# What is the heat flow rate [ ]?

"What is the heat flow rate [...]?"

## Homework Statement

The problem is:
"A copper tube (length, 3.0 m; inner diameter, 1.500 cm; outer diameter, 1.700 cm) extends across a 3.0-m long vat of rapidly circulating water maintained at 20 °C. Live steam at 100 °C passes through the tube.

(a) What is the heat flow rate from the steam into the vat?

(b) How much steam is condensed each minute?

(For copper, ##k_T## = 1.0 cal/s ⋅ cm ⋅ °C.)
"

The solution is attached as TheSolution.jpg.

## Homework Equations

##ΔQ/Δt = ####k_T## ##A## ##ΔT/L##

## The Attempt at a Solution

When looking at the provided solution, I am, basically, confused about the assumptions made and the geometries involved.

To elaborate, I am confused as to why the solution deemed it was necessary to approximate the tube wall as a flat sheet. (Also, how exactly does a tube WALL look like? - I wasn't able to find anything useful on Google Images.)

Also, it is probably because I am having trouble visualizing the geometries in this problem but, I don't see why the inner and outer surface areas are being averaged.

Lastly, I don't see why we're assuming the hypothetical plate has a thickness of 0.100 cm.

Any help in understanding how this problem is done would be greatly appreciated!

#### Attachments

• TheSolution.jpg
26.4 KB · Views: 332

## Answers and Replies

TSny
Homework Helper
Gold Member
See if this image helps.

#### Attachments

• cyl.gif
8.9 KB · Views: 409
gracy
haruspex
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I don't see why we're assuming the hypothetical plate has a thickness of 0.100 cm.
The thickness of the wall is the difference between the inner and outer radii, which will be half the difference between the inner and outer diameters.
It is perfectly feasible to treat the wall as an annulus (instead of as a flat sheet) and solve the problem analytically. http://en.wikibooks.org/wiki/Heat_Transfer/Conduction#A_hollow_cylinder.
(Since the temperature gradient through the wall changes along the length of the tube, there is still a slight approximation in assuming the heat flow is purely radial.)
The problem statement gives no guide on how accurate the answer needs to be, so I don't know how you were expected to know you could make the flat plate approximation.

Chestermiller
Mentor
The thickness of the wall is the difference between the inner and outer radii, which will be half the difference between the inner and outer diameters.
It is perfectly feasible to treat the wall as an annulus (instead of as a flat sheet) and solve the problem analytically. http://en.wikibooks.org/wiki/Heat_Transfer/Conduction#A_hollow_cylinder.
(Since the temperature gradient through the wall changes along the length of the tube, there is still a slight approximation in assuming the heat flow is purely radial.)
The problem statement gives no guide on how accurate the answer needs to be, so I don't know how you were expected to know you could make the flat plate approximation.

The link your provide really describes how it should be done precisely. The OP can compare the results from the "exact" solution with the approximate solution provided by the thin wall approximation.

Incidentally, the problem description implies that the temperature gradient through the wall does not change with axial location. The steam temperature will always be 100 C, and the outside wall temperature will always be 20 C.

Chestermiller
Mentor
The exact solution is given by $$Q=2πkL\frac{ΔT}{\ln(\frac{r_o}{r_i})}$$
where ro is the outer radius and ri is the inner radius. Now let's see how this compares with the thin wall approximation is. Let us rewrite the ratio of the outer radius to the inner radius as follows:
$$\frac{r_o}{r_i}=\frac{(r_o+r_i)+(r_o-r_i)}{(r_o+r_i)-(r_o-r_i)}=\frac{1+x}{1-x}$$
where $$x=\frac{(r_o-r_i)}{(r_o+r_i)}$$
So, $$\ln(\frac{r_o}{r_i})=\ln(1+x)-\ln(1-x)$$
Now, if we expand the ln terms in taylor series about x = 0, we obtain:
$$\ln(1+x)=x+...$$
$$\ln(1-x)=-x+...$$
so $$\ln(\frac{r_o}{r_i})=2x+...=2\frac{(r_o-r_i)}{(r_o+r_i)}+...$$
Substituting this into our original equation and retaining only the first term in the series, we obtain:
$$Q=πkL(r_o+r_i)\frac{ΔT}{(r_o-r_i)}$$
This is the same as the thin wall approximation.

