Loaded test question!!!

[david luchini]

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:

I've got my problems with SI also, but it may have saved you from a fatal flaw. A "pound" is not a unit of FORCE, it is a unit of MASS.

Force is mass x acceleration. The english units measure of force is the poundal, which is equal to 1 lb x 1ft/s?.

And height x density doesn't give pressure. HxD would provide an result in units of lb(mass)/ft?. Pressure should have a unit of lb(mass)/(ft x s?).

If you replace pound with poundal in your example above, I think you'll find you don't get where you're trying to go.

 

[markstg]

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'm with you Smart$, and go with A (pressure). Since in the context of the test questions, this is an instrument calibration question for measuring level in a tank/vessel.

To measure the level in a tank/vessel, you measure the pressure in the tank and use the formula Pressure Measured = Height (above pressure tap) x density (of the product being measure) divided by the density of water at Standard Conditions. P = H x Dprod/Dwater. Measure pressure and knowing both density of product and density of water you get H the level being measure. This is how a DeltaP transmitter is set up to measure level.

Since they didn't have Density of water in the formula, I guess that's why they said part of the equation.

In real life I get the Specific Gravity of the liquid, and use the Formula
P = H x (SG). As an example: for a product of a SG of .5 with a pressure reading of 100 inches of water gives a product level of 50 inches.

 

[Chamuit]

I'd go with A.

Pressure is the amount of force acting on an area.

 

[Smart $]

I've got my problems with SI also, but it may have saved you from a fatal flaw. A "pound" is not a unit of FORCE, it is a unit of MASS.

Force is mass x acceleration. The english units measure of force is the poundal, which is equal to 1 lb x 1ft/s?.

And height x density doesn't give pressure. HxD would provide an result in units of lb(mass)/ft?. Pressure should have a unit of lb(mass)/(ft x s?).

If you replace pound with poundal in your example above, I think you'll find you don't get where you're trying to go.

Well, I'll save some typing () by providing the following link...

http://en.wikipedia.org/wiki/Pound-force

 

[LarryFine]

My best guess is A.

I agree with you.

I agree with A as the answer...

I'd agree that there looking for A.

It's gotta be 'A'.

I'm with you Smart$, and go with A (pressure).

I'd go with A.

I'd say the A's have it.

 

[Smart $]

I'd say the A's have it.

Aye'd say it appears so...

...and that's some mighty Fine polling there, Larry

 

[Smart $]

I'd say the A's have it.

BUT... (wouldn't you just know it )


...here's the really trippy part. The question states "Height times density is part of the equation..."

Now if we happen to be working in density units of lb/ft? and convert to lb/in?, plus we accept that a one pound mass exerts a one pound gravitational force, then height times density is the complete equation for pressure (with lb/in?, or more familiarly psi, as the unit).

Would this not eliminate A as the correct answer?

 

[david luchini]

Well, I'll save some typing () by providing the following link...

http://en.wikipedia.org/wiki/Pound-force

Nice link, but it doesn't address my point. The POUND is not a unit of force, it is a unit of mass.

The poundal is a unit of force 1pdl = lb(mass) x ft/s?.

Or if you prefer, you can use another unit of force, the Pound-force = 32.174lb(mass) x ft/s?. (Or 1lbf=32.174pdl)

In your equation you said:

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...

If you had the maximum force on the bottom of the tank in pound-force, and divided that by the (maximum height in ft x density in lb/ft?), you would get: (lb(mass) x ft/s?) / (ft x lb(mass)/ft?) = ft?/s?.

I don't think a value of cubic feet per seconds squared will yield the maximum bottom of tank area.

 

[david luchini]

...here's the really trippy part. The question states "Height times density is part of the equation..."

Now if we happen to be working in density units of lb/ft? and convert to lb/in?, plus we accept that a one pound mass exerts a one pound gravitational force, then height times density is the complete equation for pressure (with lb/in?, or more familiarly psi, as the unit).

Would this not eliminate A as the correct answer?

This illustrates what I was trying to point out about pounds being mass not force. No, height times density is not the the complete equation for pressure (with lb/in?, or more familiarly psi, as the unit).

If you take density in lb(mass)/in? and multiply by height in inches, you get lb(mass)/in?. The pressure familiarly known as psi is lb(force)/in?. They are not the same.

 

[markstg]

David, you make a good point that there is a difference between lbf and lbm but the full equation for pressure excerted on a column of liquid is

P=D*H*g/gc where P is pressure in lbf/ft2
D is density in lbm/ft3
H is height of liquid in ft.
g is gravitational acceleration at the point of measurement with units of
feet/sec2, and at sea level has a value of 32.174.
gc is a constant that relates the units of force, mass, length and time. In the units we are using gc = 31.174 (lbm-ft)/(lbf-sec2).

And this still makes A (pressure) the best answer.

