4.4 Level Calculations

4.4 Level Calculations

NCEES Measurement 1.I lists hydrostatic pressure/DP, density, elevation/suppression, composition, and pressure and temperature compensation. Every hydrostatic level loop is a manometer. In US customary units the PE-working constant is

P(psi) = 0.433 × SG × h(ft)

(that 0.433 psi/ft is 62.4 lb/ft³ divided by 144 in²/ft² for water). DP level is the same statement written as P_HP − P_LP. The art is knowing what sits on each tap at 0% and at 100%.

Open tank versus closed tank

Open tank. HP (or the single pressure tap) is at the bottom nozzle. LP is atmosphere. Indicated head is only the liquid column. If you mount the transmitter below the 0% tap, a constant extra liquid head sits on HP even when the tank is “empty” at the tap. That constant is a suppressed zero: LRV is a positive pressure, URV is that pressure plus the span head.

Closed tank, dry leg. HP at the bottom, LP in the vapor space, LP leg dry (gas-filled). If vapor density is negligible, DP ≈ 0.433 × SG_liquid × h, same as an open tank. The shortcut dies on high-pressure light hydrocarbons and on steam drums: the vapor column is not zero, and a dry leg that condenses becomes an accidental wet leg.

Closed tank, wet leg. LP leg is filled with a seal fluid of SG_seal (glycol/water, oil, etc.) to a height H_leg. At 0% process level (liquid at the HP tap) the LP head is still there, so DP is negative. You elevate the zero: LRV is that negative DP, URV is the less-negative DP at 100%. The span is still the process hydrostatic 0.433 × SG_process × H_taps, not the seal-fluid head. The seal-fluid head shifts both ends of the range by the same amount if H_leg is fixed.

Worked example — wet-leg span and zeros

A closed tank has taps 10.00 ft apart. Process SG = 0.80. The LP wet leg is filled with water, SG_seal = 1.00, and the transmitter sits at the HP tap elevation so the wet-leg height is 10.00 ft. Use 0.433 psi/ft.

At 0% (liquid at the HP tap): DP = 0.433 × (0 − 1.00 × 10.00) = −4.33 psid.

At 100% (10.00 ft of process): HP equivalent water column = 0.80 × 10.00 = 8.00 ft DP = 0.433 × (8.00 − 10.00) = −0.866 psid.

Calibrate LRV = −4.33 psid, URV = −0.866 psid. Span = 3.46 psi = 0.433 × 0.80 × 10.00. That is an elevated zero (both ends negative; 0% is the more negative number).

Now the suppressed-zero twin: an open water tank, 12.00 ft span, transmitter 3.00 ft below the 0% tap.

At 0%: P = 0.433 × 1.00 × 3.00 = 1.30 psig (suppress this constant). At 100%: P = 0.433 × 15.00 = 6.50 psig. Calibrate 1.30 to 6.50 psig. Span is still 0.433 × 12.00 = 5.20 psi — the extra 3 ft is on both ends.

If operations fills the wet leg with SG 1.10 glycol instead of water, every DP number shifts by 0.433 × 0.10 × 10.00 = 0.433 psi more negative. Recalibrate. Do not leave the old LRV on a heavier seal.

Interface level

When a heavy phase (SG_h) and a light phase (SG_l) both cover the taps, with interface height h above the HP tap and tap distance H,

DP / 0.433 = SG_h × h + SG_l × (H − h)

so

h = (DP/0.433 − SG_l × H) / (SG_h − SG_l).

Numeric check: H = 8.00 ft, brine SG_h = 1.10, oil SG_l = 0.80.

All light (h = 0): DP = 0.433 × 0.80 × 8.00 = 2.77 psid. All heavy (h = 8): DP = 0.433 × 1.10 × 8.00 = 3.81 psid. Interface at 4.00 ft: DP = 0.433 × (1.10 × 4 + 0.80 × 4) = 0.433 × 7.60 = 3.29 psid.

You cannot use a single “tank SG.” You also cannot put 0% at “empty tank” unless the light phase still wets the upper tap — the formula assumes both phases remain in the span. Radar and magnetostrictive interface loops have their own dielectric and density assumptions; hydrostatic interface is a two-SG manometer.

Density, composition, and P-T compensation

Hydrostatic level is inferred inventory. If composition or temperature moves SG, the same true height produces a different DP. Hot hydrocarbon tanks and blending vessels need a density input (from T, from a densitometer, or from a compensated transmitter) or they will “change level” on a sunny afternoon. Pressure compensation on liquids is usually negligible. On vapors it is not optional when you care about the dry-leg assumption.

Exam trap: vapor-space density is not zero

A dry-leg closed tank really measures

DP = ρ_L g h + ρ_V g (H − h) = ρ_V g H + (ρ_L − ρ_V) g h.

If you compute h = DP / (ρ_L g) you have ignored ρ_V. On an atmospheric water tank, ρ_V is air and the error is tiny. On a high-pressure propane sphere or a steam drum, ρ_V is a serious fraction of ρ_L. Using the tank liquid SG alone mis-maps the span; the honest liquid gradient is (ρ_L − ρ_V). A related trap is a “dry” LP leg that fills with condensate: you are now on wet-leg math with an unknown, drifting H_leg.