Level Calibration Range

I posted this before in another thread so here it is again.

If that vessel is non pressurised,then no need to go for 2 legs..

Just multiply the specific gravity(g) of medium with maximum height of the medium(h), that is the range for ur transmitter....

The original question involved two legs. Why provide an answer to a question that wasn't asked?

That's OK provided one can neglect the changes in barometric pressure caused by local weather variations. In a small vessel, it can introduce a large error.

Hi, Romer. hope this helps you:

As I understand the calibration problem, you have a vessel with the measuring leg always full of liquid: 100% DEA, 100% HC or something in between.

Also, from your drawing, the physical distance between high pressure and low pressure instrument taps is 105". For ease of calculation, I'm gonna use 100".

In order to obtain calibration range, we need the HC specific gravity. You don't give it, so let's say it's 0.75 (some sort of light gas oil?).

O.K., here goes:

0% (zero) of transmitter is 100" X 0.75 = 75 inches equivalent water column.

This happens when the distance between taps is full of the lighter liquid: HC.

100% (URV) of transmitter is 100" X 1.05 = 105" WC. }

This results when the measuring leg is full of the heavier liquid: DEA.

Transmitter span is 30" WC.

With this calibration, 0 to 100% of transmitter gives interface level from lower (hi pressure) tap to higher (low pressure) tap.

For an interface height of H", transmitter % lecture will be:

% = (Hx1.05 + (100 - H)x0.75 -75) x 100 / 30

You can check it out with several examples of interface height.

-------------------------------------------------------------------------------

Things to watch out for: 1-) Emulsification of the two liquids in the interface region = imprecision 2-) very similar specific gravities in the upper and lower liquids = very small calibration range.

Nestor,

Thank you by your help. We calibrated the trasnmitter with your calculation and is working well.

Regards,

Romer.

Hi,

Just remember that if the 105" is the range, the differential pressure corresponding to that is multiplied by the density, with consistent units. The density of the DEA is Density of water x the Sp. Gr. Therefore expect the 105" to be:

Diff. Pressure = H, in Ft * D #/Ft^3 * S.G. = press diff , in psf or divide by 144 to get psi.

Diff. Pressure = (105"/12")ft*(62.4 #/ft^3)* 1.05 = 573.3 psfd, = 3.98 psid

Here' some shortcuts

1. Convert Differential Head of Fluid to equivalent Head of Water by multiplying w/ S.G.

Ex. Diff Head = 105" DEA = 105" * 1.05 = 110.25" water = 9.19 Ft Water

2. Divide Diff Head in ft of water by 2.31 to get psid. ( 2.31 =144/62.4)

Ex. Diff Pressure = 9.19/2.31 = 3.98 psid

The reasons I gave all the above workout is that you might be looking for answers in different units.

Summary:

Given: 105 " DEA fluid, S.G. = 1.05

Diffential Head in INCHES Water = 110.25 Inches of Water

Diffential Head in Ft Water = 9.19 Ft

Differential Pressure, psi =3.98 psi

Good Luck