Basic Rocket Science Q & A

[ckiki lwai]

 
I hope the mods find this new Q & A allright

Suppose you traveled from Earth to Mars using a Hohmann transfer orbit.
How do you calculate the time you have to wait to get another chance to fly back to Earth again using a Hohmann transfer orbit?

Edit: Sorry didn't read James Lowe1 post in the section intro thread...
I can't remove it anymore, delete it when you get the chance.

[Spacenick]

It's not strictly Orbital mechanics, but I dind't find a thread that fitted the topic better. Can someone describe to me how a tank pressure fed rocket operates, the thing I don't understand is how the tank pressure can push fuel into the chamber even though the chamber pressure must be many times higher than the tank pressure?

[zeke01]

Simple: For a tank pressure fed rocket you increase the pressure in the tanks until it's higher than the combustion chamber pressure.   To obtain high efficiency this would require thick walled tanks and pipes which makes it a very heavy, albeit very simple, propulsion system when compared to other types of rocket cycles.

Most efficient, chemical rocket engines nowadays use pumps to boost fluid pressures from low-pressure tanks to values required for efficient combustion, reducing the amount of high pressure (thick & heavy) plumbing.

zeke

[Jim]

I hope the mods find this new Q & A allright

Suppose you traveled from Earth to Mars using a Hohmann transfer orbit.
How do you calculate the time you have to wait to get another chance to fly back to Earth again using a Hohmann transfer orbit?

It has to do with knowing the where the planets are in their orbits and how long it takes for them to get into the right positions.  It is every 26 months for earth to mars

[Jim]

It's not strictly Orbital mechanics, but I dind't find a thread that fitted the topic better. Can someone describe to me how a tank pressure fed rocket operates, the thing I don't understand is how the tank pressure can push fuel into the chamber even though the chamber pressure must be many times higher than the tank pressure?

The tank pressure is always higher than the combustion chamber for a pressure fed system

[Spacenick]

I guess this means that the heat from the combustion is used to augment the tank pressure, right? Else one wouldn't really need the combustion, or am I wrong with this.

[William Barton]

I guess this means that the heat from the combustion is used to augment the tank pressure, right? Else one wouldn't really need the combustion, or am I wrong with this.

The combution chamber he's talking about is the rocket engine. The tank pressure has to be higher than the chamber pressure or the engine will starve (or worse, blow back through the valves and feedlines, creating a nice bomb).

[Spacenick]

Yeah, I'm talking about exactly the same chamber it's the one LH2 and LOX get pumped into to combust kicking the gas out through the nozzle producing the thrust.
The thing is if the tank pressure were higher than the pressure in the chamber WITHOUT using energy from the combustion, one wouldn't need the combustion and could let the fuel flow directly to the nozzle, so my question was how the pressure is augmented so that it is higher than the chamber pressure.

[Jim]

Yeah, I'm talking about exactly the same chamber it's the one LH2 and LOX get pumped into to combust kicking the gas out through the nozzle producing the thrust.
The thing is if the tank pressure were higher than the pressure in the chamber WITHOUT using energy from the combustion, one wouldn't need the combustion and could let the fuel flow directly to the nozzle, so my question was how the pressure is augmented so that it is higher than the chamber pressure.

The pressure isn't augmented.  It is by design higher in the tank than combustion chamber. 

combustion adds energy and it provides the bulk of the  increase the velocity of the exhaust.

If you have trouble grasping this, then I bet a jet engine throws you into a tizzy.   Need to read a basic rocket engine book

[Spacenick]

Well i think I got it, the important thing in the chamber is not the increase of pressure due to combustion but the high temperature of the exhaust leading to an extreme increase in volume and particle speed (which in itself isn't anything else than temperature). This in turn makes the exhaust exit the nozzle at extremly high speeds (the higher the better). Not really part of the physics we did in school physics though^^

[Antares]

Rocket Propulsion Elements by Sutton is the standard.  I advise the 6th edition.  There was a 7th updated by Biblarz, but it was full of errors.

Fundamentals of Astrodynamics by Bate, Mueller and White (known in classes as BMW) is an excellent, cheap text on orbital mechanics.

[jabe]

Rocket Propulsion Elements by Sutton is the standard.  I advise the 6th edition.  There was a 7th updated by Biblarz, but it was full of errors.

Fundamentals of Astrodynamics by Bate, Mueller and White (known in classes as BMW) is an excellent, cheap text on orbital mechanics.

