# Energy needed for achieving this mass flow rate for a rocket

• I
• Basimalasi
In summary, a steam rocket requires 3,400 MJ/sec to achieve the same thrust and mass flow rate as NASA's Delta 2 rocket. The combustion of the rocket fuel is highly efficient, but you need to provide even more power to maintain a constant flow.

#### Basimalasi

Hey,
Suppose you have an enclosed vessel that is full of water. You heat it up then you open a valve, and that will create enough thrust for it to act as a rocket.

How much energy is needed for a steam rocket to get heated enough to achieve a mass flow rate of 2430 Kg/s.
Mass of water= 256 Tons.
total mass of the rocket with the propellant is 282 Tons.

This is by no means a HEP or Nuclear Physics question.
that mass flow rate you are giving will erase your water in just 2minutes... but nevermind...

it has the exact area of this rocket engine. But I don't know how to calculate the exit area of the nozzle
Rocketdyne RS-27A
Propellants: LOX/RP-1 Kerosene
O/F Ratio: 2.245
Dry Weight: 2,528 lbs
T/W Ratio (sl): 79.11
T/W Ratio (vac): 93.75
Length: 149 inches
Diameter:
67 inches
Expansion Area Ratio (ε = Ae/At): 12

Shouldn't the problem also state the speed of the expelled water / steam as well as the mass flow to determine the power, in which case the total energy would be the power times the time it takes to empty the water from the rocket?

The required temperatures would instantly evaporate your rocket.

rcgldr said:
Shouldn't the problem also state the speed of the expelled water / steam as well as the mass flow to determine the power, in which case the total energy would be the power times the time it takes to empty the water from the rocket?
It's hard to determine the flow of an unsteady two-phase fluid

Basimalasi said:
How much energy is needed for a steam rocket to get heated enough to achieve a mass flow rate of 2430 Kg/s.
Mass of water= 256 Tons.
Just pulling a number out of the air for order of magnitude, but if it starts hot and pressurized at 300 C, the enthalpy of vaporization is 1400 kJ/kg. So 2,430 kg/sec would need 3,400 MJ/sec. For comparison, that's about the typical heat output of a commercial nuclear reactor.

As mfb says, you can't make the starting mixture hot enough to vaporize all of the water upon release without it being thousands of degrees.

imagine we had a meta-material that can withstand such temp.

You cannot maintain a constant flow by heating it in advance, flow will decrease over time.

Flow depends on the unspecified size of the narrowest point of the nozzle, and also on the shape.

mfb said:
You cannot maintain a constant flow by heating it in advance, flow will decrease over time.

Flow depends on the unspecified size of the narrowest point of the nozzle, and also on the shape.
it will be heated with a 400 MW power plant then a high insulation layer will contain it

400 MW for 2430 Kg/s gives a maximal exhaust velocity of 570 m/s. That is really, really bad for a rocket. You won't go to space with such a tiny exhaust velocity.

What do you actually want to do?

I just want to see how much power should I provide to a steam rocket as heat to make it have a same thrust and mass flow rate of NASA's Delta 2 rocket

Why don't you look up how much power the combustion of the rocket fuel releases?

Those powers are typically in the range of tens of gigawatts.

But even if the power differs those two systems have completely different reactions. one is combustion and the other is straight expansion because of ;pressure deference,Maybe the combustion has less efficiency ? :/
Also I'm kind mixed up with what exactly I am looking for. Why do we care about the power input instead of energy. Can't I determine the input energy and then see how much time will it take a 300MW to reach that energy ?

The combustion is highly efficient (you heat the material with the chemical reaction - how could this be inefficient).
To improve on that, you have to deliver even more power.
Basimalasi said:
Why do we care about the power input instead of energy. Can't I determine the input energy and then see how much time will it take a 300MW to reach that energy ?
To launch a rocket, you need some minimal mass flow - you have to counter gravity. Rockets typically start with thrust of about 1.5 to 2 times their weight, which means they waste 1/2 to 2/3 of their initial thrust already. Going down more makes it worse.

mfb said:
The combustion is highly efficient (you heat the material with the chemical reaction - how could this be inefficient).
To improve on that, you have to deliver even more power.To launch a rocket, you need some minimal mass flow - you have to counter gravity. Rockets typically start with thrust of about 1.5 to 2 times their weight, which means they waste 1/2 to 2/3 of their initial thrust already. Going down more makes it worse.

What I mean is can we treat the steam rocket as a thermal battery. we store a huge amount of energy through heating and then launch. The power plant charges the rocket once not constantly. for example if a rocket a mass flow rate of 2,430 kg/sec would require 1000 Gjoules of input energy then we bring a 400MW power plant to charge it (heat it) for ⇒ Time= Energy/Power= 1000 Gjoules/400MW = 3333.333 seconds.

What did I miss here?

The result is essentially identical to a huge flywheel. You spin it up so that the rim is at the selected exhaust velocity and then unwind a thin strip from one side. Both solutions fail for essentially the same reason. You have to be able to contain the immense pressure.

Basimalasi said:
What I mean is can we treat the steam rocket as a thermal battery. we store a huge amount of energy through heating and then launch.
As discussed before, storing the energy as heat in the rocket does not work, the rocket would evaporate.

if we hypothetically had a material that would withstand the high pressure/temperature .would it work?

Sure. If you have indestructible materials that do not conduct heat at all, fill them with incredibly hot plasma at high pressure, let the plasma out, done.

The speed of the plasma ejection would still be a factor in the energy required. Consider the case of the rocket nosed into a massive object with no atmosphere, so that all of energy ends up in the ejected plume of plasma. Then it becomes clear that the speed of that plume is a key factor in the energy output (1/2 m v^2).

Awesome. I'm going to include this in my graduation project. can you help me find a specific equation that could solve for the required energy input for creating that plasma

I don't see the point, as this indestructible material for the container does not exist.

Anyway, for a given exhaust velocity, see above for the minimal energy per mass.

I will make a plasma container like those used in nuclear reactors and I will launch that rocket, You'll see ;)
Thanks for you help!

Nuclear reactors work at way lower temperatures and pressures, and they are way more massive than rockets.