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Transport processes – Part 2a

Ron ZevenhovenÅbo Akademi University

Thermal and Flow Engineering / Värme- och strömningstekniktel. 3223 ; ron.zevenhoven@abo.fi

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Page 58 added 22.1.2018

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Thermal diffusivity

α = λ /(ρꞏcp)

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more general: T=T*

more general: θ = (T -T*)/(T0 -T*)

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Orthogonality

etc. , 0)sin(30,)sin(0,)sin(

etc. , 1)sin(3 1,)sin( and

0,1,2,3...m integer for

πππ

xπ)msin(π)m(

x½xdxπ)m

(cos

/π/π

)(π)m(

xπ)m

sin(π)m(

xdxπ)m

(cos

cA

)Axsin(x½dx)Ax(cos½½dx)Ax(cos

cA

)Axsin(dx)Ax(cos

cxsinxdxcos

cxcosxdxsin

:Note

m

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EXAMPLET

rans

port

pro

cess

es(T

RP

)

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EXAMPLE

This can be since the concrete has an 8x higher heat capacity ρ∙cp, i.e. enthalpy / volume.

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more general: h(T-T*)

more general: θ = (T -T*)/(T0 -T*)

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Note: µ0 = 0

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Separation of variables – example /1

• Q: A steel plate at 900°C is cooled by spraying 40°C water on one side of it. This gives convective heat transfer with constantheat transfer coefficient h = 5000 W/(m2.K). The other side of the plate may be consideredthermally insulated.

• For a plate with thickness d = 4 mm, calculatethe temperature on both plate surfaces 5 seconds after the spray cooling has started.

• For the steel, assume conductivity λ=20 W/(m.K) and thermal diffusivity a = 6ꞏ10-6

m2/s

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Separation of variables – example /2

• A: For the Biot number: – Bi = 5000ꞏ0.004/20 = 1;

– First eigenvalue µ1 = 0.860 (from Figure 2.2)

• Using only the first eigenvalue:– @ x=d : T(x=d) = 40+860ꞏ0.73ꞏexp(-0.28ꞏt)

– This gives T = 195°C @ t = 5 s

– @ x=0 : T(x=0) = 40+860ꞏ1.12ꞏexp(-0.28ꞏt)

– This gives T = 277°C @ t = 5 s

• It is readily seen that the second eigenvalue can be neglected.

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= Fourier number, Fo

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The functions Yk(x) are in practice (in the field addressed by this course) not needed.

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Bessel functions data

• Source: Introduction to Thermal and Fluid Engineering by Deborah Kaminski and Michael Jensen2005 by John Wiley & Sons, Inc.

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Using Bessel functions – example /1

• Q: A cylindrical column with diameter d = 0.05 m is initially at T = T0 when at time t = 0 suddenly the surface temperature is brought to T = 0 (with respect to somereference temperature).

• Similar to the case for a plane surface, determine the time until the centretemperature Tc is equalised to 0.05 = (Tc-0)/(T0-0)

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Using Bessel functions – example /2• A: For long times use the first eigenvalue,

with n = 0 this gives (see p. 37)

µ0 = the first zero of J0(..), which is 2.405, and with J1(2.405) = 0.519 for r = 0: 0.05 = 1.60ꞏexp(-0.0037ꞏt)

• This gives the result t = 937 s, which is ~ 2x faster than a plate with d= 0.05 m

• The heat flux (W/m2) can be calculated using-λꞏdT/dr and differentiated Bessel functions

)exp()()( 2

2000

0100

2

R

taµ

R

rµJ

µJµTT

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Separation of variablessimplification Fo > 0.2

(”long times”)

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1-dimensional transient conduction /1

• Using separation of variables, convectivecooling/heating (see above) by a medium flow at temperature Tflow with Bi = h.Lchar/k, with convectiveheat transfer coefficient h (Wm2.K), characteristiclength scale Lchar (m) and material conductivity k (W/m.K), gives for dimensionless time τ = Fo > 0.2, using only the first eigenvalue λ1:

rrλ

)rrλ

sin(

)τλexp(CTT

T)t,r(T

)r

rλ(J)τλexp(C

TT

T)t,r(T

)L

xλcos()τλexp(C

TT

T)t,x(T

flowstart

flow

flowstart

flow

flowstart

flow

:sphere

:cylinder

:wall plane

see tabelised dataon next page forfirst eigenvalue λ1

and constant C1

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1 -dimensionaltransient

conduction /2

• Source: Introduction to Thermal and Fluid Engineering by Deborah Kaminski and Michael Jensen2005 by John Wiley & Sons, Inc.

Wall with thickness 2L

Cylinder with radius r0

Sphere with radius r0

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)(1

)( pFp

dFt

o T

rans

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es(T

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More general: T1

T1

-T1 +

+ T1 / p

T1

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+ T1 / p

T1 + (T0 - T1)ꞏerfc(..)

+ T1 / pp. 5112.

