H3 Conduction 2

(3) Tube wall with the BC of the 1 st kind •Heat Equation:

1 d    dT    r   = 0 r  dr    dr   •BCs: (1st kind)

T ( r 1 ) = T s1  T ( r 2 ) = T s 2

  r   ln   + T s1 T ( r ) = ln (r 1 r 2 )  r 1  T s1 − T s 2

  r   •or  T ( r ) = ln   + T s 2 ln (r 1 r 2 )  r 2   T s1 − T s 2

•Heat Equation:

1 d    dT    r   = 0 r  dr    dr  

T ( r 1 ) = T s1 •(1st kind) T ( r  ) = T  s2  2 •BCs:

• Solution: 1 d    dT    r   = 0 r  dr    dr  

r 

dT  dr 

Using the two BC’s, C  =

= constant, C  T s1 − T s 2

ln (r 1 r 2 )

dT  dr 

=

 D = T s1 −

C  r 

T  = C ln r + D

(T s1 − T s 2 )ln r 1 ln (r 1 r 2 )

Thus,

  r   T s1 − T s 2   r   ln  + T s1 •or  T ( r ) = ln   + T s 2 T ( r ) = ln (r 1 r 2 )  r 1  ln (r 1 r 2 )  r 2   T s1 − T s 2

• Heat flux & heat transfer rate: qr ′′ = − k 

dT  dr 

= − k 

C  r 

=−

k (T s1 − T s 2 ) r ln (r 1 r 2 )

~

1 r 

 kC   ′ ′ qr  = 2π  rLqr  = −2π  rL  ~ constant   r    qr  =

qr  =

2π   Lk (T s1 − T s 2 ) ln (r 2 r 1 )

~ (area)

2π  (r 2 − r 1 ) L

k 

ln (r 2 r 1 )

(r 2 − r 1 )

k  

∆T 

(T s1 − T s 2 )

area, 2π  r  L where r  =

 A L

Standard form for q r

r 2 − r 1

ln (r 2 r 1 )

•“logarithmic mean radius” r   L

•Heat Equation:

1 d    dT    r   = 0 r  dr    dr   •BCs: (1st kind)

T ( r 1 ) = T s1  T ( r 2 ) = T s 2

- What does the form of the heat equation tell us about the variation of the heat flux qr ′′ with r  in the wall? - How does the transfer rate qr  vary with r  ? - Is the foregoing conclusion consistent with the energy conservation requirement?

- What does the form of the heat equation tell us about the variation of qr ′′ with r  in the wall? - How does the heat transfer rate qr  vary with r  ? 1 d    dT    r   = 0 r  dr    dr   qr ′′ = − k 

dT  dr 

~

kC  r 

r 

dT  dr 

= constant  = C

kC     qr  = 2π  rLqr ′′ ~ 2π  rL  ~ constant   r   

 Heat flux depends on r, but the heat flow rate is independent of r. - Is the foregoing conclusion consistent with the energy conservation requirement?

•* Summary: •Temperature profile:

  r   ln   + T s1 T ( r ) = ln (r 1 r 2 )  r 1  T s1 − T s 2

  r   ln  + T s 2 T ( r ) = ln (r 1 r 2 )  r 2   T s1 − T s 2

•Heat Flux :

•Heat Transfer Rate:

•where

qr ′′ = −k 

dT  dr 

=

k  r ln (r 2 r 1 )

qr  = 2π  rLqr ′′ =  A L  A L = 2π  r  L L •and

(T s1 − T s 2 )

k 

(r 2 − r 1 ) r  =

(T s1 − T s 2 )

r 2 − r 1

ln (r 2 r 1 )

•“ LMR”

(4) Tube wall with the BC of the 3rd kind

•Heat Equation:

1 d    dT    r   = 0 r  dr    dr  

T  1 and T  2 are prescribed. ∞

∞

But T s1 and T s2 are not known a priori.

