Chapter 13
13.1
500
13.2
2Btu
h in. F
(2 ) 6 1 1 25.13 25.13 Btu
1 1 25.13 25.13
3 in. h °F


Now there is convection through right end
[k(3)] =
2
2
25.13 25.13 0 0
Btu
1 (2 )
25.13 25.13 0 1
kip in. °F





[k(3)] =
25.13 25.13
25.13 37.70



Now q* = 0, Q = 0
(1)
{}f
=
(2)
{}f
= 0
(3)
{}f
= h T A
0
1
= (1) (0°F)
22
0
1
=
0
0
Assemble equations and boundary condition t1 = 200°F
1
2
200 F
25.13 25.13 0 0
50.26 25.13 0






t
t
1
0
x
F
2
3
4
50.26 25.13 0
50.26 25.13
Symmetry 37.70








t
t
t
=
13.3
xx
AK
L
2Btu
h in. F
(2in. ) (3 )
30 in.
1
5
Btu
h °F
h = 0
6
hPL
= 0, hA = 0, h T P L = 0
[k(1)] =
11
1
11
5
= [k(2)]
(1)
{}
q
f
=
0
L
q* [N]T dx =
30
0(0.1 )
Lx
1X
L
X
L
dx
= 0.1
2
2
6
3
Btu
h
L
L
=
15
30
(2)
{}
q
f
=
15 45
30 45
=
60
75
Heat flow
1
15
F
1
1
1 1 0 50 F
t
+ 75 =
1
5
(– t2 + t3) (2)
1
5
5
5
1
Solve for F1 using Equation (A) above
15 + F1 =
1
5
(t1 – t2)
1
11
2
1
00
(4)
{}f
= qA
0
1
=
2
2
0
W
500 (0.1m ) 1
m


 
0
1
{f }’s = 0
Assemble global equations with t1 = 500°C and t4 = 100°C
0
0
50 50 0 0
50 10 10 0



1
2
500 C
t
t
13.8 A composite wall is shown below. For element 1, let Kxx = 5
W
mC
, for element 2 let
Kxx = 10
W
mC
, for element 3 let Kxx = 15
W
mC
. The left end has a heat source of 600 W
[k(1)] =
1 –1 5 –5
(0.1)(5)
0.1 –1 1 –5 5
[k(2)] =
1 –1 10 –10
(0.1)(10)
0.1 –1 1 –10 10
[k(3)] =
1 –1 15 –15
(0.1)(15)
0.1 –1 1 –15 15
(2) (3) (1) 1
1
2
3
44
600 5 –5 0 0
0–5 15 –10 0
00 –10 25 –15
0 0 –15 15
x
t
t
t
Ft
2
0.1 0.1 10 m
13.9
Find T1 T2, T3, T4, Q (heat transfer through the double pane) (use A = 1 cm2)
1 1 1 0
g
AK hA
2W
mK 2
1m 0.80 1 1 1 0
W
200 200 K
[k(2)] =
11
11
A
AK
L
=
2W
mK
1m 0.025 11
11
0.01m
=
2.5 2.5 W
2.5 2.5 K
1 1 0 0
g
AK hA
2W
mK 2
1m 0.80 1 1 0 0
W
508
1
2
3
4
287.5 14.5 C
287.3 14.2 C
265.1 7.88 C
264.8 8.15 C
TK
TK
TK
TK
13.10
(1)
xx
W
K 0.20 mC

(2)
xx
W
K 0.038 mC

(3)
xx
W
K 0.12 mC

xx
hend
1 1 2 1
AK hpL
1 1 1 2
L6
Assumptions Determine temperatures at inner/outer surfaces and at interfaces. Also,
determine heat transfer through wall.
A = 1 m2
(1) 1 1 1 0
1(0.20)
K hA
1 1 0 0
0.025

(1)
2
8 8 1 0 18 8 W
K 10(1)
8 8 0 0 8 8 m


(2)
2
11 0.42 0.42
1(0.038) W
K11
0.09 m
0.42 0.42







0


(3)
1 1 0 0 9.6 9.6
1(0.12) W

1

400

{fh}lt end = hL T∞L A
1
0



=
200
0



F = Kt
1
2
3
4
t
200 18 8 0 0
t
0 8 8.42 0.42 0
t
0 0 0.42 10.02 9.6
t
400 0 0 9.6 29.6









