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No, it doesn't mean that the photons don't go any further. It means that when the temperature gradient inside the star reaches a threshold, the gas becomes convectively unstable. Heat is transferred more efficiently by moving parcels of gas than by transferring photons from hotter regions to cooler regions. So, the photons continue to diffuse outwards, but ...


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Don't think about individual single photons bouncing around. The energy (heat) diffuses slowly, exchanging energy with neighboring particles by various means (smashing into each other and radiation that only makes it as far as the next particle before being absorbed). Different photons are emitted and absorbed over and over, as well as physical collisions ...


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A formula such as this (being a analog to parallel electrical resistors, as pointed out in the comments) can get a little more complicated when different areas are involved. However, I think that the author of your formula restricted the discussion to one single heat transfer area in order to avoid dealing with that. The heat transfer rate originally starts ...


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Other small effects that you may be ignoring are heat conduction to the air and temperature drop along the pipe. If the fluid is a gas, its temperature may drop by some small amount due to pressure drop along the pipe. The fluid temperature will also drop due to the heat transfer that you calculate. An old reference for these effects is Chapter 8 of ...


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The rate of temperature change will be the power per unit mass times the specific heat. So if you have a certain mass of water $M$ flowing per second, at a velocity $v$, losing $\Delta P$ pressure per second, then work done is $v\Delta P A$ and $A = \frac{M}{\rho v}$ . Then with a heat capacity $c$ (about 4.2 kJ/kg/K for water), and the relationship between ...



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