Calculation req for N2 to move slurry down 6" line.

[v8landy]

Hi

I am looking for assiatance to put some figures to a design requirement to move recovey bottoms slurry down a 6" line approx 30m.

We have 10bar G N2 onsite. Viscosity of slurry changes, but is general quite thick (figures to come).

Basic question is does it sound possible, and would I be able to show some sort of calculation?

Thanks

 

[quark]

Yes, what you have to do is a pressure drop calculation. If the slurry is finally pushed into atmosphere (or an atmospheric tank) then the required N2 pressure(ideally) is the pressure drop itself. Volumetric requirement will be equal to the tank+piping volume, at the given pressure.

 

[v8landy]

Thanks Quark

I would be push from on tank to another which would be atm tank. (? over push up or via top?)

So what sort of info would be required?
Tank vols
Pipe vols/lenghts/bends etc
Fluid info (viscosity, desity etc)

Thanks

 

[quark]

When you transfer it from top then there is no accidental backflow. I would like to go with a dip pipe and an impingement plate for the transfer as it is a slurry.

Yes, that information is required (+add valve details)

 

[v8landy]

I prefer top feed for the same reasons. Dip leg is possible, but plate is not due to no access.

 

[quark]

Slury data viz, particle size and volume concentration is required. Don't go for dip tube when you can't use an impingement plate.

 

[v8landy]

Can anyone assist me with the calcuations?

 

[v8landy]

OK

I have calculated that I would need 24 Bar, but I need verfication of my calcualtion, and if indeed it is correct.

SG = 1.051
Tank A = 10m3 (2m high) I want to move this at about 3m^3/hr.
Tank B = 125m3 (14m high)
Approx distance from A to B = 100 m with a few 90's and valves. Open to atms

I started of using Bernoulli's equation to find V2

V2= sqrt(2*g*h1) = 6.26 m/s

I then calculated Re number and got 329
With this I calculated The Friction Coefficient - ? = 0.194
I then calculated ? = ? g the specfic weight = 10.31
I then calculated pressure loss = ploss = ? (l / dh) (? v2 / 2)

friction coefficent 0.194454017 666.6666667
l 100
dh 0.15 18918
v2 36
? 1051

2452454.068 pressure loss = pressure loss (Pa, N/m2)

bar 24.52454068

Can anyone help or confirm please

 

[CMA010]

24 bar over a 100m 6" line and a 12 m height difference seems high, very high, assuming 3 m3/hr. 3 m3/hr gives you a velocity of 0.05 m/s.

 

[v8landy]

Thanks, I have recalculated on 0.9 m/s and get 5.8 Bar.

Think I am there now. Although pressure in theory is less than 10 bar, I still have my doubts that it will work.

I suspect that the nitrogen will blow a hole though the material in the pipe i.e rat hole.

I think Progressive cavity pump is the way.

 

[CMA010]

0.9 m/s corresponds to a flowrate of 60 m3/hr, your tank would be empty in 6 minutes max. What is the visosity you used?

I'm not familiar with the term rat hole, what's that?

 

[v8landy]

Hi

I have now lowered velocity to 0.345 m/s and got 3.47 Bar(g)

I have used 5000 centi poise for viscosity

Term Rat Hole, is a hole when material stops flowing but hole is created.

i.e in pipe nitrogen blows a hole through the middle and material around outside can not move.

 

[v8landy]

CMA010, please can you show me how you are calculating flowrate m^3/hr from m/s from a known tank volume?

Thanks

 

[CMA010]

You want to "push" the slurry down a 6" line at 3 m3/hr =>

v = Q/A
= 3/(PI/4*(6*0.0245)^2)
= 164.5 m/hr
= 0.05 m/s

0.345 m/s gives you 22 m3/hr (Q = v*A).

Assuming newtonian, normal liquid flow, 100 m 6" pipe, 4 x 90 oC bend, 4 plug valves you would get 65 m3/hr (dynamic pressure loss of 8.5 bar and static pressure loss of 1.2 bar). Bare in mind that this is based on "ideal" conditions, i doubt these apply to your slurry.

If you apply N2 from the top and the liquid outlet is at the bottom i wouldn't expect a rat hole.

If i were you i would ask around at work if someone has a copy of Crane Technical Paper 410, "Flow of Fluids through Valves, Fittings and Pipe", that you can use. Furthermore the Piping Handbook of McGraw-Hill has a nice section on slurries.

 

[v8landy]

Hi CMA010

Thanks for that equation.

Have you just guessed at a flow of 65 m^3/hr? or do you have some data for that?

Still unsure about the rat hole effect, but think you are correct that if I apply steady constand pressure to top of liquid in tank it will push down pipe.

 

[CMA010]

The 65 m3/hr was calculated, based on the assumptions mentioned earlier. If you obtain a copy a Crane Technical paper 410 you can calculate the flow based on your pressure drop with the 2K method.

It's quite common to unload trucks or to empty process lines with nitrogen.