Transport processes Part 2a Ron Zevenhoven bo Akademi University - - PDF document

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Transport processes – Part 2a

Ron Zevenhoven Åbo Akademi University Thermal and Flow Engineering / Värme- och strömningsteknik

• tel. 3223 ; ron.zevenhoven@abo.fi

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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. , ) sin(3 0, ) sin( 0, ) sin( etc. , 1 ) sin(3 1, ) sin( and 0,1,2,3... m integer for                                   

    

      

π π π x π ) m sin( π ) m ( x ½ xdx π ) m ( cos / π / π ) ( π ) m ( x π ) m sin( π ) m ( xdx π ) m ( cos c A ) Ax sin( x ½ dx ) Ax ( cos ½ ½ dx ) Ax ( cos c A ) Ax sin( dx ) Ax ( cos c x sin xdx cos c x cos xdx sin : Note

m

                             

           

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EXAMPLE

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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 constant heat transfer coefficient h = 5000 W/(m2.K). The other side of the plate may be considered thermally insulated.

• For a plate with thickness d = 4 mm, calculate

the 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 Jensen 2005 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 some reference temperature).

• Similar to the case for a plane surface,

determine the time until the centre temperature 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 2 1

2 R t a µ R r µ J µ J µ T T       

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Separation of variables simplification Fo > 0.2

(”long times”)

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1-dimensional transient conduction /1

• Using separation of variables, convective

cooling/heating (see above) by a medium flow at temperature Tflow with Bi = h.Lchar/k, with convective heat transfer coefficient h (Wm2.K), characteristic length scale Lchar (m) and material conductivity k (W/m.K), gives for dimensionless time τ = Fo > 0.2, using only the first eigenvalue λ1:

                

               r r λ ) r r λ sin( ) τ λ exp( C T T T ) t , r ( T ) r r λ ( J ) τ λ exp( C T T T ) t , r ( T ) L x λ cos( ) τ λ exp( C T T T ) t , x ( T

flow start flow flow start flow flow start flow

: sphere : cylinder : wall plane

see tabelised data

• n next page for

first eigenvalue λ1 and constant C1

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1 -dimensional transient conduction /2

• Source: Introduction to Thermal and

Fluid Engineering by Deborah Kaminski and Michael Jensen 2005 by John Wiley & Sons, Inc.

Wall with thickness 2L Cylinder with radius r0 Sphere with radius r0

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) ( 1 ) ( p F p d F

t

• 

  

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More general: T1 T1

• T1 +

+ T1 / p T1

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+ T1 / p T1 + (T0 - T1)ꞏerfc(..) + T1 / p

• p. 51

12.

T1 / p (T0 - T1)

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p t

• r

p dt t e

• pt

1 1 £ : , 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 >> 0.2

• r simply

Fo > 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 2 1 with simplified be can 2

2 2 2 2 2 2 2 2 2 2

r T r r T r r r T r T r T r r T r T r r dr T dT r dr r dT T d r T r T r r 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 function erf(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 L erfc T ) t , x ( T                                                            

     

t . exp t π . ) t , L x ( Φ at L exp t π c λρ T ) t , L x ( 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 heat diffusivity 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 - 2

Inside a wall with thickness 2L, heat is generated as a result of an electric

• current. 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 temperature profile in the wall :

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) (

2 2

  x q dx T d 

2 2 2

      dx d

1

• cos

cos 1

• r

cos cos                                                              

L L L L L L

q L q x T T q L q x 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 57/58 2a

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Laplace transform – why?

While Fourier transform ”unravels” a function of time f(t) into a series of cosines and sines, i.e. oscillations, Laplace transform ”unravels” a function of time f(t) into expontial functions F(p): identified as poles Ai of F(p), with intensity ai. Fourier transform identifies periodic trends, Laplace transform identifies exponential behaviour: suitable for analysing, for example, response to a sudden change.

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