1. Heat loss from a house
2. Heat loss from a pipe
= 1.6810-5 cal/cm2-s-K
q = 1.6810-5*(18 + 14) = 5.3910-4 cal/cm2-s
3. Wind chill
The wind chill heat loss is
When no wind blowing, the heat loss is
Chapter 20 Heat transfer page 20-2
0
6
8
10
02468
v
1/2
s
– T
db
)
-3/4
4. Energy balance
t
U + v2
2 = –·v
U + v2
2 – ·q + (v·g) – ·pv – ·[·v]
First we expand the left term and yield
t
2 + v·
2 +
2
t + ·v = – ·q + (v·g) – ·pv – ·[·v]
But from equation of continuity,
t + ·v = 0
Therefore,
Chapter 20 Heat transfer page 20-3
By expanding the left term of (3), we have
L =
U
t + v·U = U
t – U
t + v·U
5. Thermal conductivity measurement using hot wire
From Fourier field equation in cylindrical coordinates,
T
2T
T(t1) – T(t2) = IE
4kL ln
t2
=> k = IE
4L ln(t1/t2)
T(t1) – T(t2)
6. Spinning of polymer melt
In the rubbery region (the core),
T1
2T1
7. Hollow fiber radiator
The overall heat transfer coefficient U will be the sum of 3 resistances:
kd
D = 1.62
Dl
by analogy,
d2v
1/3