# Thread: 2 rods placed end to end,heat eq. Have an idea i'm wrong!

1. ## 2 rods placed end to end,heat eq. Have an idea i'm wrong!

The question :Two rods L1 and L2 of different materials( hence different thermal conductivities) and different cross-sectional areas,are joined at x=a. The temperature is continuous, i.e $u(a^{+},t)=u(a^{-},t)$
And NO HEAT ENERGY IS LOST AT a, so all heat energy that flows from L1 flows into L2.

? What equation represents the condition that no energy is lost at a ?

My thoughts : Equate the rate at which heat leaves L1 to the rate at which heat enters L2. ???
In the derivation of the heat equation we find the net rate at which energy flows through a section of the rod. Energy flows in by $E^{\prime}_{in} = -\lambda(x)A(x)u_{x}(a,t)$ and energy flows out by $E^{\prime}_{out} =-\lambda(x)A(x)u_{x}(b,t)$
Where A(x) is the area of a cross section of the rod, $\lambda$ is the thermal conductivity, a and b points on the rod with b>a.
If all heat leaving L1 goes to L2,then across the join $E^{\prime}_{in}=E^{\prime}_{out}$ ?? (not sure here) and $u_{x}(a,t)=u_{x}(b,t)$ ??
which gives me $-\lambda_{1}(x)A_{1}(x)u_{x}=-\lambda_{2}(x)A_{2}(x)u_{x}$ and so $-\lambda_{1}(x)A_{1}=-\lambda_{2}(x)A_{2}$ ??

Is this full of holes?! There is a concept or thought that i don't understand, what am i missing? How do i find an equation to represent that no heat is lost at a? I know that it should be an equation in $-\lambda_{1}(x),-\lambda_{2}(x),A_{1}\,and\,A_{2}$ from the question that follows.

2. Originally Posted by punkstart
The question :Two rods L1 and L2 of different materials( hence different thermal conductivities) and different cross-sectional areas,are joined at x=a. The temperature is continuous, i.e
And NO HEAT ENERGY IS LOST AT a, so all heat energy that flows from L1 flows into L2.

? What equation represents the condition that no energy is lost at a ?

My thoughts : Equate the rate at which heat leaves L1 to the rate at which heat enters L2. ???
In the derivation of the heat equation we find the net rate at which energy flows through a section of the rod. Energy flows in by and energy flows out by
Where A(x) is the area of a cross section of the rod, is the thermal conductivity, a and b points on the rod with b>a.
If all heat leaving L1 goes to L2,then across the join ?? (not sure here) and ??
which gives me and so ??
Looks good to me!

Is this full of holes?! There is a concept or thought that i don't understand, what am i missing? How do i find an equation to represent that no heat is lost at a? I know that it should be an equation in from the question that follows.

3. Originally Posted by punkstart
The question :Two rods L1 and L2 of different materials( hence different thermal conductivities) and different cross-sectional areas,are joined at x=a. The temperature is continuous, i.e $u(a^{+},t)=u(a^{-},t)$
And NO HEAT ENERGY IS LOST AT a, so all heat energy that flows from L1 flows into L2.

? What equation represents the condition that no energy is lost at a ?

My thoughts : Equate the rate at which heat leaves L1 to the rate at which heat enters L2. ???
In the derivation of the heat equation we find the net rate at which energy flows through a section of the rod. Energy flows in by $E^{\prime}_{in} = -\lambda(x)A(x)u_{x}(a,t)$ and energy flows out by $E^{\prime}_{out} =-\lambda(x)A(x)u_{x}(b,t)$
Where A(x) is the area of a cross section of the rod, $\lambda$ is the thermal conductivity, a and b points on the rod with b>a.
If all heat leaving L1 goes to L2,then across the join $E^{\prime}_{in}=E^{\prime}_{out}$ ?? (not sure here) and $u_{x}(a,t)=u_{x}(b,t)$ ??
which gives me $-\lambda_{1}(x)A_{1}(x)u_{x}=-\lambda_{2}(x)A_{2}(x)u_{x}$ and so $-\lambda_{1}(x)A_{1}=-\lambda_{2}(x)A_{2}$ ??

Is this full of holes?! There is a concept or thought that i don't understand, what am i missing? How do i find an equation to represent that no heat is lost at a? I know that it should be an equation in $-\lambda_{1}(x),-\lambda_{2}(x),A_{1}\,and\,A_{2}$ from the question that follows.
Wht you have is fine except one thing.

If you let $u_1$ and $u_2$ be the temperatures in each section then at the join ( $x = a$) you have that both the temp and heat fluxes are continuous, ${\it i.e}.$

(1) $u_1 = u_2$

(2) $\lambda_1(x) A_1(x) \frac{\partial u_1}{\partial x} = \lambda_2(x) A_2(x) \frac{\partial u_2}{\partial x}$.

The derivatives don't cancel.

4. ## the question that follows...

the next question says that i should assume $u_{x}$ is continuous, then gives me some values in order to solve for $\lambda_{2}$ . Using ,
can i say $\frac{\partial u_{1}}{\partial x}=\frac{\partial u_{2}}{\partial x}= u_{x}$ because $u_{x}$ is continuous?

5. Originally Posted by punkstart
the next question says that i should assume $u_{x}$ is continuous, then gives me some values in order to solve for $\lambda_{2}$ . Using ,
can i say $\frac{\partial u_{1}}{\partial x}=\frac{\partial u_{2}}{\partial x}= u_{x}$ because $u_{x}$ is continuous?
Now you can b/c it says the derivatives are continuous across the interface.