found an  equation that figures wire melting current/time relationship.  It appears that the relationship between time and melting is linear.  So, twice the amps, half the time sort of thing.  I can't vouch for the below, but it looks good at a glance, except I don't follow the "Area" number.  I believe it is intended for sizing the wire in a fuse.

Example: 16 gauge copper wire: Tmelt = 1083, Area = 2581 circ mil, Time = 5 sec, Tamb = 25

E= Area in CM (of the wire)
B = Tmelt (of the wire)- Tamb in deg. C
D = 234-Tambient in deg. C
T= time in seconds.
So, E = 2581, B= 1058, D=209, T=5
Then
Ifuse = E* SQRT {<LOG[(B/D)+1]>/(T*33)}
Ifuse = 2581* SQRT {<LOG[(1058/210)+1]>/165}
Ifuse = 2581* SQRT {<LOG(6.04)>/165}
Ifuse = 2581* SQRT {0.781/165}
Ifuse = 2581* SQRT {.00473}
Ifuse = 2581* 0.0688
Ifuse = 178A

6 hours ago, apostolakisl said:

It appears I need two of them since my main power feed from the meter splits to two panels, a 150 and a 200 amp.  The main panel leviton model is considerably less expensive than the sub-panel model.

With only one main panel protector, then an incoming surge will go completely to earth via that path.  The other one not necessary.  But since these things are so inexpensive, then one in each would simply mean even better conductors to earth.  Simply better.

Sub-panel protectors must have an additional protector circuit between neutral and ground.  So that would make it more expensive.  It must also shunt to ground.  But sub-panel neutral and ground are separate.  A main panel protector has neutral and ground already connected together.. That extra protector circuit in a sub-panel protector connects neutral to ground.

3 hours ago, apostolakisl said:

found an  equation that figures wire melting current/time relationship.  It appears that the relationship between time and melting is linear.  So, twice the amps, half the time sort of thing.  I can't vouch for the below, but it looks good at a glance, except I don't follow the "Area" number.  I believe it is intended for sizing the wire in a fuse.

Example: 16 gauge copper wire: Tmelt = 1083, Area = 2581 circ mil, Time = 5 sec, Tamb = 25

E= Area in CM (of the wire)
B = Tmelt (of the wire)- Tamb in deg. C
D = 234-Tambient in deg. C
T= time in seconds.
So, E = 2581, B= 1058, D=209, T=5
Then
Ifuse = E* SQRT {<LOG[(B/D)+1]>/(T*33)}
Ifuse = 2581* SQRT {<LOG[(1058/210)+1]>/165}
Ifuse = 2581* SQRT {<LOG(6.04)>/165}
Ifuse = 2581* SQRT {0.781/165}
Ifuse = 2581* SQRT {.00473}
Ifuse = 2581* 0.0688
Ifuse = 178A

In the electrical distribution field we have experienced that copper conductors can be taken to a low red hot glow on the pole lines and yet  still perform to deliver energy to customers, when occasional crisis situations occur.. With aluminum conductors the resistance to temperature coefficient is exponential and it will evaporate suddenly if abused with  overloads. Each metal or alloy type has it's own thermal co-efficient. This makes metal evaporation specs with high currents very hard to predict exactly.

Conductor damage curves are not linear and overcurrent protection is calibrated or sized to match these curves.

 

Nice - a nerd electrical e-peen contest. You guys are your own best friends. enjoy.