Mod-02 Problem Formulation 2009

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© 2006, 2009 F. A. Kulacki Conduction Heat TransferME8341-02-1

Problem Formulation

in Conduction Heat

Transfer

Formulation of general laws


Overview

• Determined andundetermined problems

• Lumped formulation

•

Integral formulation• Differential formulation



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Determined and

undetermined

problems


Example 1: Steady one-dimensional

flow of an inviscid fluid in an adiabatic

diffuser

• Enthalpy is conserved.

• Flow is isentropic.

• Apply massconservation.

• Apply the linearmomentum equation(Newton’s 2nd Law ofMotion).

2

A

u

V

p

ρ

1

A

u

V

p

ρ

CV

V δ



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Example 1 (continued)

2

A

u

V

p

ρ

1

A

u

V

p

ρ

CV

V δ

( )

+=

⋅

+−= ∫∫

ρ

Ω ρ ρ

pu

dx

d

d nV p

E CV

CVS

0

0

Energy conservation

( ) ( )

( )

( )

( ) ( )

( )[ ]V AV pAdx

d

d nV V n p

d nV V d n

AV dx

d

AV AV

d nV

CV

CVS

CV

CVS

CV

CVS

CV

CVS

x x x

cv

CVS

ρ

Ω ρ Ω

Ω ρ Ω

ρ

ρ ρ

Ω ρ

∆

+=

⋅−−=

⋅−⋅ℑ=

=

+−=

⋅−=

∫∫ ∫∫

∫∫ ∫∫

∫∫ +

0

0

0

0

0

0

Mass conservation

Linear momentum


Example 1 (continued)

2

A

u

V

p

ρ

1

A

u

V

p

ρ

CV

V δ

+=

ρ

10 pd du

Incompressible fluid – Density is constant, and du = 0.The problem is completely determined and theequations can be solved.

Compressible fluid – The problem is undeterminedand cannot be solved. An equation of state is theadditional condition that is needed. Pressure is afunction of density, p =p(ρρρρ,u), e.g., p = ρρρρRT, and theproblem can be solved.



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Example 2. First law of

thermodynamics

x

z

y

( )t E

For a lumped formulation, the FirstLaw of Thermodynamics is written,

Γ &&& +−= W Qdt

dE

( )t E

t

( )it E

Problem is undetermined unlessrelations between heat transfer,

work, and energy generation withtime or another variable are known.


Lumped formulation

of general laws



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Lumped system configuration

with control volume

Control volume

System

Initial

State,B1,B’

Final

State,

B2, B’’.

iii ,m ,b ϑ ∆∆ Note: “b” is the specific value of the quantityof interest (“property/mass”).


Control volume, σσσσ

System

InitialState,B1,B’

FinalState,B2, B’’.

iii ,m ,b ϑ ∆∆

B Bmb B B

B B B

ii

system

′′=+′=

−=

2

1

12

∆

∆

∑

∑

∑

−=

−=

∆−∆=∆

∆=∆−∆

σ

σσ

σ

σ

i

ii

i

ii

system

i

iisystem

systemii

wbdt

dBdt

dmb

dt

dB

dt

dB

mbBB

BmbB

Combine system and controlvolume relations to get the

transformation between the two.

This expression is the general transformation

from system to control volume formulation.



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Conservation of mass

∑−==

===

=

i

iwdt

dm

dt

dm

m

Bb ,m B

dt

dm

σ 0

1

0Control volume, σσσσ

System

InitialState,B1,B’

FinalState,B2, B’’.

iii ,m ,b ϑ ∆∆


Conservation of energy – First

law of thermodynamics

nuclchem

by

by

cycle cycle

U U mgzmV

U E

PqW Q E

W Q E

W Q

++++=

−≡−=

−=

=∫ ∫

2

2

&&&

δ δ ∆

δ δ

S1

S2



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Conservation of energy

esd

i

ii

by

PPPP

Pqwedt

dE

eb , E B

PqW Q E

++=

−=−

==

−≡−=

∑σ

&&&


System

σ iP

σ eP σ d P

σ sP

( )

i ,eee

ss

id d

PPP

PP

PPP

ϑ ∆σ

σ

σ

+=

=

+=Include in internalEnergy flow acrossThe CVS.

Becomes usual expressionFor “flow work” underAssumption of normal forcesAt CVS, i.e., no shear at CVS.