Thanks for the tip...
I recently bought the "BMW" book and love it..Now i need to find the sixth edition by sutton..

cheers
jb
BTW how can the 7th be so full of errors??

[Danny Dot]

It's not strictly Orbital mechanics, but I dind't find a thread that fitted the topic better. Can someone describe to me how a tank pressure fed rocket operates, the thing I don't understand is how the tank pressure can push fuel into the chamber even though the chamber pressure must be many times higher than the tank pressure?

The pressure feed engines the shuttle uses on orbit have a chamber pressure of about 125 psi and tank pressures of about 300 psi.  I was very surprised when I first learned a rocket engine can operate with such low pressure.  The key to efficiency is high temperature, not high pressure.

[Danny Dot]

Well i think I got it, the important thing in the chamber is not the increase of pressure due to combustion but the high temperature of the exhaust leading to an extreme increase in volume and particle speed (which in itself isn't anything else than temperature). This in turn makes the exhaust exit the nozzle at extremly high speeds (the higher the better). Not really part of the physics we did in school physics though^^

You are correct.  It is the temperature that causes the exhaust velocity to be high.

[antonioe]

There were numerous hints at the true role of the chemical reaction (call it "combustion" if you wish) in a rocket engine; let me summarize what was spread over the previous comments:

The true core of a rocket engine is the nozzle, or, more precisely the convergent-divergent nozzle, a frighteningly simple but also frighteningly subtle machine invented by Gustav de Laval around the turn of the (nineteenth) century.  A Laval nozzle converts gas temperature and pressure into gas velocity in an extremely efficient (lossless) way.  Indeed, the process is very close to "adiabatic", i.e. there is almost no exchange of heat between the gas and anything else during its transit through the nozzle (well, there is some; it looks like a lot to the nozzle, but it looks like very little to the gas).

As heat and pressure are converted into velocity, the temperature (as well as the pressure) of the gas decreases.  If we had injected propellant into the nozzle at tank temperature (either because the tanks are at high pressure, as in a pressure-fed engine, or because we mechanically compressed it with a turbopump, in which case it will be a bit hotter than the tanks due to the compression), it will be mightily cold by the time it expands towards ambient pressure.  Cold also means dense, so the speed will not be very high, since the gas has decreased its volume a lot.

Enter the "combustion"; it takes the (not very hot) compressed gas and heats it up so that, when it cools down through the nozzle, it still remains at a high temperature and low density.  The chain of events could be conceived as follows:

1.- Pump or tank pressure causes the gas to want to leave through the nozzle, towards the area of lower density "out there".

2.- Chemical reaction increases the energy of the gas in the form of thermal energy.

3.- The fluid-dynamics of the nozzle accelerate the gas leaving the nozzle at the expense of pressure and temperature, therefore converting (in a most efficient way) thermal energy into mechanical (gas velocity) energy.

The main effect of the pressure at the beginning of the nozzle is to increase the "thrust density" of the device: high (chamber) pressure engines are small for the amount of propellant that goes through them in a seconds, therefore have high thrust for their size; conversely, low pressure engines have low thrust.  Sometimes (e.g. when you are lifting off vertically against 1 g) you want a substantial amount of thrust.  Other times (e.g., while delicately approaching the ISS) you do NOT want a lot of thrust.  This explains why we have both high pressure AND low pressure engines.

The efficiency of this process, when the expansion is well matched to the outside pressure, is remarkable: it is practically at the limit of the Carnot efficiency between the combustion chamber temperature and the outside conditions.  This leaves the chemistry of the reaction (the energy it adds to a unit mass of propellant) as the physical limit of the performance of any chemical rocket engine.

Society has been so spoiled by Moore's law of technological rate of improvement in electronics that they simply cannot believe that the asymptote of technological improvement in rocketry was achieved in 1970, with no hope of significant improvements...

I check my edition of Sutton; I have two: a third edition (1963 - I graduated in 1972) and a fifth (1986 - the year I left MIT to join Orbital).  According to the bookstore markings, I paid $28.95 for the thrid edition - new -  and $45.95 for the fifth, also new... *SIGH*)

[JWag]

There were numerous hints at the true role of the chemical reaction (call it "combustion" if you wish) in a rocket engine; let me summarize what was spread over the previous comments:

<snip>

Thank you for writing that.

[BriA]

I hope the mods find this new Q & A allright

Suppose you traveled from Earth to Mars using a Hohmann transfer orbit.
How do you calculate the time you have to wait to get another chance to fly back to Earth again using a Hohmann transfer orbit?

