Chapter 4, Problem 4.11P

To determine

(a)

The heat flux for each radiator material.

Answer to Problem 4.11P

The heat flux for the brass radiator is $24\times {10}^6\text{W}/{\text{m}}^{\text{2}}$ and for the aluminum radiator the heat flux is $49.4\times {10}^6\text{W}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Given:

The thickness of each radiator is $0.5\text{mm}$.

The temperature of engine fluid is $400\text{K}$.

The ambient air temperature is $300\text{K}$.

Formula used:

The expression heat flux of radiator is given by,

$q=-k\frac{dT}{dx}$ …… (I)

Here, $q$ is the amount of heat flux, $\frac{dT}{dx}$ is the temperature gradient and $k$ is the thermal conductivity.

The temperature gradient is given by,

$\frac{dT}{dx}=\frac{T_{\text{a}}-T_{\text{f}}}{dx}$ …… (II)

Here, $T_{\text{a}}$ is the ambient air temperature and $T_{\text{f}}$ is the temperature of fluid and $dx$ is the wall thickness.

Substitute $\left(\frac{T_{\text{a}}-T_{\text{f}}}{dx}\right)$ for $\frac{dT}{dx}$ in equation (I).

$q=-k\left(\frac{T_{\text{a}}-T_{\text{f}}}{dx}\right)$ …… (III)

Calculation:

The heat flux of aluminum radiator is calculated as,

Substitute $247\text{W}/\text{m}\cdot \text{K}$ for $k$, $300\text{K}$ for $T_{\text{a}}$, $400\text{K}$ for $T_{\text{f}}$ and $0.5\text{mm}$ for $dx$ in equation (III).

$\begin{array}{c}{q}_{\text{Al}}=-\left(247\text{W}/\text{m}\cdot \text{K}\right)\left(\frac{300\text{K}-400\text{K}}{0.5\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}}\right)\\ =-\left(247\text{W}/\text{m}\cdot \text{K}\right)\left(-200,000\text{m}\cdot \text{K}\right)\\ =49.4\times {10}^6\text{W}/{\text{m}}^{\text{2}}\end{array}$

Here, $q_{\text{Al}}$ is the heat flux for aluminum.

The heat flux of brass radiator is calculated as,

Substitute $120\text{W}/\text{m}\cdot \text{K}$ for $k$, $300\text{K}$ for $T_{\text{a}}$, $400\text{K}$ for $T_{\text{f}}$ and $0.5\text{mm}$ for $dx$ in equation (III).

$\begin{array}{c}{q}_{\text{Br}}=-\left(120\text{W}/\text{m}\cdot \text{K}\right)\left(\frac{300\text{K}-400\text{K}}{0.5\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}}\right)\\ =-\left(120\text{W}/\text{m}\cdot \text{K}\right)\left(-200,000\text{m}\cdot \text{K}\right)\\ =24\times {10}^6\text{W}/{\text{m}}^{\text{2}}\end{array}$

Here, $q_{\text{Br}}$ is the heat flux for Brass.

Conclusion:

Therefore, the heat flux for the brass radiator is $24\times {10}^6\text{W}/{\text{m}}^{\text{2}}$ and for the aluminum radiator the heat flux is $49.4\times {10}^6\text{W}/{\text{m}}^{\text{2}}$.

To determine

(b)

The estimate of relative area and the relative weight of changing from brass radiator to an aluminum radiator.

Answer to Problem 4.11P

The relative area of aluminum is $49\%$ of the area of brass and the weight of brass is $6.31$ times the weight of aluminum.

Explanation of Solution

Given:

The heat transfer rate is the same for brass and aluminum.

The brass is $30$ weight percent zinc and $70$ weight percent copper.

The density of brass is $8.4\text{g}/{\text{cm}}^{\text{3}}$.

Formula used:

The relation for rate of heat transfer in aluminum is given by,

${\left(\frac{dQ}{dt}\right)}_{\text{Al}}=A_{\text{Al}}{q}_{\text{Al}}$ …… (IV)

Here, ${\left(\frac{dQ}{dt}\right)}_{\text{Al}}$ is the rate of heat transfer in aluminum and $A_{\text{Al}}$ is the area of aluminum radiator.

