Okay I tried to relate kinetic energy of the sword with the temperature needed to boil blood (which would mean that the blood boils off the sword as it's cutting through the victim, and hence no blood on the sword). I don't know if I did this correctly, as I couldn't find a nice way to relate kinetic energy and temperature, but I found that the sword would have to be going 409 m/s (or 914 mph).
Edit: Here are my equations used...
$<KE> = <\frac{1}{2}mv^2> = \frac{3}{2}kT$
Which is average kinetic energy equals the average of one half the mass of an iron atom times its velocity squared. Then that equals three halves of the Boltzmann constant times temperature. I used:
- $m$ = 9.27E-26 kg (atomic mass of iron (since swords are mostly iron))
- $k$ = 1.38E-23 J/K (Boltzmann constant)
- $T$ = 374 K (temperature at which water boils plus a little bit to account for the salt in blood which raises its boiling point to about 374 K)
Plug everything in and solve for $v$. That yields $v$ = 409 m/s. To see what that compares with in terms of the sword as a whole's kinetic energy, I used the kinetic energy equation again, only this time using this velocity times the sword's mass ($m$ = 1.4 kg, mass of a katana). I got 117,000 Joules of energy. Which feels kind of low to me - that's like a car hitting you on the road - but maybe I'm not relating these equations correctly.