How Does Power Affect Force in Space Travel?

[norrrbit]

I'm getting over 5kW, which is a lot more reasonable. Check your numbers.

Ooops, you're right - that makes 5,3 kW.

If it was resting on the ground, it'd take zero power.

Obviously, that's why I have cut the text out, but you were faster :)

It doesn't take any energy to support something. It's only all the air you're moving. And with infinite area, you don't need blades. You can just make it a parachute, and it will be enough.

Then, what constrains engineers from making much bigger rotors, as it'd reduce power, i.e. energy consumption for hover flight - blades' mass and strength? New lighter materials will make rotors bigger and hover flight more economical?

 

[norrrbit]

Assuming a reflector at the bottom of the saucer, the force F= 2*dp/dt. And p for photon is E/c. So P = dE/dt = cdp/dt = (1/2)Fc.

Ok. So I understand (after small tutorial from wikipedia) that p is photons momentum and P=.5*c*mass*g. Thus finally I've got confirmed that Power is linearly dependent on Force... ;-). Just joking. I understand that it holds only for this specific experiment.

 

[norrrbit]

The thing that I learned most from this thread is that it seems one can't talk about power, force, etc. generally but only in specified context (as Gerenuk underlined).
Thank you K^2 and all you people for your effort. You've helped me a lot!

 

[Nagakumari]

Hi,
Can anyone tell me where to find solutions for the problems in Mechanics of Materials (Gere and Timoshenko) book.
That would be greatly appreciated.
Thank you.
Nagakumari

 

[rcgldr]

So you want to view the engine as doing negative work on the propellant?

It's only negative work (decrease in kinetic energy) if rocket speed is > .5 V_e (exit velocity relative to rocket). It's zero work on propellent if rocket speed = .5 V_e (just a sign change of exhaust velocity), and positive work (increase in kinetic energy) on propellent if rocket speed < .5 V_e.

In the case of the hovering object, no work is done on the object. In the case of a helicopter, work is done on the air, and the amount of work done depends on rotor efficiency. Rotor length is a big contributor to efficiency, but there are practical limits.

If you had a slug that snugly inside a friction free cylinder with an open top, then the slug would descend until the pressure of the trapped air supported the slug. Once a stable state is achieved, no work is done.

 

[K^2]

Then, what constrains engineers from making much bigger rotors, as it'd reduce power, i.e. energy consumption for hover flight - blades' mass and strength? New lighter materials will make rotors bigger and hover flight more economical?

Yes, it's mostly strength and added weight. And yes, with new materials, all sorts of aircraft have been becoming more efficient. Look at the modern sail planes.

It's only negative work (decrease in kinetic energy) if rocket speed is > .5 Ve

Well, yeah. Naturally. Alright, that works. But I still wouldn't call it the most intuitive explanation.

 

[Anamitra]

We turn on the engines which work with constant power: P - i.e. energy intake from the batteries is constant through time. To my intuition the ship should be now accelerating with constant accelaration: A. Then force acting on the ship should also be constant: F=ship_mass*A.

It would be convenient to use the formula:

F=ship_mass*A + (d/dt of ship_mass)*u
F=mA+(dm/dt)*u, [m is the mass of the spacecraft ]
P/u =mA+(dm/dt)*u ; [since P=F*u]
P= muA +(dm/dt)*u^2
[dm/dt is a negative quantity and we take it to be constant]
If P is constant(as assumed by norrbit)

We have,
dP/dt =0

Therefore,
m{(du/dt)*A+u*(dA/dt)}+uA*(dm/dt)+2u du/dt *dm/dt=0
[Noting that dm/dt is a constant ,we write dm/dt=-k;k>0]
mA^2 + mu (dA/dt)-uAk -2uAk=0
If A is constant as assumed by norrbit,

We have,
mA^2=3uAk
mA=3uk
A=3uk/m

Now m decreases with time and hence u has to decrease if A is to be a constant[and if k is constant]. Then again if u decreases A cannot be a constant
[if k is assumed to decrease we have to repeat the differentiation right from the beginning( since dm/dt= constant has simplified the differentiation that I have done). We cannot apply the simple formula A=3uk/m in this case]]
Therefore A cannot be a constant in this case if power is taken to be a constant[and if k is constant].