Conservation of energy

Include in the internalenergy flow for the CV

i

i

ii

qqq

Pqwedt

dE

ϑ ∆σ

σ

+=

−=−∑


System

σ iP

σ eP σ d P

σ sP

Combining terms for q and P,

( )σ σ

σ

ρ esd i

ii

PPPqw p

edt

dE ++−+

+= ∑

0ih=



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Integral formulation

of general laws


System and fixed control

volume at t

x

z

y

System at time t

Fixed controlVolume, σσσσ

Flow field,

V(x,t)

Properties and the velocity field are specified in time and space.

The system can move and deform in time and space with V(x,t)



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volume at t + ∆∆∆∆t

x

System at time t+∆∆∆∆T

V

n

z

y

Fixed control

Volume, σσσσ

x


V

n

System at time t+∆∆∆∆T

2

3

z

y

Fixed control

Volume, σσσσ

x


volume at t + ∆∆∆∆t

1

Determine change of B in

system regarding regions

1 to 3 via finite calculus.



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Determine change of B insystem regarding regions

1 to 3 via finite calculus.

Then take the limit as ∆∆∆∆t →→→→ 0.

( ) ( )

( ) ( ) ( ) ( )

t

t B

t

t t B

t

t Bt t B

dt

dB

t

B B B B

dt

dB

system

t t t

system

∆∆

∆

∆

∆

∆

∆

2311

2131

−+

+−+

≈

+−+≈

+

V

n

2

3

z

y

1

x

( ) ( ) ( )ϑ

∂

ρ∂=ϑρ=

=

∆

−∆+∫ ∫ →∆

dt

bdb

dt

d

dt

dB

t

tBttBlim

CVCVCV

11

0t

( ) ( ) ∫∫ ∫∫ ∫∫ σσσ

→∆σ⋅ρ=σ⋅ρ−σ⋅ρ=

∆

−∆

∆+ dnVbdnVbdnVbttB

tttBlim

in,out,

23

0t

Rate of change of B in the CV

Rate of change of B in the CV due to material flow at the boundary, σσσσ


( ) ( ) ( ) ( )

( ) ( )CVsystem

2311

tsystem

dnVbdtb

dtdB

t

tB

t

ttB

t

tBttBlim

dt

dB

σ⋅ρ+ϑ∂ρ∂=

∆−

∆

∆++

∆

−∆+=

∫∫∫ ∫∫

→∆

V

n

2

3

z

y

1

x

Combine terms to obtain the

Relation between the rate ofchange of B in the system andcontrol volume.

This development is heuristic, or“physically based”. The resultis the same as for the Reynolds

Transport Theorem.

This result is used to produce the general laws as in the case of thelumped formulation (mass, energy, entropy).



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( )( )

CVsystem

CVsystem

dnVdt

0dtdm

1b,mB

dnVbdt

b

dt

dB

σ⋅ρ+ϑ∂ρ∂==

==

σ⋅ρ+ϑ

∂

ρ∂=

∫∫∫ ∫∫

∫∫∫ ∫∫

System mass stays the same following fluid motion in

the assumed flow field.




( )( )

( )CVsystem

CVsystem

dnVedt

e

dt

de

eb,EB

dnVbdt

b

dt

dB

σ⋅ρ+ϑ

∂

ρ∂=

==

σ⋅ρ+ϑ

∂

ρ∂=

∫∫∫ ∫∫

∫∫∫ ∫∫

Begin by consider the system statement of the first law of thermodynamics.

Note that this “general law” remains valid for stationary systems and systems

that follow fluid motion.

systemsystemsystem dt

W

dt

Q

dt

dE

δ−

δ=



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( )

∫∫

∫∫∫ ∫∫

σ

σ⋅−=

δ

δ−

δ=

σ⋅ρ+ϑ

∂ρ∂

dnqdt

Q

dtW

dtQdnVed

te

system

systemsystemCV

Here the heat flux, q, is introduced

at the control surface. In the limit

of ∆∆∆∆t →→→→ 0, the integration approachesthat for the system derivative following

fluid motion.

