^^yeah, i'm having a similiar problem. but if you only have 7 terms, how are you solving for 5 pi groups. that would only be 4.


that is what he went over today. that the states of the model and the prototype must be equal. like Pi group 1 of the model equals pi group 1 of the prototype

10/19/2005 10:55:38 PM

my big question is i dont know which variables to make the pressure distribution a function of to start with...

10/20/2005 12:38:16 AM

i emailed him about it and he said for someone to bring it up in class tomorrow since it isn't due till monday

10/20/2005 3:01:34 PM

awesome, i was wondering if he mentioned it and i didnt hear that the due date was changed, cause i thought it was odd that we didnt finish the note/examples on wednesday

thanks for emailing him, i got caught up with history shit these past couple hours

i ended up getting something similar as you, i ended up using the same starting parameters and was a little off

anyways we'll find out in the morning if i'm not too hungover

10/20/2005 4:55:20 PM

was it me or did anyone else not really understand his explination this morning

can one of you help me out and explain it a little better

thanks

10/21/2005 11:27:41 AM

so has anyone figured out how to get the velocity answer he has for 4? cause i'm gettin like 9.2 at best and dont really know what else to do.

10/22/2005 4:13:34 PM

yeah, i had like 9.26 and that is what matt had and gould said that was fine. only like 3% off . . . prolly due to using diff numbers

10/22/2005 8:14:53 PM

For number 1, does anyone know what it means to put V in simplest form? I have the pi groups and put Pi group 1 as a function of Pi group 2, i just dont understand the second part of that question.

10/23/2005 12:42:03 PM

I thought that was all we had to do for 1 and 2

can anyone explain how to do number 4? I don't know where to begin with it. I got all the others i think though.

thanks

10/23/2005 1:05:37 PM

^ ditto

10/23/2005 1:15:30 PM

for #5, the vortex shedding frequency would just have units of 1/time? or would it be a force, or would there be frequency and a force?

10/23/2005 3:53:29 PM

example # 7.5 & 7.6 in the book are kinda like number 4... but i can't get my numbers to come out quite right

10/23/2005 4:09:31 PM

1/time

just like frequency cycles PER second

10/23/2005 4:35:06 PM

still confused on number 4, oh well guess i'll work on my lab report for a while and maybe it will come to me by some weird miracle

10/23/2005 5:35:27 PM

what did you guys get for your 3 pi groups?

and what do we use for the cavatiation number?

thanks

10/23/2005 6:28:00 PM

for 4 i didnt solve for pi groups, i just used the chart in the book. i mean he said there is no sense for us to continously solve for these groups when they are formally known and proven already. with that in mind, i only got 9.2 for the velocity and my cavitation pressure was way off from waht he has. so i dunno what to do.

10/23/2005 6:42:37 PM

i only had 2 pi groups, reynolds, and i used F rho V and L to come up with a pi group of F/rho V^2L^2

I'm so confused about cavitation though....why do they give you both the cavitation number and coeffcient?

10/23/2005 7:17:01 PM

i just used cavitation and got close like the others, all of my answers are like 3-5% off

oh well, shitty problem

i'm saying screw it, i got all the others

10/23/2005 7:21:57 PM

did you just do .5=2(P-Pvac)/rhoVwater^2?

10/23/2005 7:25:32 PM

oh yea i got 4! i didnt get his answers cause he apparently used different values but i got similar answers which means they have to be right.

for part a. use the formula for reynold's number and find your velocity of the model. should be around 9.2 or so.

for part b. use the cavitation formula in the chart in the book AND the pressure coefficient formula. solve the cavitation formula for P. solve the pressure coefficient formula for Pinfinity. the cavitation pressure will fit into the pressure coefficient formula by the way it is solved. Pinfinity works out to be 82.59kPa.

for part c. use F/(roe*V^2*L^2). solve for your Fm/Fp ratio. should be around .27

[Edited on October 23, 2005 at 9:24 PM. Reason : .]

10/23/2005 9:17:07 PM

Can anybody explain how the D works in #3?

10/23/2005 9:47:04 PM

it's just telling you all the parameters you need to use

so, w, h, and then you need rho, V, D, and mu

10/23/2005 10:36:28 PM

bttt for chap 8 homework

i'm confused as shit on number 1, so this isnt a good start

anyone started it yet?

10/30/2005 11:43:51 AM

I got number 1, anyone have a clue on number 2?

10/30/2005 12:37:46 PM

number 2 i got the first part which is easy, but the other two i'm confused on, can you post how to do number 1 for me

thanks

10/30/2005 4:11:34 PM

For 1, just use the umax equation we derived in example 8.2 =-R^2/4*mu(dP/dx). dp/dx is the pressure change over the length of the needle, and u get the pressure using the force/plunger area. once u get umax, get uaverage by dividing umax by 2, since umin is 0. then just do flowrate=uaveraqe*(needle area). im tryin to figure out how to integrate number 3 right now if anyone has a clue.

10/30/2005 8:26:08 PM

havent figured 3 out totally, cant quite get the integration right either

thanks for the help on 1

if i get 3 tomorrow i'll post

10/30/2005 9:27:53 PM

ok i need to ask if someone can do me a huge favor

i lost the first half of chp 8 notes with the 3 examples and his additions, can someone take a pic of theirs and send it to me

i'm fucked if i dont have those examples to look at

10/31/2005 9:26:05 AM

any help on 8.4? i got part b thinkin was solving for part a but now i have no clue how to get part a. i'm kinda jumping around so i only have 8.1 and 8.5 done.

