Chapter2_Conduction Heat Transfer (1)

Chapter Two Conduction Heat Transfer

Fourier’s Law of Heat Conduction Heat Flux (Wm-2)

q  x =

Q x  A

= −k 

dT  dx

Heat Transfer (W) Per unit Area (m2) (W/m2)

Thermal Conductivity  (Wm-!-) dT  "  " tem#erature difference (!) d x  "  " distance across section (m)

Q x

= −k A

dT  dx

• • •

Heat is conducted in the direction of decreasin$ tem#erature Thus% the tem#erature $radient is ne$ative &hen heat is conducted in 've x-direction -ve si$n to ensure that the heat transfer in 've x-direction is 've uantity

Hot face

Cold face

Hot  face temp.

 Q x = − k A

dT  dx

Temp. gradient

Cold face temp.

Conduction  Through

a Flat Slab or Wall  Through a Hollow Cylinder  Through a Hollow Sphere  Through Solids in Series  Plane

walls in series  Multilayer Cylinders  Combined

Convection and Conduction and Overall Coefficients  Conduction with nternal Heat !eneration

Plane wall From Fouriers *a&+

dT   Q x = − k A dx

 x 2

∫ 

 x1

T 1

T 2

Q x 

T 2

∫ 

Q  x dx = − kAdT  T 1

Q  x [ x2 −  x1 ] = − k A[T 2 − T 1 ] [T 1 − T 2 ]  Q x = k A  x 2 −  x1 Template copyright 2005

 x1

 x2

* T 1> T 2

Q  x  x 2

= −k A

dT  dx T  at  x2

dT      (  x ) dx = − ∫  k A  dx ∫    dx    x T  at  x 1

4/12/15

Q

1

Template copyright 2005

 T 1 − T 2   The rate of heat transfer :Q x = k A    x2 −  x1  Note that

Thermal Resistance

T 1 > T 2

R T!:

"con#$ction resistance%

Therefore 4/12/15

R TH  &

∆ x /k  '

Template copyright 2005

T 1

 R TH T 2

Calculate the heat loss per square meter of surface area for an insulating wall composed of 25.4 mm thick !er insulating !oard" where the inside temperature is #52.$ % and the outside temperature is 2&$.1 %.

(rom 'pp '.) the thermal con#$cti*ity of fiber ins$lating boar# is 0.04+ ,/m.-. The thicness 21&0.0254 m: q

k 

 A

=

=

 x2

−

 x1

(T 1

−

T 2 )

105.1 W/m 2



)52. "nsi#e%

23.1 "$tsi#e%

25.4 mm on fiber ins$lating boar#

0.048 =

0.0254

(352.7

−

297.1)

Conduction Through a Hollow Cylinder Consider the hollow clinder with !n inside r!di"s o# r 1 r 1

q

r 2

T 1

q

T 1

 R

T 2

T 2

 A = 2π  rL where

r 2 ln( ) r 1  R = 2π  k L

$

q

=

 Alm

=

 Alm

q=

k Al m  A2

−

T 1

−

r 2

T 2

−

r 1

 R =

 A1

ln( A2 /  A1 ) ( 2π   Lr 2 2π   Lr 1 ) 2π   Lr 2 ln( ) 2π   Lr 1 −

=

T 1 − T 2 ln( r 2 / r 1 ) / 2π k L

=

T 1 − T 2  R

r 2 − r 1 k Alm

=

ln( r 2 / r 1 ) 2π  k L

 Length of Tubing Tubing for Cooling Coil 

% thic&'w!lled clindric!l t"in o# h!rd r"er h!*in !n inside r!di"s o# 5mm !nd !n o"tside r!di"s o# 20 mm is ein "sed !s ! tem+or!r coolin coil in ! !th. ,ce w!ter is #lowin r!+idl inside !nd the inside w!ll tem+er!t"re is 274.9-. he o"tside s"r#!ce tem+er!t"re is 297.1-$ % tot!l o# 14.5W m"st e remo*ed #rom the  !th  coolin coil. ow m!n m o# t"in !re needed

rom %++endi %.3$ k     0.151 W/m.-  r 1 = r 2 =

q=

5

= 0.005 m

1000 2 1000

T 1 − T 2 ( 2π  k L) ln( r 2 / r 1 )

= 0.02 m

he c!lc"l!tion will e done #irst #or ! lenth o# 1 m t"in  A2

 Alm

=

q

kAlm

=

=

−

 A1

ln( A2 /  A1 ) T 1 r 2

−

T 2

−

r 1

=

=

0.1257

−

0.0314

=

ln(0.1257 / 0.0314)

0.151(0.082)(

274.9 0.02

−

−

0.08 m 2

297.1

0.005

)

