This is the 6th heat transfer class of my simulated course of Heat Transfer I

1. 6th class of Heat Transfer I
Thursday, January 30, 2014:
From textbook: J.P. Holman’s Heat
Transfer, 7th edition:
Fins (Section 2-10);
Thermal Contact Resistance
(Section 2-11)

2. Fins: Motorcycle Engine

3. Example Problem (fins): 2-67
Lcorrected = L + t/2 = 0.0064 m + 0.0016 m/2=
0.0072 m
r1 = 0.025 m /2 = 0.0125 m
r2c = r1 + Lc = 0.0125 m + 0.0072 m = 0.0197 m
r2c / r1 = 0.0197 m/ 0.0125 m = 1.5760
Am = t (r2c - r1 )= (0.0016 m)( 0.0197 m – 0.0125 m)
Am = 1.1520 x 10-5 m2
Lc
3/2 ( h/ (k Am ) ½ =

4. Example Problem (fins): 2-67
(concluded)
= (0.0072)3/2 (23/[(200)(1.1520 x 10-5 )])1/2 =
=0.0610
From Figure 2-12 in textbook: η = 98% = 0.98
qmaximum = 2π (r2c
2 – r1
2 ) h (To - T∞ )
= 2π [(0.0197)2 – (0.0125)2 ](23)(150 – 15)
= 4.5230 W
qactual = (4.5230 W)(0.98) = 4.4325 W
The heat lost by the fin is 4.43 Watts.

5. Example of Thermal Contact
Resistance (cylindrical electronic
component inside of heat sink)

6. Example Problem (thermal contact) 2-
9
Two 3.0 cm diameter 304 stainless-steel bars, 10
cm long, have ground surfaces and are exposed
to air with a surface roughness of about 1
micrometer. If the surfaces are pressed together
with a pressure of 50 atm and the two-bar
combination is exposed to an overall
temperature difference of 100 degrees Celsius,
calculate the axial heat flow and temperature
drop across the contact surface.

7. Solution Example Problem (thermal
contact) 2-9
See page 28, Equation 2-4: Heat flow = thermal
potential difference/thermal resistance, or
q = ∆Toverall /∑Rth
From page 34, Eqn. 2-12,
Rth= ∆x/kA (for conduction resistance), Rc =
1/hc A (for the contact resistance)
but there are three resistances in this system: one
conduction resistance in each bar, and one contact
resistance (between the two bars)

8. Solution Example Problem (thermal
contact) 2-9, continued
For the bars: Rth= ∆x/kA = (10 cm)(1m/100
cm)/(16.3 W/m•°C)(π((3.0 cm/2)(1m/100 cm))2 )=
8.679 °C/W.
The contact resistance: Rc = 1/hc A =
(5.28 (10-4 ) m2 ∙ °C/W)/(π((3.0 cm/2)(1m/100 cm))2
= 0.747 °C/W.
The total thermal resistance is then:
∑Rth = (2)8.679+0.747 = 18.105 °C/W.

9. Solution Example Problem (thermal
contact) 2-9, concluded
The overall heat flow is:
q = ∆Toverall /∑Rth = (100 deg C)/(18.105 deg C/W) =
5.52 W
The temperature drop across the contact surface
is: ∆Tc= (Rc /∑Rth ) ∆Toverall =
[(0.747 °C/W)/ (18.105 deg C/W )](100 deg C)
= 4.13 deg C

10. Questions?
References:
Slide 3: Retrieved on Sunday, August 27, 2017:
https://www.comsol.com/blogs/thermal-
contact-resistance-simulation/
Assignment: Read Sections 3-1, 3-2, and 3-3