PLAN Catapult Development Thread, News, etc.


taxiya

Brigadier
Registered Member
that sounds right, but than it is unfair to compare this 90% with the steam. The 5% of steam is measured from the reactor's steam outlet (thermal energy) to the shuttle on the launch track. A fair comparison should measure EMLS from the same point, that is certainly not 90%. I don't know if that would be 25% as you calculated, but you are probably close to the truth.
After realizing how you understood the "121/484MJ each", I think 25% would be wrong even if we agree with the definition of efficiency (measuring point).

Regardless the actual figure, the overall efficiency of EM launcher is much higher than steam by the same measurement, there is no doubt about it.
 

gelgoog

Captain
Registered Member
90% efficiency is typically the conversion efficiency of motion to electricity and vice-versa. Think an electric motor or alternator.
I would roughly estimate then than it has 10% loss from the EMALS itself. Linear electric motor.
Another 10% loss from electric to flywheels storage. Another 10% loss from flywheel storage back to electric again.
There might also be some (insignificant) losses from the flywheels with friction and electric transmission losses in the circuits.
The nuclear reactor power plant might be like 45% efficient at converting thermal power to electricity at best.
It depends on reactor core temperature but it is more or less the same in most LWRs. So that is another 55% loss.

Are you guys comparing the catapult performance of the Ford against the Nimitz class? You should notice both carrier families use different reactors. The Nimitz class uses 2x 550 MWt A4W reactors and the Ford class uses 2x 700 MWt A1B reactors.
 
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nlalyst

Junior Member
Registered Member
I think you misunderstood it.

The launch energy for one track is maximum 121MJ. This is provided by the four rotors together. Each rotor (flywheel/disc alternator) can store up to 121MJ at 6000RPM. However, due to the fact that flywheel's power density is lower than requirement, the 121MJ can not be released during the 2-3 seconds launch time frame. So four such rotors are put together to provide that 121MJ. At the end of a launch, each rotor will have 75% energy left unused.
Can you clarify how this fits with the 12 motor-generator flywheel sets that can supply energy to all four catapults?
 

Intrepid

Captain
Can you clarify how this fits with the 12 motor-generator flywheel sets that can supply energy to all four catapults?
The Ford class is designed to have three catapults in operation. They can switch between the catapults, so that the unused catapult is changing.
 

nlalyst

Junior Member
Registered Member
The Ford class is designed to have three catapults in operation. They can switch between the catapults, so that the unused catapult is changing.
Thanks. That helps clarify a part of the picture.

What do you make of the conflicting sources that on one side claim that each flywheel stores 121 MJ of energy and the official General Atomics statements that each of the 12 EMALS motor-generator stores 60 MJ of energy and has a 60 MW output of electricity ?

Source:
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Matheus S

New Member
Registered Member
Thanks. That helps clarify a part of the picture.

What do you make of the conflicting sources that on one side claim that each flywheel stores 121 MJ of energy and the official General Atomics statements that each of the 12 EMALS motor-generator stores 60 MJ of energy and has a 60 MW output of electricity ?

Source:
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One of my doubts.
 

Matheus S

New Member
Registered Member
After realizing how you understood the "121/484MJ each", I think 25% would be wrong even if we agree with the definition of efficiency (measuring point).

Regardless the actual figure, the overall efficiency of EM launcher is much higher than steam by the same measurement, there is no doubt about it.
But that's the question. The point I'm raising is the overall efficiency of EMALS, sources claim it would be around 90%, well above 5% for the steam catapult. But this information is quite divergent between the various sources when you look at the entire EMALS system and all the storage subsystems. I know that EMALS is more efficient than a steam catapult, but I want to estimate how much.
 

Matheus S

New Member
Registered Member
90% efficiency is typically the conversion efficiency of motion to electricity and vice-versa. Think an electric motor or alternator.
I would roughly estimate then than it has 10% loss from the EMALS itself. Linear electric motor.
Another 10% loss from electric to flywheels storage. Another 10% loss from flywheel storage back to electric again.
There might also be some (insignificant) losses from the flywheels with friction and electric transmission losses in the circuits.
The nuclear reactor power plant might be like 45% efficient at converting thermal power to electricity at best.
It depends on reactor core temperature but it is more or less the same in most LWRs. So that is another 55% loss.

Are you guys comparing the catapult performance of the Ford against the Nimitz class? You should notice both carrier families use different reactors. The Nimitz class uses 2x 550 MWt A4W reactors and the Ford class uses 2x 700 MWt A1B reactors.
Doing these calculations with the appropriate percentages of loss of efficiency, there is no more than 20% overall efficiency of EMALS.
 

nlalyst

Junior Member
Registered Member
But that's the question. The point I'm raising is the overall efficiency of EMALS, sources claim it would be around 90%, well above 5% for the steam catapult. But this information is quite divergent between the various sources when you look at the entire EMALS system and all the storage subsystems. I know that EMALS is more efficient than a steam catapult, but I want to estimate how much.
I think a good way to compare efficiency would be to find out the amount of steam needed to launch an aircraft via EMALS, since we know the amount needed to do it via the steam catapults.
 

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