Jump to content

Delta V calculation for multistage liquid fueled rockets with SRB boosters


Recommended Posts

Hello! I have recently gotten into calculating the delta V of all my rockets, just for fun :P, and have come across a problem. I have some rockets where the first stage is a liquid fueled engine and 2 SRB's. The basic equation assumes that all of the boosters burn out at the same time. However, in this case, the SRB's burn out first, while the liquid-fueled portion of the stage continues burning. Is there any perfect equation for this calculation?

Link to comment
Share on other sites

You just need to consider how much fuel the center stage used until the SRBs are expent,  wich is easy since the fuel consumption rate is constant.  Just multiply the fuel consumption (in mass)  of the center engine and burn time of SRBs. 

 

Link to comment
Share on other sites

30 minutes ago, Spricigo said:

You just need to consider how much fuel the center stage used until the SRBs are expent,  wich is easy since the fuel consumption rate is constant.  Just multiply the fuel consumption (in mass)  of the center engine and burn time of SRBs. 

 

Can you make an equation? I am only a freshman in high school, and know the basic equation, but I am not sure how to do this without an equation to plug the numbers into... :blush:

Link to comment
Share on other sites

You must consider the SRBs + liquid engine as a separate stage from the liquid engine alone.

From thrust, Isp and solid fuel mass, calculate the SRB burn time:

t = m * Isp * g0 / F

Where t is time in seconds, m is solid fuel mass in tons, Isp is specific impulse of the SRB in seconds, g0 is 9.80665m/s2, and F is the SRB thrust in kN

Using that time, calculate how much fuel will be consumed by the liquid fuel engine in the same time (same relationship as above, just rearranged):

m = F * t / (Isp * g0)

Where m is the mass of LF+O consumed in tons, t is the time from the first equation, F is the liquid engine thrust in kN, Isp is the specific impulse of the liquid engine in seconds, and g0 is 9.80665m/s2.

Now you can combine the fuel masses and Isp values as you would do for a stage which burns out simultaneously (which it is, in effect), then compute the remaining fuel for the liquid engine alone as a separate stage.

Link to comment
Share on other sites

2 minutes ago, Red Iron Crown said:

You must consider the SRBs + liquid engine as a separate stage from the liquid engine alone.

From thrust, Isp and solid fuel mass, calculate the SRB burn time:

t = m * Isp * g0 / F

Where t is time in seconds, m is solid fuel mass in tons, Isp is specific impulse of the SRB in seconds, g0 is 9.80665m/s2, and F is the SRB thrust in kN

Using that time, calculate how much fuel will be consumed by the liquid fuel engine in the same time (same relationship as above, just rearranged):

m = F * t / (Isp * g0)

Where m is the mass of LF+O consumed in tons, t is the time from the first equation, F is the liquid engine thrust in kN, Isp is the specific impulse of the liquid engine in seconds, and g0 is 9.80665m/s2.

Now you can combine the fuel masses and Isp values as you would do for a stage which burns out simultaneously (which it is, in effect), then compute the remaining fuel for the liquid engine alone as a separate stage.

Ok! Thank you to everyone!!! This really helped. 

Link to comment
Share on other sites

The wiki have some stats for SRBs and LiquidFuelEngines,  (I'm not sure if accurate for current KSP version) .  So you can also use the given burn  time for the SRB (e.g. 23, 7s for hammers)  and fuel/s for LF engines (e.g. 12,747/s for swivel*)   to figure out how much fuel your LF engine used in this time (23,7*12,747=302,10).

From the stats of the Oscar-B fuel tank we have 18units of LF + 22 oxidizer weight 0,2t,  so one unit of LF plus the needed oxidizer weight 0,011t   (in your example 302,10*0,011=3,36t)

*I'm not sure but looks like the wiki list it in units of liquid fuel per second. (too lazy to check).  

Link to comment
Share on other sites

On 1/16/2017 at 4:17 PM, Red Iron Crown said:

You must consider the SRBs + liquid engine as a separate stage from the liquid engine alone.

