Has anyone worked on the HW for 310. If so do you have any idea on how to do the last problem he assigned.

1/22/2006 5:09:19 PM

havent gotten that far yet.

What did you get for the other ones

1/23/2006 1:54:49 PM

other MEs can help you if you post the question itself

1/23/2006 4:31:56 PM

what kind of answers r u guys gettin for this HW set?

1/24/2006 6:54:10 PM

the answers are on the book's website as noted on the front cover

go to http://www.wiley.com/college/incropera

[Edited on January 24, 2006 at 7:01 PM. Reason : R]

1/24/2006 7:00:49 PM

where exactly do u see answers on that site u linked? i just clicked every subsection on the left and there is nothing.

1/24/2006 7:12:29 PM

if it's still the 5th edition, here are the answers:
http://bcs.wiley.com/he-bcs/Books?action=index&itemId=0471386502&bcsId=1737

if they're on the 6th edition now, i don't think they have the answers

1/24/2006 7:24:33 PM

5th edition --> Students --> End of chapter Answers

^or that

1/24/2006 7:48:17 PM

anyone actually get 70? i worked the part a like it was done in the problem session and i didnt get the right answer. not sure what's wrong.

1/25/2006 2:12:31 PM

i got it, but i didnt go to the problem session

1/25/2006 4:08:33 PM

how were u able to get it? mainly interested in part a; the # i get is too high and i dont know what i'm doin wrong.

1/25/2006 4:13:50 PM

im about to leave but you can probably figure it out with this

.9*700 - q"rad - q"conv = q"u

remember to conver the radiation temperatures into Kelvin, i would have gotten it the first time if i had done that, oh well now i know better

if you need more help ill be back in a few hours

1/25/2006 4:44:26 PM

ok that's the equation i used. that radiation exchange term is messing me up i think; i dont know what to do with that. i have my Tglass = 30 and Tair = 25 but i dont know what to do with the -10degreesC term.

1/25/2006 5:19:24 PM

the -10 is used for the radiation with the sky (convert to kelvin)

the 25 is used for the convection (leave in celcius)

1/25/2006 5:24:12 PM

^ yeah what he said

basicly, i think it is saying that the solar panel will continue to operate properly when the surrounding air is -10*C; the air inbetween the glass and the absorber is 25C, and the glass is 30C

remember to do radiation in Kelvin, I spent so long on this problem and i was swearing at it, then i was like, damnit, and made it in kelvin and it work the same way i had originally done it, damn

1/25/2006 7:20:06 PM

how do you do part b for number 70?

I'm not sure what it is asking and what to do with the flow rate.

1/26/2006 12:05:34 AM

use your flow rate to find the change in Temp

im cant remember if i saw this in the book or in the notes but you want to use the formula that is

dudot=mdot*Cp*dT
and dudot is measure in W
so to get dudot, take your q"u*A =>(W/m^2)*m^2

and you end up with A*q"u=mdot*Cp*dT

solve for dT

1/26/2006 12:39:09 AM

bttt for this hw, I missed the problem session

2/1/2006 9:01:17 PM

2.13

a) how do you solve dT/dx = -(1/k*Ao)*q(x)*exp(-ax) ? you could integrate but there seems to be twop functionms dependent on x: q(x) & exp(-ax).

2/1/2006 10:53:59 PM

qx just means the heat flow rate in the x direction right?

not that its a function of x

2/1/2006 11:08:53 PM

well in the problem they write it as q_x(x), so it seems like heat rate is a function of x

I am also not sure what to do in this problem, maybe multiply by dx, so you get q(x)dx = -kAoe^(ax)dT

and then integrate?

2/1/2006 11:24:20 PM

ehhhhh

n/m

[Edited on February 2, 2006 at 1:10 AM. Reason : /]

2/2/2006 1:08:22 AM

yeah, i'm having trouble with homework 4. i couldnt make it to the problem session.

how do you get started on 3.4 (b) ??

[Edited on February 14, 2006 at 9:45 PM. Reason : n/m]

2/14/2006 9:22:27 PM

>.<

[Edited on February 14, 2006 at 10:49 PM. Reason : .]

2/14/2006 10:32:39 PM

bttt. i have lab during the problem session so i can never go

can anyone tell me how to find the total heat loss from the pipe in problem 3.52? i got Ts,a and Ts,b by just finding the total heat transfer through both materials by themselves, but how do you get the total heat loss?

also, what would the thermal circuit look like? thanks

2/21/2006 8:00:22 PM

total heat loss is just qa + qb

thermal circuit has RA and RB in parallell

2/21/2006 10:15:13 PM

that's what i thought, but im not getting the same answer that's on the solutions website

2/21/2006 10:39:40 PM

hey nerdchick, what'd you get for Th when you solved qh = qi + qh? for 3.46

2/22/2006 6:59:46 PM

I didn't bother to solve for Th, I was like fuck that

2/22/2006 7:54:08 PM

i'm going to do the same

2/22/2006 8:09:53 PM