Posted by Morris Dovey on June 10, 2008, 11:18 am
If I pour a U-shaped tube, say, half full of a fluid (of density d) and
stopper the left "leg", I have a volume V0 of air trapped at temperature
T0 and ambient pressure P0.
If I (magically) warm the trapped air from T0 to T1, its volume will
expand from V0 to V1.
I know how to calculate V1 if the expansion is isobaric (takes place at
constant pressure).
However, as the expansion takes place fluid is displaced from the left
side of the tube to the right side, which should cause an increase in
pressure in the trapped air. In other words, the expansion /can't/ take
place at constant pressure.
*What I tried...*
I calculated the expansion as if it was isobaric. Then I noted that the
right column would contain 2(V1 - V0) more fluid than the left column,
and reasoned that this would amount to a mass difference of 2d(V1 - V0)
and (from a long-ago physics course) since F=ma, there would be a force
of 2dg(V1 - V0) pushing down on the right column (and so pushing up on
the trapped air in the left column).
IIRC, pressure is force divided by area - so the trapped air should be
at a pressure equal to the original pressure P0 plus the pressure caused
by the upward-displaced water. In other words:
P1 = P0 + 2dg(V1 - V0)/(cross-sectional area of U-tube), where g is
acceleration due to gravity (roughly 9.8 m/s² or 32 ft/s²)
*My head began to hurt when...*
I realized that the increase in pressure would cause a reduction in the
volume of the trapped air, which means that the calculated right side
displacement wasn't quite right, which means that the adjusted pressure
calculation was off...
...and it was (suddenly) time to look for some knowledgeable help.
I'm guessing that I might be able to do an iterative calculation, but
suspect that this kind of calculation has been done often enough that
someone must have produced a way to directly calculate V1 and P1.
Anyone know how this kind of problem is solved?
--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/
Posted by daestrom on June 11, 2008, 8:27 pm
> If I pour a U-shaped tube, say, half full of a fluid (of density d) and
> stopper the left "leg", I have a volume V0 of air trapped at temperature
> T0 and ambient pressure P0.
> If I (magically) warm the trapped air from T0 to T1, its volume will
> expand from V0 to V1.
> I know how to calculate V1 if the expansion is isobaric (takes place at
> constant pressure).
> However, as the expansion takes place fluid is displaced from the left
> side of the tube to the right side, which should cause an increase in
> pressure in the trapped air. In other words, the expansion /can't/ take
> place at constant pressure.
> *What I tried...*
> I calculated the expansion as if it was isobaric. Then I noted that the
> right column would contain 2(V1 - V0) more fluid than the left column, and
> reasoned that this would amount to a mass difference of 2d(V1 - V0) and
> (from a long-ago physics course) since F=ma, there would be a force of
> 2dg(V1 - V0) pushing down on the right column (and so pushing up on the
> trapped air in the left column).
> IIRC, pressure is force divided by area - so the trapped air should be at
> a pressure equal to the original pressure P0 plus the pressure caused by
> the upward-displaced water. In other words:
> P1 = P0 + 2dg(V1 - V0)/(cross-sectional area of U-tube), where g is
> acceleration due to gravity (roughly 9.8 m/s² or 32 ft/s²)
> *My head began to hurt when...*
Let's ignore changes in vapor pressure.
Start with the level of liquid in the two columns equal and mark the level.
Let 'X' be the height of the liquid above the mark on the open end of the
U-tube. I also assume the two vertical legs have the same cross-sectional
area (same diameter tubes)
The initial pressure in the closed end if the level in the two columns is
equal initially is the same pressure as the open end, atmospheric pressure.
Therefore P0 = Patm. (1)
After heating, we want to know 'X'. The pressure in the enclosed chamber
is:
P1 = Patm + 2*X*rho*ga
Where:
X is measured height above initial mark
rho density of liquid (in appropriate units to match X)
ga acceleration of gravity (if using 'English' units you also divide by
g-sub-c so this can be treated as '1')
The initial volume is just V0. The volume after things have heated and the
water displaced is simply:
V1 = V0 + X*A
X is measured height above mark on open side (or below mark on this side of
tube, it is the same if the diameters are the same)
A is cross-sectional area of tube in the region of the water.
Finally, the ideal gas law tells us:
P1*V1 / T1 = P0*V0 / T0
P1*V1 = P0*V0*T1 / T0
Subtitution:
(Patm+2*X*rho*ga)*(V0+X*A) = P0*V0*T1/T0
And since P0=Patm...
