Fluid pressure question!

  • Thread starter joe98
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  • #1
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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
 

Answers and Replies

  • #2
Delphi51
Homework Helper
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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.
 
  • #3
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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
 
  • #4
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,073
The pressure would have to be as high as it was with cream but water is [strike]less[/strike] more dense,
Cream floats on the surface of milk. :smile:
 
  • #5
NascentOxygen
Staff Emeritus
Science Advisor
9,244
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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?
 
  • #6
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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 cant seem to derive an equation to relate the display reading (in %) to the
sensor pressure?

Any ideas?
 
  • #7
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,073
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.
 
  • #8
27
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wouldnt the overall pressure difference=850x1000x(12-0.6)=94962Pa

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

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