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Calc req to confirm N2 will push slurry down 6"
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Source：Internet Author：Unknow Pubdate：2008-05-06
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v8landy (Chemical)
10 Jan 08 8:06
Hi
I am looking from so advice on which equation(s) I would use to calculate if it would be possible to move X slurry down a 6" line (100m long) with Nitrogen (10barg avalible) into a tank 125m3 and 12m high.
Many thanks
James
dcasto (Chemical)
10 Jan 08 14:18
you need density of slurry. At 40 feet a gravity of 1, it will take 17 psig. At a gravity of 5, its 85 psig. This ignores friction losses, put at a very slow rate or static, that zero.
v8landy (Chemical)
10 Jan 08 14:47
Hi thanks for that start.
sg is approx 1.08 (alter depending on how distilation ends)
v8landy (Chemical)
10 Jan 08 14:52
字串2
Dcasto
Forgot to ask how did you come by this info? Can you provide me with details?
Thanks
dcasto (Chemical)
11 Jan 08 14:57
Potential energy = mass time height. Pressure = hieght of fluid times density. Density = 1.08 x 62.4 = 67.39 lbs/ft^3 40 feet times 67.39 = 2696 pounds/ft^2 = 18.7 psi.
v8landy (Chemical)
12 Jan 08 11:59
I am getting there, please have Patience with me!
first question is where did you get the 62.4 from? is this the height of my tank and you have typed/converted it wrong (12m)?
Second, does the diamter of the pipe not become a factor?
So equations are
1 Potential energy = mass x height
2 Pressure = height of fluid x density
3 density = SG x height
字串8
v8landy (Chemical)
12 Jan 08 12:50
thinking again about it the figure you are using for 40 = my 12 tank height?
So I still need to calculate the pressure drop along the 6" pipe of approx 100m long (+ taking into account and bends)
jjwamp (Petroleum)
14 Jan 08 19:26
62.4 is lbs/ft^3 of water * SG = density of fluid you are dealing with.
I am not sure the exact configuration but my guess is basically you have a pipeline full of seawater or some other fluid that you wish to remove from the pipeline to do some work on it.
I generally work in english notation but you can convert back.
Basically
10 bar = 145 psi. 12 meters = 40 ft. 1.08 SG = 1.08 * 8.33lbs/gal = 9.0 lbs/gal
Equation in oilfield units is a simple equation used often in drilling.
.052 * Fluid Wt (in lbs/gal)* hieght (ft) 字串8
.052 * 9.0 * 40 = 18.72 psi.
With 145 psi you should have more than enough neglecting friction and losses in bends ect.
HOWEVER
In practice this is not going to be that easy unless you can pump N2 at a high rate. At low rate the N2 will quickly flush some of the fluid out and then will channel on top on the remaining fluid in the pipe leaving a considerable amount left in the pipe.. The PRACTICAL determining factor of whether you can evacuate the pipe (assuming the pipe is lieing more or less flat level) is going to be based more on how fast you can pump the N2 and of course the faster you pump the more friction will be caused. This is not going to be a simple X*Y/Z calculation.
jjwamp (Petroleum)
14 Jan 08 20:13
By the way .052 is a conversion factor is all.
john1964 posted in another thread the proof below if you want to know where it comes from:
Normal formation pressure is equal to the hydrostatic pressure of the fluid extending from the surface to the subsurface formation. Add surface pressure if any. 字串5
Hydrostatic pressure psi= Density X0.052 X true vertical depth;
Hydrostatic pressure is defined as the pressure due to unit weight & the vertical height of a column of fluid. Since pressure is measured in psi & depth in feet, it is convenient to convert mud weights from pounds per gallon ( PPG) to pressure gradient psi/ft & the conversion factor is 0.052.
The conversion factor ( 0.052) is derived as follows-
A cubic feet contains 7.48 USG; A fluid weighing 1 ppg is equivalent to 7.48 lbs/cu.ft;
The pressure exerted by one foot of that fluid over the area of the base would be 7.48 lbs / 144 sq inches = 0.052 psi.
v8landy (Chemical)
15 Jan 08 4:15
JJwamp many thank for your very detailed info.
A couple of question though
1 what is the 8.33 lbs/gal?
You calculation apears only to be for the vertical height I have to push the fluid. 字串3
The basic picture of the opperation would be
Vessel X (approx 10m^3) on ground level would be pressurised with N2 to push fluid Y along pipe appox 6" dia. Total lengh of pipe would be approx 100m with an intial vertical leg of approx 3 m followed by approx 80m horizontal (3 or 4 90/swept bends). it would then reach the side of the 12m (40ft) tank and therfore already be approx 3m off the ground.
I was then planning to blow into the top of the vessel.
Would the horizontal section just be calculated using the D'Arcy Weisbach equation for staight pipe? I calculated a nominal figure of appox 0.35 Bar.
Other options I am considering are to pump it over, but I would still need to blow the line clear, so this information is still valid.
We have plenty of N2, other option would be to use steam, we have lots of that.
v8landy (Chemical)
15 Jan 08 6:12
I have worked out what the 8.33 lbs/gal = density of water
字串5
v8landy (Chemical)
15 Jan 08 8:24
Would viscosity not be a factor that is required???
v8landy (Chemical)
16 Jan 08 10:53
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 字串1
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
字串4
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