- #1
Ojizo88
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[SOLVED] Heat equilibrium between two different rods
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).
[tex]H=-kA[/tex] [tex]\frac{\Delta T}{\Delta x}[/tex]
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)
[tex](-k_{a} A (T_{h} - T_{e})) / l_{a} = (-k_{b} A (T_{e} - T_{c})) / l_{b}[/tex] , with [tex]T_{e}[/tex] being the point in the middle and [tex]T_{h}[/tex] and [tex]T_{c}[/tex] 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...
[tex](-k_{a} T_{h} + k_{a} T_{e}) / l_{a} = (-k_{b} T_{e} + k_{b} T_{c}) / l_{b}[/tex]
solving for [tex]T_{e}[/tex] you get...
[tex]k_{a} T_{h} l_{b} + k_{b} T_{c} l_{a} = k_{b} T_{e} l_{a} + k_{a} T_{e} l_{b}[/tex]
Which then goes to...
[tex]T_{e} = (k_{a} T_{h} l_{b} + k_{b} T_{c} l_{a}) / (k_{b} l_{a} + k_{a} l_{b})[/tex]
However, when I put in the values I have, I get an answer HIGHER than the original [tex]T_{h}[/tex]!
I don't get what I am doing wrong here.
I have a problem after this where I have to solve for [tex]T_{e}[/tex] when 3 rods come together but I clearly can't do that one properly if I can't solve this one first!
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
[tex]H=-kA[/tex] [tex]\frac{\Delta T}{\Delta x}[/tex]
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)
[tex](-k_{a} A (T_{h} - T_{e})) / l_{a} = (-k_{b} A (T_{e} - T_{c})) / l_{b}[/tex] , with [tex]T_{e}[/tex] being the point in the middle and [tex]T_{h}[/tex] and [tex]T_{c}[/tex] 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...
[tex](-k_{a} T_{h} + k_{a} T_{e}) / l_{a} = (-k_{b} T_{e} + k_{b} T_{c}) / l_{b}[/tex]
solving for [tex]T_{e}[/tex] you get...
[tex]k_{a} T_{h} l_{b} + k_{b} T_{c} l_{a} = k_{b} T_{e} l_{a} + k_{a} T_{e} l_{b}[/tex]
Which then goes to...
[tex]T_{e} = (k_{a} T_{h} l_{b} + k_{b} T_{c} l_{a}) / (k_{b} l_{a} + k_{a} l_{b})[/tex]
However, when I put in the values I have, I get an answer HIGHER than the original [tex]T_{h}[/tex]!
I don't get what I am doing wrong here.
I have a problem after this where I have to solve for [tex]T_{e}[/tex] when 3 rods come together but I clearly can't do that one properly if I can't solve this one first!