Integral Relation for a Control Volume (Part 1)

Department of Mechanical Engineering MEHB223 Mechanics of Fluids 1: Lecture 5: Integral Relations for CV (Part 1)

Lecture 5 Integral Relation for a Control Volume (Part 1)


Integral Relation for CV : Continuity Equation

1


Lecture Summary • Flow Rates and Discharge • Control Volume Analysis : General Equation • Mass Conservation Equation (MCE)


5.1 Introduction

•

u mo ons are govern y conserva on laws ‐ mass, momentum and energy conservation

• Basic concepts for analyzing fluid motion have been described in last lecture

• General control volume analysis • Application to conservation of mass

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5.2. Flow Rates and Discharge Volume Flow Rate :

•

∀

m3 / s

dA u

Af

Velocit •

• Volume flow rate across dA :

δ∀ =

δ∀ uδt δ A = = uδ A δt δt

•

• Total Volume flow rate :

ro ile

∀ = ∑ uδ A = A f

∫ udA

A f

• Note - Variation of u with A (equation of velocity profile ) needed to evaluate the volume low rate.


5.2. Flow Rates and Discharge Volume Flow Rate : Example 1 : Air low between 2 arallel lates 80 mm a art. The ollowin velocities 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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5.2. Flow Rates and Discharge Volume Flow Rate : (Practical Application)

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



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Average velocity for every j-th sector

U

=

1

π R

2π R

2

∫ ∫ 0

0

v (r ,α ) r dr dα

Flow Rate in eight sectors of measurement cross-section

Q*

=

n

∑ ( π n R

2 pj

⋅ Us j )

j =1

[m3 / s]


5.2. Flow Rates and Discharge Volume Flow Rate : •

∀=

_

_

_

∫ u dA = u ∫ dA = u A

f

A f

A f

Mean velocit : •

_

u=

∀ A f

=

1

∫ udA

A f A f

The uniform value of velocity which corresponds to the same volume flow rate as in real low

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5.2. Flow Rates and Discharge Mass Flow Rate : Definition : -

•

k / s

m •

δm = •

m=

• δm δ∀ = ρ = ρδ ∀ = ρuδ A δt δt

∑ ρuδ A = ∫ ρudA A f

Incompressible flow : -

A f

•

m=ρ

u

=ρ

A f

u

=ρ

•

A f

Incompressible and uniform flow : •

_

m = ρu

_

_

∑ δ A = ρ u ∫ dA = ρ u A f A f

A f


5.2. Flow Rates and Discharge •

X [X / s]

Flow Rate of any extensive property :

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.

δX = xδm •

•

δ X = xδ m = xρudA Thus the flow rate :

•

X=

∑

A f

xρudA =

∫ xρudA

A f

To evaluate requires to know variation of u, ρ and x with A f

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5.2. Flow Rates and Discharge 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 : -

∫ xρudA = x ∫ ρudA

A f

A f

∫ xρudA X x= = ρ udA m ∫ •

_

A f

•

A f


5.2. Flow Rates and Discharge Cylindrical Coordinates : Commonly used in engineering to describe axi-symmetric flow eg. Flow in circular pipe. Problem becomes 2D. Elemental area, dA

δA = 2πr δr The flow rates for cylindrical axi-symmetric flow are : •

∀ = 2π ∫ urdr A f

•

m = 2π

∫ ρurdr

A f

•

∫

X = 2π xρurdr A f

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5.2. Flow Rates and Discharge Example of extensive properties flow-rate : Extensive property, X

•

Kinetic Energy, KE

KE =

u2 /2

∫ ρu

u2

A f

•

Momentum in x direction : Mx

u x

Momentum in y direction : My

u y

Enthalpy, H

h

Equation for axi-symmetric flow

General flow rate equation

Value per unit mass, x

x

2

•

dA

∫

KE = πρ ru 3 dr A f

•

=

x

2 x

A f

•

My

A f

= ∫ ρuu y dA

M y = 0

∫ ρuhdA

H = 2 πρ ruhdr

•

A f

•

H=

A f

•

∫

A f


5.2. Flow Rates and Discharge The velocity profile for a pipe flow is given by :

r ⎞ = 1 − ⎛ ⎜ ⎟ uo ⎝ R ⎠ u

2

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

• Mass flow rate • x- momentum flow rate • Mean value of x-momentum

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5.2. Flow Rates and Discharge 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


5.3. General 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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5.3. General Control Volume Equation General equation derivation : 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.

