Monday, March 30, 2015

Power plant electrical capacity on the Gerald R. Ford-class aircraft carrier

How big is the reactor on the  Gerald R. Ford aircraft carrier?

By the wikipedia the size of the power plant is 300 MW electrical output from each reactor.
Article from the Wiki:

The new reactor for the CVN 21 class overcomes many of the shortfalls of the Nimitz-class reactor and is an enabler for many of the other technologies and improvements planned for the new class. Two Bechtel A1B nuclear reactors will be installed on each Ford-class carrier, with each A1B reactor capable of producing 300 MW of electricity.

It referencing  the next two pages:
 NUCLEAR MARINE PROPULSION __  M. Ragheb __ 3/5/2015
So what can we get from this informations?
First link doesn't give details about the reactor size, second giving same data, but it is not celar:

The Gerald Ford-class carriers have A1B reactors reported to be 240-300 MW each, but running a ship which is entirely electrical, including an electromagnetic aircraft launch system. The reactors are two to three times as powerful as the A4W units in Nimitz-class.

So, let see the A4W power plant on the Nimitz.
 The only ships to use these nuclear reactors are the Nimitz class supercarriers, which have two reactors each rated at 550 megawatts. These generate enough steam to produce an unspecified electricity supply plus 140,000 shaft horsepower (140 MW).[2]

Just a small correction, 140 000 HP equal to 104 MW :)
It saying "unspecified amount of electricity"

Another interesting information from the wiki :
"The biggest problems facing the Nimitz class are the limited electrical power generation capability and the upgrade-driven increase in ship weight and erosion of the center-of-gravity margin needed to maintain ship stability."

Logical, considering that if they retrofit the ship with new generators then they have to install the turbine + generator onto the upper levels of the ship,and that has negative effect on the stability.

This is the conventional copper winding,  but we can't forget that the US military poured money into an interesting project back int the 00's:

It is the American Superconductor Company.

There is not so much reference on this page, but this company started as a direct military supplier to the Pentagon .
They supplying hight temperate super conductor generators, with integrated cooling system.

High temperature means above the boiling point of liquid nitrogen, and it is high compared to the boiling point of Helium :)

Interesting, but this business removed all reference to the navy generators.


-- FORD_wiki FORD_realistic NIMITZ_wiki
Thermal power 1 reactor 900 300 300
useful power power 1 reactor 300 MWe 100 MWe 100 MW shaft
efficiency 33% 33% 33%

A 300 MWe reactor is a big deal, that is the low side of a typical 70s land based nuclear reactor.
A reactor of that size means big trouble in the case of a nuclear accident.

So, the efficiency of a typical steam power plant is around 33%.

It can be higher, but that means more mass, so lets target that as realistic number.

Before the HMSC removed all document about the marine propulsion the test generator size was 50MW .
It sounds like a good compromise for different ship size,and its should mean four generators on the Ford.

What a coincidence, the Ford has four shaft .
See the differences between the Nimitz and Ford:
Propeller and shaft of the Ford
Propeller and shaft of the Bush 

Not so much different. No smell of electrical propulsion.  :)

Interesting, the geometry is the same.
So, put together.
50 MW generator from AMSC , same shaft geometry on both carrier ( not hanging pods with motors like here ) .

From a redundancy standpoint a reactor-generator-motor design simple,and easy to move everything around the ship, to balance it, but it make everything more error proof.

So the ship propulsion should look like an 50MWe generator, with HTSC wired rotor from the AMSC , connected to a 66MWe steam turbine, and the engine combination is connected to the shaft of and to the propeller.
Probably there is a spare , 50MW generator with a turbine somewhere , without a propeller.

It should means that the centre of  gravity is low, the redundancy is high,and the ship has enough energy to drive the catapult.


That is the main electrical consumer on the ship.
They have two options.
Capacitors or big enough generators.

On a steam catapult they can store steam in boilers.
See the catapult energy requirement.

Max mass : 40000 kg , max speed 140 knots - > 259 km/hour.  (source wiki )

Probably it is two extreme, but say that the machine is capable to launch a 40 tons plan with 259 km/hour .
259km/hour = 259000 m/hour = 71.9m/seconds
E=40000*71.9*71.9=206,78 MJoule

The length of the catapult is 99 meter.
Based on this the required acceleration to launch a 40 tons air plane by 71.9 m/sec speed :
a= v2 /(2*s) 
a= 71,92 /(2*99)=26.1 m/seconds2

let see how much time does it takes to accelerate the airplane from 0 to 71.9 m/sec.
t=71.9/26.1=2.75 seconds

So, divide the required energy by the time to get the necessary power plant size the provide the required energy:
P.plant=206,78 MJoule/2.75 sec=75.19 Mw


Of course there is huge loss between the airplane and the steam turbine , but it means that a 200 MWe power plant should be enough to launch an airplane in every 2,75 seconds,and leave enough energy for other systems of the ship.

Interesting, isn't it?

old brochure from the AMSC, that describing the HTSC motor

Update again:
Superconductor Motor for Navy Passes Full-Power Test
U.S. Navy Orders 36.5-Megawatt Superconducting Propulsion Motor

Seems like the motor is not 50 MW, but 36,5 MW.

That gives 4*36,5= 146 MW electricity , still enough for the catapult.   :)


  1. Out of curiosity (I'm trying to create a realistic as possible ship )lets say there's a ship with 15 5"railguns with expected shell speed of 2438.400m how big would the power plant of that ship need to be?

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  3. Yes, but lets not forget that the plane's are on full afterburner when they are being catapulted, so they are most certainly contributing to their own acceleration.