Rocket Propulsion Flight Time

[jtaz612]

The height that a model rocket launched from Earth can reach can be estemated by assuming that the burn time is short compared to the total flight time, so for most of the flight the rocket is in free-fall. (This estimate neglects the burm time in calculations of both time and displacement. ) For a model rocket of specific impulse Isp = 100 s, mass ratio mo/mf = 1.2 , and initial thrust to weight ratio τ o = 5 . Estimate (a) the height the rocket can reach, and (b) the total flight time. (c) Justify the assumption used in the estimates by comparing the flight time from part (b) to the time it takes for the fuel to be spent.

Some useful equations:
The specific impulse of a rocket propellant is defined as Isp = Fth / (Rg) , where Fth is the thrust of propellant, g is the magnitude of free fall accelaration at the surface of earth, and R is the rate at which the propellant is burned.

I have not shown any work because I am lost as to how to start the problem. If more of an attempt is needed let me know and i'll post what I've done so far. Thankyou.

 

[denverdoc]

well this will be a pretty crude "proof":

but total impulse, I=Isp*g*Mp where Mp=mass of propellant.
and I=deltaP=Mrocket*Vel(burnout), from mass fraction we know
(Mprop+Mrocket)/Mrocket=1.2

hence Mass Prop=.2Mrocket, subbing,

100*9.8*.2=Vb=196m/s. from kinematics, altitude,h, from burnout to apogee assuming G is constant, and no drag

2(g)h=Vb^2 =1960m, the flight time would of course be 196=a*t=20 seconds.

any thoughts re the burn time of the propellant?

 

[m2003]

I would approach this problem by using the equations provided and applying the principles of rocket propulsion and free-fall. Here is how I would solve for the height and flight time:

(a) To estimate the height the rocket can reach, we can use the equation for the specific impulse: Isp = Fth / (Rg). Rearranging this equation, we get Fth = Isp * Rg. We can then use this value for Fth in the equation for thrust to weight ratio: τ o = Fth / mg, where m is the mass of the rocket. Rearranging this equation, we get m = Fth / (τ o * g). We can now use this value for m in the equation for mass ratio: mo/mf = 1.2. Rearranging this equation, we get mf = mo/1.2. We can now use this value for mf in the equation for specific impulse: Isp = Fth / (Rg). Rearranging this equation, we get R = Fth / (Isp * g). Finally, we can use this value for R in the equation for height: h = (Isp * g * ln(mo/mf)) / (g * R). Plugging in our values, we get h = (100 s * 9.8 m/s^2 * ln(1/1.2)) / (9.8 m/s^2 * 3.75 s^-1) = 12.87 meters. Therefore, the estimated height the rocket can reach is 12.87 meters.

(b) To estimate the total flight time, we can use the equation for free-fall distance: h = 1/2 * g * t^2. Rearranging this equation, we get t = √(2h/g). Plugging in our value for h from part (a), we get t = √(2 * 12.87 m / 9.8 m/s^2) = 1.27 seconds. Therefore, the estimated total flight time is 1.27 seconds.

(c) Justifying the assumption used in the estimates, we can compare the flight time from part (b) to the time it takes for the fuel to be spent. Since the burn time is assumed to be short compared to the total flight time, we can estimate the burn time to be much less