First order ordinary differential equation

[KillerZ]

Homework Statement

I haven't done ODEs in a few years and I am trying to do this equation:

m_{Hg}C_{p,Hg}\frac{dT_{Hg}}{dt} = Q

Q = hA(T_{air} - T_{Hg})

T_{Hg}(t = 0) = 20

I need to find T_{Hg}(t=590)

Homework Equations

The Attempt at a Solution

h, A, m_{Hg}, C_{p,Hg}, T_{air} are all constants

\frac{dT_{Hg}}{dt} = \frac{hA(T_{air} - T_{Hg})}{m_{Hg}C_{p,Hg}}

\frac{m_{Hg}C_{p,Hg}}{hA(T_{air} - T_{Hg})} dT_{Hg} = dt

\int\frac{m_{Hg}C_{p,Hg}}{hA(T_{air} - T_{Hg})} dT_{Hg} = \int dt

\frac{-m_{Hg}C_{p,Hg}}{hA}\int\frac{1}{(-T_{air} + T_{Hg})} dT_{Hg} = \int dt

\frac{-m_{Hg}C_{p,Hg}}{hA} ln|T_{Hg} - T_{air}| = t + c

ln|T_{Hg} - T_{air}| = \frac{t + c}{\frac{-m_{Hg}C_{p,Hg}}{hA}}

T_{Hg} - T_{air} = e^{t/\frac{-m_{Hg}C_{p,Hg}}{hA}}e^{c/\frac{-m_{Hg}C_{p,Hg}}{hA}}

e^{c/\frac{-m_{Hg}C_{p,Hg}}{hA}} = c

T_{Hg} - T_{air} = ce^{t/\frac{-m_{Hg}C_{p,Hg}}{hA}}

T_{Hg} = ce^{t/\frac{-m_{Hg}C_{p,Hg}}{hA}} + T_{air}

T_{Hg}(t = 0) = 20

T_{air} = -7

h = 3.9

A = 0.00176

C_{p,HG} = 139.0908

m_{Hg} = 0.05

20 = ce^{0/\frac{-(0.05)(139.0908)}{(3.9)(0.00176)}} + (-7)

c = 27

T_{Hg} = 27e^{t/\frac{-m_{Hg}C_{p,Hg}}{hA}} + T_{air}

T_{Hg}(t=590)

T_{Hg}(t=590) = 27e^{590/\frac{-(0.05)(139.0908)}{(3.9)(0.00176)}} + (-7)

T_{Hg}(t=590) = 8.0822

 

[LCKurtz]

Man, you must like typing subscripts. Why not call your dependent variable T and write the equation in the form

\frac{dT}{dt} = k(T_a - T)

in the first place and put the constants in for k at the end? It would be much easier to read, not to mention typing it all in. Anyway, it looks like you solved it OK, with the caveat that I didn't check all the details.

 

[hunt_mat]

I think you made an error in using your initial condition (using the notation of LCKurtz) you had:
<br /> -\ln |T_{a}-T|=kt+c<br />
Now if T(0)=T_{0}, then I would use that at this point here to tell you that -\ln |T_{a}-T_{0}|=c and your equation becomes:
<br /> -\ln |T_{a}-T|=kt-\ln |T_{a}-T_{0}|<br />
From here I think it is just algebra to obtain the correct answer.

 

[KillerZ]

Man, you must like typing subscripts. Why not call your dependent variable T and write the equation in the form

\frac{dT}{dt} = k(T_a - T)

in the first place and put the constants in for k at the end? It would be much easier to read, not to mention typing it all in. Anyway, it looks like you solved it OK, with the caveat that I didn't check all the details.

OK doing it with the substitutions:

k = \frac{hA}{m_{Hg}C_{p,Hg}}

and

T = T_{Hg}

\frac{dT}{dt} = k(T_a - T)

\frac{1}{k(T_a - T)}dT = dt

\frac{-1}{k}\int\frac{1}{(-T_a + T)}dT = \int dt

\frac{-1}{k}ln|-T_a + T| = t + c

ln|-T_a + T| = -kt + c

-T_a + T = e^{-kt + c}

T = ce^{-kt} + T_a

Then initial conditions T(t=0) = 20

20 = ce^{-k(0)} + (-7)

27 = c

T = 27e^{-kt} + T_a

 

[hunt_mat]

This line is incorrect:
<br /> T=ce^{-kt}+T_{a}<br />