Author Topic: Electric/chemical Energy to kinetic energy conversion  (Read 135889 times)

Offline jdbenner

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #15 on: June 16, 2006, 05:28:02 PM »
Don is certainly correct but keep in mind that “most efficient” refers to mass expenditure, rather than energy expenditure.  The most efficient energy wise is when the exhaust velocity equals the rockets velocity, at any given time (so that the exhaust has zero residual kinetic energy).  Unfortunately This leads to a few limitations:
1 Chemical energy densities are suboptimum for launch from earth. 
2 It is difficult to efficiently vary exhaust velocity.
3 If pushed to the extreme of zero exhaust velocity it will literally take an infinite mass of propellant to get to any velocity what so ever.

The optimum, energy wise, fixed exhaust velocity is approximately 0.666… times the rockets final velocity. 
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Offline DonPMitchell

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #16 on: June 16, 2006, 09:47:36 PM »
Can you elaborate?  Why is it ever better to have a lower specific impulse?
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Offline ijuin

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #17 on: June 17, 2006, 01:37:24 AM »
jdbenner:

Motion is all RELATIVE. That means that the exhaust velocity relative to the vehicle velocity is all that matters, not the exhaust velocity relative to an outside frame of reference (such as the Earth). The only time where it does matter is inside an atmosphere, where high-thrust solid motors are more efficient for liftoff than the lower-thrust-higher-impulse hydrolox engines because of gravity, friction, and pressure concerns.

Offline jdbenner

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #18 on: June 17, 2006, 04:13:49 PM »
Velocity is speed in a given direction.
Momentum is mass times velocity. I = mv
Kinetic Energy is one half mass times speed squared. KE = .5mv^2

Doubling the velocity of a given mass doubles momentum but quadruples energy. 
Doubling mass for a given velocity doubles momentum but only doubles energy.   

Let us solve simplified one-dimensional propulsion problems Newtonian physics and assuming all motion to be absolute relative to the launch site and thrust is forward (away from the launch sight).
   
V = Velocity of rocket relative to the reference point.

c = XV = Velocity of expelled propellant relative the rocket

v = |c-V| = The speed of the expelled propellant relative to the reference point.

E = .5mc^2 = Energy imparted to a given mass (m) of propellant to expel it at a velocity of c.

ke = .5mv^2 = Kinetic energy remaining in a given mass (m) of expelled propellant.
 
KE = E – ke = The kinetic energy imparted to the rocket by a given mass (m) of propellant.

We wish to maximize KE/E.
By the law of conservation of energy one is the maximum value of KE/E.

KE/E = 1

(E – ke)/E = 1

Set ke equal to zero

(E-0)/E = 1

To set ke equal to zero we must set v equal to zero.

ke = .5mv^2

ke = .5m(0)^2 = 0

To set v to zero we must set c equal to V


v = |c-V|
v = 0 = c-V

c = V

So the maximum energy transfer occurs when the propellant is expelled just fast enough to lose all speed relative to the rockets point of origin (the fixed reference point).  If any velocity remains in either direction there will be kinetic energy associated with that velocity.



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Offline jdbenner

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #19 on: June 17, 2006, 04:24:09 PM »
jdbenner:

Motion is all RELATIVE. That means that the exhaust velocity relative to the vehicle velocity is all that matters, not the exhaust velocity relative to an outside frame of reference (such as the Earth). The only time where it does matter is inside an atmosphere, where high-thrust solid motors are more efficient for liftoff than the lower-thrust-higher-impulse hydrolox engines because of gravity, friction, and pressure concerns.

It is true that motion is relative.  But, relative to what?
If The rocket launches from Earth then the rocket and any propellant in it must have some velocity and kinetic energy relative to Earth.  The point where the rocket and propellant originate is a valid point of reference, since all propellant must be accelerated relative to that point, and any velocity relative to that point, remaining in the exhaust, must have been imparted to the propellant when it was in the rocket.
« Last Edit: June 17, 2006, 04:58:26 PM by jdbenner »
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Offline ijuin

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #20 on: June 18, 2006, 03:52:13 AM »
Exhaust should be computed relative to the engine that expels it, not relative to an external observer. You get the same amount of delta-v from a given fuel mass and ISP regardless of the initial velocity.

