Maths assistance required.

[ripcrow]

I have asked this question on the physics forum but can’t get a decent response. I want to calculate how fast water vapour travels within a vacuum and how to calculate its boundary layer. The scenario is as follows. A sealed vacuum system comprising of two tanks connected by a pipe and and a tap in between the two tanks. In one tank water is heated to 40 degrees Celsius while the other tank remains empty but is cooled. The hot tank has a vacuum pressure of -27 in hg applied while the cold tank has a vacuum of -29 in hg applied. How fast will the water vapour move between the two pressure differential. 

To give greater clarity -27 in hg is roughly the pressure at which 40 degrees Celsius water will vaporise at and -29 in hg is the pressure that 1 degree water will vapourise. I’d imagine that the tap will act as a nozzle and the water vapour would lose temperature as it passes through the nozzle  but I don’t know how much by. The idea is based on a cryophorus system. Can’t paste link but can be found with a quick search. 

[racket]

LOL......this is different :-)

You'll need to get back to absolute pressures , standard atmosphere is 29.92 in hg , so your tanks are at 2.92 in hg absolute and at 0.92 inhg absolute , 2.92 inhg = 1.449 psia and 0.92 = 0.452 psia so a pretty decent Pressure Ratio, its gunna go fast .

Basically you have wet saturated steam at 40 deg C , same as last stage/s of a steam turbine emptying into a Condenser at 0.5 psia

I might have to think on this a bit more

Cheers
John

[ripcrow]

Absolutely love the way you grasp the situation racket. You wouldn’t believe the comments I got on the physics forum just for asking the question. The pressure differential being a negative is complicated to calculate.

[racket]

LOL..............those sort of low pressures are encountered at very high altitudes , ~53,000 and 95,000 feet , so not really talking about "vacuums" rather just low pressures , but with a "high" 3.2 :1 Pressure Ratio between the two chambers.

I'll have a look through an old steam engine book I have to see if theres info

Cheers
John

[racket]

OK , at 1 psia steam forms at 101.7deg F -38.7C so our water is gunna boil as soon as the valve is opened and the pressure drops , but we don't know how quickly the pressure in the two tanks will "equalise" , but assuming we only want the initial velocity then we can assume that high pressure ratio exists , but theres air in those tanks that will exit before the 40C water at ~1.5psia starts to boil and send both steam and air through the "nozzle" .

LOL......this is getting complicated , no wonder the physics guys gave up :-(

Now as soon as the ~1.5psia pressure drops and the water starts to boil , the water temp will want to drop as the latent heat of evaporation kicks in unless you quickly add those BTUs .

Theres too many variable here to be able to workout any sensible numbers , with a very small "nozzle" its just gunna be airflow and we can workout the velocity , using the "normal" ways, you'll probably get a choked sonic nozzle velocity of ~1,000 ft/sec

More thinking required

[ripcrow]

Have you ever seen a cryophorus system. It’s an enclosed system with a hot side and cold side. From some lessons on YouTube the condensation rate causes the vacuum to increase ( lower air pressure ). Provided the heat can be kept up to the hot side and the heat can be removed from the cold side the volume in the cold side continues to be lower then the volume in the hot side. I’ll try another explanation. If 1 cubic meter of water is evaporated as that one cubic meter of water is condensed the volume is less therefore the air pressure is lower. Because the air pressure is lower the hot side will have lower pressure so the remaining water will boil at a lower pressure which means less heat is required to boil the water and less heat is required to be removed from the cold side.  This video describes the dew point and how pressure acts on evaporation and condensation rates. At the end of the video it shows that the rate of evaporation is less then the rate of condensation and the air would be supersaturated after it passes through a nozzle that was maintaining the pressure differential. That’s my interpretation of it youtu.be/BqFVtlQa-2w

[racket]

LOL......all academic , but what does it mean ??

Why does a Sterling engine come to mind en.wikipedia.org/wiki/Stirling_engine

[ripcrow]

I was thinking more of a turbine. Enthalpy is 2206 joules per gram of water. And that means that 2206 joules per gram has to be removed to convert the vapour back to water and allow the vacuum to persist. With turbine efficiency being 90% or above that would mean that 1985 joules are available to be harnessed per gram of water.Water to steam ratio is 1700 times at 100 degrees and at 1 atm.
I have been trying to work out what volume and mass and at what speed is available in such a system. The density of steam at 100 degrees is .5976 kg per cubic metre at 1 atm while volume is 1.6733 cubic metres per kilogram.
The density of water vapour at 40 degrees and 55.3 torr is 0.051243 kg per cubic metre and volume is 19.515 cubic metres per kilogram.
If my interpretation is correct the volume difference is huge but the volume doesn’t have much mass. It’s been along time since I played with jet engines due to work and life but as a thought experiment how do these densities and volume differ to the mass flow going through a jet turbine. Is the speed close to the speed of flow in a jet turbine and hiw much difference is there in volumes and densities.

[racket]

You've lost me :-(

If theres a pressure ratio across a turbine stage and theres mass flow then it should turn out power , but with such low densities the mass flow will be very small unless the equipment is huge , and with small mass flows the power is going to be minimal .

[ripcrow]

Just did some equations
Wasn’t sure if Bernoulli’s equations were relevant but I used it anyway. With 40 degree water at 7372 pascals and the cold side at 1 degree and 653 pascal the speed should be 1188 ft per second or 1303 kph with a pipe of internal diameter of 5 mm the flow volume is 2.84 cubic metres per second with a mass of 0.1455 kilogram per second.

[ripcrow]

Jan 31, 2021 1:21:03 GMT -5 racket said:

You've lost me :-(

If theres a pressure ratio across a turbine stage and theres mass flow then it should turn out power , but with such low densities the mass flow will be very small unless the equipment is huge , and with small mass flows the power is going to be minimal .

What mass flow and densities and what speeds do you use on the jet turbine 

[racket]

You might need to redo your volume calculation again , its a "tad" optimistic

[ripcrow]

Just tried calculating force. Mass is 0.145530 kg per second so I used the speed as 362.1054 metres per second for acceleration. Not sure if that was the right way to do it. Gives a force of 52.69 Newton’s.

[ripcrow]

Jan 31, 2021 2:25:00 GMT -5 racket said:

You might need to redo your volume calculation again , its a "tad" optimistic

I used Bernoulli’s equation. Where did I go wrong 

[racket]

Volume = area X speed ................5 mm dia = 19.635 sq mms , divide by 1,000,000 (sq mms/sq meter) to get area in sq meters , then multiply by 362 to get cubic meters/sec