10/31/2005 11:25:36 AM

i'm still having problems with number 1

for umax=-R^2/4*mu(dP/dx), is the R the reynolds number i forget, and i dont have my examples to look at anymore

also i'm getting 113.something for dp/dx is that what you guys got?

for 8.4 if you got the 41.37 for b then just multiply by gravity to get the answer on a mass basis instead of weight

10/31/2005 11:49:41 AM

R is the radius of the plunger. i was using diameter by mistake and then caught my mistake later on. (dp/dx) i got (442097/.05). it all worked out once i got the Radius/diameter thing straightened out.

10/31/2005 12:00:40 PM

ok i got all of them except parts b, c on 2 and i'm fucking up the integration on 3 still

grrr

10/31/2005 12:28:57 PM

anyone get 2 or 3?

10/31/2005 5:11:34 PM

dang i can't even get the 1st one

i followed all of these methods, and ended up getting .878 m^3/s

which is no where close, that's not a good sign i can't even get the first one, haha

my final calculation was Vol = (7939.876)(pi)(0.00015m)^2

[Edited on October 31, 2005 at 6:30 PM. Reason : .]

10/31/2005 6:28:00 PM

for #4
part a:
use eq. 5.73 assuming no work from pump or turbine. Divide each term by mass and solve for E_mech_dot/m_dot (mechanical head loss) to get the head loss per unit mass.
Also don't forget that the pressure is given as gage.
part b:
use the same equation and divide each term by m_dot*g

10/31/2005 11:26:16 PM

i'm still not getting number 2

he said in the email: You must use the relationship for wall shear stress and dp/dx that we
developed using conservation of x-momentum (i.e Ex. 8.2 in notes) for this
problem. Thus you do not need velocity.

can someone post that relationship, i lost my first half of notes and dont have that anymore, maybe i can figure it out if i have that equation

also how would i go about finding the head loss in meters

thanks

11/1/2005 12:16:14 PM

number 1 is still screwing me up....here are my values maybe someone can clear something up for me:
(dp/dx)=442097/.05
mu=5*(1.002e-3)
which gives me a umax value of (15883.725)
uavg=7941.86

so I get a huge number when I plug this in. If anybody can help I would appreciate it.

11/1/2005 6:42:43 PM

anyone get 3 right?

or 2c

[Edited on November 1, 2005 at 6:49 PM. Reason : ]

11/1/2005 6:49:13 PM

Uavg should be 1.24345,

make sure you are using the correct equation... i think you are leaving out R^2 and 4*mu

[Edited on November 1, 2005 at 6:59 PM. Reason : ...]

11/1/2005 6:58:33 PM

here is what i am doing:

[(.006^2)/(4*(5*1.002e-3))]*(dp/dx)...is this not the right formula, because it still comes out huge

11/1/2005 7:40:19 PM

i gave up on 3, its fucking impossible to understand

and the only thing i cant get is c on number 2

11/1/2005 8:06:30 PM

should be about 2.something...

check your parenthesis, maybe something is messed up, but your equation is the same one i used.

[Edited on November 1, 2005 at 8:29 PM. Reason : ...]

11/1/2005 8:27:23 PM

For #4, are we allowed to use eq. 5-73, it says in the problem to use eq. 5-77

11/1/2005 8:34:26 PM

I got number 3....here is some hints to it....i have only done number 1 besides this though....so i may need help on the others

dQ=2pi r u dr
Q= 2 pi Umax integral (r*(1-r/R)^1/n) dr from 0 to R

x=1-r/R
dx/dr=-1/R and dr=-Rdx
so
R-r=xR when r=0, x=1
R-xR=R(1-x) when r=R, x=0

So
Q= 2pi Uman integral 0 to 1 (R*(1-x)x^1/n(-Rdx)
=2piR^2 Umax ((xn/n+1)-(x^2*n)/(2n+1) from 0 to 1

and u=Q/pi R^2

that gives it to you

11/1/2005 8:39:26 PM

5.77 and 5.73 are the same... only falirly different by definition

11/1/2005 8:55:58 PM

hey can somebody send me the Ch 8 notes? the ones with the examples already worked in them... I lost mine and now I will be fucked if I can't find them.... just PM me or like send them to my email jmalexa2@ncsu.edu

11/1/2005 11:42:25 PM

anyone still having problems with number 1, use the radius for the needle, not the tube

11/2/2005 12:24:13 AM

bttt for homework 9

i'm getting 661psi for number two for some reason

also for number 3, how do you factor in the efficiency of the pump

11/9/2005 12:34:27 PM

I haven't worked 3 out yet but do you just multiply the efficiency by power? similar to thermo probs?

11/9/2005 1:04:42 PM

thats what i dont know

on number 5, i've got my table like this:

3in 3.068in 4.334ft/s 66633 .005 .0215 8.8005 26.1956ft
2.5in 2.469in 6.692ft/s 82798 .007 .0215 10.936 63.4823ft
2in 2.067in 9.548ft/s 98900 .008 .0215 13.062 131.32ft

this doesnt seem right if they have to be less than 20ft right?

could someone explain if i'm not doing something right or what

11/9/2005 5:53:07 PM