cheme 150a spring 2003 final

8/6/2019 ChemE 150A Spring 2003 Final

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Exam Physical Transport Phenomena II (tn375)24 March 2003 9.00 12.00 hr

Give your answers in symbolic notation (unless numerical values are specifically asked).

Start each (main) exercise on a separate page.

Put your name on each page. Specify your student number on the first page.

1 boiling egg

An egg at room temperature (20C ) is submerged in boiling water at t = 0. We assume thatthe egg is perfectly spherical with radius R, that flow does not play a role in this problem, that

the water temperature remains constant at 100

C , and that the egg has homogeneous mediumproperties. The temperature diffusivity of the egg is a.

a) Give the equation that governs the non-stationary temperature distribution inside the egg.Argue which terms can be neglected. Specify the initial conditions and the boundary conditions.

b) Rewrite the equations into dimensionless form.

c) Solve the full non-stationary problem by making use of separation of variables. Derive theequation for the eigenvalues. What are the values of the first three eigenvalues?

d) After some time t1 the solution is assumed to be accurately represented by the first term ofthe series obtained in c). Which dimensionless number must be considered to substantiate thisassumption? Give an expression for t1.

e) After some time tc the center temperature of the egg exceeds a given temperature Tc (onehas obtained a hard-boiled egg). Assuming tc > t1, give an expression for tc.


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2 Creeping flow between two concentric spheres

A viscous fluid flows in the space between two concentric spheres with radii R (for the outersphere) and R (for the inner sphere) as depicted in the graph below.

Outside the regions close to the entrance and exit (the regions near A and B enclosed by thesolid angle 2), the velocity components vr and v are assumed to be zero, and v = v(r, ). Theincompressible and stationary flow between the two concentric spheres has a very low Reynoldsnumber so that creeping flow may be assumed. The pressure drop between entrance and exitis p. Effects of gravity can be neglected.

a. By using the continuity equation, show that the velocity component v can be written as

v = f(r)/ sin .

b. Show that the differential equation for f(r) is given by

0 = 1

r

p

+

r2 sin

d

dr

r2

df

dr

, (1)

where is the dynamic viscosity.

c. Solve this differential equation for f(r). Hint:1

sin xdx =

1

2ln

1 cos x

1 + cos x

. (2)

d. Derive an expression for the volumetric flow rate Q [m3/s].


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3 solid wall with a heat source

Consider a solid wall (material constants , cp, ) with thickness L, which contains an internaluniform heat source q [W/m3]. At x = L there is a convective heat transfer boundary condition(heat transfer coefficient , ambient temperature Ta). At x = 0 a constant heat flux

w isapplied. The problem can be considered one-dimensional.

Analytical

a) Calculate the stationary temperature distribution inside the wall.

Numerical

The problem is discretised by introducing N cell centered nodes T1, T2, . . . , T N. The grid pointsare uniformly distributed with grid spacing x = L/N. The virtual cells T0 and TN+1 (seefigure below) are used for a straightforward implementation of the boundary conditions.

T0 T1 T2 . . . TN TN+1x

x = 0 x = L

b) Specify the numerical implementation of both boundary conditions using T0 and TN+1.

c) Take N = 2, q = 4W/m3, w = 2W/m2, = 1W/mK, = 1W/m2K and L = 2m,

Ta = 300K. Specify the full system of equations in the stationary situation.

d) Solve the system of c) to find the numerical prediction of the stationary temperature distri-bution. Calculate the temperature at the boundaries. Compare your answers with the analyticalresults obtained in a).


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Answers Exam Physical Transport Phenomena II (tn375)24 March 2003 9.00 12.00 hr

1 boiling egg

a) In spherical coordinates the differential equation becomes:

T

t=

a

r2

r

r2

T

r

(1)

The other terms vanish due to spherical symmetry. Boundary conditions t 0

T(R, t) = T1 = 373K (2)

T

rr=0

= 0 (3)

Initial conditionT(r, t = 0) = To = 293K r R

b) With

=T T1To T1

=r

R =

at

R2

we obtain

t=

1

2

2

with boundary conditions t 0

(1, t) = 0 (4)

=0

= 0 (5)

Initial condition(, t = 0) = 1 1

c) Try = F()G().1

F

F

=1

G

1

2

2

G

= 2

HenceF() = exp(2)

G follows from1

2

2

G

+ 2G = 0

General solution

G() = Asin()

+ B

cos()

Second solution does not behave physically at = 0, hence B = 0. Boundary condition (1, t) =0, implies G(1) = 0, the eigenvalues therefore follow from

sin() = 0


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The eigenvalues aren = n n = 1, 2, . . .