The rocket equation is as follows:

(delta-v) = (exhaust velocity) * ln(initial mass / final mass)

Note the complete lack of reference to initial or final velocity.

Offline jdbenner

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #21 on: June 18, 2006, 10:39:34 AM »
First off, delta-v measures a change in velocity, which implies motion relative to some point.  An object in linear motion must move relative to something besides it self.

Secondly, delta-v is a velocity change not an energy change.  True energy changes, but to know how much it changed you need to know both the delta-v (velocity change) and either the initial or the final velocity.

The reason a given velocity change is not enough to compute an energy change, is that energy is calculated from the square of the velocity.

Kinetic energy calculation
 J/Kg = .5v^2

Delta-v 1000m/s
Initial velocity zero m/s, change in kinetic energy in J/Kg = .5(1000) ^2 = 500,000J/Kg
Initial velocity +1000 m/s, change in kinetic energy in J/Kg = .5(2000) ^2 - .5(1000) ^2
= 1,500,000J/Kg
Initial velocity -1000m/s, change in kinetic energy in J/Kg = -500,000J/Kg
Initial velocity +10^5m/s, change in kinetic energy = .5(101,000) ^2- .5(100,000) ^2
= 100,500,000J/Kg

Energy is measured in Force acting over a distance like, Newton Meters, or in pound feet.
So to calculate mechanical energy we must know both force and the distance over which it acts.

Let us assume a one Newton thrust rocket motor with one second duration attached to a large mass so that we can simplify by neglecting acceleration in our calculation.
If our rocket moves 10m/s then we can calculate the energy imparted by our rocket motor as 10m/s*1s*1N = 10 Nm = 10J at a power of 10W
However if our rocket moves at 1,000m/s then our calculation becomes:
1,000m/s*1s*1N = 1,000Nm = 1,000J at a power of one KW


According to the law of conservation of energy, energy can not be either created or destroyed.  So where did the extra energy come from?  The answer is that it came from the kinetic energy in the propellant.

The kinetic energy in the propellant in the tank of a rocket moving at 1,000m/s relative to our fixed point is 5*10^5J/Kg 
Let us assume that we have a rocket with an exhaust velocity of 500m/s
That means that it will take 125,000J/Kg to accelerate the propellant 500m/s relative to the rocket. But that propellant was moving at 1,000m/s relative to our fixed point so its initial kinetic energy was 5*10^5J/Kg.
The expelled propellant is now moving at 500m/s relative to our fixed reference point so it now has 125,000J/Kg kinetic energy.  So our propellant gave up (5*10^5-1.25*10^5) = 3.75*10^5J/Kg of its kinetic energy, which is three times the energy expended in expelling the propellant.

P.S.
I first learned about optimum energy exhaust velocities from one of Dr Robert Forward’s books, unfortunately I forget which book but I remember the topic was discussed in a chapter on antimatter rockets.
Joshua D. Benner Associate in Arts and Sciences in General Science

Offline Bob B.

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #22 on: June 18, 2006, 12:01:23 PM »
It doesn’t matter to from what reference the velocity is measured.  Let’s say we have a rocket with a mass of 10,000 kg and moving 8,000 m/s relative to some external reference point.  Its momentum and kinetic energy are,

P = mv = 10,000*8,000 = 8E+7 kg-m/s

KE = 0.5mv^2 = 0.5*10,000*8,000^2 = 3.2E+11 J.

