How to Convert Lbm/ft^3 to kg/m^3?

[LT72884]

Homework Statement

The pressure drop in a duct is to be measured by a differential oil manometer. If the differential height between the two fluid columns is 5.7 inches and the density of oil is 41 lbm/ft^3, what is the pressure drop in the duct in mmHg? Report your answer to 1 decimal place.

Homework Equations

P=pgh
760mmHg - 101.325kPa=1atm = 760torr

The Attempt at a Solution

I understand how to do it, but I am stuck at converting Lbm/ft^3 to kg/m3
I used google to get that 41Lbm/ft3 is the same as 656.74kg/m3 and with that i get the correct answer after calculations. However, i don't want to use google for all my conversions. Please help me see how to convert Lbm/ft3 to kg/m3

thank you:)
Good Day

 

[Wrichik Basu]

Do it like this: $$\frac{1 \, lbm}{1 \, ft^3} = \frac{0.454\, kg}{(0.305)^3 \, m^3}$$ Calculate the value of $\frac{0.454}{(0.305)^3 }$and put it there. You have your conversion.

 

[LT72884]

Awesome:) the 0.305 is what unit? Is that a foot per meter as in 1 foot = 0.305m? I just want to make sure i get the numbers right. and 454 g per pound, hmm. my book nor homework lecture notes do not have those numbers. Well, i have them now haha. onward and upward. ANything else i should know about this problem?

thanks

 

[Wrichik Basu]

$1 ft = 0.305 m$ and $1 lbm = 0.454kg$. I had to take those values from Google, as I am not used to the FPS system.

Just keep in mind that whenever you have some conversion of this type, change each unit to the required unit, put it in the expression, raise it to the necessary power, and you have your answer.

And by the way, if such conversions come often in your work, you may like to memorise them rather than derive them every time. It will save time.

 

[Chestermiller]

I would never have done this the way that the others have done it here. Here is my approach:
$$P=\frac{\rho g h}{g_c}$$where $$g_c=32.2\ \frac{lb_m}{lb_f}\frac{ft}{sec^2}$$
So, in this problem: $$P=\frac{(41)(32.2)\left(\frac{5.7}{12}\right)}{32.2}=19.48\ \frac{lb_f}{ft^2} =0.1352\ psi$$
1 atm = 14.7 psi
So, P = 0.0092 atm = 6.99 mm Hg

I guess sometimes it helps to have a lifetime of experience with Imperial units.

 

[LT72884]

@Chestermiller Yup, there is no way i would have known to do it that way haha. i have seen the gc number before but never used it.

How did you go from 19.48lbf to 0.13psi?

@Wrichik Basu Yeah, I am going to have to mezmorize those numbers. they were not in my book. To be honest, i like metric better than what we use here in the USA.

thanks to all:)

 

[Chestermiller]

@Chestermiller Yup, there is no way i would have known to do it that way haha. i have seen the gc number before but never used it.

How did you go from 19.48lbf to 0.13psi?

That's $19.48 \frac{lb_f}{ft^2}$, not $19.48\ lb_f$. There are 144 in^2 in a square ft, so if I divide by 144, I get $0.13 \frac{lb_f}{in^2}=0.13\ psi$.

 

[LT72884]

That's $19.48 \frac{lb_f}{ft^2}$, not $19.48\ lb_f$. There are 144 in^2 in a square ft, so if I divide by 144, I get $0.13 \frac{lb_f}{in^2}=0.13\ psi$.

OH MY GOODNESS. I do not know why i didnt catch that hahahahaha.

thanks for helping me see the obvious:)

 

[scottdave]

You may find my Insight article useful. https://www.physicsforums.com/insights/make-units-work/

 

[LT72884]

You may find my Insight article useful. https://www.physicsforums.com/insights/make-units-work/

thanks:) that's a good read