I've got a question which is more related to DEMO I guess than ITER, because the latter is not meant to be 'operational'. However, let's say ITER *is* a Q=10 device (that's the goal;
https://www.iter.org/sci/Goals) so even if the breeder blanket technology is not present, should it not be a viable device for proving the thermal power can be extracted?
There appears to be a conflict in the goals where it is designed to operate for 400 to 600s, but it will take around 600s for the US proposed lithium lead breeder blanket to melt. Once the pulse stops, it'll freeze again and no thermal power can be circulated out of it?
So my questions are; if ITER is to be a Q>10 device and its design was taken forward as a power source, and a viable breeder blanket put around it, how would that breeder blanket work if it takes as long to melt the blanket coolant than the predicted pulse length? If the time to the next pulse was short, then obviously it could continue with then-melted coolant, so the next question is; how long would it take between pulses to reset a tokamak for the next pulse? I've not seen repetition rate mentioned as a goal for ITER.
Here's a calculation on how long the blanket takes to come up to operating temperature, please let me know if it is in error;
ITER's goals are 500MW for up to 600s
ITER will have a surface area of around 700m^2 (assume R=6m and r=3m)
Lithium has a specific heat capacity of 24.86J/mol/K, density of 0.534g/cm3, melting point of 180C and heat of fusion of 3kJ/mol. [Data from
https://en.wikipedia.org/wiki/Lithium]
So, to take one cc of lithium from room temperature (say 20C) to a suitable temperature for a steam cycle, say 500C, would take; (480K * 24.86J/mol/K * 0.534g/(7g/mol)) + (0.534g/(7g/mol) * 3kJ/mol) = 910J + 228J = 1150J.
Lead has a specific heat capacity of 26.65J/mol/K, density of 11.34g/cm3, melting point of 600C and heat of fusion of 4.77kJ/mol. [Data from
https://en.wikipedia.org/wiki/Lead]
So, to take one cc of lead from room temperature (say 20C) to a suitable temperature for a steam cycle, say 500C, would take; (480K * 26.65J/mol/K * 11.34g/(207g/mol)) + (11.34g/(207g/mol) * 4.77kJ/mol) = 700J + 261J = 961J.
Details of the US test blanket module for ITER are given here;
http://www.telegrid.enea.it/Conferenze/SOFE05/DATA/05_11.PDF
It is a 41.3cm thick blanket of lithium lead eutectic at 17% Li and 83% lead.
This would therefore be approx 41.3cm * 700m^2 = 2.9E8 cm3, so 49E6cm3 of lithium and 240E6cm3 of lead.
Total energy to heat this blanket up, if it were to surround the ITER vessel, would be (49E6cm3 * 1150J/cm3) + (240E6cm3 * 961J/cm3) = 56E9J + 230E9J = 286E9J.
So, the total run time for ITER to melt its lithium lead blanket to molten and provide coolant and thermal output would be 286E9J/500MW = 572s.