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the area of the slab, the
temperature difference,
Δ
T
, between the back and the front; and
inversely proportional to
Setting:
A rectangular slab of thickness
Δ
x
and with an area
A
.
The front side of the slab is at a
temperature
T
;
the back side has a somewhat
different temperature,
T+
Δ
T
.
We are trying to calculate the heat
flow rate, the amount of heat flowing
through the slab per unit time,
H =
Δ
Q/
Δ
t
.
the thickness of the slab,
Δ
x
.
H
should also somehow depend on properties of the material of the slabE
We expect
H
to be proportional to
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View Full Document Bringing all the parts together:
H =
Δ
Q/
Δ
t
– heatflow rate is measured in Joules/second,
J/s
, or Watts,
W
.
Thermal conductivity,
k
, is measured in
W/(m
⋅
K)
.
x
T
kA
H
Δ
Δ
−
=
The coefficient
k
reflects specific
properties of the material of the
slab and is called
thermal
conductivity
Thermal conductivities of different materials.
Best heat conductor – Copper; use it when you build a heat sink, as a
material for pipes in your cooling system, a radiator.
Worst heat conductors are the best insulating materials – air,
fiberglass (layers in the walls of houses in cold regions), styrofoam
(cups for your hot coffee).
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View Full DocumentHeatflow rate equation
Continuing the analogy: electric resistance,
R
, is analogous to
)
/(
kA
x
T
x
T
kA
H
Δ
Δ
−
=
Δ
Δ
−
=
Is similar to the Ohm’s law:
R
V
I
=
The current,
I =
Δ
q/
Δ
t
,
amount of charge per unit time,
is analogous to the heatflow rate,
H =
Δ
Q/
Δ
t.
The voltage,
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This note was uploaded on 02/05/2011 for the course BILD 2 taught by Professor Schroeder during the Spring '08 term at UCSD.
 Spring '08
 SCHROEDER

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