need help determining height for elevated tank

[kahlilj]

i plan to install an above-ground, lube oil storage tank. the client wants this tank to be elevated high enough to tie into an existing oil header line without using a pump (i.e. gravity-fed).

The setup is: tank will be located outside; approximately 40 feet of 1" diameter piping will be used to connect to the existing oil line; tie-in to the existing 1" diameter oil line is ~2 feet above grade; Current oil consumption (flow rate) is a maximumn of 8 gallons per day; the oil used has a density is 7.3 lbs/gallon; assume coldest temp will be ~50-60°F;

I tried sizing the minimum elevation for gravity fed flow using bernoulli's equation and came up with an elevation which does not seem right to me - ~3 feet.

can someone direct me to steps in determining what the minimum tank height should be?

 

[lilliput1]

Check with equipment manufacturer. It may not be allowed to pipe the overhead tank direct to the unit without a "lube oil day tank" because the oil level in the day tank is required not to be above a certain level (for fuel feed, it is not allowed to be above the injector) to not cause flow syphonage into the equipment when the unit is off.

Elevation of main tank should be such that the day tank is not overpressurized (beyond tank rated working pressure). Also elevation should be such that pressure drop through piping including entry, exit, fittings & valves at the maximum gpm draw would be at least equal to the minimum elevation available when the lube oil is at its minimum level in the tank. You can use data from Cameron Hydraulic Data for water then apply density correction. Do trial & error solution and add safety factor.

Address how the lube oil tank will be filled.

Pipe lube oil supply and return to the lube oil day tank. Lube oil pump should be able to lift oil from the lube oil day tank.

 

[lilliput1]

I forgot to add: Note the lube oil pump gpm or gph would be much more than the lube oil consumption which is net loss, not circulated lube oil.

 

[kahlilj]

thanks lilliput. a daytank will not be installed nor needed. instead the tank will feed into a 1" header pipe. the new tank is planned to be 2000 gallons, although we do not have a design for it yet.

what i really need help on is how to go about setting up the eqn to determine what the outlet elevation from the tank should be to reach the 8gpm flowrate.

currently they use a 55 gallon drum on an elevated stand to flow oil into this 1" header. i assumed energy equilibrium between this setup & the planned 2000 gallon tank setup. the eqn i used was:

?1÷? + v1÷2gc + z1 + hA = ?2÷? + v2÷2gc + z2 + hE + hf

assumed:
?1÷? = ?2÷? (because both tanks are @ atmospheric pressure)
hE = 0 because there is no pump involved

then determined hF =0.3 ft and solved for z2.

should i have made one side of the eqn = 0 and then solved for z2?

 

[kahlilj]

sorry, i didn't know know that the greek characters i tried to use in my previous message would not process correctly. the bernoulli eqn should be:

p1÷rho + v1÷2gc + z1 + hA = p2÷rho + v2÷2gc + z2 + hE + hf

kahlilj

 

[quark]

Your procedure is correct if you correctly calculated the frictional losses in the pipe and considered the correct pressure in the existing pipeline against which you have to discharge.

Lilliput gave you good suggestions.

See the thread378-116505

Regards,

 

[kahlilj]

thanks quark. lilliput did have good considerations which i have taken into account. the day tank issue i think will be addressed by the individual units (3 total) that the header line will feed. each unit has its own 300 gallon reservoir which, in effect, can act as a day tank. (the header line feeds this reservoir - not the units directly.)

my other concern/question is which approach should i take to solve for z2:

1) equilibrium between existing drum tank & new 2000 gallon tank:
p1÷rho + (v1)^2÷2gc + z1 + hA = p2÷rho + (v2)^2÷2gc + z2 + hE + hf

OR

2) just calc total head of new tank alone:
H = p2÷rho + (v2)^2÷2gc + z2 + hE + hf

also thanks for the thread link. very helpful!

 

[quark]

It depends upon what you consider as datum. If you consider the exit point as datum then pressure and velocity heads on the LHS and static head in RHS of the equation become zero.

So Z1 = P2/[Ρ]+v2[sp]2[/sup]/2gc+hf (I would like you explain hA and hE)

Regards,

 

[sailoday28]

How will you maintain level control in the 2000 gallon tank?
Noting the input of Quark, any swings in 2000 tank level will then impact system pressures.

 

[kahlilj]

quark, i would like to consider the entry point (or tie-in) to the existing header as my datum or reference. the other terms in the eqn are:
hf = frictional head = ~0.65 ft
hA = for hydraulic losses (turbine) = zero
hE = for hydraulic losses (pump) = zero

so if i go with existing header (2' above grade) as my datum then assume equilibrium at this point? if so, with the header line not being directly open to atmosphere, would pressure head still be zero? what would eqn look like if choosing existing header tie-in as my datum?

sailoday, level "control" needed is more or less just knowing the amount of oil remaining in the tank. so an indicator (without signal or control) is all that is needed.. level will be manually monitored.

 

[quark]

If oil is not already flowing in the existing line, then you need not consider any pressure. Your datum will be the exit into the existing day tanks and you should also consider frictional drop in the already existing line. But oil flows from some other source then the actual pressure term should come into picture and you should consider the tapping point as datum.

What Sailoday says is correct. Static head of oil changes as the level in the tank and it is better to design the system based on low level in the tank. But there may be overflow issues in day tank due to increased flowrate and that is why gravity flow of fuel oils from main storage tank to day tank is not allowed here in India.

Regards,