Thus the flow rate :

To evaluate requires to know variation of u,and x with Af

udAxmxX ==

==

f

f

AA

udAxudAxX


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Flow Rate of any extensive property :




Mean value, x , of any extensive property X is defined as the

uniform value of X which would give the same flow rate as the

real flow .

Define as : - =ff A

_

A

udAxudAx

=

=

m

X

udA

udAx

x

f

f

A

A_

Cylindrical Coordinates :




Commonly used in engineering to describe axi-symmetric flow

eg. Flow in circular pipe. Problem becomes 2D.

rr2A =Elemental area, dA

The flow rates for cylindrical axi-symmetric flow are : -

=

fA

urdr2 =

fA

urdr2m =

fA

urdrx2X


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Example of extensive properties flow-rate :




Extensive

property, X

Value per unit

mass, x

General flow

rate equation

Equation for

axi-symmetric

flow

Kinetic Energy,

KE

Momentum in x

u2/2 =

fA

2

dA2

uuKE

=

=

fA

3drruKE

2

direction : Mx

Momentum in ydirection : My

Enthalpy, H

ux

uy

h

fA

xx

=

fA

yy dAuuM

=

fA

uhdAH

fA

x

0My=

=

fA

ruhdr2H

-




-

The velocity profile for a pipe flow is given by :

where uo is the maximum velocity, R the radius of the pipe. Determine :

2

o R

r1

u

u

=

Mass flow rate

x- momentum flow rate

Mean value of x-momentum


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Example 3 : -




Air enters a square duct at section 1 with the velocity distribution as shown.

The velocity varies in y direction only. Determine : -

the volume flow rate

the mean velocity

the mass flow rate if the density is 1.3 kg/m3

55..33.. General Control Volume EquationGeneral Control Volume Equation

General equation derivation : -



Basic laws (e.g. Mass conservation, Newtons 2nd law and 1st

law of thermo) are applicable and formulated in term of

SYSTEM i.e using LAGRANGIAN View point

Thus for EULERIAN viewpoint (i.e. with CV fixed in space)

need to be reformulated.

Recall :

Control Volume (CV) : Any volume of constant shape, size,

position and orientation with respect to an observer. Can be

finite size or infinitesimal element

Control surface (CS) - The surface of CV. Mater as well as heat

and work can cross the control surface.


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Consider a CV of with multiple inflows and outflows.

Consider what happen to the CS and the system boundary (S)

initially and after fluid has entered and leave the CV.

1,iX

CV boundary at

time t and t+dt

2,iX1,oX

2,oX

CS boundary at

time tCS boundary at

time t+dt

X





Consider CS and S at time t (initially) and t+dt (final)

Time, t

Xc,t Xs,t

Time,

t+dt

Xc,t+dt XS,t+dt

Control Volume, C System, S


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General equation derivation : -55..33.. General Control Volume EquationGeneral Control Volume Equation



At timet (initially) :

At time t+dt (final) :

Rate of change of X in CV :

t,St,C XX =

1,o1,o2,i1,idtt,Sdtt,C XXXXXX ++= ++

t

XX

t

X t,Cdtt,CC

=

+

oi

S XXt

X

+

=

-

or for non-uniform flow :oi

SC XXdt

dX

dt

dX

+=

+=outflow_All A

o

lowinf_All A

iS

CV oi

uxdAuxdAdt

dXxd

dt

d

Steady Flow CV equation : -




For steady flow - properties do not change with time at all

points in the CV

Thus :

Hence, for steady flow equation becomes : -

0dt

dXC =

Note that dXs/dt is related to the system based physical law to

other quantities. Eg. When X = momentum, dXs/dt is related to

external force by the Newtons 2nd law

ioS XX

dt = = lowinf_All A ioutflow_All A oS

io

uxdAuxdAdtor


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Example 1 : -Water flows in the pipe system as given in figure below. At C, the pipe

5.5.4.4. The Mass Conservation Equation (MCE)The Mass Conservation Equation (MCE)



branch into 2 parts. The flow rate divides between the branch such that the

discharge at D is twice that at E. Determine : -

The volume flow rate and mean velocity at A

The volume flow rate at B

The diameter of pipe CD

The men velocity in pipe CE.

Example 2 : -




cm jet o water issues rom a cm iameter tan , as s own. ssume

that the velocity in the jet is (2gh) m/s. How long will it take for the water

surface in the tank to drop from ho=3m to hf=30cm ?


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Example 3 : -




T e open tan s own as a constant in ow isc arge o m s. -m

diameter drain provides a variable outflow Vout= (2gh) m/s. What is the

equilibrium height heq of the liquid in the tank ?

Example 4 : -

Water drain out of a trough as shown. The




angle with the vertical of the sloping sides is

, and the distance between the parallel sidesis B. The width of the trough is Wo+2htan,where h is the distance from the trough

bottom. The velocity of the water issuing from

the opening in the bottom of the trough is

e ual to Ve=2 h . The area o the waterstream at the bottom is Ae. Derive anexpression for the time to drain to depth h in

terms of h/ho, Wo/ho, tan and Aeg0.5/(ho1.5B),where ho is the original depth. Find the time

to drain to one half the original depth for

Wo/ho=0.2, =30o, Aeg0.5/(ho1.5B)=0.01s-1


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End ofEnd of Lecture 5Lecture 5

integral relation for a control volume (part 1)

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