I@Gyro Gearloose
INote for OP
+solved @user1 @user2...
Dwow
Dinteresting
DI = -KAdt/dx
Dbut dt isn't constant
DI will be dq/dt
Dim not getting the equations
OYum. I started doing it and got a second order DE
Got general solution for one side. Need to solve for other side, input conditions
Got general solution for one side. Need to solve for other side, input conditions
DIm getting three d wale terms
DdT/dx in the RHS
dq/dt in the LHS
dq/dt in the LHS
OLet me try from the beginning actually
Dcan't find how to relate all these
OWait you took dT/dx into consideration? The rod has negligible heat capacity so I assumed its temperature can be ignored.
OHuh
OYeah, then I messed up
Dthe source and sink are at different temperatures so the heat flow should be through the rod as well
OYup tup
O*yup
OCrap I did all of this wrong
D
OGot this and all
OCurve which like charging capacitor in RC circuit for comparison.
OWhich makes sense.
Dtime and temperature ko leke differential format banana hai
but idk what to with the dx
but idk what to with the dx
OSince it should asymptote
Dinteresting
Di see
O$Sa$ is $m{a}s_{a}$
TOpt
Dyeah It does
it'll attain a steady state when T1 = T2
it'll attain a steady state when T1 = T2
OTa = Tb.
T1 and T2 are initial values
T1 and T2 are initial values
OI didn't want to get confused
Doh haan
Dyesyes
DI can't get the equations to work
OWait, won't that temp be (T1+T2)/2 ?
Dwhy so?
OJust checking.
ONvm it depends on initial difference
Di see
OIT ALL HECKING COLLAPSED
OPRACTICALLY EVERYTHING CANCELLED
OFinal answer I got was exponential decay
O$\delta T(t) = (T_2 - T1)e^{-\frac{KA(m{1}s{1}+m{2}s{2})}{m{1}s{1}m{2}s_{2}l}t}$
TOpt
OAll conditions are satisfied
OAfaik
O@hardcoreisdead pls confirm
OBecause I didn't take temp.grad. into consideration
