Imperial Star Destroyer power generation

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Picard578
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Imperial Star Destroyer power generation

Post by Picard578 » Mon Jul 23, 2018 2:18 pm

https://starwars.wikia.com/wiki/Imperia ... _Destroyer
The ship however was not suited towards planetary atmospheres, in which full power was required to stay aloft. Any interruption in the power supply in atmosphere could be catastrophic to the vessel, despite all vital equipment being shielded.[17] The primary power generator on board the ship was large enough relative to the rest of the craft to protrude with a ventral bulb.[5]
What does this say about ISD's power generation capability?

So, by using Darkstar's figures of 34,767,000 / 69,534,000 t, and assuming basically zero altitude (we see assault ships hovering very low in AotC). Now, I had to find formula on 'net, but:

Ship mass: 69.534.000.000 kg
F = ma = 69.534.000.000 kg * 10 m/s^2 = 695.340.000.000 N = 695,34 MW.

So basically minimum power generation is 348 MW (using lower mass figure). Of course, this means that it would be unable to accelerate away from the planet.

Does this look right?

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Re: Imperial Star Destroyer power generation

Post by Mike DiCenso » Thu Jul 26, 2018 2:37 pm

I assume you went intentionally with the low-end estimate for an ISD rather than the upper end number, but then? However, since you went with the upper estimate mass in newtons. At 10 m/s, assuming the low-end mass of 37,767 million tons, F = 347,670,000,000 N or 3.4767e+11 J or 347.6 terajoules/s as the low-end is what I get.
-Mike

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Re: Imperial Star Destroyer power generation

Post by Sothis » Thu Jul 26, 2018 4:48 pm

I was wondering (this Thread made me ponder it) what other ships in the SWverse can truly challenge an ISD?

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2046
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Re: Imperial Star Destroyer power generation

Post by 2046 » Sun Jul 29, 2018 5:20 am

The idea an ISD would need to be running at full power just to "stay aloft" seems… quite odd. For instance, how does it then escape the atmosphere?

Then again, my ISD power generation page does feature a conclusion of 400-500TW max, which is awfully close to Mike's result. That said, however, anti-gravs aren't treated like super-efficient rockets in Star Wars, which is what's suggested by such calculations. If anything, the fact that things are left floating all the time suggests a technology much more kind to the batteries.

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Re: Imperial Star Destroyer power generation

Post by Iscander » Sat Aug 11, 2018 3:13 pm

Unfortunately thrust(force) and power are not so easy to interchange.

I don’t disagree with 2049’s assessment that treating Star Wars repulser lift as rocket is incorrect, but lacking more information on the efficiency of their anti-gravity engines or baring someone actual inventing anti-gravity this is probably the best method we have of approximating it with tech we understand. And we can at least draw and equivalence.

Even give that though it does come with its own problems, like introducing more unknowns into the equation. Notably Specific Impulse, or effective Specific Impulse, since the actual lifting mechanism is in question.

Knowns and assumptions:
Mass = 40 Million tons, treating as metric tonnes (Disney Star Wars comics)
Local gravity = 1 gee = 9.81 m/s^2

For **** Thrust = F = m * a = ṁ_dot * v_exhaust ****
m_dot is the mass flow rate and v_exhaust is the leaving velocity of the thruster exhaust
Isp is related to exhaust velocity by **** I_sp = v_exhaust / g_0 **** so once we select an Isp value, we can solve for m_dot from the previous equation. Then solve for Energy with **** KE = 1/2 m_dot * v_exhaust^2****

For a range of Isp (Specific Impulse) values
Isp1 = 1 sec (effectively moving full own mass each second with engines) 1.92E+12 watts or 460 Tonnes/sec
Isp2 = 21,400 sec (Dual stage ion thruster) 4.11E+16 watts or 9.8 MT/sec *
Isp3 = 30,570,000 sec (Photon rocket) 5.88E+19 watts or 14 GT/sec
I will rule out number 3 out of hand baring incredibly poor efficiency, since the energy to launch from surface to LEO is only 60MJ/kg (573.6 MT) for a 40 Million-ton ISD.