(P0+2*X*rho*ga)*(V0+X*A) = P0*V0*T1/T0
Solve this for X
P0*V0+2*X*rho*ga*V0 + P0*X*A+2*X^2*rho*ga*A = P0*V0*T1/T0
(and then a miracle occurs...)
aX^2+bX+C = 0
X=-b +-sqrt(b^2-4ac) / 2a
where:
a = 2*rho*ga*A
b=(2*rho*ga*V0+P0*A)
c=P0*V0*(1-T1/T0)
Believe it or not, you can plug this into a spreadsheet like I did with a
few extra cells and it solves pretty nicely. Set separate cells for V0, A,
T0 and T1. Then set cells equal to the formulas for 'a', 'b', and 'c'.
Finally, set a cell to the formula for 'X'. Then you can play around with
different values for V0, A, T0 and T1 with ease.
I tried a 1 inch diameter tube with the enclosed portion 10 inches long (V0
= 7.85 cu in). With an initial temperature of 530 R (70F) and a final
temperature of 630 R (170F), I got a value for X of 1.78 inches.
Making the temperature change larger, or the inital volume V0 larger by a
taller enclosed portion gave me a larger 'stroke'. Using a wider diameter
tube made no difference.
daestrom
Posted by Morris Dovey on June 11, 2008, 8:52 pm
daestrom wrote:
> Let's ignore changes in vapor pressure.
>
> Start with the level of liquid in the two columns equal and mark the
> level. Let 'X' be the height of the liquid above the mark on the open
> end of the U-tube. I also assume the two vertical legs have the same
> cross-sectional area (same diameter tubes)
>
> The initial pressure in the closed end if the level in the two columns
> is equal initially is the same pressure as the open end, atmospheric
> pressure. Therefore P0 = Patm. (1)
>
> After heating, we want to know 'X'. The pressure in the enclosed
> chamber is:
>
> P1 = Patm + 2*X*rho*ga
> Where:
> X is measured height above initial mark
> rho density of liquid (in appropriate units to match X)
> ga acceleration of gravity (if using 'English' units you also divide by
> g-sub-c so this can be treated as '1')
>
> The initial volume is just V0. The volume after things have heated and
> the water displaced is simply:
>
> V1 = V0 + X*A
> X is measured height above mark on open side (or below mark on this side
> of tube, it is the same if the diameters are the same)
> A is cross-sectional area of tube in the region of the water.
>
> Finally, the ideal gas law tells us:
>
> P1*V1 / T1 = P0*V0 / T0
> P1*V1 = P0*V0*T1 / T0
>
> Subtitution:
>
> (Patm+2*X*rho*ga)*(V0+X*A) = P0*V0*T1/T0
>
> And since P0=Patm...
> (P0+2*X*rho*ga)*(V0+X*A) = P0*V0*T1/T0
>
> Solve this for X
> P0*V0+2*X*rho*ga*V0 + P0*X*A+2*X^2*rho*ga*A = P0*V0*T1/T0
>
> (and then a miracle occurs...)
>
> aX^2+bX+C = 0
>
> X=-b +-sqrt(b^2-4ac) / 2a
>
> where:
> a = 2*rho*ga*A
> b=(2*rho*ga*V0+P0*A)
> c=P0*V0*(1-T1/T0)
>
> Believe it or not, you can plug this into a spreadsheet like I did with
> a few extra cells and it solves pretty nicely. Set separate cells for
> V0, A, T0 and T1. Then set cells equal to the formulas for 'a', 'b',
> and 'c'. Finally, set a cell to the formula for 'X'. Then you can play
> around with different values for V0, A, T0 and T1 with ease.
>
> I tried a 1 inch diameter tube with the enclosed portion 10 inches long
> (V0 = 7.85 cu in). With an initial temperature of 530 R (70F) and a
> final temperature of 630 R (170F), I got a value for X of 1.78 inches.
>
> Making the temperature change larger, or the inital volume V0 larger by
> a taller enclosed portion gave me a larger 'stroke'. Using a wider
> diameter tube made no difference.
First class! Many thanks.
--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/
Posted by daestrom on June 13, 2008, 8:07 pm
>-> I tried a 1 inch diameter tube with the enclosed portion 10 inches long
>(V0
> -> = 7.85 cu in). With an initial temperature of 530 R (70F) and a final
> -> temperature of 630 R (170F), I got a value for X of 1.78 inches.
