Loaded test question!!!

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Smart $

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Ran across this question on sample test.

6. Height times density is part of the equation for​


A. pressure
B. volume
C. force
D. specific gravity
My best guess is A. What is throwing me is the "part of the equation" phrase, as I see all the answers could be the result of the "complete" equation :roll:


Looking for others opinions... :)
 

mcclary's electrical

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Location
VA
Ran across this question on sample test.


My best guess is A. What is throwing me is the "part of the equation" phrase, as I see all the answers could be the result of the "complete" equation :roll:


Looking for others opinions... :)

I agree with you. Although not exactly a "loaded", maybe just poorly written. You could come up with an equasions to insert height*density for everyone of those answers
 

charlie b

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Atmospheric pressure is also a player. Consider a cylinder of water within a larger body of water, with the liquid at rest. Let the cylinder have a cross sectional area of A, and a height of H. Let the atmospheric pressure at the surface be P. Let the specific weight (related to density by a factor of gravitational acceleration) of the fluid by G. The pressure felt at the bottom of the cylinder would be given by P + GH.

My reference is Eshbach's Handbook of Engineering Fundamentals.
 

cadpoint

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Durham, NC
Well I guess I don't understand why these types questions are on an electrical test? Maybe these are control work type questions.
I'll just assume that this is a mountain region of this great land.

I don't think it's poorly written, but more of a compound question, maybe even the dreaded loaded question. Once one understands exactly all of what the question is asking and what's involved to answer it.

SDSU.edu

I'll go out on the limb and say "D". :D
 

winnie

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Springfield, MA, USA
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Electric motor research
I agree with A as the answer...and since pressure*area=force, I agree that 'C' should also be considered an equally valid answer.

B and D are harder; I think that you would have 'non-simplified' equations (eg density * height / height) in order to make those work.

-Jon
 

Smart $

Esteemed Member
Location
Ohio
Atmospheric pressure is also a player. Consider a cylinder of water within a larger body of water, with the liquid at rest. Let the cylinder have a cross sectional area of A, and a height of H. Let the atmospheric pressure at the surface be P. Let the specific weight (related to density by a factor of gravitational acceleration) of the fluid by G. The pressure felt at the bottom of the cylinder would be given by P + GH.

My reference is Eshbach's Handbook of Engineering Fundamentals.
Intersting :cool:
 

Smart $

Esteemed Member
Location
Ohio
Well I guess I don't understand why these types questions are on an electrical test? Maybe these are control work type questions.
I'll just assume that this is a mountain region of this great land.

I don't think it's poorly written, but more of a compound question, maybe even the dreaded loaded question. Once one understands exactly all of what the question is asking and what's involved to answer it.

SDSU.edu

I'll go out on the limb and say "D". :D
It is a prep test for instrumentation calibration.
 

Smart $

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Location
Ohio
I agree with A as the answer...and since pressure*area=force, I agree that 'C' should also be considered an equally valid answer.

B and D are harder; I think that you would have 'non-simplified' equations (eg density * height / height) in order to make those work.

-Jon
I agree B is definitely harder, but I think not impossible (haven't really given it my full attention at this point)

Regarding D, since specific gravity is just a ratio of densities, one being water, we can use the principle mentioned in the second paragraph of the following linked article to determine an unknown fluid's specific gravity.

http://www2.emersonprocess.com/siteadmincenter/PM Rosemount Documents/00816-0100-3208.pdf
 

david luchini

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Engineer
I'd agree that there looking for A. Pressure: Height x Density x Acceleration = Pressure.

You probably could go for C. Force: Height x Density x Acceleration x Area = Force. But since Height x Area = Volume, I don't see why you'd introduce height instead of volume.

I think B. Volume is not realistic since you'd have Height x Density x Area / Density. So density cancels itself out, why introduce it into the equation?
 

Smart $

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Location
Ohio
I'd agree that there looking for A. Pressure: Height x Density x Acceleration = Pressure.

You probably could go for C. Force: Height x Density x Acceleration x Area = Force. But since Height x Area = Volume, I don't see why you'd introduce height instead of volume.

I think B. Volume is not realistic since you'd have Height x Density x Area / Density. So density cancels itself out, why introduce it into the equation?
I agree B is the least likely, but I was thinking what if you were given the maximum force on the bottom of a tank, the liquid's density and maximum height and asked to determine the minimum volume of the tank?
 
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david luchini

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I agree B is the least likely, but I was thinking what if you were given the maximum force on the bottom of a tank, the liquid's density and maximum height and asked to determine the minimum volume of the tank?

That's interesting. Off the top of my head, knowing the force at the bottom of the tank, the liquid density and the height of the liquid in the tank, you still wouldn't be able to find the Volume of the liquid in the tank. I don't think its enough information, but I haven't tried to work out the equation.

Let us know if you figure it out.
 

KentAT

Senior Member
Location
Northeastern PA
I agree B is the least likely, but I was thinking what if you were given the maximum force on the bottom of a tank, the liquid's density and maximum height and asked to determine the minimum volume of the tank?

Also used for volume calcs. Height (of fluid column) x density (of fluid in the column) x horizontal area of tank or vessel bottom (in square feet) x desired conversion factor (to give you gallons, liters, etc) gives you volume of fluid in tank or vessel.

In the instrument calibration world, Height and Density can give you pressure, which can then be used to find volume if desired. Most level transmitters use this concept to output the 4-20mA analog signal to the control (PLC, etc) which displays and controls as either 0-100% level, x gallons, or both.

Now, if you have a horizontal cylindrical pressure vessel, you must apply another algorithm to account for the changing volume for the given heights - starts off minimal at the bottom of the "circle", grows to max volume change in the middle, back to minimal at the top.

kent
 
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Smart $

Esteemed Member
Location
Ohio
That's interesting. Off the top of my head, knowing the force at the bottom of the tank, the liquid density and the height of the liquid in the tank, you still wouldn't be able to find the Volume of the liquid in the tank. I don't think its enough information, but I haven't tried to work out the equation.

Let us know if you figure it out.
While it may not agree with SI unit fanatics :)roll:) maximum force on the bottom of the tank can be in pounds. Pounds divided by pressure (maximum height in ft ? density in lbs/ft?, such as 62.43lb/ft? for water) will yield the maximum bottom of tank area... and we already have the maximum height. Multiply those two together and you have the minimum volume the tank can be to achieve the maximum load... I think :roll::D
 

Smart $

Esteemed Member
Location
Ohio
...

Now, if you have a horizontal cylindrical pressure vessel, you must apply another algorithm to account for the changing volume for the given heights - starts off minimal at the bottom of the "circle", grows to max volume change in the middle, back to minimal at the top.

kent
Any vertical profile which exhibits varying horizontal cross-sectional areas at different heights would require an algorithm of some sort. I think I got that right ;)
 
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