Let’s say the rocket expels 1,000 kg of propellant at –3,000 m/s relative to the rocket.  To make the math simple, let’s say the propellant is expelled instantaneously rather than over a period of time.  Relative to the original reference point, the expelled propellant is moving 5,000 m/s; therefore its momentum and kinetic energy are,

P = mv = 1,000*5,000 = 5E+6 kg-m/s

KE = 0.5mv^2 = 0.5*1,000*5,000^2 = 1.25E+10 J.

Since momentum is conserved the, the final momentum of the rocket must be the initial momentum less that of the propellant,

P = 8E+7 – 5E+6 = 7.5E+7 kg-m/s.

Therefore the rocket’s final velocity is,

v = P/m = 7.5E+7/(10,000–1,000) = 8,333.3 m/s

and the change in velocity is,

delta-V = 8,333.3 – 8,000 = 333.3 m/s.

The rocket’s kinetic energy is now,

KE = 0.5mv^2 = 0.5*9,000*8,333.3^2 = 3.125E+11 J.

The total final kinetic energy is that of the expelled propellant plus rocket,

KE = 1.25E+10 + 3.125E+11 = 3.25E+11 J

thus the change in kinetic energy is,

delta-KE = 3.25E+11 – 3.2E+11 = 5E+9 J.


Let’s now perform the exact same calculation but measuring velocities from a reference point moving along with the rocket at its initial velocity.  The rocket’s initial velocity in this reference frame is zero, thus its momentum and kinetic energy are both zero.

In relation to our new reference point, the expelled propellant is moving –3,000 m/s, thus its momentum and velocity are,

P = mv = 1,000*(–3,000) = –3E+6 kg-m/s

KE = 0.5mv^2 = 0.5*1,000*(–3,000)^2 = 4.5E+9 J.

The final momentum of the rocket is,

P = 0 – (–1 3E+6) = 3E+6 kg-m/s

therefore it’s final velocity is,

v = P/m = 3E+6/(10,000–1,000) = 333.3 m/s.

Since the initial velocity was zero, the delta-v is also 333.3 m/s, which is the same change in velocity as determined in the first calculation.

Let’s now calculate the kinetic energy of the rocket,

KE = 0.5mv^2 = 0.5*9,000*333.3^2 = 5E+8 J

thus the total kinetic energy of propellant plus rocket is,

KE = 4.5E+9 + 5E+8 = 5E+9 J.

And since we started out with zero kinetic energy, 5E+9 J is also represents delta-KE.  We therefore see the exact same change in energy as in the previous calculation.  It made no difference from which reference point we measured the velocities.

Offline jdbenner

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #23 on: June 18, 2006, 12:19:36 PM »
The law of conservation of momentum states, for every action there is an equal but opposite reaction.

 Rockets are reaction motors so any momentum imparted to a rocket must be the inverse of the momentum imparted to the propellant.  After some propellant is expended the remainder with the rocket has the inverse of the momentum imparted to the propellant.  Thus the rocket and its remaining propellant have grater than zero momentum relative to the initial inertial reference frame. If the rockets exhaust is to slow, some of that imparted momentum from the initial propellant expenditure remains with the exhaust expended later.  That is to say, the exhaust is moving forward relative to the initial inertial frame of reference. 

To move any thing forward requires moving something backwards, and requires expending energy.  Our purpose is to move the rocket forward, not to move the propellant forward, thus setting the exhaust velocity to low wastes both propellant and energy.

Setting the exhaust velocity to high saves propellant; because momentum is mass times velocity, higher velocity give more momentum per unit mass, and we certainly are not leaving any forward momentum in the exhaust, since it is still moving backwards relative the initial inertial frame of reference.  However, setting the exhaust velocity to high wastes energy because the exhaust is moving relative to the initial inertial frame of reference it contains kinetic energy imparted by the rocket motor, since the purpose of the rocket motor is to impart motion to the rocket, rather then to impart it to the exhaust, the motion of the exhaust relative to the initial inertial frame of reference, means that unnecessary motion was imparted to the exhaust, and is wasteful of energy.
« Last Edit: June 18, 2006, 03:21:25 PM by jdbenner »
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Offline jdbenner

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #24 on: June 18, 2006, 01:04:49 PM »
  Let’s say we have a rocket with a mass of 10,000 kg and moving 8,000 m/s relative to some external reference point.  Its momentum and kinetic energy are,

P = mv = 10,000*8,000 = 8E+7 kg-m/s

KE = 0.5mv^2 = 0.5*10,000*8,000^2 = 3.2E+11 J.

