• jc2009
In summary: For an insulated boundary you say that the heat flux must be zero, which is the case for infinite wall thickness. This is what is meant by an insulated boundary. For a fixed temperature you would say that the temperature gradient is zero, and this is what is meant by u_x=0. You can get the same situation for a fixed flux by making the wall thick enough.A different boundary condition is for a fixed temperature. This means that the temperature at the wall is fixed. This is true for a very thin wall and this is the way to interpret the boundary condition u=constant. For this case you must use a very thin wall to meet the boundary condition. This is not the case in this problem.best regards,coom
jc2009
Problem 1 : Find the steady state temperature of a laterally insulated rod subject to the following conditions. Length of the rod 20, left end insulated and right end held at 100

I am not sure if i fully understand this problem, is this asking me to list the Boundary value conditions ? like , u(x,0) = 10 , u(x,20) = 100, u_x (0) = 0 ?
or a function ?

Problem 2 : The vertical displacements of a membrane were found to be given by the function u(x,y,t) = cos(15t) sin(3x) cos(4y) , What is the corresponding wave's speed c ?

for this problem first i need to take the derivative u'(x,y,t) but with respect to what?

jc2009 said:
Problem 1 : Find the steady state temperature of a laterally insulated rod subject to the following conditions. Length of the rod 20, left end insulated and right end held at 100

I am not sure if i fully understand this problem, is this asking me to list the Boundary value conditions ? like , u(x,0) = 10 , u(x,20) = 100, u_x (0) = 0 ?
or a function ?

You need to assume that the solution can be written as the sum of the steady state and the transient solution. This can be written as:
$$u(x,t)=v(x)+w(x,t)$$
In which v the steady state solution and w the transient is. Putting this in the PDE of heat you get two equations, one of which is a function of v and this can be easily solved. The equation is now:
$$\frac{d^2v}{dx^2}=0$$
Using the boundary conditions gives you for the steady state solution:
$$v(x)=100$$
A constant for the whole bar, which is correct because no heat can be added or extracted due to the insulation. A constant must arrise.

jc2009 said:
Problem 2 : The vertical displacements of a membrane were found to be given by the function u(x,y,t) = cos(15t) sin(3x) cos(4y) , What is the corresponding wave's speed c ?

for this problem first i need to take the derivative u'(x,y,t) but with respect to what?

Substitute the solution in the wave equation:
$$\frac{\partial^2u}{\partial t^2}=c^2 \cdot \left[\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2} \right]$$
You can extract the velocity c from this.

coomast

coomast said:
You need to assume that the solution can be written as the sum of the steady state and the transient solution. This can be written as:
$$u(x,t)=v(x)+w(x,t)$$
In which v the steady state solution and w the transient is. Putting this in the PDE of heat you get two equations, one of which is a function of v and this can be easily solved. The equation is now:
$$\frac{d^2v}{dx^2}=0$$
Using the boundary conditions gives you for the steady state solution:
$$v(x)=100$$
A constant for the whole bar, which is correct because no heat can be added or extracted due to the insulation. A constant must arrise.

coomast

My question is how come v(x) = 100, if you integrate v'' , you'll have to get a constant times x + constant , why just 100 ?

Hello jc2009,

Indeed the solution is v(x)=A*x+B. The two boundary conditions are on the left side (x=0)
$$\frac{dv}{dx}=0$$
and on the right hand side (x=20)
$$v=100$$

This gives you v=100 for the complete bar after applying these conditions.

best regards,

coomast

coomast said:
Hello jc2009,

Indeed the solution is v(x)=A*x+B. The two boundary conditions are on the left side (x=0)
$$\frac{dv}{dx}=0$$
and on the right hand side (x=20)
$$v=100$$

This gives you v=100 for the complete bar after applying these conditions.

best regards,

coomast

if you solve B=0 then
v(20)=A20+b =100
A=5

then v(x) = 5x

how did you get 100?

