Rockets

[RobertPaulson]

I don't know if this is right place for this question, but it looked like the best place to put it;

A rocket has an initial mass of 7x10[sup:3po4rn87]4[/sup:3po4rn87]kg and on firing burns its fuel at rate 250kgs[sup:3po4rn87]-1[/sup:3po4rn87]. The exhaust velocity is 2.5x10[sup:3po4rn87]3[/sup:3po4rn87]ms[sup:3po4rn87]-1[/sup:3po4rn87].

Q) If the rocket is pointing vertically up and starts stationary on the launch pad how long must the rocket engines fire before the rocket lifts off? (You may use, without repeating its derivation, the formula derived for N2 applied to rockets).

I understand what the question is asking, and can input all the data; but I don't know at what point is the rocket defined to have 'lifted off'. Is there a formula for this or is it at a certain time/velocity/distance?

 

[galactus]

Subhotosh is the engineer, but there may be something wrong with the rocket design. It's probably me, though.

The thrust is \(\displaystyle \left(250 \;\ \frac{kg}{s}\right)\left(2500 \;\ \frac{m}{s}\right)=625,000 \;\ N\)

The initial acceraltion off the pad is \(\displaystyle F-ma\Rightarrow \;\ 625,000-ma\Rightarrow \;\ 625,000-70,000(9.8)=-61,000 \;\ N\)

\(\displaystyle \frac{-61,000}{70,000}=\frac{-61}{70}\approx -0.87 \;\ \frac{m}{s^{2}}\)

The acceleration is negative.

Like I said, it is probably me, but it would appear the rocket has a flawed design. Was that part of the problem statement, by chance?.

I am rusty on my physics. Maybe SK or someone else can confirm or deny my assertion.

 

[wjm11]

at what point is the rocket defined to have 'lifted off'

The weight of the rocket is continuously decreasing as the engines burn. Lift off occurs when the weight (W = mg) is equal to the thrust. That is the "break even" point. At all times after that, the weight is less than the thrust, and the rocket is accelerating upward.

 

[RobertPaulson]

Subhotosh is the engineer, but there may be something wrong with the rocket design. It's probably me, though.

The thrust is \(\displaystyle \left(250 \;\ \frac{kg}{s}\right)\left(2500 \;\ \frac{m}{s}\right)=625,000 \;\ N\)

The initial acceraltion off the pad is \(\displaystyle F-ma\Rightarrow \;\ 625,000-ma\Rightarrow \;\ 625,000-70,000(9.8)=-61,000 \;\ N\)

\(\displaystyle \frac{-61,000}{70,000}=\frac{-61}{70}\approx -0.87 \;\ \frac{m}{s^{2}}\)

The acceleration is negative.

Like I said, it is probably me, but it would appear the rocket has a flawed design. Was that part of the problem statement, by chance?.

I am rusty on my physics. Maybe SK or someone else can confirm or deny my assertion.

I'm pretty sure your assertion is correct, as part b) asks What is wrong with the design of this rocket?
Due to this design flaw, does this mean that the rocket won't ever take off? Or that it will but only after a while?

at what point is the rocket defined to have 'lifted off'

The weight of the rocket is continuously decreasing as the engines burn. Lift off occurs when the weight (W = mg) is equal to the thrust. That is the "break even" point. At all times after that, the weight is less than the thrust, and the rocket is accelerating upward.

Thanks, and i'm assuming i have to take into account the loss of mass due to fuel burn?
EDIT: Sorry, :S didn't read the first sentence, my bad

 

[RobertPaulson]

Thanks to your help i think i may have just done it;

'Lift off' will happen when W=Thrust , which means that at lift off the mass will have to be about 6.37x10[sup:cfasz9lc]4[/sup:cfasz9lc]

Meaning that the rocket has to burn 7x10[sup:cfasz9lc]4[/sup:cfasz9lc] - 6.37x10[sup:cfasz9lc]4[/sup:cfasz9lc] = 6.29x10[sup:cfasz9lc]3[/sup:cfasz9lc]kg to achieve this.

Therefore at a rate of 250kgs[sup:cfasz9lc]-1[/sup:cfasz9lc] the rocket can burn this mass of fuel in about 25 seconds.

Which leaves part b) as the design is flawed as either the rocket is carrying too much fuel to begin with or the thrust is not high enough.

Amirite?

On a side note, in real life would the fuel burn rate decrease over time as a result of decreased pressure or would proper rockets take that into account?

 

[wjm11]

Therefore at a rate of 250kgs-1 the rocket can burn this mass of fuel in about 25 seconds.

Which leaves part b) as the design is flawed as either the rocket is carrying too much fuel to begin with or the thrust is not high enough.

You are correct.

I believe the rocket engines are designed to deliver constant thrust. That is controlled by the fuel pumps in the case of liquid fuel engines.

In solid fuel engines, you light them and they go. The burn rate depends on the surface area of the burn of the solid fuel. Cracks in the solid fuel lead to burn taking place in those cracks, resulting in exploding boosters (as I recall).

 

[RobertPaulson]

Cheers brah, i love learning cool physics stuff

 

[Deleted member 4993]

at what point is the rocket defined to have 'lifted off'

The weight of the rocket is continuously decreasing as the engines burn. Lift off occurs when the weight (W = mg) is equal to the thrust. That is the "break even" point. At all times after that, the weight is less than the thrust, and the rocket is accelerating upward.

That is exactly correct. Before Goddard, engineers did not believe that rockets (with respectable pay-load) can be pushed pass escape velocity - exactly for that mistake.

 

[TchrWill]

A rocket has an initial mass of 7x10^4kg and on firing burns its fuel at rate 250kgs^-1. The exhaust velocity is 2.5x103ms^-1.

Q) If the rocket is pointing vertically up and starts stationary on the launch pad how long must the rocket engines fire before the rocket lifts off? (You may use, without repeating its derivation, the formula derived for N2 applied to rockets).

The thrust of the rocket is defined by F = Vex(r)/g where F = the thrust in kilograms, Vex = the exhaust velocity of the gases, r = the propelant consumption rate and g = the acceleration due to gravity. Therefore, F = [2.5(10^3)kg/sec. x 250(kg/sec)]/9.8m/sec. = 63,775 kg(f).

The rocket cannot rise until the magnitude of the thrust exceeds the weight of the rocket or the thrust to weight ratio is greater than 1.

This will happen when 7(10)^4 - 63,775 = 6,225kg of propellant has been consumed or, at 6,225/250 = 24.9 sec.

Realistically, a rocket would never be designed to sit on the launch and burn propellant in this manner. Ignition thrust is always greater than ignition weight If a fixed rocket design is being used, the design liftoff weight of the rocket must be limited to that which derives from a target liftoff thrust to weight ratio, F/W, which derives from a design trajectory profile having the target F/W ratio. The Apollo Saturn V rocket had a liftoff F/W ratio of ~7,500,000/6,400,000 = 1.17, an extremely low F/W.