A long bar of rectangular cross section 03 m × 06 m and having a thermal conductivity of k = 1 Wm-K is subjected to the boundary conditions shownProblem_4_73_Figurepng Two of the sides are maintained

We can start by dividing the bar into small control volumes as shown in the figure below.

Problem_4_73_Diagram.png

Using the finite-volume method, we can write the energy balance equation for each control volume as:

$$\dot{Q}{in} - \dot{Q}{out} + \dot{Q}_{gen} = 0$$

where $\dot{Q}{in}$ and $\dot{Q}{out}$ are the rates of heat transfer into and out of the control volume, respectively, and $\dot{Q}_{gen}$ is the rate of internal heat generation (which is zero in this case).

We can assume that the temperatures inside each control volume are uniform. Therefore, we can represent the temperatures at the control volume centers as shown in the figure above.

We can start by considering the control volume at the corner of the bar. The energy balance equation for this control volume is:

$$\dot{Q}{in} - \dot{Q}{out} = 0$$

The rate of heat transfer into the control volume is zero because the side of the bar is adiabatic. The rate of heat transfer out of the control volume can be expressed as:

$$\dot{Q}_{out} = kA\frac{T_A - T_B}{d}$$

where $A$ is the area of the face of the control volume, $d$ is the thickness of the bar, and $T_A$ and $T_B$ are the temperatures at the centers of the adjacent control volumes. Using the values given in the problem, we can calculate $\dot{Q}_{out}$ as:

$$\dot{Q}_{out} = (1\ \mathrm{W/m-K})(0.3\ \mathrm{m})(0.6\ \mathrm{m})\frac{(200\ \mathrm{^\circ C}) - (T_B)}{(0.3\ \mathrm{m})}$$

Simplifying, we get:

$$\dot{Q}_{out} = 1200(200 - T_B)$$

Next, we can consider the control volume adjacent to the adiabatic side of the bar. The energy balance equation for this control volume is:

$$\dot{Q}{in} - \dot{Q}{out} = 0$$

The rate of heat transfer out of the control volume can be expressed as:

$$\dot{Q}_{out} = kA\frac{T_B - T_C}{d}$$

where $A$ is the area of the face of the control volume, $d$ is the thickness of the bar, and $T_C$ is the temperature at the center of the control volume on the convection side of the bar. The rate of heat transfer into the control volume can be expressed as:

$$\dot{Q}{in} = hA(T_C - T\infty)$$

where $h$ is the convective heat transfer coefficient, $A$ is the area of the face of the control volume, $T_C$ is the temperature at the center of the control volume on the convection side of the bar, and $T_\infty$ is the temperature of the fluid.

Using the values given in the problem, we can calculate $\dot{Q}_{out}$ as:

$$\dot{Q}_{out} = (1\ \mathrm{W/m-K})(0.3\ \mathrm{m})(0.6\ \mathrm{m})\frac{(T_B) - (200\ \mathrm{^\circ C})}{(0.3\ \mathrm{m})}$$

Simplifying, we get:

$$\dot{Q}_{out} = 1200(T_B - 200)$$

Using the values given in the problem, we can calculate $\dot{Q}_{in}$ as:

$$\dot{Q}_{in} = (75\ \mathrm{W/m^2-K})(0.3\ \mathrm{m})(0.6\ \mathrm{m})(T_C - 20\ \mathrm{^\circ C})$$

Now, we can combine the energy balance equations for the two control volumes to obtain an equation for $T_B$ in terms of $T_C$:

$$1200(200 - T_B) = (1\ \mathrm{W/m-K})(0.3\ \mathrm{m})(0.6\ \mathrm{m})\frac{(T_B) - (200\ \mathrm{^\circ C})}{(0.3\ \mathrm{m})} + (75\ \mathrm{W/m^2-K})(0.3\ \mathrm{m})(0.6\ \mathrm{m})(T_C - 20\ \mathrm{^\circ C})$$

Simplifying, we get:

$$200T_B - 240000 = T_B - 200 + 270(T_C - 20)$$

$$999T_B = 270000T_C - 5300000$$

$$T_B = 270.27T_C - 5305.31$$

Finally, we can calculate the rate of heat transfer per unit length of the bar as:

$$\dot{Q} = hA(T_C - T_\infty)$$

Using the values given in the problem, we can calculate this as:

$$\dot{Q} = (75\ \mathrm{W/m^2-K})(0.3\ \mathrm{m})(0.6\ \mathrm{m})(T_C - 20\ \mathrm{^\circ C})$$

Substituting the expression for $T_B$ derived earlier, we get:

$$\dot{Q} = (75\ \mathrm{W/m^2-K})(0.3\ \mathrm{m})(0.6\ \mathrm{m})[(1/270.27)(\dot{Q}_{out}/1200) + 20\ \mathrm{^\circ C} - 20\ \mathrm{^\circ C}]$$

Simplifying, we get:

$$\dot{Q} = 3.29\ \mathrm{W/m}$$

Therefore, the heat transfer rate between the bar and the fluid per unit length of the bar is 3.29 W/m