Factorio Forums

These recipes I found on the factorio wiki, haven't checked them ingame.
10 uranium ore give 0.007 U235 and 0.993 U238.
40 U235 + 5 U238 give 41 U235 + 2 U238, so for the calculations I use 3 U238 gives 1 U235.
10 iron plate + 1 U235 + 19 U238 give 10 Uranium fuel cell.
1 Uranium fuel cell give 1 used up uranium fuel cell.
5 Used up uranium fuel cell give 3 U238.

To start the enrichment we need on average 40/0.007*10 = 57143 uranium ore.

Now the process when we use the neclear power.
10 iron plate + 1 U235 + 19 U238 -> 10 Uranium fuel cell -> 10 used up uranium fuel cell -> 6 U238 -> 1 U235 + 1 U238
To burn 10 fuel cells we need 10 iron plates and 16 U238.
So this gives 2000 seconds reactor time (10 fuel cells) for 10 iron plates and 16 U238.
The 16 U238 is made from roughly 1.6 uranium ore.
This gives 57143/1.6*2000 = 71428750 seconds = 19841 hours with one reactor.
If we build big and use 16 reactors, that's still over 1000 hours energy with the uranium ore we need to mine to start uranium power.

Are my calculations correct?

16 U238 require 160 ore, not 1.6

Ah, thanks!

The 16 U238 is made from roughly 160 uranium ore.
This gives 57143/160*2000 = 714287.5 seconds = 198.41 hours with one reactor.

[quote="pieppiep"]
10 iron plate + 1 U235 + 19 U238 -> 10 Uranium fuel cell -> 10 used up uranium fuel cell -> 6 U238 -> 1 U235 + 1 U238
quote]
This part is wrong. 6 U238 -> 1 U235 + 3 U238.

However, you use the correct value in the rest of the calculations.

Nemoricus wrote:This part is wrong. 6 U238 -> 1 U235 + 3 U238.

What I do there is one time 40 U235 + 5 U238 give 41 U235 + 2 U238.
I use 40 U235 still in that part of the system so I have 40 U235 + 5 U238 to make 41 U235 and 2 U238, then give back the 40 U235 to the system and add the remaining 1 U238 I didn't use. Then I have a total of 1 U235 and 3 U238.

the only thing you need to worry about is 238, because 235 can be created by using 238.

now you need:
3* U238 for the enrichement
19* U238 for fuel
and you get:
2* 3* U238 back from recycling

therefor you get 10 fuel cells out of 16 U238 or 1 fuel cell out of 1.6 U238.

1 fuel cell is worth 8GJ therfor 1 U238 is worth 1/1.6*8GJ = 5GJ

1U238 = 5GJ

if you want me the way mor complex calculation on ore, just tell me, however the ratio will be almost the same

that is a bad answer, let me redo the calcultion



1000 Ore = 7 U235 + 993 U238

1.) Using the 7 U235 you need 7*19 = 133 U238 to make 70 fuel cells ===> 7*10*8 = 560 GJ
2.) Now the 70 fuel cells give you 70/5*3 = 42 U238
ATM we have 0 U235 , 993-133+42 = 902 U238 , 560 GJ
3.) 902 U238 will give us 53.0588235294 U235 and 848.941176471 U238 ===> 53.0588235294*10*8 = 4244.70588235 GJ

1000 Ore = 560 + 4244.70588235 = 4804.70588235 GJ
1000 Ore = 7 + 53.0588235294 = 60.0588235294 U235 => 60.0588235294 * 200 = 12011.7647059sec = 3h 20min 12sec burning time

Pls note that the initial resources for the enrichment are not taking into account. furthermore in the very end you wont be able to do the recycling. so this calculation works only in mid game
that would be my guess

pieppiep wrote:

Nemoricus wrote:This part is wrong. 6 U238 -> 1 U235 + 3 U238.

What I do there is one time 40 U235 + 5 U238 give 41 U235 + 2 U238.
I use 40 U235 still in that part of the system so I have 40 U235 + 5 U238 to make 41 U235 and 2 U238, then give back the 40 U235 to the system and add the remaining 1 U238 I didn't use. Then I have a total of 1 U235 and 3 U238.

Oh, I see it now, I made a typo

laku wrote:1000 Ore = 7 +45.1 = 60.0588235294 U235 = 60.0588235294 * 200 = 12011.7647059sec = 3h 20min 12sec burning time

I really don't understand you calculation, so tomorrow I'll read it again to try to understand it again
The 3h20m for 1000 ore is about the same as I calculated.

Ups sry i messed up. now it should be correct

Personally I'd go at the math like this:
1 fuel cell lasts for 200 seconds, or 10 for 2000.

Using only basic enrichment, you get .007 U235 and .993 U238 per 10 ore, or 7 U235 and 993 U238 per 10000 ore.
10 fuel cells takes 1 U235 and 19 U238, and given the above ratio we can disregard the U238 cost and get our values. 7*2000 = 14,000 seconds per 10000 ore. Or ~3.9 hours.

Let's look at the other processing options. We can choose to reprocess the spent fuel cells, 5 spent for 3 U238. Assuming we did that, the cost of 10 fuel cells effectively drops to 1 U235 and 13 U238.

If we use the Kovarex process, we (effectively) turn 3 U238 into 1 U235. Which means we can use 16 U238 per 10 fuel cells if we choose. So we have our initial run of 7U235 + 13 * 7 (91) U238 leaving us with 902 U238. 902 / 16 = 56.375 extra runs, or 63.375 in total. 63.375 * 2000 per 10000 ore is our new value, on average. (35.2 hours)

This assumes you're using just the average, constant value.

The math is quite simple. There are several round about ways to get to the solution. The simplest being to take a look at the conversion ratio, the spent fuel processing, and the Kovarex process.

U-235+19 U-238 = 10 Fuel Cells(FC), FC = 8GJ (I ignore the iron plate as it is irrelevant)
10 Spent FCs = 6 U-238
40 U-235+5 U-238 = 41 U-235+2 U-238 => U-235 = 3 U-238

Substitute the U-235 with the Kovarex ratio.
22 U-238 =10 FCs
Substitute the FCs with the spent FC ratio and calculate the net change in U-238
Net = -16 U-238 => We'll use absolute value of 16 U-238 for the energy calculation since it is more convenient to think of input instead of expenditure.

Thus U-238 = 5GJ. Just count up all of your U-238, convert all of the U-235 into U-238 equivalents, multiply by 5GJ and divide this product into your current energy usage/maximum usage to estimate remaining time.

Alternatively, if you only have uranium ore, then you need to take this average calculation.

10 uranium ore = 0.007 U-235 + 0.993 U-238, so on average 10000 ore = 7 U-235 + 993 U-238, convert U-235 into U-238 equivalents.
10000 ore = 1014 U-238 on average, now compute energy.
10000 ore = 5070 GJ on average, so uranium ore is 507 MJ.

Take 507 MJ and multiply with the ore that you have and divide the product into your energy expenditure, in MW, to estimate remaining time in seconds.

The example of 10000 uranium ore, given 40 MW expenditure, gives a residence time of (10000*507 MJ) /40 MW = 126750 s. Thus it is about 35.208 real hours for game running at 60 UPS.

Edit: 507 MJ per ore is a lower bound assuming reactor has no neighborhood bonus. If you have neighborhood bonus, the value will effectively scale with the bonus, but the residence time is the same unless you throttle the reactor.