Heat equilibrium between two different rods

[Ojizo88]

[SOLVED] Heat equilibrium between two different rods

Homework Statement

A rod 1.3 m long consists of a 0.8 m length of Aluminium joined end-to-end to a 0.5 m length of brass. The free end of the aluminium section is maintained at 150 oC and the free end of the brass piece is maintained at 20 oC. No heat is lost through the sides of the rod. In the steady state, what is the temperature of the point where the two metals are joined ? (Thermal conductivity of brass k_b = 109 W/mK and Thermal conductivity of Aluminum k_a = 237 W/mK).

Homework Equations

H=-kA \frac{\Delta T}{\Delta x}

The Attempt at a Solution

Well, you know in steady state that there is no net heat flow so H_aluminum = H_brass. So you have...(I give up for tying to figure this formatting out for now so please forgive me)

(-k_{a} A (T_{h} - T_{e})) / l_{a} = (-k_{b} A (T_{e} - T_{c})) / l_{b} , with T_{e} being the point in the middle and T_{h} and T_{c} being the hold and cold ends respectively.

Since the cross sectional area of the two are the same, those will cancel and youre left with...

(-k_{a} T_{h} + k_{a} T_{e}) / l_{a} = (-k_{b} T_{e} + k_{b} T_{c}) / l_{b}

solving for T_{e} you get...

k_{a} T_{h} l_{b} + k_{b} T_{c} l_{a} = k_{b} T_{e} l_{a} + k_{a} T_{e} l_{b}
Which then goes to...
T_{e} = (k_{a} T_{h} l_{b} + k_{b} T_{c} l_{a}) / (k_{b} l_{a} + k_{a} l_{b})

However, when I put in the values I have, I get an answer HIGHER than the original T_{h}!

I don't get what I am doing wrong here.

I have a problem after this where I have to solve for T_{e} when 3 rods come together but I clearly can't do that one properly if I can't solve this one first!

 

[Ojizo88]

Little bump before bed. Maybe someone can figure out what I am doing incorrectly :(

 

[Shooting Star]

You are doing incorrect arithmetics, that's all. Recheck you calc. The ans should be 95 C.

 

[Ojizo88]

You are doing incorrect arithmetics, that's all. Recheck you calc. The ans should be 95 C.

Well is the algebra at least correct? Because I've gone over that time and time again and can't find anything wrong. Maybe I am losing the forest in the threes though...

 

[Shooting Star]

The algebra is correct. Do the arithmetic very carefully once.

 

[Shooting Star]

I forgot to mention this. You had written "Well, you know in steady state that there is no net heat flow..." That is not correct. In a steady state, heat does flow, but the temp at every point remains constant.

 

[katchum]

There is one thing I want to add here:

Your first equation is wrong in terms of + and -.

When your x goes up, your T will drop. This means you should say:

-ka.A.(-(Th-Te))/la = - kb.A.(-(Te-Tc))/lb

 

[Ojizo88]

I forgot to mention this. You had written "Well, you know in steady state that there is no net heat flow..." That is not correct. In a steady state, heat does flow, but the temp at every point remains constant.

Firstly, thank you for the assistance. I found out where the wrong step was. I figured it was something I was doing wrong algebraically.

Secondly, I am sorry if I worded my statement improperly. I was trying to imply the fact that at the middle point, there is no NET heat flow that will cause a temperature change. It has come to equilibrium. I apologize for any confusion this may have caused.