Next, we consider the work term, where displacement, shaft and electricalwork are formally present. In addition, work done at the surface against

viscous forces (shear) is included. Thus the work term is re-written as,

τ

δ+

δ+

δ+

δ=

δ

dtW

dtW

dtW

dtW

dtW

esdsystem




τσ

σσ

τ

τ

δ+ϑ′′′−+

δ+σ⋅ρ

ρ−=

δ

ϑ′′′−=

σ⋅ρ

ρ−=σ⋅=

−−−−=

δ+

δ+

δ+

δ=

δ

∫∫∫ ∫∫

∫∫∫

∫∫ ∫∫

dt

Wdu

dt

WdnV

p

dt

W

duP

dnVp

dnVpP

PPPP

dt

W

dt

W

dt

W

dt

W

dt

W

CVssystem

CV

e

d

esd

esdsystem

(Displacement work)

(Electrical work drawn to system; generation term)



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( )( )

( ) ( ) ϑ′′′+−−σ⋅−=σ⋅ρ+ϑ∂

ρ∂

ϑ′′′+−−σ⋅ρ

ρ−σ⋅−=σ⋅ρ+ϑ

∂

ρ∂

δ+ϑ′′′−+

δ+σ⋅ρ

ρ

−=

δ

∫∫∫ ∫∫ ∫∫ ∫∫∫

∫∫∫ ∫∫ ∫∫ ∫∫ ∫∫∫

∫∫∫ ∫∫

τ

σσ

τ

σσσ

τσ

duPPdnqdnVhdt

e

duPPdnVp

dnqdnVedt

e

dtWdu

dtWdnVp

dtW

CV

s

0

CV

CV

s

CV

CVssystem

Combine all terms and substitute into the first law statement to obtain

the integral energy balance.


Second law of thermodynamics

T

qS

T

QS

0T

Q

21

2121

cycle

≥

δ≥∆

≤δ

−

−−

∫

&

( ) ( )

( )( ) ∫∫∫ ∫∫ ∫∫ ∫∫∫

∫∫ ∫∫ ∫∫∫

∫∫

ϑ′′′+σ⋅

−=σ⋅ρ+ϑ

∂

ρ∂

σ⋅

−≥σ⋅ρ+ϑ

∂ρ∂=

σ⋅

−=

σσ

σσ

σ

CVCV

CVsystem

dsdnT

qdnVsd

t

s

dnT

qdnVsdt

s

dt

dS

dnT

q

T

q

Begin with the Clausius inequality andapply it to a process in the usual way.

the heat flux is evaluated at the surfaceof the C.V.

We then apply the system derivative anduse the result to relate system changes

to that in the C.V.



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Differential

formulation of

general laws


Methods

• Use integral formulation with thedivergence theorem – Reduces integral expression to a P.D.E.

• Use differential volume elementswith mean value theorems of the

calculus – Produces a P.D.E. from general principles.



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( )

( ) Vdt

d0,Vt0

dVt

0

dnVdt

0

CV

CV

⋅∇ρ+ρ

=ρ⋅∇+∂

ρ∂=

ϑ

ρ⋅∇+

∂

ρ∂=

σ⋅ρ+ϑ∂ρ∂=

∫∫∫

∫∫∫ ∫∫


Conservation of energy – First law

of thermodynamics

( ) ( )

( )[ ]

( ) uVpqdt

de

duVpq

duPPdnVpdnqddt

de

duPPdnqdnVhdt

e

CV

CV

s

CV

CV

s

0

CV

′′′=⋅∇+⋅∇+ρ

ϑ′′′−⋅∇+⋅∇−=

ϑ′′′+−−σ⋅−σ⋅−=ϑρ

ϑ′′′+−−σ⋅−=σ⋅ρ+ϑ∂

ρ∂

∫∫∫

∫∫∫ ∫∫ ∫∫ ∫∫∫

∫∫∫ ∫∫ ∫∫ ∫∫∫

τ

σσ

τ

σσ

Here we specialize the

energy balance to ideal fluids with no

shaft work.

eVt

e

dt

de∇⋅+

∂

∂=

(rearranging and combining)



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( )

upVqdt

de

.Const

uVpqdt

de

′′′=∇⋅+⋅∇+ρ

=ρ

′′′=⋅∇+⋅∇+ρ

From this form for constantdensity, we can get two

important results: themechanical energy equation

and the thermal energy

equation.

The general form of theconservation of energy

equation, or first law ofthermodynamics, for adifferential control volume.


Conservation of energy – The

mechanical energy equation

• Absence of thermal, chemical, and nucleareffects

• Constant density.

