1-D Heat Distribution in Long Cylinder

[Stomachbuzz]

Hi there,

I've got a Heat Transfer problem that I can't seem to get right.

I will list all information in the problem without 'interpreting' it:
Givens: 100 cm long cylinder filled with hot water, and constantly heated to maintain the water at 100°C. Heat is transferred via conduction from inner to outer surface, and convection from there. Ignoring heat losses from top and bottom of cylinder.
InnerR=50cm, OuterR=62cm
Water temp = 100°C
Outer Temp = 80°C
Ambient Temp = 20°C
h = 10W/m^2 * K
k = 20W/m* K

"When operated steadily, the conduction heat transfer from outer surface to ambient air is equal to the conduction through the wall from inside surface to outside surface"

Find:
1) the heat loss from the cylinder to ambient air
2) Find T as a function of r
3) Find inner surface temperature

For 1, I used Q_{cond, cyl} = -kA(dT/dr)
Q/(2πr*L) dr = -k dT
Integrate:
Q = 2πk*L (T₁ - T₂)/(ln(r₂/r₁))

I found this method from a University .edu site, and it made sense to me.
Plugging in numbers, I get Q = 2π * (20) * (1m) (20°C/ (ln(62/50)) = 11,684 watts.
This seems a bit high considering the mild ΔT and small volume but...what do I know

so I go onto part 2:
Again, drawing from .edu site, I use (d/dr (r * dT/dr)) = 0
Integrate: r * dT/dr = c₁
Algebra: dT= c₁ * (dr/r)
Integrate: T = c₁ * ln(r/r₁) + c₂

I understand it that when r=r₁ (inner radius), the temperature would be equal to inner surface temp. Which I want to assume is 100°C, but you know where that gets me...
I know when r=r₂ T=T_outer = 80°C
I'm trying to solve for c₁ using a ΔT_max formula I found, and this is where it gets murky.

ΔT_{max, cyl} = (g⋅L²)/2k
L and k are given, but what's g, right?
Well, I figure I can just borrow Q from part 1, and divide by volume, no? Q^dot = g^dot * V
11684w / (.62² -.5²)(π)(1m) = 27,672 w/m³ =g^dot which, again, seems high but
The catch (for me at least) is trying to draw equivalence with the convection equation, Q^dot = hA(T_s - T_∞)
Using that one, Q= (10 w/m² *k)(.62m * 2π) (80-20°C)= 10 * 3.896 * 60 = 2,337 watts
Which is much different. 2,337 ≠ 11684

What's goin on here?

Thanks in advance for the help. I've found my way onto this forum for years doing homework through HS and college. I know I'm in for a treat.

Including screenshot of homework problem here. I do realize it lists both 2m and 100cm as height. I figured it was an error (conflicting) and didn't matter as long as height used is consistent.

 

[mjc123]

Your answer to 1 is wrong because you are assuming T₁ is 100°C. (I would have too, but apparently not.) Since you are asked to find it in part 3, it is presumably not trivial. You don't even have to calculate the conduction, you just use the equation Q = hA(T₂ - T_air), as you do at a later point and get (I think) the right answer, 2337 W. For part 2 you assume Q_cond is the same as this, and integrate the conduction equation from r to r₂. For part 3 you simply put in r = r₁.