 

[Smart $]

Nice link, but it doesn't address my point. The POUND is not a unit of force, it is a unit of mass.

The poundal is a unit of force 1pdl = lb(mass) x ft/s?.

Or if you prefer, you can use another unit of force, the Pound-force = 32.174lb(mass) x ft/s?. (Or 1lbf=32.174pdl)

In your equation you said:



If you had the maximum force on the bottom of the tank in pound-force, and divided that by the (maximum height in ft x density in lb/ft?), you would get: (lb(mass) x ft/s?) / (ft x lb(mass)/ft?) = ft?/s?.

I don't think a value of cubic feet per seconds squared will yield the maximum bottom of tank area.

For the time being, please quit referring to the "poundal" unit. It has its place, but it isn't here in this discussion.

Technically you are correct.

However, much as writer's invoke artistic license, long ago engineers decided to invoke something similar in the use of the pound weight unit... that is, 1 lb(weight) ≈ 1 lb(mass). Refer to the Wikipedia page I referred to above, in the section titled "Foot-pound-second systems of units" (<<<--direct link link to section), last paragraph of the section.

What this means is that your equation:

(lb(mass) x ft/s?) / (ft x lb(mass)/ft?) = ft?/s?​

...simply becomes:

(lb) / (ft x lb/ft?) = ft?​

Engineers have been using this "leap" for centuries (ok, ok, ...perhaps a slight exaggeration )

 

[Smart $]

This illustrates what I was trying to point out about pounds being mass not force. No, height times density is not the the complete equation for pressure (with lb/in?, or more familiarly psi, as the unit).

If you take density in lb(mass)/in? and multiply by height in inches, you get lb(mass)/in?. The pressure familiarly known as psi is lb(force)/in?. They are not the same.

And as pointed out in my previous post, lb(weight) ≈ lb(mass). Instead of the term weight, you can use force, if you prefer.

 

[LarryFine]

And this still makes A (pressure) the best answer.

YAY! :grin:​

 

[david luchini]

However, much as writer's invoke artistic license, long ago engineers decided to invoke something similar in the use of the pound weight unit... that is, 1 lb(weight) ≈ 1 lb(mass). Refer to the Wikipedia page I referred to above, in the section titled "Foot-pound-second systems of units" (<<<--direct link link to section), last paragraph of the section.

No, from your link above "In circumstances where there may otherwise be ambiguity, the symbols "lbf" and "lbm" and the terms "pounds-force" and "pounds-mass" can be used to distinguish." Pound-mass and pound-force are not interchangeable and must be distinguished. The problem is made worse by the term pound to refer to "weight." But "1 lb(weight) ≈ 1 lb(mass)" is not a correct statement. 1lb(weight) ≈ 1lb(force) would be acceptable.

Within your link, it shows in the F-P-S section, that Force = Mass x Acceleration, and Weight = Mass x Acceleration due to gravity. In other words Weight ≈ Force, so when you use then term pound(weight), you are specifying a force not a mass.

What this means is that your equation:

(lb(mass) x ft/s?) / (ft x lb(mass)/ft?) = ft?/s?​

...simply becomes:

(lb) / (ft x lb/ft?) = ft?​

Engineers have been using this "leap" for centuries (ok, ok, ...perhaps a slight exaggeration )

No. How can lb(mass) become "lb" in the divisor, but lb(mass)x ft/s? become "lb" in the dividend? That means lb(mass)=lb(mass) x ft/s?. That doesn't make sense.

It is clear that "pound" is used in many general terms that may have different meanings. You cannot substitute pound-mass with either pound-force or pound-weight. You cannot substitute pound-force with pound-mass. Pound-force and Pound-weight both refer to a "force."

Let me try to simply it by going back to basics:

Force = Mass x Acceleration (lb-force = lb-mass x ft/s?.)
Weight = Mass x Acceleration due to gravity (lb-weight = lb-mass x 32.174ft/s?.)
Density = Mass/Volume (lb-mass/ft?.)
Pressure = Force/Area (lb-mass/(ft x s?)) = Weight/Area (32.174 lb-mass/(ft x s?))

So back to your post 17, if you had the force at the bottom of the tank in pounds-weight and divided by (height (ft) x density (lb-mass/ft?)) you still end up with an answer of units of ft?/s?, which is not area.

 

[Smart $]

David, you make a good point that there is a difference between lbf and lbm but the full equation for pressure excerted on a column of liquid is

P=D*H*g/gc where P is pressure in lbf/ft2
D is density in lbm/ft3
H is height of liquid in ft.
g is gravitational acceleration at the point of measurement with units of
feet/sec2, and at sea level has a value of 32.174.
gc is a constant that relates the units of force, mass, length and time. In the units we are using gc = 31.174 (lbm-ft)/(lbf-sec?).

And this still makes A (pressure) the best answer.

