ChE 120B
1  1
Heat Transfer — Connection to Thermodynamics
Thermos – How much heat is removed from 1 liter of
water as it cools from 90
°
C to 80
°
C?
This is a question that we can answer via thermodynamics.
Energy conservation states that
energy in – energy out = net change in energy of system
if
UU
U
−=
Δ
HH
H
Δ
and from definitions of heat capacity
vv
fp
p
dU
mC dT
dU
mC dT
==
v
ff
ii
TT
p
UmC
d
T
HmC
d
T
Δ=
∫∫
which for constant
v
,
p
CC
gives
()
( )
v
f
ip
f
i
Um
CTT
Hm
−
−
Values of
v
,
p
depend on the material.
For an ideal gas
v
p
CCR
=+
Most liquids and solids, under normal conditions, are incompressible, or have a constant
density as a function of pressure.
This means that constant volume and constant pressure
changes are the same and
v
p
=
and
v
p
UC
d
T HC
d
T
=
Often the heat capacity of a material changes with temperature; hence, we often need an
average
p
C
for calculations.
From the 1
st
Law,
UQW
Δ
, and mainly we will be dealing
with cases where
0
W
=
.
However, back to the thermos. We know that eventually the hot
drink in the thermos will cool off.
Thermodynamics will not distinguish between a good
thermos that takes a long time to cool off and a poor thermos.
What is important is the heat
transfer rate – not just
Q
, but
Q
±
to be labeled
q
or
Q
later today – somewhat confusing),
the rate of heat/time being exchanged.
0
t
QQ
d
t
Δ
=
∫
±
Once we know
Qt
±
we can calculate
Q
if
Q
±
is constant
QQt
=
Δ
±
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1  2
The rate of heat transfer per unit area normal to the direction of the heat transfer is heat flux
2
(W/m )
Q
q
A
=
±
±
in which
A
is the heat transfer area.
In English, watts
2
is Btu/hr ft .
q
±
The heat flux can vary
with time, temperature, and position.
Example 1:
A 10 cm diameter copper ball is to be heated from 100
°
C to 150
°
C in 30 minutes.
C
u
ρ
=
8950 kg/m
3
and
p
C
=
0.395 kJ/kg
°
C.
What is: a) the total amount of heat transfer to
the ball, b) the average heat transfer rate, and c) the average heat flux at the ball surface?
a)
The total energy requirement is from thermodynamics
( )
pf i
UH
Q
m
C
T
T
Δ≅
Δ==
−
()
3
4
VV
3
p
f i
CT T
C
rT T
ρρ
π
=
−=
−
(
)
3
33
4
v=
8950 .1m
4.7kg
3
66
mr
D
ππ
⎛⎞
==
=
=
⎜⎟
⎝⎠
k
J
g
k
4.7 kg
0.395
C 150 100 C
92.6 kJ
Q
=°
−
°
=
b)
While the average rate of heat transfer is
ave
kk
92.6 J
J
s
1800 s
.0514
51.4
Q
QW
t
=
−
=
Δ
±
c)
The heat flux is defined as the heat transfer/time
⋅
area
2
ave
ave
ave
2
2
51.4
1640
/ m
.1
QQ
W
qW
AD
=
=
±±
These are average values of
Q
±
and
q
are likely not to be equal to the actual instantaneous
value
Q
±
or
q
during the process.
Example 2:
Heat loss from ducts in basement
A 5 m long section of an air heating system of a house passes through an unheated
basement.
The duct is 20 x 25 cm.
Hot air enters the duct at 200 kPa and 60
°
C at an average
velocity of 5 m/s.
The temperature of the air drops to 54
°
C through the loss of heat to the
basement.
What is the rate of heat loss?
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 Winter '10
 Zasadinski
 Heat, Heat Transfer, Joseph Fourier, heat flux

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