I feel kind of stupid now that I look at my last response about the n2. I was looking at the rest of the solution while I was feeding my 2 month old early this morning and I guess I was a "little out of it".
I tried to go back through and understand the friction loss calc since I didn't get the same answer from the Moody diagram. I will try and look at it tonight. If you have any further discussion/explanation from the solution I posted earlier, I would greatly appreciate it.
I WOULD REALLY LIKE TO PASS THIS TIME.!!!!!!!!!!!!!!!!!!!!!
owillis
owillis,
OK... here's how I'd explain the friction factor. For laminar flow, f=64/Re; it is *not* dependent on e/D (relative roughness). For fully turbulent flow, f is dependent on e/D only; it is *not* dependent on Re. Remember that flow is laminar when Re is less than 2100, is turbulent when Re is more than 4000, and is in the critical zone when between the two. The Moody Diagram (Friction Factors for Any Kind and Size of Pipe) shows this graphically: to the left of the laminar flow line, is gives the equation f=64/Re, in the turbulent zone on the right, the lines are flat which means f is not dependent on Re.
If you want to calculate Re for a pipe flowing full: Re=VD/v where V is mean velocity, D is diameter of pipe, and v is kinematic viscosity of fluid. In this case, v=0.930x10-5 ft2/sec (appendix 14.A) and you should get Re=93,440 which is turbulent. Relative roughness can be calculated by looking up e in a table (design column in appendix 17.A); e/D=0.0005ft/0.5ft=0.001.
From here, you can use the Moody Diagram: Choose the curve where e/D=0.001 and follow it to the left until you get to Re=1E5 (close enough, yeah?), and then head straight to the left until you get to the left axis and then read f. You should get an f of somewhere between 0.018 and 0.019. I think it's easier to use Appendix 17.B because it's quicker and less likely to result in a mistake. In this case, you'd quickly get f=0.0185 for Re=1E5 (or you could interpolate but it's probably not worth the added accuracy) and e/D=0.001.
Or... you could use the Swamee-Jain equation... but now you're going to be pushing the 6 minute solution!
Once you have f, it's pretty easy to use the Darcy-Weisbach formula for head loss, hL=(fL/D)*(V2/2g).
Now... I think you'll agree that the Hazen-Williams equation I used in my solution is *much* quicker. You only need to find C using Appendix 17.A. If you don't have all the forms of the Hazen-Williams equations (solved for V in terms of both R and D, Q in ft3/sec and mgd, and hL let me know and I'll try to scan my one-page summary for you.
This is long-winded, but I think you'll find it's about as far in depth as you'll need to go for the PE exam. Please give me some feedback on my explanation.