•BCs: (3rd kind)

At r  = r 1; qr  =  A1h1 (T ∞1 − T s1 )  At  x = r 2 ; qr  =  A2h2 (T s 2 − T ∞ 2 ) •  A1=2π  r 1 L (inner surface area) •  A2=2π  r 2 L (outer surface area)

•BCs: (3rd kind)

•Heat Equation:

1 d    dT    r   = 0 r  dr    dr  

At r  = r 1; qr  =  A1h1 (T ∞1 − T s1 )  At  x = r 2 ; qr  =  A2h2 (T s 2 − T ∞ 2 )

• Solution: •- The temperature profile will be same as the previous case. i.e.,

  r   ln   + T s1 T s1 and T s2 are not known a priori T ( r ) = and to be determined ln (r 1 r 2 )  r 1  T s1 − T s 2

- Recall that the “heat transfer rate” is constant . i.e.,

qr  =  A1h1 (T ∞1 − T s1 ) = 2π  r 1 Lh1 (T ∞1 − T s1 ) qr  =  A L

k 

(r 2 − r 1 )

(T s1 − T s 2 ) = 2

r  L L

π  

k 

(r 2 − r 1 )

•------------ (1)

(T s1 − T s 2 )

qr  =  A2h2 (T s 2 − T ∞ 2 ) = 2π  r 2 Lh2 (T s 2 − T ∞ 2 )

•-------- (2)

•------------ (3)

- T s1 and T s2 can be determined from equations (1)~(3) although complicated. - Temperature profile itself, however, is less interesting than the heat transfer rate, q r 

T  − T ∞ 2 qr  = ∞1 = U  A(T ∞1 − T ∞ 2 ) expression for R tot  , U, A? (A1 or A2?)  Rtot 



 Rtot  = 

1

 2π  r 1 Lh1

+

ln (r 2 r 1 ) 2π   Lk 

+

 1 1 = =  2π  r 2 Lh2  U 1 A1 U 2 A2 1

 1 r  ln (r 2 r 1 ) r   =  A1 Rtot  =  + 1 + 1  U 1 k  r 2 h2   h1 1

 r  r  ln (r 2 r 1 ) 1  =  A2 Rtot  =  2 + 2 +  U 2 k  h2   r 1h1 1

Too complicated to remember! Any easier way?

• Recall the slab problem. T ∞1

T  − T ∞ 2 qr  = ∞1 = U  A(T ∞1 − T ∞ 2 )  Rtot  where  Rtot  =

1  AU 

=

1  Ah1

+

 L  Ak 

+

h1

h2

1

T ∞ 2

 Ah2

Then, for the present problem in cylindrical geometry

 1

 Rtot  = 

 A1h1

+

(r 2 − r 1 )  A L k 

+

1 

1

1

= =   A2h2   A1U 1  A2U 2

  r 2 − r 1    L where  A L = 2π  r  L L = 2π    ln (r 2 r 1 ) 

1 U 1

=  A1 Rtot 

1 U 2

=  A2 Rtot 

U 1 : Overall heat transfer coefficient based on the inner surface U 2 : Overall heat transfer coefficient based on the outer surface

(5) Composit e Wall w ith Negligible Contact Resistance

qr  = 2π  r 1 Lh1 (T ∞1 − T s1 )

qr  = qr  = qr  =

2π   Lk  A ln(r 2 r 1 ) 2π   Lk  B ln(r 3 r 2 ) 2π   Lk C  ln(r 4 r 3 )

(T s1 − T 2 ) (T 2 − T 3 )

(T 3 − T 4 s )

qr  = 2π  r 4 Lh4 (T s 4 − T ∞ 4 )

qr  =

T ∞1 − T ∞ 4  Rtot 

= U 1 A1 (T ∞1 − T ∞ 4 ) = U 2 A2 (T ∞1 − T ∞ 4 )

 Rtot  =  Ri +  R A +  R B +  RC  +  Ro

= = =

1  Ai hi

+

(r 2 − r 1 ) (r 3 − r 2 ) (r 4 − r 3 )  A L, Ak  A

1 2π  r i Lhi 1 U i Ai

=

+

+

 A L, B k  B

ln (r 2 r 1 ) 2π   Lk  A

+

+

 A L,C k C 

ln(r 3 r 2 ) 2π   Lk  B

+

+

1  Aoho

ln (r 4 r 3 ) 2π   Lk C 

+

1 2π  r o Lho

1 U o Ao

 Note that  Rtot is a constant that is independent of position r .  But, U is tied to specification of an interface. U i : overall heat transfer coefficient based on the inner surface U o : overall heat transfer coefficient based on the outer surface