Theoretical Max Isp of a Nuclear Saltwater Rocket?

[DareMightyThingsJPL]

I've long been interested in the concept of the NSWR (Nuclear Saltwater Rocket), and their high thrust with high efficiency, what would be called a torch-drive in sci-fi. However, I haven't been able to find their theoretical maximum performance. Sure, I know that in https://path-2.narod.ru/design/base_e/nswr.pdf , they conceptualize a NSWR that can go interstellar, having an Isp of approx. 480,000 seconds. But they limit the rocket with using 90% enriched uranium. Furthermore, there may be fissionable isotopes of different elements that release more energy than U233, that could be used.  So I must ask, what would be the theoretical maximum Isp that a NSWR could achieve?

[farmerben]

The thing to do is figure out the exhaust velocity.  If the ship is mostly fuel its delta V is equal to its exhaust velocity.  It's been a while since I've looked into it but I'd spitball .05c.  

[DareMightyThingsJPL]

6 hours ago, farmerben said:

It's been a while since I've looked into it but I'd spitball .05c.  

.05 percent c, or .05 c?

[farmerben]

5%

[DareMightyThingsJPL]

Ok, Thanks. I will make the next calculations assuming an exhaust velocity of 5% c

Quote

Finding v

v = c * Z

 

c = 299,792,458 m/s

Z = 5%

= 0.05

 

v = 299,792,458 * 0.05

v = 14,989,622.9 m/s

Therefore, v  is approx. 15 million m/s

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Finding I

I = v  / g 

 

v = 14 989 622.9m/s

g = 9.80665m/s (this is Earth's Gravity)

 

I = 14,989,622.9 / 9.80665

I = 1,528,520.2286204

Therefore, I  is approx. 1.5 million seconds of Isp

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In Summation, the theoretical specific impulse of an ideal NSWR, assuming an exhaust velocity of 0.05c, is 1,528,520 s