Second Semester AY 2019-2020 · A thick-walled tube of stainless steel (k = 21.63 W/m K) with an inner diameter (ID) of 0.0254 m and an outer diameter (OD) of 0.0508 m is covered

Second Semester AY 2019-2020

Department of Chemical EngineeringUniversity of the Philippines Diliman 1

2

Determine the heat transfer across the 1 inch-thick slab shown at the right, whose thermal conductivity varies linearly with temperature according to the equation

o2 o F

Btuk 0.2 0.005T

h ft F

= +

3

Assumptions

steady state conditions

constant transfer area

isotropic material

material properties do not vary significantly within the temperature range

wall thickness is significantly smaller than

the width and height

heat convection from the bulk surroundings occurs at a significantly higher rate

4

x

y

0 1

12

1.5

22 1.5

m 61

012

−= =

−

y mx b= +

xy 6x 1.5= +

h 2y 12x 3 = = +

( ) A w h 2 12x 3 24x 6→ = = + = +

5

q dTk

A dx= −

( )( )

112 250

0 350

1 q dx k T dT

A x→ = −

( )

112 250

0 350

dx q 0.2 0.005T dT

24x 6→ = − +

+

( )

( )1

12

2502

350

0

0.2T 0.0025T q

1ln 24x 6

24

− +→ =

+

Btuq 14,182

hr=

6

q dTk

A dx= − ( )

dx q 0.2 0.005T dT

24x 6→ = − +

+

What about the temperature profile?

( ) ( )21

q ln 24x 6 0.2T 0.0025T C24

→ + = − + +

o1 1

o2 2

BCs : x 0, T 350 F

1 x , T 250 F

12

= =

= = choose one

1 C 1435→ =

( )20.0025T 0.2T 1435 591ln 24x 6+ = − +

7

CHECK o1 1

o2 2

BC1: x 0, T 350 F

1BC2 : x , T 250 F

12

= =

= =

( )2BC1: 0.0025T 0.2T 1435 591ln 24 0 6+ = − +

o430 F T

−→ =

o 350 F

2 1BC2 : 0.0025T 0.2T 1435 591ln 24 6

12

+ = − +

o330 F T

−→ =

o 250 F

8

Shell balance

Heat conduction in different geometry

Conduction in a hollow cylinder

Conduction in a hollow sphere

Summary

1x 0= 2x L=

0T

LTx

( )T x ?=

▪ steady-state conduction

▪ isotropic wall material with constant properties

▪ unidimensional heat transfer along x-direction

▪ constant heat transfer area

▪ no internal heat generation

shell

thickness x=

9

Shell balance




Summary

1x x= 2x x x= +

x

xq x xq+


▪ isotropic wall material with constant properties

▪ unidimensional heat transfer along x-direction

▪ constant heat transfer area


10

Shell balance




Summary

heat heat heat heat

in out generation accumulation

− + =

x x x

q qA A 0 0

A A+

− + =

A x x x x

q q

A A 0

x+

−

→ =

since x 0 : →d q

0dx A

− =

11

Shell balance




Summary

dTk Fourier's law

d 0

dxdx

→

− =

d q0

dx A

− =

2

2

d T k 0

dx→ − =

2

2

d T 0

dx→ = ( )

( )0

L

boundary conditions:

1 at x 0, T T

2 at x L, T T

= =

= =

1

dT C

dx→ = ( ) 1 2 T x c x c→ = +

12

Shell balance




Summary

( )0 L 1T T c 0 L − = −

( )0 1 2BC1: T c 0 c= +( ) 1 2T x c x c= +

( )L 1 2BC2 : T c L c= +

L 01

T T c

L

−→ =

( )L 00 2

T TBC1: T 0 c

L

− = +

2 0c T→ =

( ) L 00

T TT x x T

L

− = +

L 0T TdT

dx L

−→ =

L 0T TdTq kA kA

dx L

− = − = −

13

Shell balance




Summary

plane wall/slab

(cartesian)

cylinder/annulus

(cylindrical)

sphere

(spherical)

pA w h= cA 2 rL= 2sA 4 r=

14

Determine the heat transfer across the 1 inch-thick slab shown at the right, whose thermal conductivity varies linearly with temperature according to the equation

o2 o F

Btuk 0.2 0.005T

h ft F

= +

15

1x x=

2x x x= +

x

xq x xq+

x

y

h 2y 12x 3= = +

( )A x w h 24x 6= = +

16

heat heat heat heat


− + =

x x x

q qA A 0 0

A A+

− + =

A x

( ) ( )

( )x x x

q q24x 6 24x 6

A A 0

24x 6 x+

+ − +

→ =

+

since x 0 : → ( )1 d q

24x 6 024x 6 dx A

− + =

+

( )d q

24x 6 0dx A

→ + =

( )

