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University of JordanDepartment of Chemical Engineering

Dr. Ali Al-ShawabkehLecture Notes -10-

Conduction-Convection Heat TransferCritical Insulation Thickness for a Pipe:

Pipe:• A problem of interest is choosing the thickness of insulation to minimize the heat loss for a

fixed temperature difference between the inside ofthe pipe and the flowing fluid far away from the

pipe.

( )32143421

∞′′′

∞

+

−=′

R R

r hk r r

T T q

cond

π π 21

2ln 1

1

where:

1r = outer radius of the pipe (bare pipe outer radius).r = radius of the insulating material.

• Or maximizing the total thermal resistance, t R′ :( )

32143421

∞′′

+

R R

r hk r r

cond

π π 21

2ln 1

• By increasing the insulation radius, r :

− the conduction resistance ( )k r r

π 2ln 1 increases which is a positive effect (since q ′ is decreased).

− while the convection resistancer h π 2

1 decreases which is a negative effect (since q ′ is

increased).

Conclusion:there is a value for r (thickness radius) at which t R′ is maximum or q ′ is minimum.

To find this value:0=

′dr Rd t

( )0

21

2ln 1 =⎟

⎠ ⎞

⎜⎝ ⎛ +

hr k r r

dr d

π π

hk

r =

• (does this value ofhk

r = gives maximum t R′ ?)

q

×

1T

materialinsulating

k ⇒

⇒⇒⇒⇒

⇒

⇒

∞T h

1r r

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r 1r

t R′

r

cr

q ′bq ′

∗r

maxq ′min R ′

b R ′

r 1r cr

q ′bq ′

∗r

maxq ′

==′

)(22

hk r dr

Rd t ???

( ) 0221

2ln

3

2

12

2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎟ ⎠ ⎞

⎜⎝ ⎛

+= k

hhr k

r r

dr d

h

k r

π π π

(concave up), i.e., athk

r = , t R ′ is a minimum.

hk

r c = (called critical insulation radius).

• Practically, it turns out that addinginsulation in cylindrical and sphericalexposed walls can initially cause thethermal resistance to decrease, therebyincreasing the heat transfer rate becausethe outside area for convection heattransfer is getting larger. At some criticalradius, cr , the thermal resistanceincreases again and consequently the heattransfer is reduced.

• For insulation radius less than cr , the heatloss increases with increasing r and forinsulation thickness greater than cr the

heat loss decreases with increasing r .• For an insulated sphere:

hk

r c2=

Note:

− For a thick pipe, cr r 1 ,increasing the insulationthickness alwaysdecreases the heat loss

(+ve effect).− For a thin pipe, cr r 1 :1)- Increasing the insulationthickness increases the q ′ (−ve effect) as far as cr r .2)- q ′ will reach a maximumwhen cr r = ).3)- Further increase in r will decrease q ′ but it is still larger than bq ′ until ∗= r r where bqq ′=′ .4)- Further increase in r will eventually give a +ve insulation effect ( q ′ becomes less than bq ′ ).

1r cr r

maxq ′

q ′bq ′

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Example Problem:

Cooling of an electric resistorA 0.5 W, 1.5 M Ω cylindrical graphite resistor has a diameter of 1 mm and is 20 mm long; it has a

thin glass sheath and is encapsulated in micanite (crushed mica bonded by a phenolic resin). Themicanite serves both as additional electrical insulationand to increase the heat loss. It can be assumed that50% of the R I 2 heating is dissipated by combinedconvection and radiation from the outer surface of themicanite to surroundings at 300 K with h = 16W/m 2⋅K; the remainder is conducted through copperleads to a circuit board. If the conductivity of micaniteis 0.1 W/m ⋅K, what radius will give the maximumcooling effect, and what is the corresponding resistortemperature?

Assume that the resistance of the glass sheath isnegligible.

Solution:Maximum cooling effect can be attained when q is max (i.e., at cr r = ).

m00625.016

1.0 ===hk

r c

W25.05.0%50%50 2max =∗=∗= R I q

( )⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢

⎣

⎡

+

−= ∞

c

c

hr k r r

T T Lq

1ln2

1max π

( )⎟⎟

⎠ ⎞

⎜⎜

⎝ ⎛ +∗+= ∞

c

c

hr k

r r

L

qT T

1ln2

1max

π

( )⎟ ⎠ ⎞

⎜⎝ ⎛

∗+∗+=

00625.0161

1.05.025.6ln

)02.0(225.0

300 T π

K 1.370=T

q

k

⇒⇒⇒

⇒⇒

⇒⇒

∞T h

r

1r

T