~ denote the heat flow vector
Let Q
~ = cal/sec/m
|Q|
2
~ denote the heat flow vector
Let Q
~ • ~n dS is cal/sec of heat flowing through a section of surface
Q
of area dS
~ denote the heat flow vector
Let Q
RR
~ • ~n dS is cal/sec of heat flowing through the entire
H = SQ
surface S
Let u = u(x, y, z) be the temperature at point (x, y, z). The
temperature gradient points in the direction of increasing temperature
Fourier’s Law:
~ = −k∇u
Q
ZZ
ZZ
~ • ~n dS = −k
Q
H=
S
∇u • ~n dS
S
In an interior section of volume ∆V , the heat gain is proportional to the mass and the change in temperature
Heat Gain = c · ∆u · ρ ∆V
Heat Loss = −cρ∆u∆V
∂u
∆u
∆V ≈ −cρ
∆V
∆t
∂t
This is the rate that the heat is leaving one small interior section
Rate of Heat Loss = −cρ
Total rate that the heat is leaving the solid is:
ZZZ
−cρ
V
∂u
dV
∂t
ZZ
ZZZ
∇u • ~n dS =
−k
S
−cρ
V
∂u
dV
∂t
ZZ
ZZZ
∇u • ~n dS =
−k
ZZZ
−cρ
S
V
ZZZ
∇ • (∇u) dV =
−k
V
∂u
dV
∂t
−cρ
V
∂u
dv
∂t
¶
ZZZ µ
∂u
k∇2 u − cρ
dV = 0
∂t
V
k∇2 u − cρ
∂u
=0
∂t
k
∂u
= ∇2 u
∂t
cρ
∂u
k
= ∇2 u
∂t
cρ
q
Let α =
k
cρ
∂u
= α 2 ∇2 u
∂t
The Heat Equation
µ 2
¶
2
2
∂u
∂
u
∂
u
∂
u
= α2
+ 2 + 2
∂t
∂x2
∂y
∂z
In the special case where u = u(x, t), the heat equation becomes:
2
∂u
2∂ u
=α
∂t
∂t2