vent sizing 2

[401laurent]

Hello,
I'm sizing the vent for vacuum tank in condition of cold spray after hot washing, Could any one help please giving the formulas for this calculation, the aire venting and the air inflow required, in order to size the vacuum valve.
the data given is: MAV(Maximum allowable vacuum)for the tank ( psi), Cold spray temperature (oF), cold spray flow (lbs/h).
Thanks.
Laurent.

 

[mcecasf]

edit - my steam maths had an error so revisiting.

 

[mcecasf]

I've not worked out how to edit posts - should have said "my poor knowledge of compressible flow"

Sheet attached

 

[zekeman]

Looks like you're assuming an allowable pressure of about 0.5 bar, since you assumed choked flow.

Neither of the cases you did represents the true situation.


In fact you have a mixture of steam and air so I attempted this problem conservatively and got flows less than yours.

First, I assumed no airflow until the pressure inside was o.5 bar (correct me if that is wrong)


Next , using the steam tables, the initial partial pressure of steam was 8.5 psi and the air 6.2, T=185F

It is clear to me that the energy equation for water injection results in condensing steam, with only second order heat exchange from the air.

So, I followed the steam tables down to 2 psi partial pressure of steam and 5+ for the air at 129F (forgive me for using these units since the steam tables I have are in BTU/F units.

At this point I allow air to enter, but conservatively DIO NOT let it mix. to sustain the pressure
The energy equation is with good acuracy

Ww(129-25)=1070Ws
So
Ww=0.097Ws
Looking for a rate of temperature change so I can get the change in volume as the mixture cools.
From the tables, I got
for 65M^3 (1950FT^3) of mixture in cooling 3.7F I condensed 1950(1/173-1/191)=1.16lb steam
Frrom the ratio of Ww/Ws=.097, that means the amount of water needed
1.16/.097=11.95 lb cooling water
Since you are pumpinmg about
10lb/sec
we need 1.195 seconds to do it
and the rate of temperature drop becomes
3.77/1.195=3.15F/sec
The volumetric change is then
3.15/(460+126)*1950=10.4 FT^3/sec
Which is the inlow air conservatively needed to sustain the inside pressure.

Subsequent cooling is less demanding on airflow.

 

[zekeman]

Oops, big error.
Forgot to replace the steam that condensed oot
1.16*v=1.16*173=200
v= specific volume
200/1.195 sec=167 ft^3/sec
which I believe is the most consrvative.
The air referenced to the outside is
.5*167=83.5 FT^3/sec
and swamps out previous answer.

 

[zekeman]

Yet another correction (and hope the last)
I also noted in going from steam partial pressur P=2 psi to p=1.8 Psi (corresponding to delta T=3.7 deg F)
along the sturation cuurve for the cooling, the number of moles, n that must be added is obtained from
0.2*144*1950=n*R*T
R=1544 universdal constant
solving for n
n=.2*144*1950/1544/490=0.0742 moless of air which weighs
2.3 lb
At standard conditions the volume is
2.3/.078=29FT^3
and since t=1.195 sec
the flow rate is
29/1.195=24 FT^3/sec
My final answer.