Fluid Pressure Equation for 150 m3 Silo | Sensor Calibration and Display Reading

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Homework Help Overview

The discussion revolves around a fluid pressure problem involving a 150 m³ cylindrical silo used for storing cream. The problem includes deriving an equation that relates the display reading of an electronic pressure sensor to the gauge pressure at a specific height within the silo. Participants also explore the implications of using water instead of cream and how that affects the height required for the silo to read 100% on the display.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants attempt to derive a formula relating the display reading to the sensor pressure, considering the sensor's height above the base of the silo. There are discussions about incorporating the sensor height into the formula and how to adjust for different fluid densities.

Discussion Status

Several participants have provided insights and attempted to derive equations, but there is no clear consensus on the correct approach or formula. Some participants question the assumptions made regarding fluid density and the implications for the display reading. The discussion remains active with various interpretations being explored.

Contextual Notes

Participants note the challenge of deriving an equation that accurately reflects the sensor's calibration for cream while considering the effects of using water. There is also mention of the limitations of the sensor's ability to measure fluid levels below its mounting height.

joe98
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Homework Statement


A 150 m3 silo- a cylinder (12 m high) is used to store cream (density 850 kg m-3). An electronic pressure sensor (measures gauge pressure) is mounted 0.6 m up from the base of the tank, and a 0 to 100% display is used to indicate the level. It is not possible to sense
the level once it is below the sensor, but the display can be calibrated with a constant
offset to account for this extra fluid (for example, the sensor can show 10% when
there is no pressure). Derive an equation to relate the display reading (in %) to the
sensor pressure. What offset should be used on the display to account for the 0.6 m
mounting height? If water is used instead of cream, how high must the silo be in
order for the display (calibrated for cream) to read 100% when the silo is full?


Homework Equations



P=ρgh
m=ρv


The Attempt at a Solution


Pressure(at the base of cylinder)=850x9.8x12=99960Pa
Pressure at sensor=1000x9.8x0.6=5880Pa

how high must the silo be in
order for the display (calibrated for cream) to read 100% when the silo is full?

99960=pgh
99960=1000x9.8xh
h=10.2meters

derive an equation
i got (9.8hx850/99960)x100

not sure of the offset but i got 0.6/12=0.05 not sure with this

any suggestion guys, much appreciated
 
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Derive an equation to relate the display reading (in %) to the
sensor pressure.
This means you need a formula to relate the display Reading (R) to the actual Pressure (P). For instance, if the sensor was at the bottom, it would be
R = P/(ρg12)*100%
You'll have to figure out how to incorporate the sensor height of 0.6 into that formula. One method is to do a couple of examples, say cream height 0.6 and cream height 10.4. Then fiddle with a linear formula until it fits.

If water is used instead of cream, how high must the silo be in
order for the display (calibrated for cream) to read 100% when the silo is full?
The pressure would have to be as high as it was with cream but water is less dense, so according to P=ρgh the height would have to be greater than 12 m to make up for the smaller density.
 
Delphi51 said:
This means you need a formula to relate the display Reading (R) to the actual Pressure (P). For instance, if the sensor was at the bottom, it would be
R = P/(ρg12)*100%

so would R=P/(pg0.6)*100
 
Delphi51 said:
The pressure would have to be as high as it was with cream but water is [strike]less[/strike] more[/color] dense,
Cream floats on the surface of milk. :smile:
 
joe98 said:
Pressure(at the base of cylinder)=850x9.8x12=99960Pa
Pressure at sensor=1000x9.8x0.6=5880Pa
Why did you use ρ of H₂O when this is offset for cream?
 
that was for a different part of the question where "If water is used instead of cream, how high must the silo be in
order for the display (calibrated for cream) to read 100% when the silo is full"

but i can't seem to derive an equation to relate the display reading (in %) to the
sensor pressure?

Any ideas?
 
joe98 said:
that was for a different part of the question where "If water is used instead of cream, how high must the silo be in
order for the display (calibrated for cream) to read 100% when the silo is full"
The nature of the fluid below the level of the probe cannot affect the display reading because the probe cannot react to it. The display reading is calibrated with that level of cream as an offset, and it remains that way for all fluids.
 
wouldnt the overall pressure difference=850x1000x(12-0.6)=94962Pa

so the reading could be 0.05+(pressure/94962)
 

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