DP level transmitter

The reading on the transmitter is affected by the density of the fluid, so the instrument must be set up on the one with the lower density in the case of immiscibles.

If level is critical, then ultrasonics or radar might be a better choice of measurement technique in this case.

Hi,

The area am confused about is the usage of this formule Diff Presure = Height x Density x Specific Gravity or is it Height x S.P. And in what pressure unit is the answer going to be. PSI, inH20,Bar, e.t.c

Is It open to atmosphere ?

Regards

Jose

Hi

The DP transmitter is installed on a closed pressurized tank. Low side ontop and High side at the bottom of the tank.

The pressure P = F_orce / A_rea as you know. Now think of a liquid column in the tank, the bottom area of that is 1 cm².

The unit of pressure depends on the way you will follow:

1- This is the illustrative one as in physics books. F_orce in Newton and P_ressure in Pascal (N/m²).

ΔP [N/m²] = h [cm] * S_pecific W_eight [N/cm³] / A_rea [cm²]

Specific Weight defined as density times gravity:

SW[N/m³] = ρ[kg/m³] * g[m/s²]

Let the tank is full of water. Then ρ = 1000 kg/m³ ( @1 bar and 4^oC !) and g = 9.8 m/s² and h=720 cm. This yields:

SW = 1000*9.8 = 9800[N/m³] = 9800*10^-6[N/cm³] Now the pressure:

ΔP = 720[cm] * 9800*10^-6 [N/cm³] = 720*9800*10^-6 [N/cm²] = 720*9800*10^-6*10⁴ [N/m2]

ΔP = 70560 pascal = 0.706 Bar = 706 mbar

2-The more practical way: We will use kg-force (kgf)as the unit of force and tecnical athmosphere (at) as the unit of pressure.

ΔP [kg/cm²] = h[cm]*ρ[g/cm³]. With ρ= 1 gr/cm³ we will find:

ΔP = 720[gf/cm²] = 720*10^-3 [kgf/cm²] = 0.72 at ( ≈ 0.7 bar)

The "bar", "at", "atm" are the pressure units those are very close each other.(Take a look at Wikipedia for pressure units).

In fact u can use the h directly as pressure unit, like ρ*7200 mmH₂O.

For both method it is essential that u must use true density value! that is the value at pressure and temperature of your tank.

Say tank is full of oil then ΔP will be 0.9*0.7 bar why i assumed ρ = 0.9 gr/cm³ for oil.

You can obtain only an avaraged value for pressure since the tank contains liquids those have different density values.

Assume you have water in the tank.

If you want to measure the level from the bottom tap to a height of 100",

then the span for callibration would be 100" Water column.

If you want to measure the same 100" of a light fluid with Sp.Gr of say 0.5 ,then the span would be 100 X 0.5 =50 " Water column

The hight U use it here is tank total hight or the hight from the location of the lower point of DP ( because most of DP lower point is higher than the bottom plate of the tank )

My question is for

Mr feridun who made v good explanation

Yes dear guest,

With the explanation(#6) i've presumed that the DP transmitter is placed lower than the bottom of tank.

regards

DP= pressure conversion facter PCF(.433PSI/Sq feet) * hight(Feet)* Specific Gravity

How can I calculate the closed tank level with 2 transmitter I mean not use dp transmitter. Thanks for your help.

James

Not a good idea to use two pressure transmitters to calculate level.

Say the tank was working at 100.0 psig and level is 50.0 " from lower tap and the specific gravity is 1.0.

The transmitter connected to top of vessell would show 100.0 Psig.

The bottom transmitter would read 100 Psig+ 50" of water= 101.8 psig.

In an ideal world(on paper) you know the level is 101.8-100 psig =50" of level.

With transmitter accuracy and other things considered the level calculation would not be accurate.

If the pressure in the top of the tank is low or a similar order of magnitude to the hydrostatic pressure then you can fit one pressure sensor in the top and one in the bottom. You will then need to subtract the two pressure readings an convert them into a fluid level which is easy to do if you know the specific gravity of the liquid.

Although you are adding together errors of two sensors if the top pressure sensor is much lower which is often the case its error will be significantly smaller than the bottom sensor error.

Also DP transmitters if not installed correctly can become inaccurate due to moisture travelling down the negative side creating a head of liquid offset which is constantly changing.

2 sensors may work out much lower cost than a DP transmitter, possibly 1/3 the price in total.

Using two sensors also cuts down on the pipe work.