FFS Battery Build

[jstaley]

Either I'm having a senior moment or am a Plank. Could someone provide an example of a Battery (TNE FFS pg 66). I'm attempting to design a TL-10 electric vehicle with a .335mw powerplant. I want to give it an 8-hour duration but cannot figure out its stats. Please provide the steps so I can write them down and not embarrass myself again.

 

[AnotherDilbert]

At TL-10 the output is 0.8 MW for one m³. You need 0.335 MW so 0.335 / 0.8 = 0.419 m³.

This battery will drain in 1 hour, but you want 8 hours: Checking Discharge Rate table, there is no output factor for 1 to 10 hours, so you simply need a battery eight times the 1 h battery, hence 8 × 0.419 = 3.35 m³.

The battery should be 3.35 m³, 3.35 × 2 = 6.7 tonnes, and 3.35 × 0.003 = MCr 0.01.

Clear enough?

 

[jstaley]

At TL-10 the output is 0.8 MW for one m³. You need 0.335 MW so 0.335 / 0.8 = 0.419 m³.

This battery will drain in 1 hour, but you want 8 hours: Checking Discharge Rate table, there is no output factor for 1 to 10 hours, so you simply need a battery eight times the 1 h battery, hence 8 × 0.419 = 3.35 m³.

The battery should be 3.35 m³, 3.35 × 2 = 6.7 tonnes, and 3.35 × 0.003 = MCr 0.01.

Clear enough?

Can't seem to wrap my head around the Discharge Rate. Self-Check
If I change the Duration to 10-hours I get the x.1 modifier
.335mw/.8 =.419m3
.419m3 x10=4.19m3
x.1 mod =.419m3
.419m3 x2=8.4t
.419m3 x0.003=MCr 0.01

Yes/No
OTOH that's a big battery for a range truck. Need to look at a fuel cell.

 

[jstaley]

Can't seem to wrap my head around the Discharge Rate. Self-Check
If I change the Duration to 10-hours I get the x.1 modifier
.335mw/.8 =.419m3
.419m3 x10=4.19m3
x.1 mod =.419m3
.419m3 x2=8.4t
.419m3 x0.003=MCr 0.01

Yes/No
OTOH that's a big battery for a range truck. Need to look at a fuel cell.

Can't seem to wrap my head around the Discharge Rate. Self-Check
If I change the Duration to 10-hours I get the x.1 modifier
.335mw/.8 =.419m3
.419m3 x10=4.19m3
x.1 mod =.419m3
.419m3 x2=8.4t
.419m3 x0.003=MCr 0.01

Yes/No
OTOH that's a big battery for a range truck. Need to look at a fuel cell.

Still screwed it up...I think it should be .8t and MCr 0.001.

 

[AnotherDilbert]

Can't seem to wrap my head around the Discharge Rate. Self-Check
If I change the Duration to 10-hours I get the x.1 modifier

Ok, 10 h is slightly simpler.

.335mw/.8 =.419m3

Yes!

.419m3 x10=4.19m3
x.1 mod =.419m3

No, you did the 10 h compensation twice.

The 10 h output rate is 0.1, so a .419 m³ battery would deliver 0.419 × 0.8 × 0.1 = 0.0335 MW.
To get the size of the needed battery you have to reverse that calculation: You need 0.335 MW for 10 h, each m³ delivers 0.8 MW for 1 h or 0.8 × 0.1 = 0.08 MW for 10 h, so you need 0.335 / 0.8 / 0.1 = 4.19 m³ of battery.

.419m3 x2=8.4t
.419m3 x0.003=MCr 0.01

0.419 × 2 = 0.838 tonnes.

OTOH that's a big battery for a range truck. Need to look at a fuel cell.

The battery is calculated for full throttle, you can probably get by with a smaller battery, but still very large.
A fuel cell is probably a better alternative.



A TL-10 fuel cell has an output of 0.5 MW for each m³.
You need 0.335 MW, so 0.335 / 0.5 = 0.67 m³ with a mass of 0.67 tonnes and a cost of MCr 0.013.
Fuel consumption is 0.3 kl/h = 0.3 m³/h for each MW, so 0.3 × 0.335 MW × 10 h = 1 m³ for 10 h.

 

[jstaley]

Ok, 10 h is slightly simpler.


Yes!


No, you did the 10 h compensation twice.

The 10 h output rate is 0.1, so a .419 m³ battery would deliver 0.419 × 0.8 × 0.1 = 0.0335 MW.
To get the size of the needed battery you have to reverse that calculation: You need 0.335 MW for 10 h, each m³ delivers 0.8 MW for 1 h or 0.8 × 0.1 = 0.08 MW for 10 h, so you need 0.335 / 0.8 / 0.1 = 4.19 m³ of battery.


0.419 × 2 = 0.838 tonnes.


The battery is calculated for full throttle, you can probably get by with a smaller battery, but still very large.
A fuel cell is probably a better alternative.



A TL-10 fuel cell has an output of 0.5 MW for each m³.
You need 0.335 MW, so 0.335 / 0.5 = 0.67 m³ with a mass of 0.67 tonnes and a cost of MCr 0.013.
Fuel consumption is 0.3 kl/h = 0.3 m³/h for each MW, so 0.3 × 0.335 MW × 10 h = 1 m³ for 10 h.

Thank You, makes sense now. I guess I just needed to see an example. Off to try my hand at BattleDress-12.