Question

This equation from problem 2 is needed to solve problem 3.

please skip C

problem 2 temperature field. Problem 3 (Fourier's Law of Heat Flow) From a temperature field, we can use Fourier's Law to compute the heat transfer rate (which is a vector field): Where k is a constant known as the thermal conductivity. For the purposes of this exercise, we can assume that k is 1. We wish to apply Fourier's Law to the temperature field given in Problem 2 to find the heat transfer in that rectangular plate. Answer the following: a. Based on the unofficial rule of heat transfer (i.e. heat flows from hot stuff the cold stuff), briefly describe how you expect heat to flow in our rectangular plate. Which corner will the heat go to? b. Compute the gradient of this temperature field. c. Plot the gradient as a field of arrows, like we did in class. d. Based on your gradient field, which direction will heat flow?

Answer 1

Qa

Assuming cartesian coordinates,

+Y axis +X axis Plate

Based on the given vector field,

=-k(

$\overrightarrow{q}= -k(\frac{\partial T}{\partial x}\widehat{i}+\frac{\partial T }{\partial y}\widehat{j})$ ............................................................( as given in question)

where $\widehat{i}$ and $\widehat{j}$ are unit vectors along x and y axes.

and Temperature field

we get

$\overrightarrow{q}= (-6x\widehat{i}+6y\widehat{j})$ units.,

where $\widehat{i}$ and $\widehat{j}$ are unit vectors along x and y axes.

Therefore this shows that, heat is transferred in negative x direction along the x axis, and in positive y axis along the Y axis.

Lower Temperature Higher Temperature Lower Temperature ty Heat Transfer direction along x axis. Heat Transfer direction along Y axs. +X -X Higher Temperature -y

As depicted above using a schematic diagram.

Heat will go to the top and the left edges of the plate. Since no temperature variation along the z axis.

Qb

Gradient of temperature is given by

$gradT=(\frac{\partial T }{\partial x}\widehat{i}+\frac{\partial T }{\partial y}\widehat{j})$

this operation on

gives,

$grad T=6x\widehat{i}-6y\widehat{j}$ (ans)

where $\widehat{i}$ and $\widehat{j}$ are unit vectors along x and y axes.

Qd

Since Heat transfer rate as a vector field is a vector quantity we can take Cartesian components of this vector along x and y axes.

Along x axis,

Based on the gradient field, Temperature is increasing along the positive x direction with respect to the x coordinates as

_ 622 OT

This means heat is transferred from positive x end of the plate to the negative x end of the plate along x axis.

Similarly

Along y axis,

$\frac{\partial T}{\partial y}=-6y\widehat{i}$

Temperature is decreasing in the +y direction, this implies heat transfer is from negative y direction to the positive y direction.