T1 / p

(T0 - T1)

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ptor

pdt

t

eo

pt

11£:

,1

(T0 - T1)

(T0 - T1)

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T1

T1

T1

T0

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-T1 / α

+ T1 / p

(T0 - T1)

(T0 - T1)

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+ T1 / p

+ T1 / p

+ T1

=1 – x + x2 – x3 + x4 …

(T0 - T1)

(T0 - T1)

(T0 - T1)

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!

Fo >> 0.2or simplyFo > 0.2

(T0 - T1) (T0 - T1)

(T0 - T1)

+ T1

+ T1

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q = √ p / a

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Transformation simplification

term 1 to reduced terms 2

21

withsimplified be can 2

2

2

2

2

2

2

2

2

2

2

r

T

rr

T

rr

r

T

r

T

r

Tr

r

Tr

Tr

r

drTdTrdrr

dTT

d

rTr

T

rr

T

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Using Laplace transform – example /1

• Q: Reconsider the gypsum + steel wall (p. 28), thickness 0.05m + 0.05 m, now for short times,

i.e. Fo = at/d2 < 0.2, i.e. t < 1250 s ≈ 20 minutes

• T(x, t=0) = 0 (°C) L = 0.05 m; a = 4ꞏ10-7 m2/s, λ= 0.4 W/mꞏK, ρꞏcp = 1 ꞏ106 J/m3ꞏK;

• Calculate the temperature at the centre of the wall(x=0) when T(x=±L, t) = 100 (°C), after 10 s and after 4 minutes

• Give also the heat flux

Φ”heat (W/m2) at x = +L

Data error functionerf(x) = 1 – erfc(x)

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Using Laplace transform – example /2• A: For the temperature at the centre:

gives T(0,10 s)= 0, T(0, 240 s) = 14.25 (°C)Note that the second term erfc(3L/4√at) < 10-6

• For the heat flux:

gives for t = 10 s: Φ”heat = 100ꞏ113ꞏ1 = 11300 (W/m2) for t = 240s: Φ”heat = 100ꞏ23ꞏ0.997 = 2296 (W/m2)

t

.erfc)t,x(T

at

LerfcT)t,x(T

t

.exp

tπ

.)t,Lx(Φ

at

Lexp

tπ

cλρT)t,Lx(Φ

"heat

p"heat

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A classroom exercise -1

• Water is transported through a pipeline which is located at several meters below the ground surface. For a situation where the temperature of air, soil and pipeline are at T0 = 7°C at sime t = 0, followed by a sudden change to a lower air temperature T1 = -8°C, calculate how deep below the ground (in meters) the pipeline should be to avoid freezing of the water (at 0°C) after 60 hours. Use for the soil a heatdiffusivity a = 1.38∙10-7 m2/s, and assume that the water in the pipeline does not move.

• (answer : 0.178 m).

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A classroom exercise - 2Inside a wall with thickness 2L, heat is generated as a result of an electriccurrent. The amount of heat generated per unit wall lenghth, as function of the distance, x (m), from the wall centre is given by q(x) = qL∙(1+β∙(T - TL) (unit : W/m), where TL is the temperature at the wall. Assuming a steady-state situation in one dimension, x (the wall is very large in directions y and z):

a. Show that the temperature profile inside the wall can be described by

with T = TL at x = ± L ; dT/dx = 0 at x = 0

where λ is the thermal conductivity of the wall.

b. Show that with new variable θ(x) = q(x) /λ = (qL/λ)∙(1+β∙(T - TL) the differential

equation becomes

with θ = qL/λ at x = ± L ; dθ/dx = 0 at x = 0

where µ2 = qL∙β/λ .

c. Show that this is solved to give the following solutation for the temperatureprofile in the wall :

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 0)(

2

2

xqdx

Td

 02

2

2

dx

d

 

1-

cos

cos1

or

cos

cos

L

L

L

L

L

L

qL

qx

TTq

L

qx

q

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Sources used(besides course book Hanjalić et al.)

• Beek, W.J., Muttzall, K.M.K., van Heuven, J.W. ”Transport phenomena” Wiley, 2nd edition (1999)

• R.B. Bird, W.E. Stewart, E.N. Lightfoot ”Transport phenomena” Wiley, New York (1960)

• * C.J. Hoogendoorn ”Fysische Transportverschijnselen II”, TU Delft / D.U.M., the Netherlands 2nd. ed. (1985)

• * C.J. Hoogendoorn, T.H. van der Meer ”Fysische Transport-verschijnselen II”, TU Delft /VSSD, the Netherlands 3nd. ed. (1991)

• D. Kaminski, M. Jensen ”Introduction to Thermal and Fluids Engineering”, Wiley (2005)

• S.R. Turns ”Thermal – Fluid Sciences”, Cambridge Univ. Press (2006)

* Earlier versions of Hanjalić et al. book but in Dutch

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Laplace transform – why?

While Fourier transform ”unravels” a function of timef(t) into a series of cosines and sines, i.e. oscillations, Laplace transform ”unravels” a function of time f(t) intoexpontial functions F(p):

identified as poles Ai of F(p), with intensity ai.

Fourier transform identifies periodic trends, Laplacetransform identifies exponential behaviour: suitable for analysing, for example, response to a sudden change.

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