Loaded test question!!!

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Smart $

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And I never will, its a very bad leap to make.:D



Good link, as it back up what I've been trying to say. For instance, "There has been much confusion (and numerical error!) because of the differences between lbf, lbm, and slug." Or, "1 lbf = 32.174 lbm? ft/s2." Or "It is not proper to say that one lbm equals one lbf, but it is proper to say that one lbm weighs one lbf under standard earth gravity."

In other words, you cannot divide a pound-force by a pound-mass in order to cancel out the term "pounds."




I get a pressure of 624.3lbf/ft?

A force on the bottom of the tank of 62430 lbf.

An area of the bottom of the tank of 100 sf.

And I think I see what is throwing you off. The mass of the water in the tank is 62430 lbm, and the force on the bottom of the tank is 62430 lbf. So since the value of lbm and lbf is the same, they must be the same thing? Therefore, dividing my pressure (lbf) by my density (lbm/ft?) I can cancel out "pounds." But as I mention above, you can't cancel out pounds that way.
The whole point is by your method, somewhere along the line you converted mass to weight (m?g). The summation of that conversion is g/gc of 1lbf/lbm. After doing many of such calc's one considers it to be understood and omits it from the calc's using simply lb rather than lbf and lbm.

Moving on, can we now say height*density is part of an equation to determine volume. Yes, something will cancel out, either height or density, likely height, but in the equation height is necessary to determin area or pressure or force depnding on which parameters are known.

Furthermore, I did post this question in the Electrical Calculations/Engineering forum, but I do not believe the prep or real tests are going to require engineering level nomenclature for the calculations.
 

Smart $

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...After doing many of such calc's one considers it to be understood and omits it from the calc's using simply lb rather than lbf and lbm.

...
Sorta forgot how we got here...

Anyway, I believe the physics level of the testing to be one which does omit the g/gc transition, thus I'm back to thinking P = D * H is considered the complete equation and therefore not " a part of the equation" and thus eliminates "A. Pressure" from being a truly viable option for a correct answer. Now whether the test writer views it this way is anybody's guess (other than the writer and those provided with his/her answers, of course) :D
 

markstg

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.

Moving on, can we now say height*density is part of an equation to determine volume. Yes, something will cancel out, either height or density, likely height, but in the equation height is necessary to determin area or pressure or force depnding on which parameters are known.
Height * Density is part of the equation for Pressure. How do you determine Volume. From your example:
Height * Density * Units Conversion factor = pressure
10ft * 62.43lbm/ft3 * g/gc = 624.3 psig
what would you mulipy and/or dived the product of Height and Density to get volume?
 

david luchini

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The whole point is by your method, somewhere along the line you converted mass to weight (m?g). The summation of that conversion is g/gc of 1lbf/lbm. After doing many of such calc's one considers it to be understood and omits it from the calc's using simply lb rather than lbf and lbm.

No, my method does no such thing. In my method mass is mass and pressure is pressure. And dividing pounds of one by pounds of the other does not (and cannot) cancel out pounds. I don't mind if you use simply lb rather than lbf and lbm, AS LONG AS you realize when you are referring to force and when you are referring to mass. For example, the pounds in PSI refers to pounds-force, and the pounds in the density of water refers to pound-mass.

Anyway, I believe the physics level of the testing to be one which does omit the g/gc transition, thus I'm back to thinking P = D * H is considered the complete equation and therefore not " a part of the equation" and thus eliminates "A. Pressure" from being a truly viable option for a correct answer. :D

You are still making the same mistake. Density * Height is NOT a complete equation for Pressure. Pressure = Force/Area and Force = Mass * Acceleration. So an equation for Pressure would provide an answer in units of lbm * (ft/s?) / ft?. Or simplifying, Pressure would be in units of lbm / (ft * s?). Density * Height provides an answer in units of lbm / ft?. As you can see, the units for D * H are not correct for Pressure.

A complete equation of Pressure would be = Density * Height * Acceleration.

So "A. Pressure" is still probably the best answer.

From your example:
Height * Density * Units Conversion factor = pressure
10ft * 62.43lbm/ft3 * g/gc = 624.3 psig
what would you mulipy and/or dived the product of Height and Density to get volume?


markstg you are close, but it is not Height * Density * Unit Conversion Factor = Pressure (H*D*g/gc=P)

It is Height * Density * Acceleration = Pressure (H*D*g=Pressure)

For example 10ft * 62.43lbm/ft? * 32.147 ft/s? (acceleration due to gravity) = 20069.4 lbm/(ft * s?).

