- #1
Xyius
- 508
- 4
I do not want to post the entire homework problem as I understand all of it except this one part.
The problem deals with energy balance and it states that water is flowing through a nozzle. (From a bigger cross sectional area to a smaller one.) I am trying to calculate the energy of the kinetic energy but my book does something weird to me.
The problem is in English units. (Very annoying!) The kinetic energy of a fluid would be..
[tex]\frac{1}{2}mv^2=[Btu][/tex]
So the specific energy should be..
[tex]\frac{1}{2}v^2=[\frac{Btu}{lbm}][/tex]
In the book they do the following..
[tex](\frac{1}{2}v^2)(\frac{1[Btu/lbm]}{25037 [ft^2/s^2]})[/tex]
That 25037 is throwing me for a loop. Why do I need this conversion? I thought KE was already in the correct units?
My assumption is that it is similar to metric units since KE is in Joules and normally you want it in KJ. Seems like I am just ignorant of English units. Can anyone shed some light on this?
The problem deals with energy balance and it states that water is flowing through a nozzle. (From a bigger cross sectional area to a smaller one.) I am trying to calculate the energy of the kinetic energy but my book does something weird to me.
The problem is in English units. (Very annoying!) The kinetic energy of a fluid would be..
[tex]\frac{1}{2}mv^2=[Btu][/tex]
So the specific energy should be..
[tex]\frac{1}{2}v^2=[\frac{Btu}{lbm}][/tex]
In the book they do the following..
[tex](\frac{1}{2}v^2)(\frac{1[Btu/lbm]}{25037 [ft^2/s^2]})[/tex]
That 25037 is throwing me for a loop. Why do I need this conversion? I thought KE was already in the correct units?
My assumption is that it is similar to metric units since KE is in Joules and normally you want it in KJ. Seems like I am just ignorant of English units. Can anyone shed some light on this?