Radial Heat Conduction through a Cylindrical Pipe

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Homework Help Overview

The discussion revolves around a problem involving radial heat conduction through a cylindrical PVC pipe, focusing on calculating the heat transfer rate and the subsequent temperature change within the pipe. The context includes specific parameters such as the lengths, diameters, thermal conductivity, and temperature differences.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply Fourier's Law of Conduction and surface area calculations to determine the heat transfer rate. They express uncertainty regarding the units for the calculated heat transfer rate (Q) and seek clarification on the implications of their results.

Discussion Status

Participants are actively engaging with the original poster's calculations, providing guidance on deducing the units for Q and highlighting the need for additional information to compute the temperature decrease accurately. Multiple interpretations of the unit question are being explored without reaching a consensus.

Contextual Notes

There is a noted lack of information regarding the contents of the pipe, which is necessary for calculating the temperature change over time. The original poster's inexperience with this type of calculation is acknowledged, indicating a learning process in progress.

Thisbe Schwer
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Homework Statement


A 1.75m long PVC pipe with a thermal conductivity of 0.19 W/mK has an internal diameter of 3mm and an external diameter of 5.5mm. Inner temperature is 298K and outer temperature is 273K. Calculate the heat transfer rate through the pipe and thus the decrease in the inner temperature after one minute.

Homework Equations


Fourier's Law of Conduction: Q = -kA(dT/dr)
Surface area of cylinder: A = 2πrL

The Attempt at a Solution


I've got the formula Q = [k2πL(Ti - To)] / [In(r2/r1)] and substituted in the values:
Q = [0.19W/mK * 2π * 1.75m * 25K] / [In(1.8333)]
Q = 16.625/0.6061...
Q = 27.428 (3dp)
But I'm a complete newbie to this kind of calculation, and I'm not sure what the unit for Q is? So I'm kind of stuck at this part of the question and not sure where to go from here.
Sorry for the silly question, but any help would really be appreciated![/B]
 
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Thisbe Schwer said:
not sure what the unit for Q is
You can deduce it by treating the units in your expression (K, m, W) as algebraic variables and cancelling as appropriate.
Thisbe Schwer said:
thus the decrease in the inner temperature after one minute.
You need more information to compute this, such as what is in the pipe.
 
So would the units for Q be W/m?
 
Thisbe Schwer said:
So would the units for Q be W/m?
You had
Thisbe Schwer said:
0.19W/mK * 2π * 1.75m * 25K
Leaving out everything except the units:
(W/mK) * m * K. What does that reduce to?
 

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