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Inst Tools > Blog > Practical Questions > Calculate Differential Pressure Sensed by Level Transmitter

Calculate Differential Pressure Sensed by Level Transmitter

Last updated: April 5, 2019 12:25 pm
Editorial Staff
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Calculate Differential Pressure Sensed by Level Transmitter

Question :

Calculate the differential pressure sensed by the level transmitter at three different water levels in this boiler steam-drum level measurement system: The measurable points are 0%, 50%, and 100% of transmitter range.

Drum Level differential pressure transmitter

Assume a density for (hot) boiler drum water of 36 lb/ft3 , a density for steam in the drum of 7 lb/ft3 , and a density for (warm) water in the “wet leg” of 61.8 lb/ft3 . If the pressure at the “low” (L) side of the transmitter is greater than the pressure at the “high” (H) side, be sure to express the differential pressure quantity as a negative number.

Credit will be given for correctly calculating each of the differential pressures:

  • (6 points) Transmitter ∆P at 0% water level     = _______”W.C.
  • (6 points) Transmitter ∆P at 50% water level   = _______”W.C.
  • (6 points) Transmitter ∆P at 100% water level = _______”W.C.

Answer :

  1. (6 points) Transmitter ∆P at 0% water level = -38.832 ”W.C.
  2. (6 points) Transmitter ∆P at 50% water level = -31.864 ”W.C.
  3. (6 points) Transmitter ∆P at 100% water level = -24.896 ”W.C.

Also Read : Boiler Drum Level Measurement Principle

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1 Comment
  • quangdinh says:
    June 15, 2022 at 11:15 pm

    Instrument Range:
    LV =[ [( 9″ x 36 + 40″ x 7) – 49″ x 61.8] x 0.0005787037] / 0.036127292000084 lbs/in3
    = – 38.83 ” WC at standard condition
    or LV = – 1.40289 psi
    UV = [ [( 39″ x 36 + 10″ x 7) – 49″ x 61.8] x 0.0005787037] / 0.036127292000084 lbs/in3
    =-24.896″ WC at standard condition
    or UV = -0.89942129054 psi
    Density water at 0.036127292000084 lbs/in3 standard conditions
    Controller Range: -15″ to 15″ WC

    Reply

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