Question

How do I integrate this equation? http://imgur.com/f7gCS

Answer 1

would you just integrate both sides and then figure out the carnage?

something akin to expanding is my first thought

Answer 2

the right side might become hat/Mcp ... but i got no idea what those letters are spose to represent

Answer 3

1/(T-Tw) can be written as (T-Tw)^(-1)... now i believe Tw is a constant..

so integral of any (ax+b)^n with respect to x is [(ax+b)^(n+1)]/[a(n+1)] plus constant.. which wont be present as the integration is definite..

Answer 4

I have the solution, however I just don't know the steps in between. I am a bit rusty on my integration. Here's the solution if it helps.

Answer 5

lol, well I got the right side :) except for a - prolly

Answer 6

Sorry Tw is just temperature of water T0 is initial temperature

Answer 7

\[\int_{0}^{t}dt\to t-0=t\]

Answer 8

i cant make out what the left side is spose to "be" in order to even try to get it to look like the answer

Answer 9

Me neither, it's driving me insane. I think it's just confusing purely because there's notation rather than numbers.

Answer 10

yeah, its like it was pulled out of thin air and we are spose to simply "know" what it is :)

Answer 11

is Tw constant? and T variable?

Answer 12

or both variable and subnotated from some evil purpose?

Answer 13

u = t-tw

du = 1 -(t'w+tw') ???

i cant even make out what to u sub even if we could u sub

Answer 14

Yes Tw is constant

Answer 15

then maybe:

u = (T-K)

du = 1 dT

du = dT

du/u int up to ln(T-K).

ln(T-K) - ln(To-K) = ln((T-Tw)/(To-Tw))

Answer 16

looks almost doable

Answer 17

-2 1-3 -1(3-1) 2
--- = --- = ------ = --
-4 1-5 -1(5-1) 4

i spose that good

Answer 18

\[\int_{T_o}^{T}\frac{dT}{T-T_w}=-\frac{ha}{Mc_p}\int_{0}^{t}dt\]
\[\Large\left.ln(T-T_w)\right|^{T}_{T_o}=-\frac{ha}{Mc_p}t|^{t}_{0}\]
\[\Large ln(T-T_w)-ln(T_o-T_w)=-\frac{ha}{Mc_p}(t-0)\]
and then it plays out from there

Answer 19

Would it not be at the 3rd step \[\ln (T-T _{w})-\ln(T _{0}-T)\] because we have to sub in the limits?

Answer 20

You sub into "T" ... Tw is a constant; so think of it like say "4"

Answer 21

assume the limits to be numbers like 3 and 5

ln(T-"4") from 3 to 5 is just:

ln(5-"4") - ln(3-"4") for example

Answer 22

lol, I am idiot, makes perfect sense. Seriously need to brush up on integration!!! Thank you for the help, I was putting out my hair for a long time!

Answer 23

:) yw, and good luck