ur not as bad off as u might think . . .

look at your units, for the pressure from the oil and from the water u have something with s^2

use this conversion to get lbf /ft^2

lb/(ft*s^2) = 0.0310809502 lb/ft^2

[Edited on September 11, 2005 at 9:26 PM. Reason : but yeah, the first one fried my brain . . . how do u do #2]

9/11/2005 9:22:58 PM

fucking a

thanks dude


i can actually help with #2:

you're going to have:
deltaP=p1-p2=(rho(meriam)-rho(water))*g*l

stems from: p1 + rho(water)*g*(a+l) - rho(meriam)*g*l-rho(water)*g*a=p2
the value a = distance from top of meriam blue on p2 side to bulb, but it cancels out of above equation so its not incredibly important

should end up with (1750-1000)*9.81*.02=147.1 Pa

9/11/2005 9:32:53 PM

only one i have no clue on is #5 now, i think we need more dimensions to find the volume of water

bah

9/11/2005 9:40:05 PM

where do i get the rho of the meriam blue?

9/11/2005 9:46:50 PM

i think that too.....


does that mean you know how to do number 6????

9/11/2005 9:47:13 PM

rho is sg time 1000 kg/mcubed, he emailed us the sg

9/11/2005 9:47:59 PM

Gould gave it to us in an email or in class or something, can't remember. It's 1.75.

1.75(1000kg/m^3) to be exact

[Edited on September 11, 2005 at 10:01 PM. Reason : clarified]

9/11/2005 9:54:03 PM

#6:

using the FBD he gave us in class, once you have the forces found and the distances they are away from the axis', plus into a moment equation to find the Fb value, i must have one of my distance values off by i think .2 meters cause i'm off of his answer by like 10kN

hope this helps a little bit...

[Edited on September 11, 2005 at 9:57 PM. Reason : ..]

9/11/2005 9:56:40 PM

I didn't use that wierd obscure conversion factor... instead found in the book where it had lbf/32.2lbm ft/s^2 and just used that at the end. It was close enough to the answer he had... 1080.9lbf to be precise.

I'm still not getting something right about 4. How do you get the area to plug into the equ F=rho*g*hc*A?
Or is that wrong?

9/11/2005 9:57:00 PM

for #4:

the area should be the width of the gate that they gave you times the length of the gate, which should be 2.828 something, just 2sqrt(2)

make sure on the right side you have your hc, yc, yp values origined at that free surface, not the free surface on the left (higher one)

9/11/2005 9:58:51 PM

so how does one go about 3.7?

9/11/2005 10:50:34 PM

Is anyone else still working on this?

9/11/2005 11:20:07 PM

yea me..still cant get 4

9/11/2005 11:34:35 PM

what is your problem with it? do you have any of the last 3?

9/11/2005 11:37:04 PM

i've been so busy worryin about 1 and 4 i havent even attempted the last 3. luckily i figured out 1 only cause of the damn conversion factor. i did everythin right except convert right which fucked me up. with 4, i cant get my overall force to work out. i have 2 resultant forces and the moment arms, summing about point B and including the Fa but my Fa is too big when i calculate it.

9/11/2005 11:40:36 PM

hmm. what are your two resultant forces and how far from B are they?

9/11/2005 11:41:57 PM

Fr1 = 333148 and its 1.2597 from B
Fr2 = 111049 and its .93999 from B

[Edited on September 11, 2005 at 11:53 PM. Reason : wrong #]

9/11/2005 11:46:31 PM

i have Fr2=110857, but the others are close enough that they shouldn't matter.

Not sure how you have it set up, but the free surface is at the top of the gate, so hc2 should =1m. so rho*g*hc2*A = 1000*9.8*1*11.312


did that help in any way?


[Edited on September 12, 2005 at 12:06 AM. Reason : ]

9/11/2005 11:53:54 PM

i redid my numbers and now i have:

Fr1=332573, distance from B=1.4039
Fr2=110858, distance from B=.94267

i said 110858(.94267)+332573(1.4039)-Fa(2.828)=0

my Fa turns out to be like 50kn

9/12/2005 12:08:01 AM

#5 or #6 anyone? I'm beating myself up over these two problems. Any tips would be appreciated.

9/12/2005 1:33:19 AM

when did he say the solutions would be up? i checked the board like an hour ago and only the first two sets were up...

9/12/2005 3:53:30 PM

and what was he saying at the beginning of class today? Damn bus came 20 min late.

9/12/2005 3:58:40 PM

when i asked him when the solutions would be posted he said by lunchtime today. so much for that.

9/12/2005 4:47:09 PM

I was planning on going over there later and taking a picture of them.

So if they are up, and the pics are halfway decent then I will put em up on here later.

9/12/2005 4:55:37 PM

haha thats what i was gonna do

9/12/2005 4:56:28 PM

when u goin? i was plannin to around 8 or so. but if ur goin b4 then i wont bother.