15.2W  

−

The negati*e sign in#icates that the heat flow is from r2 on the o$tsi#e to r1 on the insi#e. ince 15.2 , is remo*e# for 1 m length the nee#e# length is:

length

=

14.5W   15.2W   / m

=

0.94m

Conduction Through a Hollow Sphere r 2

q

r 1

T 1

 R

T 2

T 1

q T 2

 A = 4π  r 2

T 1 − T 2 T 1 − T 2 = q=  R (1 / r 1 − 1 / r 2 ) / 4π  k 

(

1

−

1

r 1 r 2 where  R = 4π  k 

)

CONDUCTON TH!OU"H SOLDS N S#!#S

Plane $alls in Series  A T 1

C 

B

q=

T 2 T 3

q

T 1 − T 4 ∆ x A ∆ x B ∆ xC  + + k  A A k  B A k C  A

=

T 1 − T 4

− = T 1 T 4 ∑ R  R A + R B + RC 

T 4

where

q

T 1

 R A

T 2

∑ R =

 R B

Electrical analogy 

∆ x A

k  A A

T 3

+

∆ x B

k  B A

+

 RC 

∆ xC 

k C  A

T 4

 EXAMPLE 4.3-1 Heat Flo Through an !n"ulate# $all of a Col#  %oo&

% cold'stor!e room is constr"cted o# !n inner l!er o# 12.7 mm o#  +ine$ ! middle l!er o# 101. mm o# cor& o!rd$ !nd !n o"ter l!er o# 7.2mm o# concrete. he w!ll s"r#!ce tem+er!t"re is 255.4- inside the cold room !nd 297.1- !t the o"tside s"r#!ce o# the concrete. 6se cond"cti*ities #rom %++endi %.3 #or +ine$ 0.151 #or cor& o!rd$ 0.0433 !nd #or concrete$ 0.72 W/m.-. C!lc"l!te the he!t loss in W #or 1 m2  !nd the tem+er!t"re !t the inter#!ce etween the wood !nd cor&  o!rd.

he resist!nce #or e!ch m!teri!l !re  RC  =  R B =  R A =

∆ x A k  A A

∆ x B k  B A

∆ xC  k C  A

= = =

7.2 × 10

−3

= 0.100

0.72 101 . × 10

−3

0.0433 12.7 × 10 0.151

−3

= 2.34

= 0.0841

q=

= = =

T 1 − T 4 T 1 − T 4 = ∆ x A ∆ x B ∆ xC   R A +  R B + RC  + + k  A A k  B A k C  A 255 255.4 − 297 297.1 0.084 + 2.34 + 0.1 0.1 ' 41.7 2.530 530 − 1.48 W

o c!lc"l!te the tem+er!t"re T 2$ q=

T 1 − T 2  R A

− 1.48 =

255.4 − T 2

0.084 084 T 2 = 25.79 - 

%ultilayer Cylinders r 4

r 3

q

r 2

 A C   B

q=

ln( r 2 / r 1 ) 2π  k  A L

=

+

T 1 − T 4  R A +  R B +  R C 

r 1

T 1

T 2

T 1 − T 4 ln( r 3 / r 2 ) 2π  k  B L

=

T 3

+

T 4

ln( r 4 / r 3 ) 2π  k C  L

T 1 − T 4

∑  R

∑ R =

ln( r 2 / r 1 ) 2π  k  A L

+

ln( r 3 / r 2 ) 2π  k  B L

+

ln( r 4 / r 3 ) 2π  k C  L

 EXAMPLE 4.3-' Heat Lo"" fro& an !n"ulate# Pi(e

% thic&'w!lled t"e o# st!inless steel (%) h!*in ! k      21.3 W/m.& with dimensions o# 0.0254m , !nd 0.0508m  is co*ered with ! 0.0254m l!er o# !sestos (:) ins"l!tion$ k    0.2423 W/m.&. he inside w!ll tem+er!t"re o# the +i+e is 811- !nd the o"tside s"r#!ce o# the ins"l!tion is !t 310.8-. or ! 0.305m lenth o# +i+e$ c!lc"l!te the he!t loss !nd !lso the tem+er!t"re !t the inter#!ce etween the met!l !nd the ins"l!tion.

q=

T 1 − T 3  R A + R B

he resist!nces !re

 R A =

ln( r 2 / r 1 ) 2π  k  A L ln(

=

=

ln( d 2 / d 1 ) 2π  k  A L

0.0254

)

0.0127 2π  (21.3)(0.305 305)

-/W = 0.0173 -/W

 R B

=

ln( r 2 / r 1 ) 2π  k  B L ln(

=

=

ln( d 2 / d 1 ) 2π  k  B L

0.101

)

0.0508 2π  (0.2423)(0.0.305)

= 1.493 -/W

he he!t tr!ns#er r!te is

q=

q=

T 1 − T 3  R A + R B 811 811 − 310 310.8

0.0173 + 1.493 493 = 331 331.7 W

q=

T 1 − T 2  R A

331 331.7 =

811 811 − T 2

0.0173 T 2 = 805 805.5 -