( )

o

o


1 at x 0, T 350 F

1 2 at x , T 250 F

12

= =

= =

17

( )d q

24x 6 0dx A

+ =

( )

( )

o

o


1 at x 0, T 350 F

1 2 at x , T 250 F

12

= =

= =

( ) 1

q24x 6 C

A

+ =

1Cq dT k

A dx 24x 6→ = − =

+( ) 1

dx 0.2 0.005T dT C

24x 6→ − + =

+

( ) ( )2 21

6 0.2 350 250 0.0025 350 250 C ln

8

− + − = −

( )21 2 0.2T 0.0025T C ln 24x 6 C→ + = − + +

( ) ( ) ( )2

1 2BC1: 0.2 350 0.0025 350 C ln 6 C+ = − +

( ) ( ) ( )2

1 2BC2: 0.2 250 0.0025 250 C ln 8 C+ = − +

1C 591=

( ) ( ) ( )2

2BC1: 0.2 350 0.0025 350 591ln 6 C + = − + →2C 1435=

( )20.2T 0.0025T 591ln 24x 6 1435+ = − + +

Using shell balance, determine the heat transfer rate across the frustum shown at the right, whose thermal conductivity is measured to be 27 Btu/hrftF.

18

Shell balance




Summary

19

Shell balance




Summary p1A p2A

p1 p2

am

A AA

2

+=

am xA

T q k

→ = −

p2

p1

A1

A

arithmetic mean:

20

Shell balance




Summary

L

1r 2rr


▪ isotropic cylinder material with constant properties

▪ slender geometry (r << L) --> 1D heat transfer along the r-direction (radial)

▪ constant cross-section area


shell thickness r=

transfer area 2 rL=

shell volume 2 rL r=

21

Shell balance




Summary

heat heat heat heat


− + =

r r r

q qA A 0 0

A A+

− + =

2 rL r

r r r

q q2 rL 2 rL 0

A A

+

− =

r r r

q qr r

A A 0

r r+

−

→ =

since r 0 : →1 d q

r 0r dr A

− =

22

Shell balance




Summary

1 d qr 0

r dr A

− =

1 d q r 0r dr A

→ =

1

q dTk

A r r r C

d→ =−= 1C dr

dTk r

→ = −

( )

( )1 1

2 2


1 at r r , T T

2 at r r , T T

= =

= =

( )12

CT ln r C

k = − +→

d q r 0

dr A

→ =

( )11 1 2

CBC1: T ln r C

k= − +

( )12 2 2

CBC2: T ln r C

k= − +

( ) ( )12 1 2 1

CT T ln r ln r

k − = − −

( )2 11

2

1

T TC k

rln

r

−= −

23

Shell balance




Summary

( )( )

( )( )2 1

1 1 2

2

2 12 1 1

2

11

T TB C

T TC TC1: T ln r

rln

r

ln rr

lnr

−= + →

−= −

( )( )

( )( )2 1 2 1

1 1

2 2

1 1

T T T TT ln r T ln r

r rln ln

r r

− −= + −

( )11

22 1

1

rlnrT T

rT T lnr

−=

−

( )2 1

2

1

T TdT 1

dr rrln

r

−→ =

24

Shell balance




Summary

( )2 1

2

1

T Tq dT q 1k k

A dr 2 L rln

r

r

r

−= − → = −

( )( )1

2

1

2

l

T Tq

r

k 2 Lr

n

−= −

1

2 1

2 r

r r

r

−

−

2 1lm

2

1

logarithmic mean:

r r r

rln

r

−=

( ) 2 1

2 1

lm

T Tk r2 L

r r

−= −

−

lm rA

T q k

→ = −

( )2 1

2

1

T TdT 1

dr rrln

r

−=

25

Shell balance




Summary

1r2r

r


▪ isotropic spherical material with constant properties

▪ unidimensional heat transfer along r-direction


shell thickness r= 2transfer area 4 r=2shell volume 4 r r=

26

Shell balance




Summary

heat heat heat heat


− + =

r r r

q qA A 0 0

A A+

− + =

2 4 r r

2 2

r r r

q q4 r 4 r 0

A A

+

− =

2 2

r r r2

q qr r

A A 0

r r+

−

→ =

since r 0 : →2

2

1 d qr 0

r dr A

− =

27

Shell balance




Summary

2

2

1 d qr 0

r dr A

− =

2

2

1 d q r 0r dr A

→ =

2 21

q d

d r C

Tk

A rr→ = =− 1

2

C dr dT

k r→ = −

( )

( )1 1

2 2


1 at r r , T T

2 at r r , T T

= =

= =

12

C 1T C

k

r= +→

2d q r 0

dr A

→ =

11 2

1

C 1BC1: T ln C

k r= +

12 2

2

C 1BC2: T C

k r= +

12 1

2 1

C 1 1T T

k r r

− = −

( )2 11

2 1

T TC k

1 1

r r

−=

−

28

Shell balance




Summary

( ) ( )