The relation for rate of heat transfer in brass is given by,

${\left(\frac{dQ}{dt}\right)}_{\text{Br}}=A_{\text{Br}}{q}_{\text{Br}}$ …… (V)

Here, ${\left(\frac{dQ}{dt}\right)}_{\text{Br}}$ is the rate of heat transfer in brass and $A_{\text{Br}}$ is the area of brass radiator.

The heat transfer rate is same for brass and aluminum.

Equate equation (IV) and (V) to obtain expression for $A_{\text{Al}}$.

$\begin{array}{c}{A}_{\text{Al}}{q}_{\text{Al}}=A_{\text{Br}}{q}_{\text{BR}}\\ A_{\text{Al}}=\frac{A_{\text{Br}}{q}_{\text{BR}}}{q_{\text{Al}}}\end{array}$ …… (VI)

The expression for weight is given by,

$W=D\times V$ …… (VII)

Here, $W$ is the weight, $D$ is the density and $V$ is the volume.

Calculation:

The area of aluminum is calculated as,

Substitute $24\times {10}^6\text{W}/{\text{m}}^{\text{2}}$ for $q_{\text{Br}}$ and $49.4\times {10}^6\text{W}/{\text{m}}^{\text{2}}$ for $q_{\text{Al}}$ in equation (VI).

$\begin{array}{c}{A}_{\text{Al}}=\frac{A_{\text{Br}}\left(24\times {10}^6\text{W}/{\text{m}}^{\text{2}}\right)}{49.4\times {10}^6\text{W}/{\text{m}}^{\text{2}}}\\ =0.486A_{\text{Br}}\times 100\%\\ =48.6\%A_{\text{Br}}\\ \approx 49\%A_{\text{Br}}\end{array}$

The wall of thickness of both the radiators is same.

$V_{\text{Al}}=49\%V_{\text{Br}}$

Here, $V_{\text{Al}}$ is the volume of aluminum and $V_{\text{Br}}$ is the volume of Brass.

The weight of brass is calculated as,

Substitute $8.4\text{g}/{\text{cm}}^{\text{3}}$ for $D$ and $V_{\text{Br}}$ for $V$ in equation (VII).

$w_{\text{Br}}=8.4\text{g}/{\text{cm}}^{\text{3}}\left(V_{\text{Br}}\right)$ …… (VIII)

Here, $w_{\text{Br}}$ is the weight of brass and $V_{\text{Br}}$ is the volume of

The weight of aluminum is calculated as,

Substitute $2.71\text{g}/{\text{cm}}^{\text{3}}$ for $D$ and $V_{\text{Al}}$ for $V$ in equation (VII).

$w_{\text{Al}}=2.71\text{g}/{\text{cm}}^{\text{3}}\left(V_{\text{Al}}\right)$ ……. (IX)

Here, $w_{\text{Al}}$ is the weight of aluminum.

Substitute $0.49V_{\text{Br}}$ for $V_{\text{A}}$ in equation (IX).

$\begin{array}{c}{w}_{\text{Al}}=2.71\text{g}/{\text{cm}}^{\text{3}}\left(0.49V_{\text{Br}}\right)\\ =\left(1.33\text{g}/{\text{cm}}^{\text{3}}\right)V_{\text{Br}}\end{array}$ …… (X)

Divide equation (X) by (VIII).

$\begin{array}{c}\frac{w_{\text{Br}}}{w_{\text{Al}}}=\frac{8.4\text{g}/{\text{cm}}^{\text{3}}\left(V_{\text{Br}}\right)}{\left(1.33\text{g}/{\text{cm}}^{\text{3}}\right)V_{\text{Br}}}\\ w_{\text{Br}}=6.31w_{\text{Al}}\end{array}$

Conclusion:

Therefore, the relative area of aluminum is $49\%$ of the area of brass and the weight of brass is $6.31$ times the weight of aluminum.