1
2
3
4
t 18.54 C
t 16.72 C
t 17.76 C
t 19.27 C


1
(1) (1)
x xx 2
2
t18.54
1 1 1 1 W
q K 0.20 14.56
t16.72
L L 0.025 0.025 m
 
 


13.11
Unit definition
Given in problem statement (Mathcad used to solve this one)
Ltotal = 20 cm
W
mC
h1 = 50
2
W
mC
h2 = 80
2
W
mC
Tleft end = 100°C
Tinf = 20°C
dia = 0.5 cm
radius =
dia
2
A = 0.00002 m2
radius = 0.0025 m
W
0.0136 0.007 W
0.0209 0.0026 W
512
Set up equations to solve for temperature
1,1 0,0 0,1
1,0 1,1 0,0 0,1
1,0 1,1 0,0 0,1
1,0 1,1
1 2 2
2 2 3 3
3 3 4 4
44
00
0
0
00
k k k
k k k k
k k k k
kk
Guess
t2 = 75°C
Given
2 1,0 left end
3
4
5
t
t
2
3
t
t
()F K T
F
2
3
t
t
4
5
t
t
2
3
4
5
t
t
t
t
=
30.717
51.077
69.113
42.348
°C
left end
T
13.12 A tapered aluminum fin (k =
200W
m °C
) shown in Figure P13-12, has a circular cross-section
with base diameter of 1 cm and tip diameter of 0.5 cm. The base is maintained at 200°C and
W
13.13
(1) 1 –1 1 0
[] –1 1 0 0
xx
AK
k hA
L
Use unit area, A = 1 ft2
2
2
Btu
(1ft ) 0.10 1 –1 1 0
h ft °F Btu.1ft
2
(2) 1 –1 0.048 – 0.048
(1ft )(0.02)
[] 5 in. –1 1 – 0.048 0.048
12 in. (1ft)
k
2
(3)
2
1 –1 0 0
(1ft )(0.8) Btu
0.5 in. –1 1 0 1
h ft °F
12 in. (1ft)
(3) 19.2 –19.2
[]–19.2 23.2
k
2
1 1 105
Btu

A = 0.025 Kxx = 20 h = 20 L = 20
100
A
[k1] =
1 –1 0 0
–1 1 0 0
0 0 0 0
0 0 0 0
k1 =
1
xx
K
Ak
L
[k1] =
25 –25 0 0
–25 25 0 0
0 0 0 0
0 0 0 0
0 0 0 0
xx
K
0 0 0 0
0 1 –1 0
xx
K
0 0 0 0
0 25 –25 0
0 0 1 –1
0 0 –1 1
0000
0 0 0 1
[k3] = [k3c] + [k3h] [k3] =
0 0 0 0
0 0 0 0
0 0 25 –25
0 0 –25 25.5
100
0
0
0
inf
0
0
0
1
0
0
0
10
100
0
0
10
[K] =
25 –25 0 0
–25 50 –25 0
0 –25 50 –25
0 0 –25 25.5
232
228
13.16 A hot surface of a plate is cooled by attaching fins (called pin fins) to it. The surface of the
plate (left end of the fin) is at 90°C. The typical fin is 6 cm (0.06 m) long and has a cross-
1 –1 2 1
–1 1 1 2
6
xx
AK hPL
L
–6 1 –1
5 10 400
–1 1
0.02
21
10 0.006 0.02
12
6
0.1004 – 0.0998 W
[k(3)] =
0.1004 – 0.0998 W
– 0.0998 0.10045 C
1 1 0.012
10 20 C 0.006m 0.02 m W
hT PL
0.013

1
2
3
4
90 C
0.1004 – 0.0998 0 0
2(0.1004) – 0.0998 0
2(0.1004) – 0.0998
Symmetry 0.10045
t
t
t
t
1
0.012
2(0.012)



F
1
2
F
F
=
1
2
11
11
xx t
KA
t
L
(7)
11
L
13.18
See Equation (13.4.28)
Now convection from left end
10
left
h
0
13.19
[k] = [B]T [D] [B] t A +
s
h[N]T [N] ds