Scaling Arguments — The Fourier Number

Turkey Cooking

Woman's Home Companion Cook Book

Weight n weight Tune/Unit Weight in time/weight

6-10 (8) 2.1 20-25 3.110-16 (13) 2.6 18-20 2.9518-25 (21) 3.0 15-18 2.8

Suppose the whole family is getting together, and we needed to cook a 30-pound turkey. How long should we cook it?

says 1 3 nt nM

1 3t M

Why should a 30-pounder takes in 30 = 3.4 ~ 14 min/lb?

9-1

Turkey Cooking

Why do bigger turkeys take less time to cook per pound?

Turkeys are spheres? — 1st approximation

We know that 2

0

, s

s

T T re f

T T R

, everything here just depends on

dimensionless or "scaled" radius and time. Basically, to cook a turkey, we need to make sure that the interior temperature reaches a

certain value for a certain time; hence:

Since in each case, 0,a , we should have the same dimensionless time, 0

02

t

R

20R

Now this gives us the real time, which depends on R2 of the turkey, assuming all turkeys have the same α:

The mass of turkey (sphere) is: 1 3

3 or M

M R R

21 3

2 3

2 3

M M

so the actual time it takes for a turkey to cook

goes like 2 3M .

If, as in the cookbook, we want to determine the time/mass, we have to: 2 3

1 32 3

1 3 per pound

MM

M M

or M

1/ 3 .nt nM Hence, the theoretical justification of our graph, without any tough PDE's to solve. Often in engineering analysis, the dimensionless Fourier number, τ, is sufficient to scale the problem.

9-2

ChE 120B

Important Numbers on Heat Transfer

2

p

kt

C L - Fourier number — dimensionless time

ratio of convective heat transferBiot number

conductive heat transfer

hL

k

When to do a complete analysis — when not to. First, always do the simplest scaling relationship.

Transient problem – ID,

[ 0

2_ _ _ _ _ _ _ _ _ _ _ _ _ _ _

2 , ,C T

y=0 y=L

0

p

x bT T

xt x

in dimensionless form 0

T T

T T

When to use tabulated data/scaling:

Note that for transient problems in dimensionless form, the solutions are functions of

20

, char. dim.

surface

surface

T Tt

T T

Char. Dim x/L, 0/r r , etc.

And Biot# =.dim.

h

k char

So, all problems are similar and can be used to scale up when needed. What is important in the system geometry, the Fourier number and the Biot number.

9-3

ChE 120B Hamburger Cooking Scaling

For Biot# >> 1, τ = αt/L2 governs the time it takes for a given dimensionless temperature change.

Imagine – McDonalds is planning a 1/2-pounder to replace the 1/4-pounder. To use the same buns, the diameter of the burger is the same but the patty is thicker, however, the centerline temperature of patty must be at the sample temperature as before to prevent intestinal problems due to raw meat. Using the same cooking surface, how much longer will the 1/2-pounder take to cook?

Eng. Analysis:

important dimension

2

b

2

2

2

2

V R b

M V R b

for 1/4-pounders

1

1 2

1

t

b

22 2

2

t

b

but we want 0

0

x surf

surf

T Tsame

T T

so 1 2 – dimensionless time, temperature same

1 22 2

1 2

= t

b b

________ should be same so

1 22 2

1 2

t t

b b

2 12b b to account for extra meat, as R stays the same

1 2

221 1

2

t t

b b

solving we see that

1 24t t

or a 1/2 pounder will take 4 times as long to cook as a 1/4 pounder. This is likely why McDonald's puts two 1/4 pound patties to make their 1/2 pound burger.

9–4

ChE 120B

Superposition

Suppose we are faced with the following problem:

3

1

4 2

t t

t

t

At first glance, a separation of variables solution seems impossible. However, we can break the problem in to simpler sub-problems due to the linearity of the differential equation:

2 2

2 2

x

T T

y

-- Assume we have 4 solutions to the simple problems below:

We can solve each of these by separation of variables.

