It is an interesting issue, indeed.
I have nevertheless some questions.
The current will be constant?
The time up to the maximum temperature reaching will be short-time [0-2 sec.] or hours?
The water flows along the cylinder, I think.
If it is short time the phenomenon is almost adiabatic so the water does not play any function.
From IEEE-80/2000 form.37
I=Amm^2*SQRT(TCAP/time/alpha/ro/10^4*ln((Tf+Ko)/(Ti+Ko)))
alpha=0.00403
ro=2.86
alpha=0.00403 TCAP=2.56
ro=2.86 Ko=228
If the phenomenon duration will be long but still not infinite
Pinput*time=Qtemp.raise+Poutput*time
Pinput= I^2*Ro*(1+alpha*(T-To))
NOTE: alpha is defined only for 0-100 dgr.C for 700 dgr. Will be less. The formula could contain more terms relevant for an elevated temperature.
Poutput=Pconv+Prad+Pcond.
Pconv depends on water properties
Pconv=h*surf.area*(T-Ta)
h=Nu*k/D
Nu=Nusselt factor k=water thermal conductivity D=AL cylinder diameter.
Nu=C*Re^m*Pr^(1/3)
For an isothermal long horizontal cylinder, as Hilper suggests.
See:
and Forced Convection Experiments-2.doc
Re=Reynolds number
Re=w*D/niu
w=water speed [2-3 m/sec ???] niu=cinematic viscozity [for water 40 dgr.C=0.66/10^6 m^2/sec]
Pr= Prandtl number [for water 40 dgrC=4.35]
Prad=eps*sigma*(Tf^4-Ta^4)
eps[for oxided AL]=0.1-0.2 sigma=Stefan-Boltzmann constant for black body.
Tf and Ta in Kelvin dgr.
Pcond may be neglected for fluids.
Qraise.temp=Tcap*Vol*(Tf-Ti)
Vol=cylinder volume
Tcap=thermal capacity of AL [2.6 J/cm^3/dgr.C]
The calculation normally include an integration but an average calculation could be enough.