On 02/09/2011 10:22 AM, Jed Rothwell wrote: > Stephen A. Lawrence <sa...@pobox.com <mailto:sa...@pobox.com>> wrote: > > > The energy produced was apparently *exactly* what was needed to > boil away the input water -- no more, no less. > > And *that* is strange. > > > Nope. That's steam at 1 atm. It never gets any hotter than just above > boiling.
NO. Jed, I can't believe you're making this mistake! That's *exactly* like saying oxygen can't get any hotter than -183C (its boiling point) unless you raise the pressure above 1 atmosphere! There is nothing magic about water vapor -- it's just another gas, and it can exist at 1 atmosphere at any temperature above its boiling point. Increase its temperature while holding the pressure steady, and its density drops, that's all. Now, if you boil water in an /open/ boiler with a /submerged/ heating element, the temperature of the steam will never go above 100C (give or take a degree). The temperature of the steam in that case is pegged to the temperature of the water through which it must pass, and the temperature of the water is fixed at boiling, unless you close the boiler and raise the pressure. But in this case the heating element (the walls of the tube) is only submerged until the water boils. After that, the steam is in direct contact with the heating element, and no longer in close contact with liquid water, and there is nothing to keep its temperature from rising well above boiling. The geometry of the water jacket may be more complex than a simple tube but the same argument applies: Once the water has boiled away and the inner wall of the water jacket is in direct contact with the steam, the steam temperature is no longer fixed at boiling. > > >> It comes out faster with more enthalpy if the pump adds more >> energy to it. > > THAT'S THE POINT! > > If the reactor produced even a few hundred watts more than what > was needed to vaporize the water, the temperature of the steam > would have been substantially higher than boiling. > > > Nope. It would just move faster out of the end of the hose, as I said. > You have to raise the pressure to make the temperature go up. Sorry, that is completely wrong. Look, if it's moving faster out of the end of the hose, but it's the same number of moles of steam (which it *must* be, because the pumping rate is fixed), then the steam must be more "spread out", right? It must be taking up more volume per mole. Volume coming out is the integral of the flow rate, flow rate is the speed of the steam times the area of the hose opening; ergo, if it's going faster, you've got a larger volume coming out. Pressure is fixed, number of moles are fixed, and the volume has increased. What's that tell us? PV = nRT; let's solve for T. T = PV/nR 'n' is fixed, 'R' is a constant, 'P' is fixed, 'V' has increased -- so the temperature has also increased. QED. > > - Jed >