1 ,T x y =

1

0

0 0

t

2 ,T x y 2

0

0

0

t

3 ,T x y

3

0

0 0

t

4 ,T x y 4

0

0

0

t

9–5

ChE 120B

We know that T1 + T2 + T3 + T4 is also a solution of the differential equation that has boundary conditions

1 2 3 4 10, 0, 0, 0,T y x T y x T y x T y x T

2 2 3 4 2, , , ,T y b x T y b x T y b x T y b x T

and so on…..

Since each sub-solution also satisfies 2 2

2 20

T T

x y

, the sum must also.

9—6

ChE 120B

Melting a Fuse Wire (More Superposition)

Fuse design — how much current do you need before one melts — important question.

Fuses blow because the wire melts due to too much current. How to model this system?

23 3

energy1

time e

WS Ke

cm cm 2current density amps/I cm

1 13 conductivity ohm K cm

1 resistanceek

Model — Ke, K, ρ, Cp, physical parameters constant

Input — Output + Generation = Accumulation

area area vol. elvol. el

2 q - 2 q 2 S 2 e p

TrL r rL r dr r dr L r dr L C

t

lim

0 e p

rq r rq r dr TrS r C

dr dr t

gives r e p

Trq rS r C

r t

e p

k T Tr S C

r r r t

Rearranging, we get

e

p

S

ST Tr

r r t C t

1T Tr S

t r r r

What we wind up with is a non-homogeneous P.D.E. How do we solve? — first, need B.C.'s, I.C.

0 0 @ t =0 -TT T is ambient

9-7

ChE 120B Symmetry:

0 @ 0, or is finite @ 0T

r T rr

Surface: f

Tk h T T

r

Newton's law @ surface

We also need to non-dimensionalize problem.

0

f

f

T T

T T

a little different format than usual, but it is handy for the 2nd B.C.

2 standard

r t

R R

0 0

2

f fT T T TR S

R R RR

0 0

2 2

f fT T T T dS

R R

2 2

00

1

ff

p

SR SRkT T T TC

2

0

e

pp

S Rk

C TC

2

0

eS R

kT

Dimensionless rate of generation

G

9-8

ChE 120B

Dimensionless Problem Statement:

1

2G

I.C. 0, 1

B.C. ,0 = finite

B.C. 2

0

0

f

f

T TK h T T

R

B.C. 2 1 1 1

d hRBi

k

Now we need to solve our set of equations

How to think about this problem:

1. We would know how to solve if G = 0

A) Massage B.C.'s to get separable solutions. (Actually, they are already OK)

B) Use e-functions to get answer.

2. How do we go about getting the equations in this form?

A) Transient problem can't always have easy superposition solutions.

B) We can often determine the steady-state distribution — What does this tell us

about the transient problems?

Steady-state Problem

1 0@ steady stateG

G

9-9

ChE 120B

2

12

GC

11 0 because 0 at Imposing some B.C.'s

2

CGC

Then, one more integration gives:

2

24

GC

B.C#2

1 Bi

2

1 2 1 2 4 n

G GBi C

2 4 2

G GC

Bi

2

4 4 2

G G G

Bi

2 21

4

G

Bi

2 24

G Bi

Bi

Now we know what happens when t and we get steady state!

22

4

G Bi

Bi

New superposition tack for transient problems.

Let or 0

is the solution to our original problem.

9-10

ChE 120B

is the steady state solution to our original problem. Is the solution for easier than .

I hope so.

Original equation

1

G

Plug in

0*

2*1 1

2G

Now the trick is becoming clear.

We made sure that

1=G and =0

= 0 so this reduces the equation for to:

1

0

B.C.'s on

I.C. 0, 1 0, 0

1 0,

1 0,

,0 ,0 finitefinite

1 1Bi

1 1 1 1Bi

9-11

ChE 120B

1 1Bi

because

1 1Bi

Hence, for , we have a separable solution

1

0

0, 1

,0 finite

1,1Bi

Solution for is straight forward:

Let Y N to get

21/

YN

22

10 Y

Y C e Y

22 0 3 0

10

NN C J C Y N

and

2

1 2 0 3 0C e C J C Y

The easiest B.C. to apply is that is always finite @ 0.

