Steady state temperature of insulated rod

In summary, the steady state temperature of a laterally insulated rod is 10 degrees Celsius when kept at 0 degrees Celsius on the left and 100 degrees Celsius on the right.
  • #1
jc2009
14
0
Problem: Find the steady state temperature of a laterally insulated rod subject to the following conditions. Length of the rod 10, , left end kept at 0 and right end at 100

Solution: i got this equation from a book i don't know if this applied to this problem but i don't know what to do.
u(x,t) = v(x) + w(x,t) , v is the steady solution and w the transient putting this is the PDE of Heat , we get two equations , one of which is a function of v and this can be easily solved ( but i don't see what the equations are) .

From this point i don't know what to do

thanks
 
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  • #2
Hello jc2009,

The basic equation of heat conduction in 1 dimension is given as:
[tex]\frac{\partial u}{\partial t}=k \cdot \frac{\partial ^2 u}{\partial x^2}[/tex]
Now if it can be assumed that the equation can be written as the sum of a transient and steady state part as follows:
[tex]u(x,t)=v(x)+w(x,t)[/tex]
with v(x) the steady-state and w(x,t) the transient solution and u(x,t) the total solution, you can obtain two new equations by substituting this into the original PDE. This is thus:
[tex]\frac{\partial v}{\partial t}+\frac{\partial w}{\partial t}=k \cdot \left[\frac{\partial ^2 v}{\partial x^2}+ \frac{\partial ^2 w}{\partial x^2}\right][/tex]
Now the first term is zero because the steady-state is independent of time. This gives two new equations as follows:
[tex]\frac{\partial w}{\partial t}=k \cdot \frac{\partial ^2 w}{\partial x^2}[/tex]
[tex]0=\frac{\partial ^2 v}{\partial x^2}[/tex]
The second one gives you the steady-state solution, the first one the transient one. In your case the solution for the transient can be written as:
[tex]v(x)=A\cdot x + B[/tex]
Using the boundary conditions, you get:
[tex]v(x)=10\cdot x[/tex]

Hope this helps,
coomast
 
  • #3
coomast said:
Hello jc2009,

The basic equation of heat conduction in 1 dimension is given as:
[tex]\frac{\partial u}{\partial t}=k \cdot \frac{\partial ^2 u}{\partial x^2}[/tex]
Now if it can be assumed that the equation can be written as the sum of a transient and steady state part as follows:
[tex]u(x,t)=v(x)+w(x,t)[/tex]
with v(x) the steady-state and w(x,t) the transient solution and u(x,t) the total solution, you can obtain two new equations by substituting this into the original PDE. This is thus:
[tex]\frac{\partial v}{\partial t}+\frac{\partial w}{\partial t}=k \cdot \left[\frac{\partial ^2 v}{\partial x^2}+ \frac{\partial ^2 w}{\partial x^2}\right][/tex]
Now the first term is zero because the steady-state is independent of time. This gives two new equations as follows:
[tex]\frac{\partial w}{\partial t}=k \cdot \frac{\partial ^2 w}{\partial x^2}[/tex]
[tex]0=\frac{\partial ^2 v}{\partial x^2}[/tex]
The second one gives you the steady-state solution, the first one the transient one. In your case the solution for the transient can be written as:
[tex]v(x)=A\cdot x + B[/tex]
Using the boundary conditions, you get:
[tex]v(x)=10\cdot x[/tex]

Hope this helps,
coomast

excellent answer, very detailed, thank you coomast
 
  • #4
coomast said:
Hello jc2009,

The basic equation of heat conduction in 1 dimension is given as:
[tex]\frac{\partial u}{\partial t}=k \cdot \frac{\partial ^2 u}{\partial x^2}[/tex]
Now if it can be assumed that the equation can be written as the sum of a transient and steady state part as follows:
[tex]u(x,t)=v(x)+w(x,t)[/tex]
with v(x) the steady-state and w(x,t) the transient solution and u(x,t) the total solution, you can obtain two new equations by substituting this into the original PDE. This is thus:
[tex]\frac{\partial v}{\partial t}+\frac{\partial w}{\partial t}=k \cdot \left[\frac{\partial ^2 v}{\partial x^2}+ \frac{\partial ^2 w}{\partial x^2}\right][/tex]
Now the first term is zero because the steady-state is independent of time. This gives two new equations as follows:
[tex]\frac{\partial w}{\partial t}=k \cdot \frac{\partial ^2 w}{\partial x^2}[/tex]
[tex]0=\frac{\partial ^2 v}{\partial x^2}[/tex]
The second one gives you the steady-state solution, the first one the transient one. In your case the solution for the transient can be written as:
[tex]v(x)=A\cdot x + B[/tex]
Using the boundary conditions, you get:
[tex]v(x)=10\cdot x[/tex]

Hope this helps,
coomast

In this part :
The second one gives you the steady-state solution, the first one the transient one. In your case the solution for the transient can be written as:
[tex]v(x)=A\cdot x + B[/tex]
Using the boundary conditions, you get:
[tex]v(x)=10\cdot x[/tex]

I think you meant to say "THe solution for the steady state,(not the transient one) can be written as:
[tex]v(x)=A\cdot x + B[/tex]
Using the boundary conditions, you get:
[tex]v(x)=10\cdot x[/tex]

because you if you integrate 2 times the v(x) part using the boundary condition you get
u'' = 0 , Integrate [ u '' ] = u' = A
Integrate [u ' ] = Ax+B ----> v(x) = 10x , or this is the transient one ?
Also how exactly you got 10x? , is this because of the length of the rod ? , how do you know A = 10 , and B=0 in Ax+B ?
 
Last edited:
  • #5
Hello jc2009,

Indeed I meant the steady-state solution :-)

The boundary conditions are:
v=0 at x=0
v=100 at x=10

Putting this into the solution v=A*x+B gives you the system to solve:

100=A*10+B
0=A*0+B

From which you get: v=10*x

best regards,

coomast
 

Related to Steady state temperature of insulated rod

1. What is the steady state temperature of an insulated rod?

The steady state temperature of an insulated rod refers to the temperature that the rod reaches when the heat flow into the rod equals the heat flow out of the rod. In other words, the temperature of the rod stops changing and remains constant over time.

2. How is the steady state temperature of an insulated rod calculated?

The steady state temperature of an insulated rod can be calculated using the thermal conductivity of the material, the length and cross-sectional area of the rod, and the temperatures at each end of the rod. This can be done using the formula: T = (k * A * ΔT) / L, where T is the steady state temperature, k is the thermal conductivity, A is the cross-sectional area, ΔT is the temperature difference, and L is the length of the rod.

3. What factors can affect the steady state temperature of an insulated rod?

Some factors that can affect the steady state temperature of an insulated rod include the thermal conductivity of the material, the length and cross-sectional area of the rod, the temperature difference between the two ends of the rod, and the surrounding environment.

4. How does the insulation of the rod affect its steady state temperature?

The insulation of the rod plays a crucial role in determining the steady state temperature. Insulation helps to reduce heat loss from the rod, which can result in a higher steady state temperature. Similarly, insufficient insulation can lead to a lower steady state temperature as more heat is lost to the surroundings.

5. Can the steady state temperature of an insulated rod change over time?

In most cases, the steady state temperature of an insulated rod remains constant over time once it has been reached. However, certain factors such as changes in the surrounding environment or changes in the properties of the rod (e.g. due to wear and tear) can cause the steady state temperature to change. In these cases, the temperature may need to be recalculated to determine the new steady state temperature.

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