You could then apply your Unit Conversion Factor, if you wanted to convert the units to a more familiar term, such as Pounds(force)/ft?.
We know that 1lbf=32.174lbm*ft/
s?, so if we multiply the answer of 20069.4 lbm/(ft * s?) times (1lbf/(32.174lbm*ft/s?)) (which is your conversion factor gc) then we get the Pressure = 624.3lbf/ft?.
 

markstg

Senior Member
Location
Big Easy
:)
Acceleration.



markstg you are close, but it is not Height * Density * Unit Conversion Factor = Pressure (H*D*g/gc=P)

It is Height * Density * Acceleration = Pressure (H*D*g=Pressure)

For example 10ft * 62.43lbm/ft? * 32.147 ft/s? (acceleration due to gravity) = 20069.4 lbm/(ft * s?).

You could then apply your Unit Conversion Factor, if you wanted to convert the units to a more familiar term, such as Pounds(force)/ft?.
We know that 1lbf=32.174lbm*ft/
s?, so if we multiply the answer of 20069.4 lbm/(ft * s?) times (1lbf/(32.174lbm*ft/s?)) (which is your conversion factor gc) then we get the Pressure = 624.3lbf/ft?.

Ok, using g/gc and calling it a unit conversion factor is not correct, and your formula Is correct P=H*D*g, now if we are using English Engineering Units, than the forumla becomes H*D*g/gc.

And A (pressure) is still the right answer.:)
 
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Smart $

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No, my method does no such thing. In my method mass is mass and pressure is pressure. And dividing pounds of one by pounds of the other does not (and cannot) cancel out pounds. I don't mind if you use simply lb rather than lbf and lbm, AS LONG AS you realize when you are referring to force and when you are referring to mass. For example, the pounds in PSI refers to pounds-force, and the pounds in the density of water refers to pound-mass.

You are still making the same mistake. Density * Height is NOT a complete equation for Pressure. Pressure = Force/Area and Force = Mass * Acceleration. So an equation for Pressure would provide an answer in units of lbm * (ft/s?) / ft?. Or simplifying, Pressure would be in units of lbm / (ft * s?). Density * Height provides an answer in units of lbm / ft?. As you can see, the units for D * H are not correct for Pressure.

A complete equation of Pressure would be = Density * Height * Acceleration.

So "A. Pressure" is still probably the best answer.
I knew you were going to be picky about using the word "converted". Unfortunately I realized it after the allotted edit time. :roll: Perhaps "transition". You say you keep lb as mass throughout, but as soon as you bring gravity into the equation, you are making the transition. Yes I realize the difference. As I stated twice previously, you are technically correct, but when either force or pressure is given and you have to determine the other, the transition has to be made.

Regardless, that is not the point. Prior to the push to go metric here in the states, the distinction between pound mass and pound force was not made except for laboratory type scenarios and it sufficed for all but the most critical of applications. The point is that it still does.
 

Smart $

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...what would you mulipy and/or dived the product of Height and Density to get volume?
After due consideration, an equation for volume does not require height times density as part of the equation. As such, I agree "B. Volume" is an incorrect answer.
 

Smart $

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Location
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... As I stated twice previously, you are technically correct, but when either force or pressure is given and you have to determine the other, the transition has to be made.

...
Ooops! I meant to contrast force and pressure to density.
 

david luchini

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As I stated twice previously, you are technically correct, but when either force or pressure is given and you have to determine the other, the transition has to be made.

Smart, I was going to let this go because it seems that we're basically on the same page. But, since you think I'm being too technical on this question, AND I noticed that in the "Service Conductors, limited or not" poll your answer is based on a technicality (with which I agree), I thought I'd put forward this one more thought just for fun.

I see you haven't made the leap yet. :mad:

Here's why I haven't made the leap:

The leap allows that pounds-force and pounds-mass are the same, so that the equation for pressure can be Height(H) * Density(D), giving a result in "pounds per square foot" (which sound reasonable for Pressure.)

But we also know that Pressure = Force / Area, and that Force = mass * acceleration, and Density(D) = mass/ Volume. Since we live on earth, lets talk specifically about acceleration due to gravity (g).

So Pressure (P) = m*g/A = H*D.

Simplifying: m*g=A*H*D. We know that Area * Height = Volume, so,

m*g=V*D, or m*g=V*m/V.

Simplifying further: m*g=m ?!?!?!:-?

So, in making the leap, acceleration due to gravity becomes a constant of the value 1, with no units (not even 1 ft/s?.) Gravity ceases to exist and the world spins out of control.:)

That's why I haven't made the leap.;)
 

Besoeker

Senior Member
Location
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While it may not agree with SI unit fanatics :)roll:)
You should. It makes many calculations so much easier and you have little choice but to use them in everyday live for some things anyway.
Ever heard of a 0.13 HP bulb for example?

Back to the question.
Height gives you distance - it's m (metres) in SI.
Density gives you mass per unit volume kg/m^3.
Height times density thus gives kg/m^2
That's mass per unit area and part of the equation for pressure.
So maybe that answers the original question.
To get pressure, you need to multiply by the gravitational constant, 9.81 m/s^2 in this case. An acceleration as david luchini correctly noted.
That gives you pressure in Newtons/m^2.

