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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

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!

Organic Marble
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M. Çağlar TUFAN
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  • Hello! Thanka for answer @Organic Marble, but how it can be 0 seconds? :/ – M. Çağlar TUFAN May 02 '19 at 19:46
  • Alright, I'm waiting for your reply. Until you come i will try to solve other questions, there are 7 more questions except questions that asked in here. – M. Çağlar TUFAN May 02 '19 at 19:50
  • Oh, no. These questions are not for my homework, I make this for hobby but I'm aiming professional carrer about Rocket Scientist or Engineer, also Physicist. But, if it would not eligible to answer, it's not problem. – M. Çağlar TUFAN May 02 '19 at 19:57
  • Alright @Organic Marble, thanks for your helps! Have a good day! – M. Çağlar TUFAN May 02 '19 at 20:03
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    I looked at this a little bit, and I'm questioning whether equation 3.37 (the burn time equation) is really valid. The basic rocket equation is dv = Isp g0 ln(MR) so if you divide through by g0, you get dv / g0 = Isp ln(MR) and that seems to mean that equation 3.37 is always going to be zero. Can you provide the derivation of equation 3.37 from your book? The burn time equation I know requires you to know the value of the initial mass, not just a ratio, and this makes sense to me intuitively. Equation 1.21 here http://www.braeunig.us/space/propuls.htm gives a burn time, you need m0. – Organic Marble May 02 '19 at 21:38
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    Equation 3.37 is the burn time equation, not the delta v equation. I want to see the derivation of the burn time equation. – Organic Marble May 02 '19 at 22:03
  • Sorry, system doesn't let me edt $dv = -C\frac{dM}{M}$ to $\int_{v{{0}}}^{v{{f}}}dv = -C\int_{M{{0}}}^{M{{f}}}\frac{dM}{M}$ integrating and applying the limits results $v{{f}} - v{{0}} = -C(ln(M{{f}}) - ln(M{{0}})) = C ln(\frac{M{{0}}}{M{{f}}})$, here $v{{f}} - v{{0}} = \Delta v$ so, $\Delta v = C ln(\frac{M{{0}}}{M{{f}}})$ this is the derivation in my book. But, when we want to account forces on launching rocket, we should use $\Delta v = C ln(\frac{M{{0}}}{M{{f}}}) - gt{{b}}$ so that, $t{{b}} = ln(\frac{M{{0}}}{M{{f}}})I{_{sp}} - \frac{\Delta v}{g}$ for the Equation 3.37. – M. Çağlar TUFAN May 02 '19 at 22:10
  • There is a longer derivation in my book but, if I try to write it here, it can take 4-5 comments, I can edit the question with derivation if you want – M. Çağlar TUFAN May 02 '19 at 22:13
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    What book is this from? Maybe I can find it online. – Organic Marble May 02 '19 at 22:15
  • This book is "Introduction to Rocket Science and Engineering" by Taylor S. TRAVIS, get online hre – M. Çağlar TUFAN May 02 '19 at 22:23
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    I was afraid you were going to say that. I am not a fan of that book. That makes me more inclined to think that equation is wrong. But I will look into it. – Organic Marble May 02 '19 at 23:23
  • Alright, thank you. Do you have any better or more professional book recommendation? I did dive into Rocket Science with this book :P If you have a good pathway to being a good Rocket Scientist to share, I would be very happy! – M. Çağlar TUFAN May 02 '19 at 23:32
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    This very site has a great list of sources:https://space.meta.stackexchange.com/questions/249/resources-and-references-on-the-topic-of-space-exploration The best book I know is Sutton but only the 4th edition is truly great. The later edition online is not nearly as good. But still better than the Taylor one. – Organic Marble May 02 '19 at 23:34
  • This list is very useful, thank you so much. Another question, do you have any link to download 4th edition of Sutton's Rocket Propulsion Elements Book? I couldn't find it, checked 10 pages of Google, still nothing. – M. Çağlar TUFAN May 03 '19 at 00:11
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    It's not online unfortunately. – Organic Marble May 03 '19 at 00:16

2 Answers2

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  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.

Organic Marble
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    Oh, thanks! You tried so hard, i shouls sorry about this, you gave your hours by answering me. Thank you again and thank you. Have a good day and you are true teacher to me! – M. Çağlar TUFAN May 03 '19 at 00:41
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    I wouldn't do it if I didn't enjoy it. The world needs more aerospace engineers. Please run the numbers on that Apollo case in #7 with the real world data and let me know if you get the same results. – Organic Marble May 03 '19 at 00:42
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    I will do it, tomorrow and i will surely feedback you for this exercise. My eyes began to hurt, here is 3:45am, I won't forget this exercise and I will start working with book that you suggested tomorrow. Have a good day, sir! – M. Çağlar TUFAN May 03 '19 at 00:50
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    I get the same results. $I{{sp}} = \frac{I}{\Delta m{{propellant}}g}$ we get $\Delta m{{propellant}} = 29,6071kg/s$ and $v{{exhaust}} = C = gI{{sp}} = 3080,34m/s$ and using $\Delta t = \frac{M{{L}}}{E{{V}}}\cdot(1-e^{-(\frac{\Delta v}{E{{V}}})}) = 156.92s$. But I have a question, With $I{{sp}} = \frac{I{{sp}}}{\Delta v{{propellant}}}g$ we found $\Delta m{{propellant}} = 29,6071kg/s$, why this gives us mass flow rate instead of $m{{propellant final}} - m{{propellant first}}$ so change in propellant mass, I couldn't get this, I haven't solve any problem using this method. – M. Çağlar TUFAN May 03 '19 at 10:18
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    Thanks for checking my math! I calculated mass flow rate from $\dot m = T / (I_{sp} g_0)$ and multiplied this mass flow rate by the burn time (which we already knew for this real world problem) to get the delta mass. I will have to think about the equation you show. – Organic Marble May 03 '19 at 12:18
  • Hmm. You divided total thrust ($T$ or $Impulse = F\Delta t$) to $I{{sp}}g{{0}}$ same in my specific impulse equation, yep. So the $\dot{m}$ is equal to $\Delta m{_{propellant}}$ as I can see. Is it true? – M. Çağlar TUFAN May 03 '19 at 13:53
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    I got $\dot m$ = 29.63 kg/s so multiplying that flow rate by the burn time of 156.92 s gives $\Delta m$ = 4650 kg. So my $\dot m$ is not equal to my $\Delta m$ if I understand your question. – Organic Marble May 03 '19 at 14:35
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    Oh, OK. I understand now. Thank you so much @Organic Marble. I'm going to start reading book that you recommanded now. I will ask you questions more and more when I stuck about new book. Have a good day, you helped me so much! – M. Çağlar TUFAN May 03 '19 at 15:41
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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.

W H G
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  • Thanks! I used your equation on the Apollo data linked in my answer and it worked great. I have no idea why Taylor gives that bogus equation. – Organic Marble Nov 01 '23 at 23:15