Hello jc2009,

You are not applying the boundary conditions right. The left hand side is insulated, thus the derivative is 0. This is:
$$\frac{dv}{dx}=0=\frac{d}{dx}(Ax+B)=A$$
Thus A=0 and the other boundary condition gives you
$$v=100=B$$

best regards,

coomast

coomast said:
Hello jc2009,

You are not applying the boundary conditions right. The left hand side is insulated, thus the derivative is 0. This is:
$$\frac{dv}{dx}=0=\frac{d}{dx}(Ax+B)=A$$
Thus A=0 and the other boundary condition gives you
$$v=100=B$$

best regards,

coomast

$$\frac{dv}{dx}=0=$$ , is $$u_{x}$$ right
so whenever an end is insulated $$u_{x}$$ is the correct format of that boundary
THis has always caused me problems in this PDE class I am taking, in this problem $$u_{x}$$ is used because it is insulated , but in other problems where insulation is not present they anyways use $$u_{x}(0,t)=0$$ for a vertical side , so besides using $$u_{x}$$ in a rod what does exactly $$u_{x}$$ means for other problems , for example : a 10 X 20 plate ? why do they use $$u_{x}$$ for a side of a plate even when there is no insulation? why not just use u(x,t)

jc2009 said:
$$\frac{dv}{dx}=0=$$ , is $$u_{x}$$ right
so whenever an end is insulated $$u_{x}$$ is the correct format of that boundary
THis has always caused me problems in this PDE class I am taking, in this problem $$u_{x}$$ is used because it is insulated , but in other problems where insulation is not present they anyways use $$u_{x}(0,t)=0$$ for a vertical side , so besides using $$u_{x}$$ in a rod what does exactly $$u_{x}$$ means for other problems , for example : a 10 X 20 plate ? why do they use $$u_{x}$$ for a side of a plate even when there is no insulation? why not just use u(x,t)

Yes, $$u_x=\frac{\partial u}{\partial x}$$. Now stating that this is zero is indeed what is required for an insulated boundary. This comes from the law of Fourier of conductivity. What does this means? Consider a wall with a thickness d, area A, conductivity k and a temperature T1 and T2 at the two sides. Now the heat flux going through the wall is descibed by the law of Fourier as:
$$f=-k\cdot A \cdot \frac{T_1-T_2}{d}$$
or for a very thin wall, which can then be used to integrate for different situations:
$$f=-k\cdot A \cdot \frac{\partial T}{\partial x}$$
In which x the direction of heat flow, thus perpendicular to the wall sides. Looking at this equation you immediately see the influences of the different parameters. The larger the area A, the larger the flux and the thicker the wall, the smaller the flux. Also the higher the conductivity, the higher the flux as is with a higher temperature difference.

If the flux f is zero due to insulation you get:
$$0=f=-k\cdot A \cdot \frac{\partial T}{\partial x}=\frac{\partial T}{\partial x}$$
This is the boundary condition you need to use in your problems.

hope this clears some problems.

coomast

## 1. What is steady state temperature?

Steady state temperature is the temperature at which a system or object has reached a state of equilibrium, where the amount of heat entering the system is equal to the amount of heat leaving the system. This means that the temperature of the system remains constant over time.

## 2. How is steady state temperature different from equilibrium temperature?

While steady state temperature refers to a constant temperature in a system, equilibrium temperature refers to the temperature at which two objects in contact reach the same temperature. In other words, steady state temperature is a state of equilibrium, but not all equilibrium states have a steady state temperature.

## 3. What factors affect the steady state temperature of a system?

The steady state temperature of a system is affected by various factors, including the amount of heat being added or removed, the thermal conductivity of the materials involved, the size and shape of the system, and the presence of any external influences such as wind or radiation.

## 4. How is steady state temperature measured?

Steady state temperature can be measured using a thermometer or other temperature sensing device. The device should be placed in the system and allowed to reach equilibrium, at which point the temperature reading will indicate the steady state temperature.

## 5. What are some real-world applications of steady state temperature?

Steady state temperature is an important concept in many fields, including thermodynamics, engineering, and meteorology. It is used to understand and control heat transfer in systems such as buildings, engines, and electronic devices. It is also a key factor in weather forecasting and climate modeling.

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