Let’s say the rocket expels 1,000 kg of propellant at –3,000 m/s relative to the rocket.  To make the math simple, let’s say the propellant is expelled instantaneously rather than over a period of time.  Relative to the original reference point, the expelled propellant is moving 5,000 m/s; therefore its momentum and kinetic energy are,

P = mv = 1,000*5,000 = 5E+6 kg-m/s

KE = 0.5mv^2 = 0.5*1,000*5,000^2 = 1.25E+10 J.

Since momentum is conserved the, the final momentum of the rocket must be the initial momentum less that of the propellant,

P = 8E+7 – 5E+6 = 7.5E+7 kg-m/s.

Therefore the rocket’s final velocity is,

v = P/m = 7.5E+7/(10,000–1,000) = 8,333.3 m/s

and the change in velocity is,

delta-V = 8,333.3 – 8,000 = 333.3 m/s.

The rocket’s kinetic energy is now,

KE = 0.5mv^2 = 0.5*9,000*8,333.3^2 = 3.125E+11 J.




Let’s now perform the exact same calculation but measuring velocities from a reference point moving along with the rocket at its initial velocity.  The rocket’s initial velocity in this reference frame is zero, thus its momentum and kinetic energy are both zero.

In relation to our new reference point, the expelled propellant is moving –3,000 m/s, thus its momentum and velocity are,

P = mv = 1,000*(–3,000) = –3E+6 kg-m/s

KE = 0.5mv^2 = 0.5*1,000*(–3,000)^2 = 4.5E+9 J.

The final momentum of the rocket is,

P = 0 – (–1 3E+6) = 3E+6 kg-m/s

therefore it’s final velocity is,

v = P/m = 3E+6/(10,000–1,000) = 333.3 m/s.

Since the initial velocity was zero, the delta-v is also 333.3 m/s, which is the same change in velocity as determined in the first calculation.

Let’s now calculate the kinetic energy of the rocket,

KE = 0.5mv^2 = 0.5*9,000*333.3^2 = 5E+8 J
 

Forgive me Bob if I am putting words in to your mouth, but your original seamed to show that the law of conservation of energy holds in any reference frame.  But what I was trying to show was how to minimize the portion of the energy leaving with the propellant.  So I cut out the part about total energy, leaving only the part about the energy of the rocket it self.  And I believe that your calculations confirm what I was trying to say.
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Offline Bob B.

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #25 on: June 18, 2006, 02:52:06 PM »
Which reference frame you choose to observe the event is arbitrary and the velocities, momentums, and energies are all relative to whatever reference frame you choose.  You can select a reference frame that appears to maximize the energy imparted to the rocket and minimize that of the expelled propellant, or vice versa.  But no matter what reference frame you choose, the delta-v of the rocket will always be the same. 

In the first of my calculations the rocket appeared to lose energy and in the second it appeared to gain energy.  This was just the result of the arbitrary placement of the reference frame.  However, in both sets of calculations the rocket’s delta-v was 333.3 m/s.  We can repeat the calculations selecting difference reference frames, but we will always get the same change in velocity.

Minimizing the energy of the expelled propellant it just a matter of selected the right reference frame, but doing so doesn’t make the rocket go any faster.

Offline jdbenner

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #26 on: June 18, 2006, 05:23:58 PM »
Let me try a different tact. So I don’t confuse every body with the frame of reference arguments.
the frame of reference is irrelevant, except that for me, it made the explanation easer.