 

[rcgldr]

It's only negative work (decrease in kinetic energy) if rocket speed is > .5 V_e

Well, yeah. Naturally. Alright, that works. But I still wouldn't call it the most intuitive explanation.

Still switch this case to a rocket "hovering" above the moon (no atmosphere), the rocket isn't moving, so no work done on the rocket, only on the propellent expelled from the rocket. This is a similar case to the hovering helicopter (the heli doesn't move, only the air). I think it's a reasonable comparason.

 

[norrrbit]

After 4 pages of posts, I will try to recapitulate what I understood. I will use as simple and widely understood laws of physics as possible.

1. spaceship is propelled with constant drive power: P. That's because
2. it uses constant mass of propellant per second: m_p
3. which has final velocity of v_p (which I measure relatively to the ship, this velocity is constant & specific for propellant and engine construction)
4. thus propellant with m_p accelarates from 0 velocity to v_p in 1 sec which produces constant force F_p=m_p*v_p
5. F_p is counter-balanced (according to Newton's III law) by thrust F_s
6. which equals F_s=m_s*a_s (ship's mass*ship's accelaration)
7. thus F_p=F_s => m_p*v_p=m_s*a_s
8. Power is kinetic energy of a propellant's mass flow in a second: P=1/2*m_p*v_p^2
9. from 7&8 we have: P=1/2*m_s*a_s*v_p
10. because we use up the propellant, m_s drops and a_s increasesConclusions:
1. With constant power drive spaceship will accelerate faster
2. With constant power drive, thrust is also constant P=1/2*Thrust*v_p

So I was wrong in neglecting changes in ship's mass (and expecting constant acceleration of the ship - but that's why I used ionic drive - propellant's mass is realtively less important as it is more propellant-efficient), but my intuition about relation between power & thrust was not that absurd!Please correct me if I am wrong.

 

[rcgldr]

2. it uses constant mass of propellant per second: m_p

That should be "m-dot": ṁ_p, the mass flow per unit time (not mass flow per second, not important here, but the time it takes to reach terminal velocity is not needed to be known).

4. thus propellant with m_p accelarates from 0 velocity to v_p in 1 sec which produces constant force F_p=m_p*v_p

v_p is terminal velocity, and the time it takes to reach terminal velocity is not stated. The mass flow rate takes care of the time factor, so the thrust F = ṁ_p v_p.

8. Power is kinetic energy of a propellant's mass flow in a second: P=1/2*mp*vp^2

Again replace m_p with "m-dot": P=1/2 ṁ_p v_p² = 1/2 F v_p

 

[K^2]

Still switch this case to a rocket "hovering" above the moon (no atmosphere), the rocket isn't moving, so no work done on the rocket, only on the propellent expelled from the rocket. This is a similar case to the hovering helicopter (the heli doesn't move, only the air). I think it's a reasonable comparason.

Yeah, for hovering, the equations are similar. Though, mass flow rate has to be factored in separately.

So let's see...

F = mg = -\dot{m} V_p

P = -\frac{1}{2} V_p^2 \dot{m}

\dot{m} = -\frac{mg}{V_p}

P = \frac{1}{2} mgV_p

m(t) = m_0 e^{-\frac{gt}{V_p}}

P(t) = \frac{1}{2} m_0 gV_p e^{-\frac{gt}{V_p}}

Ok, maybe not so similar, but the mg term is the same.

 

[norrrbit]

That should be "m-dot": ṁ_p, the mass flow per unit time (not mass flow per second, not important here, but the time it takes to reach terminal velocity is not needed to be known).

Ok rcgldr, your remarks are understood, but I just wanted to make it as pictorial as possible for someone out of physics community.

 

[rcgldr]

That should be "m-dot": ṁ_p, the mass flow per unit time (not mass flow per second

Ok rcgldr, your remarks are understood, but I just wanted to make it as pictorial as possible for someone out of physics community.

My main concern was the impression from your previous post that the expelled propellent would accelerate to terminal velocity in 1 second. Using mass flow rate eliminates having to specify how long it takes for the propellent to reach terminal velocity.