Respuesta :
Answer:
a) [tex]div F = 27 \frac{W}{km^3}[/tex]
b) [tex]\alpha=\frac{27 W/km^3}{3}= 9\frac{W}{Km^3}[/tex]
c) [tex]T(0,0,0)=-\frac{10}{2(27000)} (0) +7605.185=7605.185[/tex]
Step-by-step explanation:
(a) Suppose that the actual heat generation is 27W/km3 What is the value of div F? div F- Include units)
For this case the value for div F correspond to the generation of heat.
[tex]div F = 27 \frac{W}{km^3}[/tex]
(b) Assume the heat flows outward symmetrically. Verify that [tex] F= \alpha r[/tex] where [tex]r=xi +yj+zk[/tex]. Find a α, (Include units.)
For this case we can satisfy this condition:
[tex]div[\alpha (xi +yj +z k)]]=\alpha(1+1+1)=3\alpha[/tex]
And since we have the value for the [tex]div F[/tex] we can find the value of [tex]\alpha[/tex] like this:
[tex]\alpha=\frac{27 W/km^3}{3}= 9\frac{W}{Km^3}[/tex]
(c) Let T (x,y,z) denote the temperature inside the earth. Heat flows according to the equation F= -k grad T where k is a constant. If T is in °C then k=27000 C/km. Assuming the earth is a sphere with radius 6400 km and surface temperature 20°C, what is the temperature at the center? 27,0 C/km.
For this case we have this:
[tex] F =-k grad T[/tex]
And grad T represent the direction of the greatest decrease related to the temperature.
So we have this equation:
[tex] 10(xi +yj+zk)=-27000 grad T[/tex]
And we can solve for grad T like this:
[tex] grad T = -\frac{10}{(27000)} (xi+yj+zk)[/tex]
Andif we integrate in order so remove the gradient on both sides we got:
[tex]T=-\frac{10}{2(27000)} (x^2 +y^2 +z^2) +C[/tex]
For our case we have the following condition:
[tex]x^2 +y^2 +z^2 = 6400 , T=20 C[/tex]
[tex] T=-\frac{1}{54000} (6400^2)+C =20[/tex]
And we can solve for C like this:
[tex] C= 20+\frac{6400^2}{5400}= 7605.185 [/tex]
So then our equation would be given by:
[tex]T=-\frac{10}{2(27000)} (x^2 +y^2 +z^2) +7605.185[/tex]
And for our case at the center we have that [tex]x^2+ y^2+ z^2 =0[/tex]
And we got:
[tex]T(0,0,0)=-\frac{10}{2(27000)} (0) +7605.185=7605.185[/tex]