It has to do with knowing the where the planets are in their orbits and how long it takes for them to get into the right positions.  It is every 26 months for earth to mars

I actually had to solve this exact problem a couple months ago.  Couldn't resist the chance to give a detailed explanation.

1)  The 26 months referred to earlier is the synodic period, which is calculated using the periods of the two planets in question.  It refers to the amount of time it takes for two planets in circular, coplanar orbits to come back to their same position relative to each other.  Granted, neither Earth nor Mars are in exactly circular orbits, nor are they coplanar.  But it's close enough for a first approximation (and for a homework problem).

2)  There is a formula to calculate the angle (I'll call it the beta angle) between the two planets that must exist to start a Hohmann transfer from Earth to Mars, and vice versa.  In order to do a Hohmann from Earth to Mars, Earth must be trailing Mars by about 44 degrees.  To do the return transfer back to Earth, Earth must be trailing Mars by about 75 degrees.

3)  In order to actually solve the problem of how long you have to wait to do a Hohmann transfer back to Earth after arriving at Mars, you must look at the angle between the planets at arrival (found using the period of time for the initial hohmann), the angle between the planets at departure (known), and the angular distance each planet covers during the interim (unknown).  You then set the interim as the variable, find the amount of angular distance each planet has to cover during the time period in question (Earth covers 2*pi plus the unknown angle, Mars covers all three angles but does not make a full revolution), divide it by the angular velocity of each planet (360 degrees/planet period), set the two times equal to each other, and solve for the unknown angle.

As an aside, another book I would recommend to anyone interested in this topic (in addition to the ones that have already been touted, which are all excellent) is "Orbital Mechanics" by Prussing and Conway.  The formulas for all the equations I mentioned above are in the tail end of Chapter 6.

[AnimatorRob]

This seems like a good place to ask this..
As I understand it, an upper stage is responsible for the final ,precise delta-v to place the payload in is correct orbit, varying it's burn time to correct for variations in first stage performance, atmospheric conditions, etc. My question is how is that accomplished with a solid fueled upper stage, for example,  a Castor 30.

[Jim]

My question is how is that accomplished with a solid fueled upper stage, for example,  a Castor 30.

By the spacecraft, another stage or  requirements that account for the errors.

The mission is planned for "excess" performance then flies t

[ckiki lwai]

I hope the mods find this new Q & A allright

Suppose you traveled from Earth to Mars using a Hohmann transfer orbit.
How do you calculate the time you have to wait to get another chance to fly back to Earth again using a Hohmann transfer orbit?

It has to do with knowing the where the planets are in their orbits and how long it takes for them to get into the right positions.  It is every 26 months for earth to mars

I actually had to solve this exact problem a couple months ago.  Couldn't resist the chance to give a detailed explanation.

1)  The 26 months referred to earlier is the synodic period, which is calculated using the periods of the two planets in question.  It refers to the amount of time it takes for two planets in circular, coplanar orbits to come back to their same position relative to each other.  Granted, neither Earth nor Mars are in exactly circular orbits, nor are they coplanar.  But it's close enough for a first approximation (and for a homework problem).

2)  There is a formula to calculate the angle (I'll call it the beta angle) between the two planets that must exist to start a Hohmann transfer from Earth to Mars, and vice versa.  In order to do a Hohmann from Earth to Mars, Earth must be trailing Mars by about 44 degrees.  To do the return transfer back to Earth, Earth must be trailing Mars by about 75 degrees.

3)  In order to actually solve the problem of how long you have to wait to do a Hohmann transfer back to Earth after arriving at Mars, you must look at the angle between the planets at arrival (found using the period of time for the initial hohmann), the angle between the planets at departure (known), and the angular distance each planet covers during the interim (unknown).  You then set the interim as the variable, find the amount of angular distance each planet has to cover during the time period in question (Earth covers 2*pi plus the unknown angle, Mars covers all three angles but does not make a full revolution), divide it by the angular velocity of each planet (360 degrees/planet period), set the two times equal to each other, and solve for the unknown angle.

As an aside, another book I would recommend to anyone interested in this topic (in addition to the ones that have already been touted, which are all excellent) is "Orbital Mechanics" by Prussing and Conway.  The formulas for all the equations I mentioned above are in the tail end of Chapter 6.

Well thanks anyway, I had the same question on my exam only with Pluto instead of Mars, but I didn't had the time to solve the question anyway. I did pass the exam tough