Consequently

(, ) =n

An exp(2n)

sin(n)

The coefficients An are to be derived from the initial condition: (, 0) = 1:

n

Ansin(n)

= 1

Multiplying by sin(m)/ and 2 (the weight function for orthogonality), integrating, and using

orthogonality, we get

Am =

1

0 sin(m)d

1

0sin2(m)d

=2(1)n+1

n

The full solution is therefore

T(r, t) = T1 (T1 To)n

2(

1)

n+1

n exp(2nat/R2)sin(n rR ) Rr

d) The relevant number is the Fourier number Fo = at/R2. The hand-outs mention Fo = 0.5 ascriterion. Hence

t1 =0.5R2

a

e) Proceeding with only the first term of the series we get for r = 0

T(0, t) = T1 2(T1 To)exp(2at/R2)

Solving T(0, tc) = Tc yields

tc =R2

2a

log

T1 ToT1 Tc

+ log 2

The interesting observation is that the egg radius is much more important than the initialtemperature (square dependence versus log dependence).

(For Tc = 80C we find tc t1. With R = 2cm and tc 300s we estimate a 3 10

7m2/s.)


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2 Creeping flow between two concentric spheres

part a

The continuity equation in spherical coordinates reads:

t+

1

r2

r

r2vr

+

1

r sin

(v sin ) +

1

r sin

(v) = 0 . (6)

Assuming stationary flow with constant density and vr = v = 0 this reduces to

(v sin ) = 0 . (7)

Integration of the above expression gives:

v =f(r)

sin . (8)

part b

In a stationary creeping flow with constant density and vr = v = 0, the momentum equationfor the -direction is given by:

0 =1

r

p

+

2v

vr2 sin2

, (9)

In spherical coordinates, the term 2v reads

2v =1

r2

r

r2

vr

+

1

r2 sin

sin

v

+

1

r2 sin2

2v2

. (10)

The last term on the right-hand side is zero because v = 0. The momentum equation then

reduces to

0 = 1

r

p

+

1

r2

r

r2

vr

+

1

r2 sin

sin

v

vr2 sin2

. (11)

With v = f(r)/ sin we find:

1

r2

r

r2

vr

=

1

r2 sin

r

r2

df

dr

(12)

1r2 sin

sin v

= f(r)

r2 sin3 (13)


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v

r2 sin2 =

f(r)

r2 sin3 (14)

Substitution in the momentum equation gives:

0 =

1

r

p

+

r2 sin

r

r2 df

dr

. (15)

part c

The momentum equation can be integrated as follows:

p

sin =

r

r

r2

df

dr

= B . (16)

Integration of the left-hand side gives:

p1 p0 = B

10

d

sin (17)

=B

2

ln

1 cos

1 + cos

(18)

=B

2

ln

1 + cos

1 cos

ln

1 cos

1 + cos

(19)

= B ln

1 + cos

1 cos

. (20)

Integration of the right-hand side gives:

r

r

r2

df

dr

= B f =

Br

2

C1r

+ C2 . (21)

The integration constants C1 and C2 follow from the no-slip boundary conditions, i.e. f = 0 atr = R and r = R:

C1 =BR2

2and C2 =

BR

2( + 1) , (22)

yielding

f =

BR

2 r

R +

R

r

( + 1)

, (23)

with

B =p

ln1+cos 1cos

. (24)

part d

The volume flux Q follows from

Q =

R

R

v(r)2rdr , (25)


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with v = f(r)/ sin . The volume flux will be independent of , so that a convenient value for can be chosen, e.g. = /2. Using the above expression for the function f(r), we find for thevolume flux

Q =

RR

BR

22

r2

R+ R ( + 1)r

dr (26)

=BR

r3

3R + Rr ( + 1)

2 r2RR

(27)

=BR3

1

6+

2

2

2+

3

6

=

BR3

6(1 )3 . (28)


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3 solid wall with heat source

a) The governing equation is

cp

tT =

2

x2T + q

with boundary conditions

xT

x=0

= w

xT

x=L

= (T(L) Ta)

Stationarity implies t

T = 0. Integration yields

T(x) = q

2x2 + bx + c

The integration constants follow b and c follow from the boundary conditions:

xT

x=0

= b = w b =

w/

and

xT

x=L

= qL + w = (q

2L2

w

L + c Ta)

Solving for c we obtain the stationary temperature distribution inside the wall:

T(x) =q

2(L2 x2) +

w

(L x) +qL

+

w

+ Ta

b) Employing the virtual cells T0 and TN+1 the implementation of the boundary conditionsreads

T1 T0

x= w

TN+1 TNx

=

TN+1 + TN

2 Ta

c) Apart from the boundary conditions we have for i = 1 . . . N :Ti1 2Ti + Ti+1

x2=

q

After substitution of the numerical values we get the following set of equations

T0 T1 = 2

T0 2T1 + T2 = 4

T1 2T2 + T3 = 4

1

2T2 +

3

2T3 = Ta


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d) Substitution of the first equation in the second, the second in the third, and so on, we obtain

T3 = Ta + 3

T2 = Ta + 9

T1 = Ta + 11

T0 = Ta + 9

The boundary temperatures are:

T(x = 0) =T0 + T1

2= Ta + 10 = 310K

T(x = L) =T2 + T3

2= Ta + 6 = 306K

which is exactly equal to the results obtained analytically. Nevertheless, the numerical predictionof the temperature distribution is not entirely accurate, as can be seen in the figure below.

0

2

4

6

8

10

12

0.5 0.5 1 1.5 2 2.5

x

cheme 150a spring 2003 final

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