In this example, we are given a temperature distribution for a wall of thickness 0.02
meters.
This wall is immersed in a fluid that has a bulk temperature of T infinity of 100 degrees
C. The given temperature distribution, or T(x), is equal to 200 degrees C plus -200
degrees C per meter X, plus 30 degrees C per meter squared X squared.
We are asked to solve for three things.
The first is the heat flux, or q'', at both faces of the wall.
The second thing we're asked for is the rate of heat storage in the wall, and finally,
we are asked to find the heat transfer coefficient, or h, of the fluid on both sides of the wall.
So let's start with the first problem.
We know that q'' is equal to -k times dT/dx, and since we are given k, we can solve for
dT/dx by taking the derivative of our original temperature distribution with respect to x.
So dT/dx is equal to -200 plus 60x.
So plugging in for x equals zero, because we're solving for q'' at x = 0, we find that
q'' is equal to 200 watts per meter squared.
Now the reason the degrees C and the kelvin cancel out is that we're dealing with temperature
differences.
So now, let's solve for the heat flux at x=0.2 meters.
So to do that we plug in the same values, but instead of plugging in x=0, we'll Plug
in x=0.2.
This is equal to 188 watts per meter squared.
So now we're going to move on to the second question; we know that heat generation, which
we'll call E gen, is equal to the energy in minus the energy out.
The energy in is just our answer for q'' at x=0, and the energy out is 188 watts per meter
squared, or the heat flux out of the wall at x=0.2.
So this means that our energy generation in the wall is 12 watts per meter squared.
Now we'll move on to question 3, which asks us, what is the heat transfer coefficient
of the fluid at both x=0 and x=0.2.
So we know that the heat flux in at x=0 has to be equal to h, our heat transfer coefficient,
times the temperature at x=0, minus the temperature of the bulk fluid.
We have our temperature distribution, which was given, and if you plug in x=0 to that
temperature distribution, you find that the temperature at x=0 is equal to 200 degrees
C. Now, solving for h, we find that h is equal to 2 watts per meter squared kelvin.
Now again, in this instance, a difference of 100 degrees C is equal to the same difference
in kelvin, which is why we can put the units of h in kelvin instead of degrees C, because
that's a more standard unit notation.
Now, let's go into the heat flux at x=0.2 meters.
We can again plug in x=0.2 into our original temperature distribution, and we find that
the temperature at x=0.2 is equal to 161.2 degrees C. Solving for h, we find that h at
x=0.2 is equal to 3.07 watts per meter squared kelvin.