For a rough confirmation of these values, I’ll look at potential energy
PE=mg∆h
Where ∆h is from the time dependent **** d = V_0 * t + 1/2 a * t^2****. For a single second, the ISP would fall 4.095m without its engines firing in a 1gee field.
Per second the ISD will need to overcome 1.92E+12 watts or 460 Tonnes/sec of ∆PE/sec.

Assuming a 100% efficient engine and no transfer losses, you need at least 460 tonnes/sec and an effective Isp of 1.

Let’s look at inefficiency real quick.
A quick web search gives me about 6% losses for transmission and distribution of electrical power in the USA. So, 94% efficient. And a NEMA premium efficiency electric motor runs from 85.5-96.2% so let’s say 90% for arguments sake.

That knocks up total power to around 544 tonnes/sec for staying aloft.

Starting to feel like Disney may have gone overboard on the Nerf bat since they took over, because GJ turbolasers are starting to look more reasonable than kT. Between this, Rebels, and the underwhelming power of the Dreadnought assault cannon in TLJ.


* Had to correct the typo 4.11E+11 watts to the correct value 4.11E+16 watts
Last edited by Iscander on Wed Aug 15, 2018 2:04 am, edited 1 time in total.

Iscander
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Re: Imperial Star Destroyer power generation

Post by Iscander » Sun Aug 12, 2018 2:58 pm

Now that is predicated pretty heavily on a very high energy efficiency since everything from ISDs to droid remotes have a repulsor lift and running out of fuel never seems to be a concern for any of them. At least while hovering.

Now it could also be true that it is a tech that doesn't scale up as efficiently and the power requirements for the ISD would be much higher.

But the above hits a lower limit, and also lays out a range of other options dependent on the engine operation.

Picard578
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Re: Imperial Star Destroyer power generation

Post by Picard578 » Fri Aug 17, 2018 11:48 am

Mike DiCenso wrote:
Thu Jul 26, 2018 2:37 pm
I assume you went intentionally with the low-end estimate for an ISD rather than the upper end number, but then? However, since you went with the upper estimate mass in newtons. At 10 m/s, assuming the low-end mass of 37,767 million tons, F = 347,670,000,000 N or 3.4767e+11 J or 347.6 terajoules/s as the low-end is what I get.
-Mike
I calculated both low (348 MW) and high (695 MW) ends.
2046 wrote:
Sun Jul 29, 2018 5:20 am
The idea an ISD would need to be running at full power just to "stay aloft" seems… quite odd. For instance, how does it then escape the atmosphere?

Then again, my ISD power generation page does feature a conclusion of 400-500TW max, which is awfully close to Mike's result. That said, however, anti-gravs aren't treated like super-efficient rockets in Star Wars, which is what's suggested by such calculations. If anything, the fact that things are left floating all the time suggests a technology much more kind to the batteries.
I screwed up a bit there. I calculated maximum power requirement of 695 MW, which compared to 400 TW (or 400.000.000 MW) is 0,0002%. If anything, power requirement I calculated might be too low.

359
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Re: Imperial Star Destroyer power generation

Post by 359 » Sun Aug 19, 2018 7:33 pm

It's an interesting note on power, however you cannot directly calculate power from thrust like that.
The units for Watts = Newtons * Meters / Seconds. There needs to be some form of displacement over time to use kinematics, not just would-be displacement, but the thing must actually move. In conventional thrusters we'd measure the displacement of the engine exhaust and use it's mass, acceleration, and displacement to calculate power for a hovering object.

A while back I did a similar analysis for acclamator-class transports here. But there I measured the motion of the ship to get engine output power on the order of 6.7 to 8.4 TW for the transport.


EDIT: Somehow missed Iscandar's excellent thrust analysis when writing this...

359
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Re: Imperial Star Destroyer power generation

Post by 359 » Sun Aug 19, 2018 7:41 pm

Iscandar wrote:For a rough confirmation of these values, I’ll look at potential energy
PE=mg∆h
Where ∆h is from the time dependent **** d = V_0 * t + 1/2 a * t^2****. For a single second, the ISP would fall 4.095m without its engines firing in a 1gee field.
Per second the ISD will need to overcome 1.92E+12 watts or 460 Tonnes/sec of ∆PE/sec.
You'd probably want to take the limit: lim(∆t->0)∆h to get an accurate value, using a whole second for the interval is going to massively inflate the result.

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