It seems more references use the same numeric value for gc as g. As such the result of g/gc is:

Seems like the result should be 1lbf/lbm...

 

[david luchini]

It seems more references use the same numeric value for gc as g. As such the result of g/gc is:

Seems like the result should be 1lbf/lbm...

Yes, the answer should be 1lbf/lbm. g is wrongly listed as 32.174lbm x ft/s?. it should be 32.174ft/s?.

 

[Smart $]

No, from your link above "In circumstances where there may otherwise be ambiguity, the symbols "lbf" and "lbm" and the terms "pounds-force" and "pounds-mass" can be used to distinguish." Pound-mass and pound-force are not interchangeable and must be distinguished. The problem is made worse by the term pound to refer to "weight." But "1 lb(weight) ≈ 1 lb(mass)" is not a correct statement. 1lb(weight) ≈ 1lb(force) would be acceptable.

Within your link, it shows in the F-P-S section, that Force = Mass x Acceleration, and Weight = Mass x Acceleration due to gravity. In other words Weight ≈ Force, so when you use then term pound(weight), you are specifying a force not a mass.



No. How can lb(mass) become "lb" in the divisor, but lb(mass)x ft/s? become "lb" in the dividend? That means lb(mass)=lb(mass) x ft/s?. That doesn't make sense.

It is clear that "pound" is used in many general terms that may have different meanings. You cannot substitute pound-mass with either pound-force or pound-weight. You cannot substitute pound-force with pound-mass. Pound-force and Pound-weight both refer to a "force."

Let me try to simply it by going back to basics:

Force = Mass x Acceleration (lb-force = lb-mass x ft/s?.)
Weight = Mass x Acceleration due to gravity (lb-weight = lb-mass x 32.174ft/s?.)
Density = Mass/Volume (lb-mass/ft?.)
Pressure = Force/Area (lb-mass/(ft x s?)) = Weight/Area (32.174 lb-mass/(ft x s?))

So back to your post 17, if you had the force at the bottom of the tank in pounds-weight and divided by (height (ft) x density (lb-mass/ft?)) you still end up with an answer of units of ft?/s?, which is not area.

I see you haven't made the leap yet.

Here's another link on the matter, approached from a "non-leaper" point of view. Scroll down to, and read sections "Comment about the gravitational conversion constant, gc" and "The relationship between force and mass units"

Again, I say you are technically correct. But for the sake of making a point (or not ), let's work an example... and I'm just making this up as I type, so I do not know the answers beforehand...

A flat bottomed, straight-vertical-sided tank contains 1000 cu. ft. of water (at STP if you prefer). The water level is 10 ft. (i.e. top to bottom)

What is the pressure at the bottom of the tank?
What is the force on the bottom of the tank?
What is the area of the bottom of the tank?

Use 62.43lb/ft? for the density of water.

 

[david luchini]

I see you haven't made the leap yet.

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

Here's another link on the matter, approached from a "non-leaper" point of view. Scroll down to, and read sections "Comment about the gravitational conversion constant, gc" and "The relationship between force and mass units"

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."

Again, I say you are technically correct. But for the sake of making a point (or not ), let's work an example... and I'm just making this up as I type, so I do not know the answers beforehand...

A flat bottomed, straight-vertical-sided tank contains 1000 cu. ft. of water (at STP if you prefer). The water level is 10 ft. (i.e. top to bottom)

What is the pressure at the bottom of the tank?
What is the force on the bottom of the tank?
What is the area of the bottom of the tank?

Use 62.43lb/ft? for the density of water.

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.

 

[winnie]

David,

I believe that you made a small error when you said 'the pound is not a unit of force, it is a unit of mass'. I don't believe that you can ignore the fact that pound has been used with both meanings.

The term 'pound' has historically been used both as a unit of force or as a unit of mass. As you have made clear, 1 pound force a different unit than 1 pound mass, so the term pound itself is ambiguous.

The most common usage that I encounter is to use pound to mean force, as in pounds per square inch, or as a unit of mass for commerce, with a 1g field assumed.

Just to be clear, I disagree only with the way you've assigned the word pound to have a specific meaning, not the physics of lbf and lbm.

-Jon

 

[david luchini]

I don't believe that you can ignore the fact that pound has been used with both meanings.

It is clear that "pound" is used in many general terms that may have different meanings. You cannot substitute pound-mass with either pound-force or pound-weight. You cannot substitute pound-force with pound-mass. Pound-force and Pound-weight both refer to a "force."

I don't believe that I have ignored the fact that pound has been used with both meanings.

The term 'pound' has historically been used both as a unit of force or as a unit of mass. As you have made clear, 1 pound force a different unit than 1 pound mass, so the term pound itself is ambiguous.

You forgot "pound" as it refers to currency.:grin:

But you are correct, I probably should have said a pound is not a unit of force as you are trying to apply it, but I think the point was clear.