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[Korbac]

Just pretend that K is a new unit of heat the universe made up.

[nenjin]

0K is okay, but it should be a bit hotter.

As in, a little hotter than "absolute stasis?"

[Culise]

Just pretend that K is a new unit of heat the universe made up.

Incidentally, if Mechwarrior 4 is to be believed, you only shutdown at 10,000K. Advances in cooling, I guess. Although that's a little suspect since it always cools down to 0K, which seems like a harmful operating temperature as well. Not to mention that 10,000K is about twice the temperature of the sun's surface.

Dem clan maths.

Degrees Kerensky, or Kerenskies for short. ^_^

[nenjin]

+2 internets to you!

[da_nang]

I'd say the temperature reader is off.

Spoiler: MATH (click to show/hide)

An average motherboard has dimensions 300 mm x 250 mm.

Let's say for the sake of comparison, the motherboard is made of water and is at 100ºC. Water has an thermal diffusivity of c = 2.338*10^-5 m²/s.

Heat equation tells us that ∂T/∂t = c(∂²T/∂²x + ∂²T/∂²y), T=(x,y,t), 0 s <= t <= 2700 s (45 min)

Initial condition:
T(x,y,0) = 378.15 K, 0 < x < 0.3, 0 < y < 0.25
T(x,y,0) = 298.15 K at the interval boundaries

Boundary condition:
T(x,0,t) =  T(x, 0.25, t) = 298.15 K for 0 <= x <= 0.3 and t >= 0
T(0,y,t) =  T(0.3, y, t) = 298.15 K for 0 <= y <= 0.25 and t >= 0


Discretize into an MxNxP mesh.
h = (x_i - x_i-1)/M
d = (y_j - y_j-1)/N
n = (t_k - t_k-1)/P

Notation: T(x_i,y_j,t_k) := T_ijk
Approximating with Crank-Nicolson method:
(∂T/∂t)_xiyjtk ~ n^-1(T_ijk - T_ij,k-1)
(∂²T/∂²x)_xiyjtk-½ ~ ½h^-2(T_i+1,jk - 2T_ijk + T_i-1,jk + T_i+1,j,k-1 - 2T_ij,k-1 + T_i-1,j,k-1)
(∂²T/∂²y)_xiyjtk-½ ~ ½d^-2(T_i,j+1,k - 2T_ijk + T_i,j-1,k + T_i,j+1,k-1 - 2T_ij,k-1 + T_i,j-1,k-1)

Insert:
n^-1(T_ijk - T_ij,k-1) = ½ch^-2(T_i+1,jk - 2T_ijk + T_i-1,jk + T_i+1,j,k-1 - 2T_ij,k-1 + T_i-1,j,k-1) + ½cd^-2(T_i,j+1,k - 2T_ijk + T_i,j-1,k + T_i,j+1,k-1 - 2T_ij,k-1 + T_i,j-1,k-1)

Let r = cnh^-2 and q = cnd^-2

2T_ijk - 2T_ij,k-1 = r(T_i+1,jk - 2T_ijk + T_i-1,jk + T_i+1,j,k-1 - 2T_ij,k-1 + T_i-1,j,k-1) + q(T_i,j+1,k - 2T_ijk + T_i,j-1,k + T_i,j+1,k-1 - 2T_ij,k-1 + T_i,j-1,k-1)

Reorder:

-rT_i-1,jk - qT_i,j-1,k + (2+2r+2q)T_ijk - rT_i+1,jk - qT_i,j+1,k= rT_i-1,j,k-1 + qT_i,j-1,k-1 + (2-2r-2q)T_ij,k-1 + rT_i+1,j,k-1 + qT_i,j+1,k-1

Let t_k = [T_1,1,k  T_2,1,k  ...  T_M-1,1,k  T_1,2,k  T_2,2,k  ...  T_M-1,2,k ... T_M-1,N-1,k]^T

Then, the PDE is solved by solving the equation At_k = Bt_k-1 + r_k from k=1 to k = P where r_k is a constant vector of boundary conditions.

t_k = A^-1(Bt_k-1 + r_k)

The numerical solution shows that even water at a lower temperature should have gone down to almost room temperature.

