Chapter 20, Problem 37P

(a)

To determine

To Calculate:The thermal current in each cube.

Answer to Problem 37P

The thermal current in copper cube is approximately $\text{0}\text{.96 kW}$ .

The thermal current in aluminum cube is approximately $\text{0}\text{.57 kW}$ .

Explanation of Solution

Given:

Thermal conductivity of copper, $K_{Cu}=40\text{1W/m}\cdot \text{K}$

Areaof cross-section of copper, $A_{cu}=9.0\times {10}^{-4}\text{m²}$

Temperature difference, $\Delta T=80\text{K}$

Thermal conductivity of aluminum, $K_{Al}=237\text{W/m}\cdot \text{K}$

Areaof cross-section of aluminum, $A_{Al}=9.0\times {10}^{-4}\text{m²}$

Formula used: The amount of heat $\left(\Delta Q\right)$ conducted per unit time $\left(\Delta t\right)$ is

  $\frac{\Delta Q}{\Delta t}\text{=}\frac{\text{k}{\rm A}\Delta {\rm T}}{L}\mathrm{..........}\left(1\right)$

Here, L the thickness of the substance. The cross-section area of the substance, $\Delta T$ the

temperature difference among the ends of the substance and k the thermal conductivity of thesubstance.

The thermal resistance is given by:

  $R=\frac{L}{kA}$

Then the equation (1) is written as

  $\frac{\Delta Q}{\Delta t}\text{=}\frac{\Delta {\rm T}}{R}$

Here, the thermal current $\text{I=}\frac{\Delta {\rm T}}{R}$

Calculation:

The thermal current within each cube, is expressed as follows:

  $I=\frac{\Delta {\rm T}kA}{\Delta x}$

The thermal resistance of each cube is:

  $R=\frac{\Delta x}{kA}$

By using the equation (2), the thermal current through the copper can be expressed as follows:

  $I=\frac{\Delta {\rm T}KCuACu}{\Delta xCu}$

  $\begin{array}{l}\text{ICu=}\frac{\left(\text{401W/m·K}\right)\left({\text{9×10}}^{\text{-4}}{\text{m}}^{\text{2}}\right)\left(\text{80K}\right)}{{\text{3×10}}^{\text{-2}}\text{ m}}\\ \text{= 962 W}\left(\frac{\text{1kW}}{\text{1000W}}\right)\\ \text{= 0}\text{.962 kW}\end{array}$

By using the equation (2), the thermal current through the Aluminum can be expressed asfollows:

  $I=\frac{\Delta {\rm T}KAlAAl}{\Delta xAl}$

  $\begin{array}{l}IAl=\frac{\left(\text{237 W/m·K}\right)\left({\text{9×10}}^{\text{-4}}\text{m²}\right)\left(\text{80K}\right)}{{\text{3×10}}^{\text{-2}}\text{ m}}\\ =\text{ 568}\text{.8 W}\left(\frac{\text{1 kW}}{\text{1000 W}}\right)\\ =\text{ 0}\text{.57 kW}\end{array}$

Hence, the thermal current in aluminum cube is approximately $\text{0}\text{.57 kW}$ .

Conclusion: The thermal current in each cube is to be calculated by using temperature difference and thermal conductivity.

(b)

To determine

To Calculate: The total thermal current.

Answer to Problem 37P

The total thermal current is $\text{1}\text{.53kW}$

Explanation of Solution

Given:Current passing through the copper $\left(I_{cu}\right)$ is $\text{962 W}$ .

Current passing through the aluminum $\left(I_{Al}\right)$ is $\text{568}\text{.8 W}$ .

Formula used:

Given that the cubes are in parallel, therefore total thermal current can be expressed as:

  ${\text{I = 1}}_{\text{Cu}}{\text{ +I}}_{\text{AI}}$

Calculation:

Substitute the values and solve:

  $\begin{array}{c}\text{I = 962 W +569W}\\ \text{= 0}\text{.962 kW+0}\text{.569 kW}\\ \text{= 1}\text{.53 kW}\end{array}$

Conclusion: Hence,the total thermal current is $\text{1}\text{.53kW}$

(c)

To determine

To Calculate:The thermal resistance of two cube combination.

Answer to Problem 37P

The equivalent thermal resistance is $\text{0}\text{.052 K/W}\text{.}$

Explanation of Solution

Given:The temperature difference $\Delta T$ is $\text{80°C}$ and thermal current (I) is $\text{1}\text{.53kW}$

Formula used:

The equivalent thermal resistance is obtained by:

  $\text{Req=}\frac{\text{ΔT}}{\text{I}}$

Where, $\Delta T$ is the temperature difference and I is the thermal current.

Calculation:

Substitute the values and solve:

  $\begin{array}{c}\text{Req=}\frac{\text{80K}}{\text{1}\text{.53kW}}\\ \text{= 52}\text{.28K/kW}\left(\frac{\text{1}\text{.0kW}}{\text{1000 W}}\right)\\ \text{=0}\text{.05228 K/W}\end{array}$

Hence, the equivalent thermal resistance is $\text{0}\text{.052 K/W}$ .

Conclusion:Equivalent thermal resistance can be calculated using temperature difference and thermal current.