0pVgzV2

1

dt

d 2 =∇⋅+

+ρ

This equation is reducible to the Bernoulli equationfor steady flows along a streamline



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Conservation of energy – The

thermal energy equation

•

Subtract the mechanical energy equation from thefull energy equation.

• Result is the thermal energy equation.

uqdt

dTc

uq

dt

du

0pVgzV2

1

dt

d

upVqdt

de

2

′′′=⋅∇+ρ

′′′=⋅∇+ρ

=∇⋅+

+ρ

′′′=∇⋅+⋅∇+ρ

Thermal energy equation for

a differential C.V.



( )( )

( )

( )

sT

q

dt

ds

dsdnT

q

dnVsdt

s

dnT

qdnVsd

t

s

dt

dS

dnT

q

T

q

CVCV

CVsystem

′′′=

⋅∇+ρ

ϑ′′′+σ⋅

−=σ⋅ρ+ϑ∂

ρ∂

σ⋅

−≥σ⋅ρ+ϑ

∂

ρ∂=

σ⋅

−=

∫∫∫ ∫∫ ∫∫ ∫∫∫

∫∫ ∫∫ ∫∫∫

∫∫

σσ

σσ

σ



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uqdt

du,

dt

du

dt

dsT

.Const

1pdduTds

sTq

dtds

′′′=⋅∇+ρ=

=ρ

ρ+=

′′′=

⋅∇+ρ ( )

T uT q

T s ′′′+∇⋅−=′′′

21

This equation governs the rate

of entropy generation owing toconduction of heat. Note that the

entropy generation rate will always begreater than zero because of the

increase of entropy principle. Thus, a

relation between the heat flux andtemperature gradient is suggested here

Appendix. Differentiation

of Multivariable functions

Partial Time Derivative

Substantial Time Derivative




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Multivariate Function

• General description of a scalar as afunction of position and time.

– f = f(x,t), where t = time

– All particles have a location at somestarting point x0. Thus, generally, x

= x(x0,t)

– The position x0 identifies a particle at

some time t0.

© 2006, 2009 F. A. Kulacki Conduction Heat Transfer39

The general flow field with

scalar transport


( )t),t,x(x 00

x

y

z

( )00 t,x

the along the particle path for The initial location of the along the particle path for

Particle path in time and space.



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Partial time derivative

•

Consider a fixed position for all time, (ξξξξ).• The change in the scalar field at this

fixed position is only with time.


x

y

z

ξ

volume in formulati e quations and appl

Many particle path of observation ove extension is to the

volume in formulati equations and appl

3t2t1t

Partial time derivative


If the scalar field is denoted by g(x,t), then for the fixed observation point the time variation of the field is,

z,y,xt

g

dt

dg

∂

∂=

i This derivative is (physically) the ti



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Substantial time derivative

• If the observation of time variation ofthe scalar field is made “following theparticle” along its path, the changeobserved owing to the “advected”motion of the observation must beincluded.

• The derivative with respect to timeunder this physical description isdenoted a Dg/Dt, or as in this course,

dg/dt when no ambiguity arises.




( )t),t,x(x 00

x

y

z

( )00 t,x

the part along the particle path for t > The initial location of the part along the particle path for t >

Particle path in time and space and observation path in time.

Along the particle path, the path on which

the time variation of the scalar field is to be observed, we have g(x,t) = g[x(x 0 ,t),t].

Note that g(x,t) is identified with a particular Starting location (x 0 ,t 0 ), but now we must

Consider x = x(x 0 ,t) where x 0 is taken as a parameter.



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( ) ( )

∂

∂+

∂

∂+

∂

∂+

∂

∂=≡

==

z

g

dt

dz

y

g

dt

dy

x

g

dt

dx

t

g

Dt

Dg

dt

dg

.timeof functionaformallyisx

.motionparticlefollowingtime.t.r.wateDifferenti

]t,t,tx[gt,xgg 0

Time variation independent

of advection.

Variation due to the spatial variation because of advection

along the particle path where time dependence of x must be included.



( ) ( )

gvt

g

x

gv

t

g

z

g

dt

dz

y

g

dt

dy

x

g

dt

dx

t

g

Dt

Dg

]t,t,tx[gt,xgg

ii

i

0

∇⋅+∂

∂=

∂

∂+

∂

∂=

∂

∂+

∂

∂+

∂

∂+

∂

∂=

==

∑



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End of Mod-02


Mod-02 Problem Formulation 2009

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