Solving Unit Conversion Problems: Changing from ft to gal/min

[maack_j]

I think this problem is straightforward , but I'm not used to these units. (Normally I use SI units)

From a physics book (U.S version ) I have this equation:

$h=10 + 4.43*Q^2$

where the unit of $h=[ft]$, and $Q=[ft^3/s]$

now the book rewrites the above equation, such that $Q=[gal/min]$ and get

$h=10 + 2.2*10^{-5}*Q^2$I just can't see what the book does to get from the first equation to the secound. Can anyone help me with this?

Best regard
J. Maack

 

[RUber]

1 Gallon is equal to 0.133681 ft³.
Does that math work out?

 

[DrClaude]

Knowing that 1 gal = 0.1337 ft³, can you do it?

[Edit: beaten by RUber...]

 

[maack_j]

Knowing that 1 gal = 0.1337 ft³, can you do it?

[Edit: beaten by RUber...]

I can't get the same result as the book with that 1 gal = 0.1337 ft³

What i do is
$4.43*(1/60^2)*0.1337$

Isn't that correct?

 

[DrClaude]

I can't get the same result as the book with that 1 gal = 0.1337 ft³

What i do is
$4.43*(1/60^2)*0.1337$

Isn't that correct?

Nope.

Personally, I don't like those kind of equations which work only for certain choices of units. So do as I do, start by figuring out what the units of the factor 4.43 have to be for the original equation to hold.

 

[maack_j]

Nope.

Personally, I don't like those kind of equations which work only for certain choices of units. So do as I do, start by figuring out what the units of the factor 4.43 have to be for the original equation to hold.

The unit for 4.43 would be $[s^2/ft^5]$ for the equation to hold.

 

[DrClaude]

The unit for 4.43 would be $[s^2/ft^5]$ for the equation to hold.

That's correct, but to simplify further calculations, keep the volume in there (and the units of h): [ft (s/ft³)²]. Now you should be able to do the conversion to [gal/s].

 

[maack_j]

That's correct, but to simplify further calculations, keep the volume in there (and the units of h): [ft (s/ft³)²]. Now you should be able to do the conversion to [gal/s].

Perfect! Tanks

 

[HallsofIvy]

First, in order for this to makes sense the different units must measure the same thing! That is the case here- both "cubic feet" and "gallons" measure volume, both "seconds" and "minutes" measure time. "cubic feet per second" and "gallons per minute" measure how fast volume is changing- perhaps how fast a liquid is moving through a pipe.

h, in the first equation is in feet and Q is in cubic feet per second. But Q is squared so $Q^2$ has units of "cubic feet squared" (or feet to the sixth power) over "seconds squared". That tells us that the "10" must have units of feet and the "4.43" must have units of "seconds squared over feet to the fifth power": $\frac{s^2}{ft^5}\times\frac{ft^6}{s^2}= ft$ which can then be added to the "10 ft" to get "h ft".

In the second equation, h is still in feet but now Q is in gallons per minute so the "$2.2*10^{-5}$" must have units of [itex]\frac{min^2 ft}{gal^2}[/tex] so that $\frac{min^2 ft}{gal^2}\times\frac{gal^2}{min^2}= ft$ again. There are about 7.48 cubic feet per gallon (I had to look that up) and 60 seconds per minute (that I did not).<br /> That gives $\frac{1}{7.48} \frac{gal}{ft^3}$ and $\frac{1}{60} \frac{min}{sec}$ so $\frac{1}{7.48^2} \frac{gal^2}{ft^6}$ and $\frac{1}{3600}\frac{min^2}{sec^2}$ So our conversion, from $\frac{s^2}{ft^5}$ to $\frac{min^2 ft}{gal^2}$ can be done by<br /> $$\left(4.43\frac{sec^2}{ft^5}\right)\times\left(\frac{1}{60}\frac{min}{sec}\right)^2\left(\frac{1}{7.48}\frac{gal}{ft^3}\right)^2= \left(4.43\right)\left(\frac{1}{3600}\right)s\left(\frac{1}{55.95}\right)= 2.18\times 10^{-5}$$<br /> <br /> That's not exactly the same as the given "$2.2\times 10^{-5}$", perhaps because of round-off error, but that's the idea.[/itex]

 

[Chestermiller]

First, in order for this to makes sense the different units must measure the same thing! That is the case here- both "cubic feet" and "gallons" measure volume, both "seconds" and "minutes" measure time. "cubic feet per second" and "gallons per minute" measure how fast volume is changing- perhaps how fast a liquid is moving through a pipe.

h, in the first equation is in feet and Q is in cubic feet per second. But Q is squared so $Q^2$ has units of "cubic feet squared" (or feet to the sixth power) over "seconds squared". That tells us that the "10" must have units of feet and the "4.43" must have units of "seconds squared over feet to the fifth power": $\frac{s^2}{ft^5}\times\frac{ft^6}{s^2}= ft$ which can then be added to the "10 ft" to get "h ft".

In the second equation, h is still in feet but now Q is in gallons per minute so the "$2.2*10^{-5}$" must have units of [itex]\frac{min^2 ft}{gal^2}[/tex] so that $\frac{min^2 ft}{gal^2}\times\frac{gal^2}{min^2}= ft$ again. There are about 7.48 cubic feet per gallon (I had to look that up) and 60 seconds per minute (that I did not).<br /> That gives $\frac{1}{7.48} \frac{gal}{ft^3}$ and $\frac{1}{60} \frac{min}{sec}$ so $\frac{1}{7.48^2} \frac{gal^2}{ft^6}$ and $\frac{1}{3600}\frac{min^2}{sec^2}$ So our conversion, from $\frac{s^2}{ft^5}$ to $\frac{min^2 ft}{gal^2}$ can be done by<br /> $$\left(4.43\frac{sec^2}{ft^5}\right)\times\left(\frac{1}{60}\frac{min}{sec}\right)^2\left(\frac{1}{7.48}\frac{gal}{ft^3}\right)^2= \left(4.43\right)\left(\frac{1}{3600}\right)s\left(\frac{1}{55.95}\right)= 2.18\times 10^{-5}$$<br /> <br /> That's not exactly the same as the given "$2.2\times 10^{-5}$", perhaps because of round-off error, but that's the idea.[/itex]

$That should be 7.48 gal/ft³, not 7.48 ft³/gal.<br /> <br /> Chet$