Question

This is part of the question needed for the answer I’ll send the actual question too but it only allows me to attach one image

Answer 1

a) We know that Fourier's law is

`Q=(k)/(x)(T_0-T)`

We can do all the manipulations to get T on one side and then discuss the units of it, so let's pass k/x dividing Q and pass T0 as minus, and correct the signal of T

`\begin{gathered} Q\cdot(x)/(k)=T_0-T \\ \\ Q\cdot(x)/(k)-T_0=-T \\ \\ T=T_0-Q(x)/(k) \end{gathered}`

Now we have the temperature in terms of x. We can see that T0 is the temperature in Kelvin, which means that Qx/k also need to be in Kelvin, considering only units we write that

`K=(W)/(m^2)\cdot(m)/(u)`

Remember that Q = W/m^2 and x = m in terms of unit, I'm assuming that the unknown unit is "u", which is the unit of "k", so let's do all the algebra do get "u" isolated.

`\begin{gathered} K=(W)/(m)\cdot(1)/(u) \\ \\ K\cdot(m)/(W)=(1)/(u) \\ \\ u=(W)/(m\cdot K) \end{gathered}`

Then the unit for k is W/mK, in Fourier's Law, "k" represents the thermal conductivity of the material.

Now let's see the integral, we have two ways to evaluate it, I'll do one more mathematical and another one more easy and quick.

As we write

`T=T_0-Q(x)/(k)`

The dT/dx would be

`\begin{gathered} (dT)/(dx)=(d)/(dx)(T_0-(Q)/(k)x) \\ \\ (dT)/(dx)=(d)/(dx)T_0-(d)/(dx)((Q)/(k)x) \\ \\ (dT)/(dx)=-(Q)/(k) \\ \end{gathered}`

See that dT/dx is a constant, now we can evaluate

`\int ^d_0((dT)/(dx))dx=-\int ^d_0(Q)/(k)dx=-(Q)/(k)(d-0)=-(Q)/(k)d`

Basically, we did the way back because we did the dT/dx and then integrated in dx. We can use it to find the temperature T when we have the dT/dx. if the plot of T and x is a line, the dT/dx would be the slope of the line and using that integral we can find the temperature using the slope.

b)

On the temperature T(d) we have

`T(d)=T_0-(Q)/(k)d`

We know that T0 = 372K, and d = 0.3m, let's plot for Q/k = 0.1, 10 and 100

We have that

`\begin{gathered} T_1=372-0.1=371.9\text{ K} \\ T_2=362-10=362\text{ K} \\ T_3=272-100=272\text{ K} \end{gathered}`

Assuming that Q is constant and k is changing, we can confirm that for higher values of k, the value of T(d) will be very close to the value of T0, and for lower values of k, T(d) will be colder than T0, that's because "k" is the thermal conductivity, so the higher conductivity means that the temperature will tend to be equal faster.

c)

Looking at table 1 and the explanation on page 1 pick the material A for the second layer. Because it has the lowest thermal conductivity of all materials. We can do the linear regression using two points, we know that

`\begin{gathered} k(100)=1.6_{} \\ k(2000)=3.1 \end{gathered}`

And we have a linear function

`k(T)=aT+b`

So let's find out a and b values, then a will be

`a=(k(2000)-k(100))/(2000-100)=(3)/(3800)\approx0.00079`

And the b value will be

`\begin{gathered} k(100)=(3)/(3800)\cdot100+b_{} \\ \\ b=k(100)-(300)/(3800)=(289)/(190)\approx1.5210 \end{gathered}`

So k(T) for the material B is

`k(T)\approx0.00079\cdot T+1.521`

For more precision use the fraction

`k(T)=(3T)/(3800)+(289)/(190)`

d)

The problem says that Q is constant for all interfaces, the means that

`\begin{gathered} Q=(k_1)/(x_1)(T_0-T_1) \\ \\ Q=\frac{k_2_{}}{x_2}(T_1-T_2) \\ \\ Q=\frac{k_3_{}}{x_3}(T_2-T_3) \end{gathered}`

Using the first and the second equation we write

`\begin{gathered} (k_1)/(x_1)(T_0-T_1)=(k_2)/(x_2)(T_1-T_2)_{} \\ \\ T_0-T_1=(x_1)/(x_2)(k_2)/(k_1)_{}(T_1-T_2) \\ \\ T_0=T_1+(x_1)/(x_2)(k_2)/(k_1)_{}(T_1-T_2) \\ \\ T_0=T_1(1+(x_1)/(x_2)(k_2)/(k_1))-T_2\cdot(x_1)/(x_2)(k_2)/(k_1) \end{gathered}`

In the same way, let's do it with the third and second equation

`\begin{gathered} (k_3)/(x_3)(T_2-T_3)=(k_2)/(x_2)(T_1-T_2) \\ \\ T_2-T_3=\frac{x_3_{}}{x_2}\cdot(k_2)/(k_3)(T_1-T_2)_{} \\ \\ T_3=T_2-(x_3)/(x_2)\cdot(k_2)/(k_3)(T_1-T_2) \\ \\ T_3=T_2(1+(x_3)/(x_2)(k_2)/(k_3))-T_1\cdot(x_3)/(x_2)(k_2)/(k_3) \end{gathered}`

Look that the equation for T0 and T3 has T1 and T2 as variables, we can write it as a matrix now