Temperature problem in fluids mechanics

PythagoreLove
Messages
19
Reaction score
0

Homework Statement


Assume the temperature of the exhaust in an exhaust pipe can be approximated by
T=T0(1+ae-bx)[1+c cos(\omegat)]
T0=100oC,
a=3,
b=0.03m-1,
c=0.05,
\omega=100 rad/s.

If the exhaust speed is a constant 3 m/s, determine the time rate of change of temperature of the fluid particle at x=0 and x=4 m when t=0.

Homework Equations


We know the equation of the temperature in function of the time and the position
T=T0(1+ae-bx)[1+c cos(\omegat)]




The Attempt at a Solution


We know that the time rate of change of temperature of the fluid particle is dT/dt

\frac{dT}{dt}=T0(1+ae-bx)[-c \omega sin(\omegat)]

When t=0, dT/dt=0... unfortunately that isn't the answer, since I don't use the exhaust speed (and have no idea how I could use it).

Thank you
PytLove
 
Physics news on Phys.org
What if you treated x and t as a variables...
 
Seems like a wonderful idea and totally worked, that problem was so different from the others I've done in fluid mechanics... I have no idea that I needed to use x(t) in my dT/dt.

Thank you Lawrence

PytLove
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top