> Which is fine in theory, but not in practice. The difference in the
> vapour pressure of water between your two temperatures is hugely
> greater than the pressure produced by a column of water 1.78 inches
> high. I'd guess it would be of the order of ten *feet* of water.
In the 'steady state'. But if you're just heating the *air* and not the
water, the water vapor pressure will change only slowly as heat transfers
into the water. Since Morris is working with his fluidyne engine, we know
the water is not being heated directly but the air only. And we know it
won't stay 'hot' for long. So assuming the water vapor pressure stays
pretty constant wouldn't be too far off.
> At high temperatures, a first approximation would ignore the gas in the
> apparatus, and consider *only* the vapour pressure of the water. The
> little program I posted here this morning gives a good approximation to
> the vapour pressure at moderate temperatures. Way up near 100C, it's a
> bit off, but only by a few percent. This could be improved by choosing
> the two initial temperatures more appropriately, for example 30 and
> 80C, instead of 10 and 40C, which the program uses. You'd have to use a
> published table to put in the two values of the VP at the two new
> temperatures.
Again, *if* you're heating the water. But I believe Morris is trying to
only heat the air. And the water temeprature changes from contact with the
air would be a couple orders of magnitude smaller in anything that cycles on
the 'many times a minute' time scale that Morris is working with.
daestrom
Posted by daestrom on June 14, 2008, 12:06 pm
>-> In the 'steady state'. But if you're just heating the *air* and not the
> -> water, the water vapor pressure will change only slowly as heat
> transfers
> -> into the water. Since Morris is working with his fluidyne engine, we
> know
> -> the water is not being heated directly but the air only. And we know
> it
> -> won't stay 'hot' for long. So assuming the water vapor pressure stays
> -> pretty constant wouldn't be too far off.
> Maybe. But only a thin film at the surface of the water has to get hot
> by contact with the air to raise the vapour pressure. It could happen
> very fast.
> And even if the vapour pressure is *constant*, that's not the same as
> it being *zero*.
> I did a few experiments with the program I wrote today, comparing the
> temperatures that would correspond to various depressions of the water
> surface, with and without vapour pressure. The results were utterly
> different. For example, with the 100-millimetre empty space, when the
> two water levels are equal, and with a depression of the level in the
> closed arm of 50 mm, if there is no vapour pressure, the temperature is
> close to 100C. But if the air (or whatever gas) in the closed space is
> saturated with water vapour, the temperature is only about 65C. That's
> a *huge* difference.
> The problem is, we really don't know what the vapour pressure is.
> Assuming that the air is saturated with water vapour represents a
> limiting case. Assuming the vapour pressure to be zero is the opposite
> limit. Reality is probably somewhere in between. But where?
> A few days ago, I suggested that it would be much better for Morris to
> use some other liquid, other than water, which has a negligible vapour
> pressure. But this may not be practicable for him.
Seems like a thin layer of oil would be very easy to do. Even household
cooking oil can work here.
> (Right now, he's probably hammering together an ark, and gathering as
> many good-looking women as possible to ride with him in it. His
> fluidyne project may be far from his mind.)
> I'm not sure what the purpose of this U-tube device is. I don't think
> much of it as a thermometer.
I believe he is thinking of the fluidyne Stirling pump that he's been
working on.
daestrom
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> stopper the left "leg", I have a volume V0 of air trapped at temperature
> T0 and ambient pressure P0.
> If I (magically) warm the trapped air from T0 to T1, its volume will
> expand from V0 to V1.
> I know how to calculate V1 if the expansion is isobaric (takes place at
> constant pressure).
> However, as the expansion takes place fluid is displaced from the left
> side of the tube to the right side, which should cause an increase in
> pressure in the trapped air. In other words, the expansion /can't/ take
> place at constant pressure.
> *What I tried...*
> I calculated the expansion as if it was isobaric. Then I noted that the
> right column would contain 2(V1 - V0) more fluid than the left column, and
> reasoned that this would amount to a mass difference of 2d(V1 - V0) and
> (from a long-ago physics course) since F=ma, there would be a force of
> 2dg(V1 - V0) pushing down on the right column (and so pushing up on the
> trapped air in the left column).
> IIRC, pressure is force divided by area - so the trapped air should be at
> a pressure equal to the original pressure P0 plus the pressure caused by
> the upward-displaced water. In other words:
> P1 = P0 + 2dg(V1 - V0)/(cross-sectional area of U-tube), where g is
> acceleration due to gravity (roughly 9.8 m/s² or 32 ft/s²)
> *My head began to hurt when...*