Some exercise questions about Tsiolkovsky's Rocket Equation and Thrust Equation

Question

I have found 6 of the question's answers, but I don't know they are correct. The last question about time of the burn I get negative result -0.42 seconds. Time could not be a negative value, so I checked my answer more than 10 times but can't get where I do mistake.

Questions

My Answers

1st question's answer is,

$$1atm = P{_{o}} = 101300Pa$$ $$F{_{nozzle}} = (P{_{e}} - P{_{o}}) \cdot A{_{e}}$$ $$F{_{nozzle}} = (101325 - 101300) \cdot 1 = 25N$$

2nd question's answer is,

$$Po = 0Pa$$ $$F{_{nozzle}} = (P{_{e}} - P{_{o}}) \cdot A{_{e}}$$ $$F{_{nozzle}} = (101325 - 0) \cdot 1 = 101325N$$

3rd question's first answer is,

$$F{_{nozzle}} = 25N, \dot{m} = 1kg/sec, v{_{e}} = 400m/sec$$ $$F{_{thrust}} = \dot{m}v{_{e}} + (P{_{e}} - P{_{o}}) \cdot A{_{e}}$$ $$F{_{thrust}} = \dot{m}v{_{e}} + F{_{nozzle}}$$ $$F{_{thrust}} = 1 \cdot 400 + 25 = 425N$$

3rd question's second answer is,

$$F{_{nozzle}} = 101325N, \dot{m} = 1kg/sec, v{_{e}} = 400m/sec$$ $$F{_{thrust}} = \dot{m}v{_{e}} + (P{_{e}} - P{_{o}}) \cdot A{_{e}}$$ $$F{_{thrust}} = \dot{m}v{_{e}} + F{_{nozzle}}$$ $$F{_{thrust}} = 1 \cdot 400 + 101325 = 101725N$$

4th question's answer is,

$$I{_{sp}} = 363sec, I = 2MN = 2000000N$$ $$C = g \cdot I{_{sp}}$$ $$C = 9,8 \cdot 363 = 3557,4m/s$$

5th question's answer is,

$$I{_{sp}} = 363sec, I = 2MN = 2000000N, C = 3557,4m/sec$$ $$F{_{thrust}} = \dot{m} \cdot C$$ $$\dot{m} = \frac{F{_{thrust}}}{C}$$ $$\dot{m} = \frac{2000000}{3557} = 562,2kg/sec$$

6th question's answer is, $$I{_{sp}} = 363sec, I = 2MN = 2000000N, C = 3557,4m/sec, \dot{m} = 562,2kg/sec, \Delta v = 7700m/sec$$ $$\frac{M{_{full}}}{M{_{empty}}} = e^{\frac{\Delta v}{C}}$$ $$\frac{M{_{full}}}{M{_{empty}}} = e^{\frac{7700}{3557,4}} = 8,7$$ $$MR = 8,7$$

7th question's answer is,

$$I{_{sp}} = 363sec, I = 2MN = 2000000N, C = 3557,4m/sec, \dot{m} = 562,2kg/sec, \Delta v = 7700m/sec, MR = 8,7$$ $$t{_{b}} = ln(\frac{M{_{full}}}{M{_{empty}}}) \cdot I{_{sp}} - \frac{\Delta v}{g}$$ $$t{_{b}} = ln(8,7) \cdot 363 - \frac{7700}{9,8} = -0,42sec??$$

Why the 7th question's answer is coming negative value as a time value? Also, are other answers are true?

Thanks!

Answer 1

1. Wikipedia gives sea level pressure as 101325 Pa, that would make the answer zero, otherwise OK, what you have is close enough.
2. Looks good to me
3. Looks good to me
4. Looks good to me
5. Looks good to me
6. Looks good to me
7. I don't like that equation. Let's check it with some real world data from here. The burn took 156.92 seconds. We can calculate mass flow from Isp and thrust, it's 29.63 kg/s. So total prop burned is 4650 kg and the mass-ratio is 1.3838. Isp is 314 so the first term in the burn time equation is 102.0. Delta-V is 999.4 m/s and g0 is 9.8, that makes the 2nd term in the burn time equation 101.9. Subtract the two terms and I get essentially zero, just like you did. Please check my numbers, it is literally an exercise for the student

I'm gonna suppress my true feeling about that equation and just say

I don't understand the derivation of that equation or the conditions under which it is supposed to apply. I recognize that the flaw may be in my analysis and not the equation.

Answer 2

For problem 3.27 not enough information is given. Assuming constant mass flow rate, the burn time is: $$\tau = \frac{R-1}{R} \frac{c}{a_0}$$ where R is mass ratio ($MR$), $c$ is exhaust velocity, $a_0$ is initial acceleration. Choose a value for $a_0$ and you can get $\tau$.

Derivation: Let $M_p$ be propellant mass, then $\tau = M_p/\dot m$. Let initial mass be $M_0$, $F$ for thrust and use $\dot m=F/c$ to get $$\tau = (M_p/M_0)(M_0 c/F) $$ Use $a_0 = F/M_0$ so $\tau = (M_p/M_0) (c/a_0)$. A little algebra on $R$ converts $(M_p/M_0)$ to the $R$ expression above.

NOTE: The equation used for 3.27 should be zero. Multiply both sides by g and $gI_{sp}$ is exhaust velocity so the first term is the ideal velocity (no losses) and the second is the ideal velocity.