To size a relief valve for a LN2 cryostat I have to know “worse case” boil-off rate when vacuum insulation failed and inner vessel with LN2 is in direct contact with room temperature air at 300K.
So, I am looking for a heat flux value (watt/m2) to LN2 inner vessel when vacuum insulation completely failed and dT = 300-77=223K.
Typical LN2 cryostat has approx. 1″ thick MLI super insulation, then the heat flux can be calculated as conduction through the 1” MLI and dT=223K. q1= ? [W/m2]
In case of No MLI (only vacuum insulation), the calculation is more difficult, since it will be air condensation on the outer surface of the inner LN2 vessel. q2 = ? [W/m2]
I can do all the calculations myself, but since it is a very typical task, then I believe it is a well known number for an expert.
CSA Archive 
Research Center Karlsruhe (Lehmann and his team) did some studies / experiments on vacuum breakdown while I was there – see for typical values for helium:
W. Lehman and G. Zahn, “Safety Aspects for LHe Cryostats and LHe Transport Containers,” ICEC7, London, 1978.
This really highly depends on the number of MLI used in the dewar. Not sure whether it covers nitrogen as well but you can get an idea about the heat loads. I believe I have the data somewhere but need to dig deeper to find it.