0 0 , soY

9-12

ChE 120B 3 1 21 0, Let BC C C C C

Now to apply the second B.C.: 1,1iB

2

0Ce J

0 1J J

and

2 2

0 1 Bi Ce J C e J

This defines 0 1 Bi Jn J

1

0

n

n

JBi

J

Similar to tan n = n

Bi

for rectangular case we dealt with before.

Applying the I.C. gives:

2

01

nn n

n

C e J

and 1 0, so

01

1 n nn

C J

To determine the nC , we recognize that are orthogonal w.r.t. , hence Multiply

by 0 mJ d and integrate from 0 to 1

1 1

0 0 010 0

1 dm n m mn

J d C J J

1 1

2

0 0

0 0

1 C dm m mJ d J

9-13

ChE 120B

1

0

01

20

0

1 m

m

m

J d

C

J d

What can we say physically about this solution now?

Where do the physical properties of the material come into the solution?

characteristic times 21

Note that for Bi << 1 — case likely to be important here, we can look at first eigenvalue: Taylor series for the Bessel functions are:

2

0 1 21

2J

1 1 2J

and inserting into the equation for eigenvalues:

1

2

2

2

12

Bi

2

1

2Bi

212Bi

12 .Bi

0 1 2

21 1

2

BiJ

Hence, simple solution is:

9—14

ChE 120B

andpC V Ahte the characteristic time pC V A

th

Same as lumped parameter model.

9-15

ChE 120B

Superposition

Consider only variations along Z direction because 1BiN r . Current passes through the

wire producing heat at a uniform volumetric rate of 3eS W cm per unit volume. Wire loses

heat from surface by Newton's Law of cooling.

Energy Balance:

Input — Output + Source = Accumulation

2 2 2 202 z z z Z Z z e pR q R q R zh T T R zS R z C T

T

reduces to: 0

2, with Fourier's Law; ,z

e p zz

q h T dtT T S C q k

R t dz

we get: 2

02

2 peCST h T

T Tz Rk k k t

02

0

let 0 ,p

T T z ktz

T L C L

then we get the dimensionless form:

222 2

2

2 peC LS LhL T

z Rk k k t

2

22

m Gz t

22

2 2, eS LhL

m GRk k

9–16

Boundary Conditions: 0, 0 1, ,0z

How to solve? — Well, let's look at steady-state — we know how to solve that.

Let z be the steady state temperature distribution. Then 0t

and

2

22

0m Gz

with 0 1 0

1 2 2sinh mz cosh mz

GC C

m

2 22 20

G GC C

m m

1 21 0 sinh m cosh mz 1

GC

m

2

1

cosh m 1

sinh m

GmC

2

1 cosh mcosh mz 1 sinh mz

sinh m

G

m

So what — let's look at a new problem:

and use our superposition principle

2

22

m Gz

2 2

2 22 2

0m m Gz z

0 by construction and

9-17

ChE 120B

2

22

mz

0, 0 0, 0 0

1, 0 1, 0 0 ,0 0 ,0z z z z

So we now have a homogeneous P.D.E. whose B.C. are of the Drichlet form — solve by

separation of variables:

Let Z z Y

2

22

Z YY m YZ Z

z

2

2 2 22

0, 0Z Y

Z M Yz

Solution is 2 2

1 2 3, sin cosm

z C e C z C z

30, 0 0C

1, 0 sin 0 1, 2,...n n

2 2

1

, e sinn m

n

z Cn n z

1

,0 sinn

z z Cn n z

1 1 2

0 0sin sin mz m z dz C m z dz

1

0 2 sin mC z m z dz

9—18