[Philii]

Spoiler (click to show/hide)

I'd say the temperature reader is off.

[spoiler=MATH]An average motherboard has dimensions 300 mm x 250 mm.

Let's say for the sake of comparison, the motherboard is made of water and is at 100ºC. Water has an thermal diffusivity of c = 2.338*10^-5 m²/s.

Heat equation tells us that ∂T/∂t = c(∂²T/∂²x + ∂²T/∂²y), T=(x,y,t), 0 s <= t <= 2700 s (45 min)

Initial condition:
T(x,y,0) = 378.15 K, 0 < x < 0.3, 0 < y < 0.25
T(x,y,0) = 298.15 K at the interval boundaries

Boundary condition:
T(x,0,t) =  T(x, 0.25, t) = 298.15 K for 0 <= x <= 0.3 and t >= 0
T(0,y,t) =  T(0.3, y, t) = 298.15 K for 0 <= y <= 0.25 and t >= 0


Discretize into an MxNxP mesh.
h = (x_i - x_i-1)/M
d = (y_j - y_j-1)/N
n = (t_k - t_k-1)/P

Notation: T(x_i,y_j,t_k) := T_ijk
Approximating with Crank-Nicolson method:
(∂T/∂t)_xiyjtk ~ n^-1(T_ijk - T_ij,k-1)
(∂²T/∂²x)_xiyjtk-½ ~ ½h^-2(T_i+1,jk - 2T_ijk + T_i-1,jk + T_i+1,j,k-1 - 2T_ij,k-1 + T_i-1,j,k-1)
(∂²T/∂²y)_xiyjtk-½ ~ ½d^-2(T_i,j+1,k - 2T_ijk + T_i,j-1,k + T_i,j+1,k-1 - 2T_ij,k-1 + T_i,j-1,k-1)

Insert:
n^-1(T_ijk - T_ij,k-1) = ½ch^-2(T_i+1,jk - 2T_ijk + T_i-1,jk + T_i+1,j,k-1 - 2T_ij,k-1 + T_i-1,j,k-1) + ½cd^-2(T_i,j+1,k - 2T_ijk + T_i,j-1,k + T_i,j+1,k-1 - 2T_ij,k-1 + T_i,j-1,k-1)

Let r = cnh^-2 and q = cnd^-2

2T_ijk - 2T_ij,k-1 = r(T_i+1,jk - 2T_ijk + T_i-1,jk + T_i+1,j,k-1 - 2T_ij,k-1 + T_i-1,j,k-1) + q(T_i,j+1,k - 2T_ijk + T_i,j-1,k + T_i,j+1,k-1 - 2T_ij,k-1 + T_i,j-1,k-1)

Reorder:

-rT_i-1,jk - qT_i,j-1,k + (2+2r+2q)T_ijk - rT_i+1,jk - qT_i,j+1,k= rT_i-1,j,k-1 + qT_i,j-1,k-1 + (2-2r-2q)T_ij,k-1 + rT_i+1,j,k-1 + qT_i,j+1,k-1

Let t_k = [T_1,1,k  T_2,1,k  ...  T_M-1,1,k  T_1,2,k  T_2,2,k  ...  T_M-1,2,k ... T_M-1,N-1,k]^T

Then, the PDE is solved by solving the equation At_k = Bt_k-1 + r_k from k=1 to k = P where r_k is a constant vector of boundary conditions.

t_k = A^-1(Bt_k-1 + r_k)

The numerical solution shows that even water at a lower temperature should have gone down to almost room temperature.
[/quote][/spoiler]
 
My computer is not explode(burn) yet.
And Speccy still show 128'C