Question

A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both the length and the diameter of this bar are doubled, the rate at which it will conduct heat between these reservoirs will be

Answer 1

Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

1. Conduction could be described by the Law of Fourierin the form: $Q=kA\frac{T_1-T_2}{L}$ where $Q$ is the rate of heat transferred  by conduction, $k$ is the thermal conductivity of the material, $T_1$ and $T_2$ are the temperatures of each heat deposit, $A$ is the cross area to the flow of heat, and ${L}$ is the distance that the flow of heat has to go.
2. For the original cylinder the Fourier's law would be: $kA_1\frac{T_1-T_2}{L_1}=25W$, and if $A_1=\frac{\pi D_{1}^{2}}{4}$, then the expression would be:$k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W$ where $D_1$ is the diameter of the original cylinder, and ${L_1}$ is the length of the original cylinder.
3. For the new cylinder, in the same fashion that for the first, Fourier's Law would be: $Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2}$,where $Q_2$ is the heat rate in the second case, $D_2$ and ${L_2$ are the new diameter and length.
4. But, $D_2=2D_1$ and $L_2=2L_1$, substituting in the expression for $Q_2$: $Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}$.
5. Rearranging: $Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1})$.
6. In the last declaration of  $Q_2$, it could be noted that the expressión inside the parenthesis is actually  $Q_1$, then:  $Q_2=\frac{2^2}{2}(25W)=50W$.
7. It should be noted, that the temperatures in the hot and cold reservoirs never change.

Answer 2

A solid cylindrical bar where length and external diameter are doubled has a heat transfer rate of 50 watts.

Let suppose that both solid cylindrical bar have the same inner diameter, but the new solid cylindrical bar is the double as long as the original cylindrical bar and the external diameter is the double as great as in the original cylindrical bar.

In addition, we assume that conductive heat transfer is one-dimensional and stable. Then, the heat transfer rate of the cylindrical bar is directly proportional to the length and inversely proportional to the natural logarithm of the diameter ratio:

$\frac{Q'}{Q} = \frac{\frac{2\cdot l}{\ln \frac{2\cdot D_{e}}{D_{i}} } }{\frac{l}{\ln \frac{D_{e}}{D_{i}} } }$

$\frac{Q'}{Q} = 2\cdot \frac{\ln \frac{D_{e}}{D_{i}} }{\ln \frac{2\cdot D_{e}}{D_{i}} }$

$\frac{Q'}{Q} = 2\cdot \frac{\ln \frac{D_{e}}{D_{i}} }{\ln \frac{D_{e}}{D_{i}}+\ln 2 }$

$Q' = 2\cdot Q \cdot \left(\frac{\ln \frac{D_{e}}{D_{i}} }{\ln \frac{D_{e}}{D_{i}} +\ln 2} \right)$

$Q' = 2\cdot Q \cdot \left(\frac{1}{1+\frac{\ln 2}{\ln \frac{D_{e}}{D_{i}} } } \right)$

Where:

• $Q$ - Original heat transfer rate.
• $Q'$ - New heat transfer rate.
• $D_{i}$ - Internal diameter.
• $D_{e}$ - External diameter.

If $D_{e} >>> D_{i}$, then the expression is reduced into this expression:

$Q' = 2\cdot Q$

If we know that $Q = 25\,W$, then $Q' = 50\,W$.

A solid cylindrical bar where length and external diameter are doubled has a heat transfer rate of 50 watts.

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