The rocket equation can be written:

(M+P)/M = e^ (dv/c)

M = empty mass of rocket
P = mass of propellant
dv = change in velocity
c = exhaust velocity
 
So let us plug in some numbers.
M = 1,000Kg
dv (delta-v) = 1,000m/s

c1 = 100m/s
c2 = 333m/s
c3 = 667m/s
c4 = 1000m/s
c5 = 10,000m/s
c6 = 100,000m/s

Rearrange the rocket equation to solve for P.

P = Me^ (dv/c) – M

Plug in values above

P1 = 1,000Kg*e^ (1,000m/s/100m/s) – 1,000Kg = 2.20*10^7Kg
P2 = 1,000Kg*e^ (1,000m/s/333m/s) – 1,000Kg = 1.91*10^4Kg
P3 = 1,000Kg*e^ (1,000m/s/667m/s) – 1,000Kg = 3478Kg
P4 = 1,000Kg*e^ (1,000m/s/1,000m/s) – 1,000Kg = 1718Kg
P5 = 1,000Kg*e^ (1,000m/s/10,000m/s) – 1,000Kg = 105Kg
P6 = 1,000Kg*e^ (1,000m/s/100,000m/s) – 1,000Kg = 10.1Kg

The final mass and delta-v remain constant, but let us look at the energy required to eject the exhaust. What is the energy expenditure?

KE = .5mv^2

KE1 = .5(P1) (c1) ^2 = .5(2.20*10^7Kg)(100m/s) ^2 = 1.10*10^11J
KE2 = .5(P2) (c2) ^2 = .5(1.91*10^4Kg) (333m/s) ^2 = 1.06*10^9J
KE3 = .5(P3) (c3) ^2 = .5(3478Kg) (667m/s) ^2 = 7.74*10^8J
KE4 = .5(P4) (c4) ^2 = .5(1718Kg) (1,000m/s) ^2 = 8.59*10^8J
KE5 = .5(P5) (c5) ^2 = .5(105Kg) (10,000m/s) ^2 = 5.25*10^9J

KE6 = .5(P6) (c6) ^2 = .5(10.1Kg) (100,000m/s) ^2 = 5.05*10^10J

As you can see, exhaust velocities can be too high as well as too low, if we are optimizing energy expenditure rather than mass expenditure.
« Last Edit: March 18, 2007, 09:22:18 AM by jdbenner »
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Offline DonPMitchell

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #27 on: June 18, 2006, 06:53:37 PM »
If I have a given fixed amount of fuel, then there is a fixed amount of chemical energy stored in the rocket.  Why would I not attempt to maximize the specific impulse of my engine to get the most delta-V for a given amount of stored energy?
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Offline Bob B.

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #28 on: June 18, 2006, 06:56:28 PM »
Let me try a different tact.
Okay, I think the new method explains your point better.


As you can see, exhaust velocities can be too high as well as too low, if we are optimizing energy expenditure rather than mass expenditure.
I can see where the energy of the exhaust gas might be a consideration for electric propulsion, since electricity is used to accelerate the gas.  The higher the gas velocity the bigger the power system must be.  Therefore one must optimize the propulsion system by considering both the propellant mass and the power requirements.

However, for chemical propulsion I see no advantage to optimizing energy expenditure.  You want the most delta-v as possible out of a given mass of propellant, and the only way to do that is by accelerating the gas to as high a velocity as is attainable.

Offline jdbenner

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #29 on: June 18, 2006, 07:54:25 PM »
However, for chemical propulsion I see no advantage to optimizing energy expenditure.  You want the most delta-v as possible out of a given mass of propellant, and the only way to do that is by accelerating the gas to as high a velocity as is attainable.


Bob you are right about chemical rockets.  The reason that higher exhaust velocities are desirable in chemical rockets is because chemical rocket exhaust velocities are somewhere between a half and a quarter of the optimised velocity for typical space applications.
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