Last edited:
Thank you all for your answers and, sorry for the late response.

I thought about the two height dimensions of the flat plate which is, technically, additional surface area introduced from the unrolling of the cylinder from which heat could escape which (slightly) reduces the amount of heat that would escape from the surface area being studied in this problem. Is that why the flat-plate approximation is an approximation? That's also why it's called the thin-wall approximation, right? Assuming that is a reason, are there more reasons why the flat-plate approximation is an approximation?

Since the temperature gradient through the wall changes along the length of the tube, there is still a slight approximation in assuming the heat flow is purely radial.
If we don't care about temperature uniformity and only care about the average rate of heat release per unit of surface area then this does not alter the calculations (even in a minute way), right?

Chestermiller:
How did you go from ##[(r_o + r_i) + (r_o – r_i)]/[(r_o + r_i) – (r_o – r_i)]## to ##(1 + x)/(1 – x)##?

Additionally, it seems to me that ##L = r_o – r_i## such that the ##L## and ##(r_o – r_i)## factors cancel out leaving ##Q = π k ΔT (r_o + r_i)## which kind of seems like ##Q = π k ΔT 2r_{average} = 2 π r_{average} k ΔT = A k ΔT## so, is ##A/L## equivalent to ##2 π r_{average}##? (The units seem to work well for that, at least.)

Also, instead of Q did you mean ΔQ/Δt, by any chance?

Chestermiller
Mentor
Thank you all for your answers and, sorry for the late response.

I thought about the two height dimensions of the flat plate which is, technically, additional surface area introduced from the unrolling of the cylinder from which heat could escape which (slightly) reduces the amount of heat that would escape from the surface area being studied in this problem. Is that why the flat-plate approximation is an approximation? That's also why it's called the thin-wall approximation, right? Assuming that is a reason, are there more reasons why the flat-plate approximation is an approximation?

If we don't care about temperature uniformity and only care about the average rate of heat release per unit of surface area then this does not alter the calculations (even in a minute way), right?

Chestermiller:
How did you go from ##[(r_o + r_i) + (r_o – r_i)]/[(r_o + r_i) – (r_o – r_i)]## to ##(1 + x)/(1 – x)##?

Additionally, it seems to me that ##L = r_o – r_i## such that the ##L## and ##(r_o – r_i)## factors cancel out leaving ##Q = π k ΔT (r_o + r_i)## which kind of seems like ##Q = π k ΔT 2r_{average} = 2 π r_{average} k ΔT = A k ΔT## so, is ##A/L## equivalent to ##2 π r_{average}##? (The units seem to work well for that, at least.)

Also, instead of Q did you mean ΔQ/Δt, by any chance?
How did you go from ##[(r_o + r_i) + (r_o – r_i)]/[(r_o + r_i) – (r_o – r_i)]## to ##(1 + x)/(1 – x)##?

Divide numerator and denominator by ##(r_o + r_i)##

Also, in the equations I wrote, L is the length of the cylinder, not the difference in radii, so L doesn't factor out.

Also, instead of Q did you mean ΔQ/Δt, by any chance?
In the equations I wrote, Q is the total rate of heat flow.

Divide numerator and denominator by ##(r_0 + r_i)##.
Also, in the equations I wrote, L is the length of the cylinder, not the difference in radii, so L doesn't factor out.
In the equations I wrote, Q is the total rate of heat flow.
Oh, now I see. :)

This is the same as the thin wall approximation.
If your general answer is the same as “the thin-wall approximation” doesn't that mean that “the thin-wall approximation” is not an approximation (assuming that the width of the flat plat approaches 0 infinitely)? Is “the thin-wall approximation” an approximation in this problem simply because we have to assign a finite width?

Lastly, why does it matter whether or not the heat is transferred in a perfectly-radial manner?

haruspex
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If your general answer is the same as “the thin-wall approximation”
It wasn't. It only reduced to thin wall approximation when Chet discarded all but the first term of the power series.