Viscosity component in eqauation

[v8landy]

Hi

I am trying to calculate the nitrogen pressure required to blow a slurry of SG 1.08 (viscosity at moment unknown, but is about the same as thick paint) down a 6" line into a tank 12 m high (2 m off ground).

Starting from a vessel of approx 10m^3 on ground level the pipe will rise to about 3m then run along the pipe bridge. Exact rout not fixed but guess will be 3 or 4 90 bends.

It will then reach the tanks, at this point I am un decided if it should go in the top or blow up the bottom. (12m high)

 

[mjpetrag]

What is the fluid level in the first tank? What is the highest level of fluid in the second tank? Is the second tank open to the atmosphere?

 

[mjpetrag]

Also, what is the total pipe length?

 

[v8landy]

Hi

Total pipe lenght would be 100 m. Undecided now about pipe diameter, but could be 3 to 6" Probably 3 or 4 90 bends and 3 or 4 valves.

Fluid in first tank (probably about 10m^3) (tank not picked) but would be max 2m of ground.

Highest level in second tank would be 12m. Pipe to tank would be on pipe racking which is about 3m off ground. So ? if to go in bottom or top.

Second tank is open to atms.

Thanks

 

[mjpetrag]

For a 6" pipe (converting everything to metrics)...

P1 + .5(rho)(v1^2) + (rho)(g)(h1) = P2 + .5(rho)(v2^2) + (rho) (g) (h1) .... becomes

(rho) (g) (h1) = .5 (rho) (v2^2)

V2 becomes about 6 m/s...which is a little high, but you can adjust accordingly

Since I don't know your viscosity, I will guess a friction factor (f) of .064 (64/Re)

Your major friction loss is

h = f * (L/D) * (V2^2/2g) ~~~ 76m head

Add about an extra meter for 90 bends & valves and an extra meter to get up to the pick rack and you need about 80m of head. To dump into the top of the tank, you need another 11 meters of head ( [12+2] - 3)...so around 90 meters of head minimum.

So the question becomes, are the friction losses due to the extra 11 meters of head to dump into the top greater or less than the back pressure created by the liquid in the second tank by dumping 1 meter from the bottom of the tank.

You should be able to solve the problem from here by calculating the pressure in the second tank at full capacity (just to be safe).


Just check to make sure the calculation is correct, I did this very quickly.


*Note* take into account the fact that dumping into the top of the tank requires longer pipe and the major friction loss needs to be recalculated for this.

 

[v8landy]

Many thanks for you help so far!.

A few questions.

1 - where has the 0.5 come from?
2 - confriming that H1 is my stating tank max head?
3 Assuming I get a viscosity or at least make a comparison to a known fluid value of "by-eye" the same viscosity, this value will become my f. If not you say 64/Re. Re = reynolds number? Calculation/equation?
4 - in final equation yuou have L and D, L being my total pipe lengh? and D being diamter of pipe?

Still needing help to get to final answer, please.

J

 

[v8landy]

Jumping ahead and using your figure of 90m of head

And using the following

Equation in oilfield units is a simple equation used often in drilling.

.052 * Fluid Wt (in lbs/gal)* hieght (ft)

0.52 * 9.0 * 295 (ft) = 135 psi or 9.5 Bar

tanking of the 11 m to go into top of tank = 8.5 Bar

Does this seam correct? or can you sugest somthing better?

 

[mjpetrag]

1 - Bernoulli's equation

2 - That is the amount of head you need to overcome friction loses due to pipe length, valves, 90s, and increase in height.

3 - Re = Reynold's Number...you should be able to find these equation's by yourself

4 - L = total pipe length, D = Pipe diameter

 

[v8landy]

I am getting lost now!

For example to get Re, I do not know the viscosity!

And Bernoulli's equation I do not know how to get the 0.5

I would like to understand it but also need the solution to get boss of back.

enzyme1434@yahoo.co.uk if this will make it easyer for you,

 

[v8landy]

Hi I have got some new data, if some can please verify.

SG is now 1.051 i.e density 1051 Kg/m^3
I am unabale to measure viscosity but I would liken it to a thick pourable honey approx 3000 cp i.e 3 Ns/m^3
Pipe lengh remains 100m with a diamter of 0.15m
Velocity m/2 I am assuming 6

I get Re number to be 315
and friction coeff to be 64/315 =0.203

Using

pressure loss = friction coeff(L/D)*(rho*V^2)/2))

I get pressure loss to = 2560000 Pa i.e 25.6 Bar

Do this look correct??????

Or again anyone a better option?

I have not taken into account the tank at 12, but even at 25.6 for a horizontal pipe this is a no go opperation.

 

[mjpetrag]

Yes...I am not sure if I am 100% correct but I got 24 Bar when i calculated it. Make sure your viscosity's correct...


Typical Viscosities (in cP at 70 deg F)
Water 1
Gasoline 8
Kerosene 10
Sulfuric Acid 25
Blood 50
Sae 10 Oil 60
Olive Oil 100
Glucose 500
Paint 500
Caster Oil 1000
Glycerine 1500
Glycerol 1500
Catsup 3000
Molasses 3000
Honey 5000
Corn Syrup 5000


If your viscosity is too high, try increasing the temperature in the line to push the material through quicker (if this is an option).

 

[v8landy]

Thanks for reply.

Did you get 24 Bar from my equations or your orginal above (90m head + working our pressure loss)???

It is not possible to heat up the slurry, plus I would rather use a pump, so glad if it is >10 barg N2. 24 is nice.

J

 

[mjpetrag]

FYI

Bernoulli's equation is

P1 + ½ ? v1^2 + ? g h1 = P2 + ½ ? v2^2 + ? g h2

P1 is assumed to be zero since we want to find a baseline of how much extra pressure we need.

V1 is assumed to be zero since the water level at the top of the tank is very slow.

P2 is considered to be a free jet so it is also 0.

H2 is at ground level so that term becomes 0. You are left with

? g h1 = ½ ? v2^2

V2 = sqrt(2*g*h1)

 

[mjpetrag]

I got 24 bar by using the new friction factor (.203) and got 242 m head (~~24 bar).

that question mark in the above post is actually the symbol for density, it didn't show up when i posted it

 

[v8landy]

Many thanks for your guidance.

 

[v8landy]

I still need to question the result becuse my pump supplier is sugesting a pressure of only 5bar. And for them to sugest that rules out pumps, so realy not in his favour to give me a low figure.

 

[v8landy]

My speadsheet does work now.

But I have to question why when I used Bernoulli's equation, and I get 6 m/s, do I need 6 m/s???

If I drop V to 1.4 m/s, pressure drops to just over 5bar.

 

[mjpetrag]

I may be wrong about that 6 m/s. That was just based on instantaneous flow from fluid pressure. Is 1.4 m/s reasonable for your application?

 

[v8landy]

Hi, the 6 m/s does not seam to have a relavence to it, so I have searched around and talked to other and a figure of about 1-3 m/s is normal for slurry/liquid.

I have used 0.9 and pressure has come out at about 5bar.

BUT I do not think I will use nitrogen to blow over for many reasons, like potential to rat hole, cost, etc.

Like look at progressive cavity pump instead.

But equations backs up my project justification.

Thanks