Chimney effect and pressure difference

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  • #1
weirdoguy
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Homework Statement:
Factory chimney has a height of ##h=22,4m##. Its side inlet opening is tightly closed with a cover whose area is ##S=1m^2##. Atmospheric air temperature is ##t_0=0^\circ C## and its pressure is ##P_0=10^5 Pa##. Calculate the value of the average air temperature in the chimney, if it is known that due to the temperature difference, the force ##F=85N## acts on the cover.
Relevant Equations:
Hydrostatic pressure; relation between density of air in different temperatures
Kind of similar question was once on polish high school ("matura") exam, and the solution that the authors gave was:
Pressure at the bottom of the chimney inside of it is ##p_{top}+\rho_1gh##, and outside ##p_{top}+\rho_2gh## where:
##p_{top}## is the atmospheric pressure at the height of ##h##,
##\rho_1## is the density of air inside the chimney,
##\rho_2## is the density of air outside.
Force is then ##F=|\rho_1-\rho_2|ghS##. But in that situation, densities were given in the problem. Here they are not. I guess we can use the equation:
##\frac{\rho_1}{\rho_2}=\frac{T_2}{T_1}##, where density in temperature of ##0^\circ C## can be found in the tables, or somewhere (it's somewhere around ##1,2kg/m^3## if I remember correctly). But how to use this ##P_0## given in the problem? At what level is it measured? In yet another exercise it was said that ##P_0## is measured at the bottom, but still using it forces me to use mollar mass of air through Clapeyron equation...

In yet another problem book I found similar exercise in which authors use coefficient of thermal expansion...

I really don't know where to go with this.
 

Answers and Replies

  • #2
haruspex
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There's PV=nRT, and n/V relates to the density. Does that help?
 
  • #3
weirdoguy
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Well, yes, I know. If I use that I get ##\frac{P}{\rho}=\frac{RT}{\mu}## and I need to use molar mass, which I have a feeling I'm not allowed to o0) I guess I don't know how to finish this exercise, since I see different options for that. So my question is what would you do? I would go with the density, without using ##P_0##, but autors gave it for a reason. So maybe I should use this molar mass. Unfortunately I don't have the book in which this exercise is, my student got that as a homework. I asked on polish forum if someone know where is it from, I would like to see what answer authors gave.
 
  • #4
haruspex
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As a thought experiment, suppose we replace the air with a gas of greater molar mass, but same pressure at top of chimney. The pressure difference would increase for the same temperature difference, yes? So there is no way without knowing the air molar mass.
 
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  • #7
Lnewqban
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@weirdoguy was trying to avoid using data not given in the problem, specifically, air's molar mass. Looking the density up in such a table should be even less desirable.
I see.
I believe that he still could use the tables as a reference to verify his density calculations.
It seems logical to assume that the pressure would be exerted on the external surface on the cover (which may be located at ground level), as normally, air in chimneys are hotter than open atmospheric air.
 
  • #8
weirdoguy
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Thank you guys for your opinion. I used molar mass and the answer is:
$$T=\frac{T_0P_0\mu ghS}{P_0\mu ghS-FRT_0}$$
 
  • #9
haruspex
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Interesting. Note that T should tend to infinity as the pressure difference on the plate approaches that between the top and bottom of the tower outside, ##\rho gh##.
Which definition of ##P_0## are you using, top or bottom?
 
  • #10
weirdoguy
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Bottom. I also did think about this issue, that is about the subtraction in the denominator. My line of thought was similar to that on the said matura exam:
$$(1)\qquad F=(\rho_0-\rho)Sgh $$
$$(2)\qquad\rho=\frac{\rho_0T_0}{T}$$
$$(3)\qquad\rho_0=\frac{P_0\mu}{RT_0}$$
Putting it all together:
$$F=\left(1-\frac{T_0}{T}\right)\frac{P_0\mu ghS}{RT_0}$$
Maybe the problem is with the second and third equation, maybe they are not compatible with each other.
 
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  • #11
haruspex
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Bottom. I also did think about this issue, that is about the subtraction in the denominator. My line of thought was similar to that on the said matura exam:
$$(1)\qquad F=(\rho_0-\rho)Sgh $$
$$(2)\qquad\rho=\frac{\rho_0T_0}{T}$$
$$(3)\qquad\rho_0=\frac{P_0\mu}{RT_0}$$
Putting it all together:
$$F=\left(1-\frac{T_0}{T}\right)\frac{P_0\mu ghS}{RT_0}$$
Maybe the problem is with the second and third equation, maybe they are not compatible with each other.
That last equation looks good in the limit as T tends to infinity. It gives
##\rho gh=F/S=\frac{P_0\mu gh}{RT_0}##
##P_0\mu=RT_0\rho##
But for some mass of air M, volume V, molar count n, ##\mu=M/n, \rho=M/V##, so this is equivalent to ##P_0V=nRT_0##.
 

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