Slightly off topic, a force of one Newton moved through a distance of one metre in one second is one Watt. Neat and tidy and it's one of the reasons I like SI.
 

Smart $

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Location
Ohio
Smart, I was going to let this go because it seems that we're basically on the same page. But, since you think I'm being too technical on this question, AND I noticed that in the "Service Conductors, limited or not" poll your answer is based on a technicality (with which I agree), I thought I'd put forward this one more thought just for fun.



Here's why I haven't made the leap:

The leap allows that pounds-force and pounds-mass are the same, so that the equation for pressure can be Height(H) * Density(D), giving a result in "pounds per square foot" (which sound reasonable for Pressure.)

But we also know that Pressure = Force / Area, and that Force = mass * acceleration, and Density(D) = mass/ Volume. Since we live on earth, lets talk specifically about acceleration due to gravity (g).

So Pressure (P) = m*g/A = H*D.

Simplifying: m*g=A*H*D. We know that Area * Height = Volume, so,

m*g=V*D, or m*g=V*m/V.

Simplifying further: m*g=m ?!?!?!:-?

So, in making the leap, acceleration due to gravity becomes a constant of the value 1, with no units (not even 1 ft/s?.) Gravity ceases to exist and the world spins out of control.:)

That's why I haven't made the leap.;)
Not a problem by me (and I'm not referring to gravity ceasing to exist and the world spinning out of control). But you can't make the leap if you keep bringing g into the equations. Just leaving g out and letting lb represent both lbf and lbm as necessary is what the leap is all about.

Here's a simple equation in which the units of measure do not cancel out: E = I R.
 
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david luchini

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But you can't make the leap if you keep bringing g into the equations.

Um, it wasn't me who brought g into the equation, it was some guy named Isaac Newton. I'm pretty sure that he was smarter than I am, so I'm gonna take his word for it.

By taking the leap (and arriving at P = H * D,) you'd calculate that the pressure on the bottom of a vessel holding a column of water that is 1" dia and 12" high to be the same whether you were at Death Valley, Denver or the top of Mt. Everest.

Of course, the pressure wouldn't be the same at those locations, but it might be "close enough" for your purposes.

But that wouldn't make P=H*D a "complete" equation for pressure, it would make it a "close enough" equation for pressure.
 

mivey

Senior Member
This thread just stuck me as funny after being absent from the site for a short while. How did I ever get by without all the fun? :grin:
 

Smart $

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Location
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Um, it wasn't me who brought g into the equation, it was some guy named Isaac Newton. I'm pretty sure that he was smarter than I am, so I'm gonna take his word for it.

By taking the leap (and arriving at P = H * D,) you'd calculate that the pressure on the bottom of a vessel holding a column of water that is 1" dia and 12" high to be the same whether you were at Death Valley, Denver or the top of Mt. Everest.

Of course, the pressure wouldn't be the same at those locations, but it might be "close enough" for your purposes.

But that wouldn't make P=H*D a "complete" equation for pressure, it would make it a "close enough" equation for pressure.
I calculate the variation of gravitational acceleration over the surface of the planet Earth to be approximately 0.6555% when adjusting only for altitude. Other factors can also vary gravitational acceleration.

I think anyone intelligent enough to make the leap would also be aware of whether local deviation from the mean would be within tolerance of the required accuracy.

Additionally, if one does not know where the measurement is going to occur, how can he adjust for any deviation?
 

Besoeker

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I calculate the variation of gravitational acceleration over the surface of the planet Earth to be approximately 0.6555% when adjusting only for altitude. Other factors can also vary gravitational acceleration.

The point is, as both david l have indicated, you need to include the gravitational constant to get from mass to force.
 

Smart $

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The point is, as both david l have indicated, you need to include the gravitational constant to get from mass to force.
And my point is that pound started out and still is a unit of weight. As such, an object weighing one pound by definition has one pound of mass and exerts one pound of gravitational force. The gravitational constant is already included by definition. It really is that simple.
 

Besoeker

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Location
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And my point is that pound started out and still is a unit of weight. As such, an object weighing one pound by definition has one pound of mass and exerts one pound of gravitational force. The gravitational constant is already included by definition. It really is that simple.

The pound or pound-mass (abbreviation: lb, lbm) is a unit of mass used in the imperial, United States customary and other systems of measurement. A number of different definitions have been used, the most common today being the international avoirdupois pound of exactly 0.45359237 kilograms.
No mention of the gravitational constant in that definition.
Do you have one that does?
 

mivey

Senior Member
No mention of the gravitational constant in that definition.
Do you have one that does?
Tweeet! Flag on the play (card if you are a soccer fan--not sure if there is a whistle because I can't hear over the vuvuzelas).

Both countries have adopted the pound as a unit of mass. That said, it is also customary to use pound to indicate force as well. We often see specs where the force is labeled with "lb" instead of "lbf". It is just the way it is and while the use of pound-mass and pound-force is encouraged, tradition continues.
 
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