2

2 12 1

11

2

1

1

2 12

1

T T 1BC1: T C

T

r1 1

r r

T 1C T

r1 1

r r−

−−= + →

= −

−

( ) ( )2 1 2 11

1

2 1 2 1

T T T T1 1T T

r r1 1 1 1

r r r r

− −= + −

− −

11

2 1

2 1

1 1

r rT T

T T 1 1

r r

− − =

− −

( )2 1

2

1 2

T TdT 1

dr r1 1

r r

−→ = −

−

29

Shell balance




Summary

( )2 1

2

2 1

T TdT 1

dr r1 1

r r

−= −

−

( )2

2 1

2 1

2 1

2

T Tq

r

dT q 1k k

A d 4 r rr r

r

r

−= − → = −

−

( ) 2 1

2 1

2

2 1

T Tq k 4

r rr r

− = − −

gm 2 1

geometric mean:

r r r=

( ) 2 1

2 1

2gm

T Tk r4

r r

−= −

−

gm rA

T q k

→ = −

( )2 1

2

1 2

2 1

T T 1

rr r

r r

−= −

−

( )2 1

2

2 1

2 1

T T 1

rr r

r r

−= −

30

Shell balance




Summary

plane wall/slab

(cartesian)

cylinder/annulus

(cylindrical)

sphere

(spherical)

am

Tq Ak

x

= −

lm

Tq Ak

r

= −

gm

Tq Ak

r

= −

2lm, gm am

1

rIf 1.5, A A A .

r ♪

A thick-walled tube of stainless steel (k = 21.63 W/mK) withan inner diameter (ID) of 0.0254 m and an outer diameter (OD)of 0.0508 m is covered with a layer of asbestos (k = 0.2423W/mK). The inside wall temperature of the pipe is 811 K andthe temperature at the interface of between the metal andinsulation is 805 K. If the temperature at the outer surface ofthe asbestos layer is 310 K, calculate the following:

a. The steady state heat loss per unit length of the tube.b. The thickness of the asbestos layer.

31

32

L

r


▪ isotropic cylinder material with constant properties

▪ slender geometry (r << L) --> 1D heat transfer along the r-direction (radial)

▪ constant cross-section area


shell thickness r=

transfer area 2 rL=

shell volume 2 rL r=

heat heat heat heat


− + =

2 rL r

r r r

q q2 rL 2 rL 0

A A

+

− =

1 d q

r 0r dr A

− =

→

1

1

r 0.0127 mT 811 K=

=

2

2

r 0.0254 m T 805 K=

=

33

1 d qr 0

r dr A

− =

1

q dT r k r C

A dr→ = − =

( )12

C T ln r C

k→ = − +

( )

( )1 1

2 2

BCs: 1 at r r , T T

2 at r r , T T

= =

= =

11

2 1 2

1

rln

rT T

T T rln

r

− =

−

lm

Tq kA

r

= −

→

2 1lm lm lm

2

1

r rA r r

rln

r

2 L, −

=

=

lm

0.0254 0.0127r 0.0183 m

0.0254ln

0.0127

−= =

( )lm

Tq k 2 r L

r

= −

( )lm

q Tk 2 r

L r

= −

( )( )

805 K 811 K2 0.0183

0

W21. m

0.0254 m 0. 127 m63

m K

− =

−

−

q W1,175 or q 1,175 W for a 1-m long tube

L m= =

34

( ) asbestosa q 1,175 W for a 1-m long tube=

steel asbestosassuming steady-state: q q=

( ) o 2 o 2

o 2o

2

r r T Tfor the insulation layer: q k 2 L

r rrln

r

− −

= − −

( )o

W 310 K 805 K1

n

W0.242,175 2

rml

0. 4

3m

025 m

K

− = −

or 0.0482 m=1r 2r

or( ) asbestosb r 0.0228 m or 22.8 mm =

A thick-walled cylindrical tubing of hard rubber (k =0.151 W/mK) having an inside radius of 5 mm and anoutside radius of 20 mm is being used as a temporarycooling coil in a bath. Ice water is flowing rapidly inside,where the wall temperature is 274.9 K. The outsidesurface temperature is 297.1 K. If a total of 52.74 kJmust be removed from the bath in one hour, how longmust be the tubing to be used as cooling coil?

35

Using shell balance, develop an expression for the steady-state temperature profile through a container cover as shownbelow. The top and bottom surfaces as well as both ends ofthe cover are well insulated.

36

Documents

Second Semester AY 2019-2020 · A thick-walled tube of stainless steel (k = 21.63 W/m K) with an inner diameter (ID) of 0.0254 m and an outer diameter (OD) of 0.0508 m is covered