I'm getting really tired of this. Peter, you didn't read, or didn't understand, what I wrote.
You don't seem to understand the fundamental point, which is that the rate of boil-off is being determined by the pump, with no feedback from the reactor. The flow rate is fixed and 100% of the water is boiled to steam. If the reactor were generating 10% more power than needed to exactly boil off the water, just where do you think that excess power would go? On 02/09/2011 11:02 AM, Peter Gluck wrote: > Jed is right, it is an open system and even if the surface of heating > is at > 300 C, the time of contact is short and the steam cannot be overheated > much. > > On Wed, Feb 9, 2011 at 5:50 PM, Stephen A. Lawrence <[email protected] > <mailto:[email protected]>> wrote: > > > > On 02/09/2011 10:22 AM, Jed Rothwell wrote: >> Stephen A. Lawrence <[email protected] <mailto:[email protected]>> 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 >> >