δ X i ,1 CV boundary at time t and t+dt

X δ X o,1

CS boundary at time t

δ X o, 2

CS boundary at time t+dt

δ X i , 2


5.3. General Control Volume Equation General equation derivation : Consider CS and S at time t (initially) and t+dt (final)

Time, t X c,t

X s,t

X c,t+dt

X S,t+dt

Control Volume, C

System, S

Time, t+dt

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5.3. General Control Volume Equation General equation derivation : XC ,t

= X S, t

At timet (initially) : X C , t + dt = X S, t + dt + δX i ,1 + δX i , 2 − δX o ,1 − δX o ,1 At time t+dt (final) : Rate of change of X in CV :

• • δX C X C,t + dt − X C,t δX S = = + ∑ Xi − ∑ Xo δt δt δt

dX C dt

=

dX S dt

•

•

+ ∑ Xi − ∑ Xo

or for non-uniform flow : d

ρxd ∀ = dt ∫

dXS

CV

dt

+

∑ ∫ ρuxdA − ∑ ∫ ρuxdA i

All _ inf low A i

o

All _ outflow A o


5.3. General Control Volume Equation Steady Flow CV equation : For steady flow - properties do not change with time at all points in the CV Thus : dX C dt

=0

Hence, for steady flow equation becomes : S

dt

•

•

= ∑ X o − ∑ Xi

S

dt

or

=

∑ ∫ ρuxdA

All _ outflow A o

o

−

∑ ∫ ρuxdA

i

All _ inf low A i

Note that dX s /dt is related to the system based physical law to other quantities. Eg. When X = momentum, dX s /dt is related to external force by the Newton’s 2nd law

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5.4. The Mass Conservation Equation (MCE) Derivation of the MCE : Let : X = mass = m General CV Equation becomes :

dm C

By the law of conservation of mass applied to system :

dm S

us t e

app e to

Or for general non-uniform flow : For constant density flow:

dt dt

s : d

dm C dt

=

•

dm S

•

+ ∑ mi − ∑ mo

dt

=0 •

=

− i

• o

∫ ρd ∀ = ∑ ∫ ρudA − ∑ ∫ ρudA

dt CV

i

All _ inf low A i

d ∀C dt

o

All _ outflow A o

•

•

= ∑ ∀i − ∑ ∀ o


5.4. 5. 4. The Mass Conservation Equation (MCE) Steady Flow MCE : for steady flow : dmc /dt = 0 •

•

∑m −∑m i

Thus the equations become :

=0

∑ ∫ ρudA − ∑ ∫ ρudA i

All _ inf low A i

•

Notes on pp ica i ity :

o

∀i −

o

=0

All _ outflow A o

•

∀o = 0

- Single fluid with or without change of phase - Mixture of fluids with or without chemical reaction - Mixture of fluids with solid particulates - If without reaction - applicable to individual species as well as total flow.

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5.4. 5. 4. The Mass Conservation Equation (MCE) Example 1 : Water flows in the pipe system as given in figure below. At C, the pipe 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.


5.4. 5. 4. The Mass Conservation Equation (MCE) 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 h f =30cm ?

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5.4. 5. 4. The Mass Conservation Equation (MCE) 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 V out = √ (2gh) m/s. What is the equilibrium height heq of the liquid in the tank ?


5.4. 5. 4. The Mass Conservation Equation (MCE) Example 4 : Water drain out of a trough as shown. The angle with the vertical of the sloping sides is α , and the distance between the parallel sides is B. The width of the trough is W o+2htanα , 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 water stream at the bottom is A e. Derive an expression for the time to drain to depth h in terms of h/ho , W o /ho , tan α and Aeg0.5 /(ho1.5 B), where ho is the original depth. Find the time to drain to one half the original depth for W o /ho=0.2, α =30o , Aeg0.5 /(ho1.5 B)=0.01s-1

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End of Lecture 5

15

Integral Relation for a Control Volume (Part 1)

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