Calling engineers: Need help with unit conversions

[Hethj7]

Well, school is back in session and sometimes the seemingly simplest things take me all night to do. (Kind of how working on the Stang can be at times).

Anyway, if anyone can help me convert seconds^2/feet^3 to gallons per minute, that would be great.

I believe there are .13368 cubic feet/ 1 gallon. But how can I get the seconds squared into minutes?

[bigred90gt]

I 'm not an engineer, but I really believe that there is more that 0.13368 cu ft in a gallon. Picture a 1'x1'x1' box and a gallon of milk. The math ain't there. Do a search on google.com for conversions and formulas, you'r bound to find something helpful there.

Caymon

[Hethj7]

I know logically that doesn't make sense, but that is correct. There actually are 7.48 gallons in 1 cubic foot. Don't ask me how....it doesn't necessarily make sense to me either, but here it is: conversions

[MEDIK418]

Are you trying to estimate flow?

[daveGT]

I think you want to convert flow numbers, right?

Converting seconds^2 / cubic feet to gallons / minute doesn't make sense - you have to convert like measurements.

It sounds like you want to convert flow numbers, from gallons per minute (gal/min) to cubic feet per second (ft^3/sec).

Therefore:

1 gallon = 0.13368 cubic feet, or 1 gal / 0.13368 c.f.

1 minute = 60 seconds, or 1 min / 60 sec.

So, convert gallons to cubic feet and then convert minutes to seconds:

(1 gal/min) * (0.13368 cf/gal) * (1 min./60 seconds) = .0022

Multiply out and cancel the units:

Answer = 0.0022 cubic feet /second = 1 gallon/minute

So...

10 gallons per minute = 10(0.0022 cf/sec) = 0.022 cf/sec


Hope this helps.

[Hethj7]

Yea, I guess I knew that my original conversion didn't really make sense. When I get home I will have to check all my units and go back through the problem. I need to end up with gal/min, but in my calculations gravity is involved, thus giving me ft/s^2. Somehow I need to get that accelaration into a velocity, although I don't recall a time frame being given for this problem.

Anyway, thanks for the replies. I don't have the problem with me right now, so I can't really add anything to this thread. Guess I will work on it some more tonight.

[The Deuce]

I don't know if this has them all, but I use this site for most of my conversion needs.

http://convert.french-property.co.uk/index.htm#length

[Hethj7]

Thanks for the link Deuce! That one I found using google is pretty good too (check out the link in my previous post).

Anyway, here was the original problem (if anyone cares at this point).

Water at 60 degrees (F) is to be pumped into a tank that is 150 feet above the pump. The pipe is 1000 feet long and 12 inches in diameter. Friction factor is .025 and neglect entrance and exit losses. Plot head pressure (in feet of water) vs. discharge flow rate, Q (gal/min).

Using the energy equation and Continuity equation Q=AV, then simplifying:

head pressure = change in elevation + (friction factor) * (length/diameter) * (4Q/(pi*d^2)) * 1/(2g)

This equation is correct, as we did it in class. I guess my main problem comes from getting value for g , 32.2 ft/s^2. I can't figure out how to get rid of the s^2 so that I can be left with just values for gal/ min.

Well, if anyone actually read through all this, sorry I bored you to death. For anyone else who may actually have an idea, I am all ears !

[jj_jonathon]

Originally posted by Hethj7
Thanks for the link Deuce! That one I found using google is pretty good too (check out the link in my previous post).

Anyway, here was the original problem (if anyone cares at this point).

Water at 60 degrees (F) is to be pumped into a tank that is 150 feet above the pump. The pipe is 1000 feet long and 12 inches in diameter. Friction factor is .025 and neglect entrance and exit losses. Plot head pressure (in feet of water) vs. discharge flow rate, Q (gal/min).

Using the energy equation and Continuity equation Q=AV, then simplifying:

head pressure = change in elevation + (friction factor) * (length/diameter) * (4Q/(pi*d^2)) * 1/(2g)

This equation is correct, as we did it in class. I guess my main problem comes from getting value for g , 32.2 ft/s^2. I can't figure out how to get rid of the s^2 so that I can be left with just values for gal/ min.

Well, if anyone actually read through all this, sorry I bored you to death. For anyone else who may actually have an idea, I am all ears !

geez...that looks like a problem i would do that mixes my calc2 class with my physics class...i hate that stuff...

friction factor...good ole u (close the the u anyways)...

youre making me wonder why im a physics major..lol..i want out!

[Hethj7]

You want out? Me too!

It makes me feel good to know that others feel the same way going through some of these courses.

I just finished my marketing degree this past December, but decided I was mad at myself for not getting my Mechanical Engineering Degree. So, here I am at 23 starting back into ME courses. I only have two years left, so that is not too bad. But, it has been 3 years or so since I have had a Calculus or Physics course .

Sometimes I am ready to just get my MBA instead and head out into the working world. But I keep telling myself this degree will be worth it once I am done. Looks like it will be a lot of late nights working problems and going nuts for the next two years !

[95mustanggt]

seems like everything is under control and I was too late!

The one good thing about metric units and that engineering calculations are much simplier!

[daveGT]

Ah, fluid dynamics. Fun stuff.

Remember, you need to calculate the total head loss (THL) from a system in units of "feet." You gotta watch your units, and write out your entire energy conservation equation based on Bernoulli's equation to avoid unit problems. Been there, friend!

Remember, the (2g) term is just a conversion term to convert your velocity, or kinetic head (v^2) to units of "feet." Likewise the pressure head (p) must be divided by gamma (density of the fluid) to get units of feet. The elevation, or potential head is already in units of feet. Your system losses, friction (hf) must also be in units of feet: hf = friction factor(length)(v^2) / (diameter)(2g) Everything can now be totaled.

You're getting mixed up when you try to combine the flow rate (Q = v * A) with Bernoulli. Write out Bernoulli, sovle for the velocity at the open end of the pipe, then calculate flow (Q). Then you can convert your flow (Q) anything you want.

Gotta love fluids! Hope this is helpful!

[Hethj7]

daveGT - Excellent write up there! Thanks for the help. I did get the problem finished today, but with little understanding of how I got the correct units. Your post clears things up some. Thanks!!

It doesn't make it any easier that this is for a mechanical systems class with a professor that assumes we have all had fluids already. I am enrolled in fluids right now though, so all this is being thrown at me all at once, although I have of course had some experience with energy equations and the like.

[MTU 50]

Fun Stuff.

In May I will graduate with my B.S. in Mech. Eng. Tech.

I took fluids two years ago and have forgot a lot of it since I haven't used it since, but I did get a 'B'.

Its not too hard, but it can be very frustrating at times when you have no idea how to do something. Of course, once you see it done the correct way, you are kicking yourself that you couldn't figure it out.

daveGT did a good job of explaining things.

[fiveohpatrol]

The one good thing about metric units and that engineering calculations are much simplier!

VERY true. I used to hate the metric system before I started studying engineering, but now I absolutely hate working with the english system.
I think it would be alot more impressive to tell someone that your car makes 223,800Watts rather than 300hp

[Hethj7]

Originally posted by MTU 50
Fun Stuff.


Its not too hard, but it can be very frustrating at times when you have no idea how to do something. Of course, once you see it done the correct way, you are kicking yourself that you couldn't figure it out.

No doubt there. I can spend hours trying to figure out something and not get anywhere. Then, once I see the solution it becomes so obvious. I wonder how I miss those little points sometimes, but it seems like it repeatedly happens. It is stuff I understand, but I just fail to apply it to the problems.