Been studyin for the past couple days for the test tomorrow, but if anyone had hw 7 or 8 I could look at, I'd be grateful. I turned in all the hws, but only having the solutions available for 5 days out of the entire semester is a pain in the ass when it comes to going back over material that wasnt understood. Also, if anyone is studying for this later tonight, let me know. I doubt many people are left after the exemptions he gave though.
4/29/2007 5:04:48 PM
fuck dude, unless you are close it isnt worth itill check and see if i have 7 and 8it was bullshit how he couldnt put that stuff online
4/29/2007 5:21:16 PM
Yeah, I thought it was pretty messed up, at least go over it in class or let me look at it for more than the 15 minute allowance. I can look at it after I'm just done with the assignment and understand what I did wrong, but now a month or two later I have nothing to look back on. If you're within I'd say 60 minutes drive of south raleigh I'm willing to come out to where ever. At this point I've studied my ass off up to this section of the notes/hw. Thanks, and just let me know if you can find them. I'll check back in a bit.
4/29/2007 5:44:53 PM
i have 7 and 8 but i dont know why 8 is wrong
4/29/2007 6:01:24 PM
Well I got a 90 on 8, but I was missing a lot of data from the excel sheet, and my analysis of the house was overcomplicated. On 7, for the cfm flow requirement problem, how did you calculate out the cfm based on square footage?
4/29/2007 6:18:34 PM
^ was that 7.2if so i did this:(15 cfm/person)(30 people) = 450 cfmand(50 people/1000 ft^2)(15 cfm/person)(500 ft^2) = 375 cfm
4/29/2007 6:51:25 PM
Yeah, 7.2. Ok, I got the first part right, but I broke the second part down completely the wrong way. Between the wrong answer and the same notes as out of the book, I couldnt tell what the right procedure was. Last question, and thanks for helpin me, on hw 6 for problem 2, what was T2, after the cooling coil. I assumed T2=8Celsius, same as the Tcoil for the problem, but after I got it back realized that would mean there would be condensate to account for.
4/29/2007 7:49:31 PM
T2 = 21.8159 C
4/29/2007 8:14:15 PM
Alright thanks man, helped me out tremendously.
4/29/2007 8:53:01 PM