If there is a significant quantity of water droplets, they will fall until all what remains is the fluid stream in the hose is vapor. Considering a stream of 10m/s, 1.5g/s out of the hose, with, 5cm2 of area, the pressure inside above 1atm the chamber is P=F/A=(1.5*10(-3)*10)/5*10(-4)=(1.5*10(-2)*10(4))/5=1.5*20=30N/m2 or and increase of 3*10(-4) atm. Hardly more than 0.1 degrees, if that much.
> You don't expect water droplets above the boiling point. The temperature of > the mixture of steam and droplets will be *at* the boiling point. > The actual boiling point inside the conduit will be slightly elevated > because of a slight increase in pressure. Rossi emphasizes that the pressure > is at atmosphere inside the reactor, but in fact it must be slightly higher, > or there would be no flow of the fluid. The pressure difference, flow rate, > and tube geometry are related by a simple formula, and reasonable estimates > indicate an elevation in the bp of a degree or so is easily plausible.
