## Displacer Level Measurement Calculations

**Problem 1 :-**

**A steel displacer of volume 0.25 cubic feet is 50% immersed in oil of relative density 0.78; **

**the density of steel = 490 lbs per cubic foot**

**the density of water = 63 × lbs per cubic foot**

**Calculate the apparent weight of the displacer.**

true weight of displacer = 0.25 × 490 = 122.5 lbs

volume of oil displaced = 0.25 × 0.5 = 0.125 cubic feet

weight of oil displaced = (0.125 × 0.78 × 63) lbs = 6.143 lbs

**∴ Apparent weight of displacer = true weight − displaced weight**

Apparent weight of displacer = 122.5 − 6.143 lbs = 116.36 lbs

**Problem 2 :-**

**A steel displacer of volume 0.1 cubic feet is to be used as a level sensor in a vessel to store water.**

** Calculate the following;**

**1:- the maximum and minimum weights of the displacer.**

**2:- the maximum range of level able to be sensed, if the diameter of the displ. is 2 inch.**

true weight of displacer = 0.1 × 490 lbs = 49 lbs

volume of water displaced = 0.1 cubic feet

weight of water displaced = 0.1 × 63 lbs = 6.3 lbs

∴ min. weight of the displacer = 49 − 6.3 lbs = 42.7 lbs

∴ max. weight of the displacer = 49 lbs

**volume of the sensor = πr ^{2}L**

Note : π = 3.142, 1 in = 1/12 ft = 0.0833ft

volume of the sensor ⇒ 0.1 ft^{3} = 3.142 × (0.0833)^{2} × L

0.1 ft^{3} = 0.0218 × L

hence, the length of displacer, L = 0.1 / 0.0218 = 4.587 ft , or 55 inches

**Problem 3 :-**

#### A steel displacer 1 metre long, and volume of 1000 CCs. is to be used to measure the interface level in an oil and water separator. The SG of oil is 0.78, the density of water is 1 kg/litre. Density of steel is 8000 kg per cubic meter.

#### Find,

#### 1:- the diameter of the displacer.

2:- the displacer weight in air, oil & water.

3:- the “0%” and “100%” weights, if the 50% interface level is set at the displacer’s mid-point, and the span is 0.5 metres.

The displacer is assumed to be totally covered with oil or water.

**volume of the sensor = πr ^{2}L**

1000 c.c.s = 3.142 × r^{2} × 100 cm

thus, we now have, r^{2} = 1000 / (3.142 x 100 ) = 3.183 cm2

∴ the radius of displacer, r = Sqrt (3.183) cm = 1.78 cm

hence, the diameter of displacer = 1.78 × 2 cm = 3.56 cm

**displacer’s weight in air = density × volume**

Note that volume unit has to be inconsistent with density unit 1 ccs = 10^{−6} m^{3}

= 8000 kgm^{−3} × (1×10^{3}/10^{6}) m^{3} = 8.0 kg

**displacer’s weight in oil = (weight in air) − (weight loss in oil)**

= 8.0 – (0.5 × 1 × 0.78) kg = 7.61 kg

**displacer’s weight in water = (weight in air) − (weight loss in water)**

= 8.0 – (0.5 × 1 × 1) kg = 7.50 kg

**at 0% interface level,** the displacer’s length will be covered 25% with water and 75% with oil. Hence, the displacer’s total weight will be made up of 75% of its weight in oil and 25% weight in water;

∴ displacer weight at 0% = (0.75 × 7.61) + (0.25 × 7.5) = 7.582 kg

**at 50%,** the displacer’s total weight will be made up of 50% weight in water, and 50% weight in oil;

∴ displacer weight at 50% = (0.5 × 7.61) + (0.5 × 7.5) = 7.555 kg

**at 100%,** the displacer’s total weight will be made up of 75% weight in water, and 25% weight in oil;

∴ displacer weight at 50% = (0.25 × 7.61) + (0.75 × 7.5) = 7.527 kg

Article Source : N. Asyiddin

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in order to find weight loss in oil and weight loss in water, 0.78×1 and 1×1 are gives density of oil and water respectively. which is multiplied with 0.5. how you got 0.5? volume of oil/water displaced will be equal to the volume of the displacer which is equal to 1000ccc right? (when the displacer is in oil or water completely, the displaced fluid volume will be equivalent to the volume of the displacer. correct me if I’m wrong). instead of 1000ccc(=0.001 m3) how you got 0.5