Heat Transfer phenomena plays an important role in many industrial and environmental problems. We will devote much time to acquire an understanding of heat transfer effects and to develop skills needed to predict heat transfer rates. In order to get familiar with the subject, please follow the following procedure: 

1. Think of one "Real World Application" of Heat Transfer (in cooking, manufacturing, climate, energy production and conversion, water heating systems, propulsion, internal combustion engines, cryogenic storage equipment, electronic cooling, air conditioning and refrigeration systems, etc.). 

2. State the physical problem. 

3. Sketch the problem with a figure. 

4.  Explain the physics of the process as you understand it.

5.  Identify the important variables, what is known and you can control (independent variables) and what is unknown (dependent variables). 

6.  Identify  the relevant physical laws that apply here. 

 Check out the Webpage http://www.howstuffworks.com/refrig.htm to see the simple explanation of how a refrigerators work. Your goal in this assignment will be to pick a simple or a complicated example and explain ( in a language  that a high school student can understand) how and why your selected "Real World Application" of Heat Transfer  works.

Problem nos (using 4th Edition)

1.11, 1.14a,1.19,1.23,1.31 and 1.33

2.9, 2.17,2.51,3.6(a) and (b),3.8(a),3.11

Design Problem:

You have only 20KW available through solar power to heat the house to 25C on a cold day. The total wall area is 200 m2 and roof area and floor areas are 50 m2 and 40 m2respectively. On a cold day outside temperature is -15C with winds blowing at 30 kmph, the convective heat transfer coefficient is 45 W/m2K for the roof and the walls and about 10 W/m2K for the floor and the inside of the house.  Select an insulation (from commercially available products), and the thickness of the insulation for the walls. You definitely need to put some double pane windows in the house as well. If the window area is about 1 m2, what adjustment you have to make to the thickness of the insulation per window. If you have to allow for 20% heat loss due to insufficient sealing between the doors and windows, would you reduce the number of windows or increase the insulation thickness ?

Answers:
2.9    2000 K/m, -2000 K/m, 2000 K/m, q"=-200Kw/m2, +200 KW/m2 and -200KW/m2
2.17a        k=15.0 W /m K. at 400K b) k=70 W/mK at 380K
2.51       q= 1.8 10^6 W/m3, (c) q conv= 1.8 x 10^5 W/m2K (d) Eout=7.77x 10^7 J/m2
3.6a         h=996 W/m2K  (b) 14.5 W/m2 K
3.8a        29.4 W
3.11 a 

3.11b L=86 mm

Due Date March 12th, 2003.

 3.24, 3.40a, 3.47,3.58,3.73a,3.105

Design of fins 

 An electric heater is made by sandwiching a heating element between two thin plates 40-cm long and 40-cm high. When operating with air at 20C the heater is to transfer 3000 W with a maximum plate temperature of 120C. Propose a suitable system of cylindrical fins. Minimum diameter of the fins should be 3mm with a minimum pitch of 4d. Weight and volume are to be kept low. A fan forces air through the heater with a velocity of 5 m/s so that the convective heat transfer coefficient to the air is 80 W/m2C. 

Answers
3.24

3.40a

3.47
a) Ts= 778.7 C.  b)Ts=1153 , Ti==778.3C c) t=0.0175m, Ti=318.2C
3.58 Will depend on the insulation material you choose
3.73a

3.105

 
 
 

5.4(a),(b),(c),5.11, 5.15,  5.41(a) and 5.72

Answers:

5.11 

5.15
 

5.41a
 

5.72

6.29 and 6.35
7.14, 7.18,7.33a 
Answers
6.29(a) 34.3 W/m²C, (b) 59 W/m²K
6.35 42.5C
7.14 55KW, 31.1KW, 90.7 KW
7.18 31.7C
Hint:

7.33a 64C

7.47(a)(b)(c), 7.62a, 7.70, 8.16a,8.26,8.30a

Answers
7.47 (a) h=235 W/m²K
7.47(b) 0.868 W  (c) 1.019 W

7.62a Hint: Use 7.57 to find h and check Bi number. If Bi number is greater than 0.1 use one term series approximation in section 5.6.2 to find the time to =15.2 s

7.70Using

find h and T. For water h=3559 W/m2K and Ts=18.7C. For Air h=20.1 and Ts=672C
8.16a Tmo =113.5C, Ts,i=60C, Ts,o= 153.5C
8.26 q=-1281W and L=15.4 m
8.30a Tm,o=61.3C, q=1519W

8.51,8.64,8.82,9.18,9.38 and 9.45a

Answers
8.51: Total heat dissipation= 100KW, heat dissipation to air is 2.67KW

8.64 Tm,o =15.7C and q=438W

8.82 L=1.8 m

9.18 qloss= 223 W, cost 43 cent a day

9.38 k=0.56 W/mK, h=7.87 W/m2K, e=0.815

9.45(a) 250W

Assignment #9

11.7, 11.14, 11.18,11.25,11.36a,11.47,11.71ab
No due  Date 

Due Date: Friday May 9th, 2003

11.59, 11.76(a)(b) and (c) , 12.10a, 12.20, 12.30a,b 13.2,13.10

11.59 A= 243 m2
11.76 (a) Tc,o =335.1K (b) L=8.11m (c) 0.998 kg/s
12.10a  2000 W/m2
12.20
 

12.30 (a) e=0.249 (b)e=0.358
13.2 Semicircular groove F1,2= 2/pi, Rectangular groove F123,4 =W/(W+2H), Vgroove F12,3 =sin(q)
13.10
(b) 

Problems 12.79,12.100a,
               13.22,13.29,13.53,13.63,13.81

Answers:

12.79 T=999K
12.100a q=1556 MW/m3 Hint: Find a for the Niobium surface which is 0.2 and e is 0.243
13.22a  q12= 13.4W, G1= 6825 W/m2
13.29 q12= -0.162 W
13.53 (a) 11995 W (b) 191W (c) 983 W, with e₂=1 it is 998W
13.63 T1= 1225K, T3= 1167K
13.81 (a) 
(b) q=1.32 W