# need help...forgotten the basics

• May 29th 2007, 06:50 PM
hks
need help...forgotten the basics
i am trying to solve this [t is time and T is temperature] to find the time t,

6450*Pi*(0.00185^2)*(0.0625)*836
dt = --------------------------------------------- dT
(1.2^2)*7.5 - 108*Pi*0.00185*0.25*T

between the limits 0 and 180degC

i have tried & got this answer, t = 3.6236 ln (10.8 - 0.156923*T)

and when solving between the limits, I am getting error because of ln (10.8-0.156923(180)) which is negative.

can someone help me:(
• May 29th 2007, 06:55 PM
ThePerfectHacker
Your equation is a mess. Type it more neatly.
• May 29th 2007, 07:14 PM
hks
sorry, hope this is more readable

dt = (3.6236)/(10.8 - 0.156923T) dT

thanks
• May 29th 2007, 07:27 PM
ThePerfectHacker
Quote:

Originally Posted by hks
sorry, hope this is more readable

dt = (3.6236)/(10.8 - 0.156923T) dT

thanks

Ignore the numbers,
Say, $a,b,c>0$ and we have:

$dt = \frac{a}{b-c\tau }d\tau$

I hate the subhuman differencial notation, let me write instead,

$\frac{dt}{d\tau}=t' = \frac{a}{b-c\tau}$

Now integrate both sides,

$\int t' d\tau = \int \frac{a}{b-c\tau} d\tau$

$t = -\frac{a}{c}\ln |b-c\tau|+C$

Now just evaluate this at $\tau = 180$ and then subtract $\tau =0$. (You can ignore the constant).

Once you do that you can substitute the true values for $a,b,c$.