saturated steam 5

[shoubaki]

This system includes six vessels , the temperature of the output steam from the final vessel is 122.4 C (if it is still steam) ,
each vessel have demister at upper portion ,


zekeman : our major problem that the output steam from the last vessel is too low as you thought,the problem encountered when increase the demand on the output , and the machine designed to produce more output

note that the industrial steam enter only the first vessel and then return to back line

 

[rmw]

So, is the effect you are asking about in the original post the first effect? Define 'purified water'. Are you distilling purified water or making purified water, and if making purified water, what are you distilling. What is the boiling point elevation (depression) of the fluid that is fed into the first effect?

rmw

 

[shoubaki]

I am distilling purified water , but about poiling point that I tried to know !! may the pressure (if its not saturated pressure ) in the first effect increase the poiling point ?, how can I determine it? the information I have mentioned in drawing I attached

 

[rmw]

I think your temperature measuring element is being wetted by carryover or ME slip, thereby cooling it below the sat temp of the vessel. Is it possible to move it?

Or... the 'purified' water is loaded with air or some other dissolved gas so that as was stated much earlier in this thread, there is a partial pressure relationship going on between the steam vapor present and the noncondensable gasses present.

rmw

 

[zekeman]

shoubaki,
This has been going on too long
By now you (we) should have figured out your problem.
The speculation on entrained gases vs pure hot water could be resolved as follows
You need to find out what is entering tank 1 and then do an energy balance to help resolve the issue.I propose the following crucial steps to get started:
1 get the temp, pressure and flow rate of the steam entering tank 1
2 get the temperature, pressure and flow rate of the water entering tank1
3 do a heat balance as follows
[(Uhw*w' +Usteam*w"]/(w'+w")= k1*Uwater2+k2*Usteam2-thermal heat transfer losses per lb of fluid per unit time
Uhw=internal energy of water (70 deg)
Usteam= internal energy high pressure steam
w' flow rate of incoming water
w" flow rate of incoming steam
k1 fraction of incoming fluids that is water
k2 fraction that is steam in tank
U water2 internal energy of assumed saturated water
Usteam2
This left hand side gives the average internal energy of the mixture in 2.
Now based on your measurements you should be able to get k1 and k2 the fractions of that energy that might be allocated ( first assume adiabatic
ignoring thermal losses from the tank)

Post you results and we can further assist.