9/12/2005 6:28:54 PM

ladies and gentleman, get ur Set#3 solutions right here:
(view the full page on screen, do not maximize..sorry for fuzzyness, could not use flash due to glare)

http://www.filefarmer.com/brianj320/solutions/solution_3_1a.JPG
http://www.filefarmer.com/brianj320/solutions/solution_3_1b.JPG
http://www.filefarmer.com/brianj320/solutions/solution_3_2.JPG
http://www.filefarmer.com/brianj320/solutions/solution_3_3.JPG
http://www.filefarmer.com/brianj320/solutions/solution_3_4a.JPG
http://www.filefarmer.com/brianj320/solutions/solution_3_4b.JPG
http://www.filefarmer.com/brianj320/solutions/solution_3_5a.JPG
http://www.filefarmer.com/brianj320/solutions/solution_3_5b.JPG
http://www.filefarmer.com/brianj320/solutions/solution_3_5c.JPG
http://www.filefarmer.com/brianj320/solutions/solution_3_6a.JPG
http://www.filefarmer.com/brianj320/solutions/solution_3_6b.JPG
http://www.filefarmer.com/brianj320/solutions/solution_3_6c.JPG
http://www.filefarmer.com/brianj320/solutions/solution_3_7.JPG

9/12/2005 9:00:20 PM

Dude, you are so awesome. Much thanks.

9/12/2005 10:16:56 PM

thanks dude

9/13/2005 12:04:29 PM

anyone start the sample test yet??

I can't figure out the last one and parts of the first one and wanted to see how my answers compared to others.

1.) @A Pabs=103.0106kPa
not sure what to do about the other points

2.) I found v=112.85299m/s, seems high to me but it could be right

3.) P1-P4= 1890.5433 lbf/ft^2= 13.129psi

4.) No clue how to do these types of problems. Hints or anything would be greatly appreciated.

9/13/2005 3:39:42 PM

just started it, i'll get back on here after i've tried all of them

seems long for a 50 min exam though

[Edited on September 13, 2005 at 4:00 PM. Reason : adfad]

9/13/2005 3:59:57 PM

can you give me a run down of how you worked number 3, its confusing the hell outta me...

9/13/2005 4:17:13 PM

I just solved #4 on the practice test and got an answer of 1003 Kn, anybody else get the same?

9/13/2005 4:43:41 PM

i just finished 1 and got a different answer.
at A, P= 101.8746 kpa
b, P= 101.8715 kpa
c, P= 105.396 kpa

9/13/2005 4:45:08 PM

i'm getting different answers from everyone else:

Pa = 101.879 : just Patm (99.9165) + rho*g*h1(20cm)


how did you get Pb?

9/13/2005 4:48:00 PM

for 3 I solved the way he solved the hw problem that was similar... not sure if it's right but that's how I did it.

(p1-p4)=(p1-p2)-(p2-p3)-(p4-p3)

P1 is point A
P4 is point B
P2 is at h = 20.3/12ft
P3 is the lowest point

then gamma*h

[Edited on September 13, 2005 at 4:48 PM. Reason : n]

9/13/2005 4:48:05 PM

for Pa for the first one isn't it h= .20m + .11m?

9/13/2005 4:49:36 PM

Did anyone actually use the gas constant for problem number 1?

My answers for 1:

A, P=101.855 kPa
B, P=101.852 kPa
C, P =105.375 kPa

Note: I used Patm = 99.898 kPa and g =9.807

9/13/2005 5:01:13 PM

i thought for Pb you have to use P=rho*R*T

is this right: cause with that i get 101.246 kPa at B

9/13/2005 5:11:11 PM

I know Dr. Gould said that we could use our book, but does anybody know if he meant just so we could look at the tables...in other words, can we use the whole book?

9/13/2005 5:13:25 PM

I'm not sure but I sure am going to need to use the center of mass and inertia equ table

I do know that he doesn't want us using the book solely to learn the material while we are taking the test

9/13/2005 5:18:22 PM

for #4, do we have to find the Fr on the vertical side of the gate as well as the angled portion to do a moment equation?

i found the Fr on the gate to be... 1470kN and yp of the gate as 4.305m

anyone else get this?

9/13/2005 5:25:45 PM

when I did # 4 I came up with the Fr=1470Kn and yp= 4.305m as well. I only used the angled portion of the wall and have Emailed him to see if this is correct. Ill post when he tells me something.

9/13/2005 5:28:38 PM

the angled gates arent bad, but we havent dealt with an angled portion and veritcal portion on the same gate, i would think that the horizontal force on the vertical portion would factor in, but the area of the pressure distributions overlap and i'm not sure how that affects it

9/13/2005 5:31:20 PM

DiscProdiG your right. I had forgotten about P=rho*R*T for a gas. I knew the gas constant was in there for a reason. My corrected answers are:

A, P=101.855 kPa
B, P=101.754 kPa
C, P =105.279 kPa

9/13/2005 5:32:17 PM

^

can you give me the equations you used and some numbers so i can see where i'm going wrong on Pc

also i'm getting Patm = 29.5 inHg * 3.387kPa/inHg = 99.9165 kPa not 99.898 like you had above

9/13/2005 5:34:36 PM

I found Patm using 29.5 inHG * 1 atm/ 29.9 in HG * 101,325 kpa/ 1 atm

is that not right?

I got it to be 99.9695kpa

9/13/2005 5:39:04 PM

i just used the conversion in the front 1 inHg = 3.387 kPa

9/13/2005 5:41:44 PM

Pc = Pb + rho*g*h


Pb = 101.754 kPa
rho = 998 kg/m^3
g = 9.807 m/s^2
h = .36 m

Pc = 105.277 kPa

9/13/2005 5:53:44 PM

Has anybody worked #2? I worked it and got 4.25m/s...did anybody else get that

I came up with an equation:

V= (hmgsin(2.15))/(uWL)

u=equals the dynamic viscosity of H2O at 20C

9/13/2005 5:55:06 PM