From thrust, Isp and solid fuel mass, calculate the SRB burn time:

t = m * Isp * g0 / F

Where t is time in seconds, m is solid fuel mass in tons, Isp is specific impulse of the SRB in seconds, g0 is 9.80665m/s2, and F is the SRB thrust in kN

Using that time, calculate how much fuel will be consumed by the liquid fuel engine in the same time (same relationship as above, just rearranged):

m = F * t / (Isp * g0)

Where m is the mass of LF+O consumed in tons, t is the time from the first equation, F is the liquid engine thrust in kN, Isp is the specific impulse of the liquid engine in seconds, and g0 is 9.80665m/s2.

Now you can combine the fuel masses and Isp values as you would do for a stage which burns out simultaneously (which it is, in effect), then compute the remaining fuel for the liquid engine alone as a separate stage.

Quick question: When gis 9.81 m/s, does that change if taking off from another body (say Eve)?

Link to comment
Share on other sites

16 minutes ago, Benjamin Kerman said:

Quick question: When gis 9.81 m/s, does that change if taking off from another body (say Eve)?

No. Ordinarily one would use exhaust velocity in meters per second instead of the Isp * g0 term in the equation, but for historical reasons* rocket engine efficiency is often given in seconds of specific impulse and uses a convenient, well-known constant to convert it to a velocity. Same deal with the Tsiolkovsky's rocket equation. So always use the Earth standard g for those two equations (TWR calculations should use local g, though).

 

* The reason is the post war collaboration between metric-using German rocket scientists and Imperial-using American ones. Rather than having to convert numbers all the time, they used seconds as the unit is the same for both systems. It's a pity, exhaust velocity is a more intuitive measure as it literally describes how fast the exhaust exits the engine, whereas relating what those seconds represent is a little more convoluted.

Link to comment
Share on other sites

On 1/16/2017 at 6:17 PM, Red Iron Crown said:

You must consider the SRBs + liquid engine as a separate stage from the liquid engine alone.

Since the SRBs and the LF-O engine have different exhaust velocities, do you not have to use the weighted average for figuring out the dV of the portion when they're firing together? 

Link to comment
Share on other sites

Just now, FullMetalMachinist said:

Since the SRBs and the LF-O engine have different exhaust velocities, do you not have to use the weighted average for figuring out the dV of the portion when they're firing together? 

What you do in this case is calculate the average Isp of the engines with the equation 7b9c847d55656373321afe88a7622fdf.png

This seems a little complicated, but all you have to do is plug in the thrust in kN into the thrust slots, and plug in the Isp of the engine into the Isp slot.

Link to comment
Share on other sites

6 minutes ago, FullMetalMachinist said:

Since the SRBs and the LF-O engine have different exhaust velocities, do you not have to use the weighted average for figuring out the dV of the portion when they're firing together? 

You do, the equation governing Isp for heterogenous engine clusters is this one:

5649bb43490708471f734294d36eb565.png

Where F and Isp are force and specific impulse for each engine (you can combine identical engines together by adding their thrusts but not their Isps). I didn't mention this as the OP stated he had already figured out how to do the math for when they burn out simultaneously, I probably should have included this for clarity.

Link to comment
Share on other sites

Just now, Red Iron Crown said:

You do, the equation governing Isp for heterogenous engine clusters is this one:

5649bb43490708471f734294d36eb565.png

Where F and Isp are force and specific impulse for each engine (you can combine identical engines together by adding their thrusts but not their Isps). I didn't mention this as the OP stated he had already figured out how to do the math for when they burn out simultaneously, I porbably should have included this for clarity.

Am I the OP? (Not exactly sure what that means)

Link to comment
Share on other sites

4 minutes ago, sardia said:

Do Mechjeb and KER accurately calculate combined stages? Sorry if the OP doesn't like these  mods.

Nah, I don't mind, but I do not know. I just don't like trusting other people to do my calculations for me. :wink:

Edited by Benjamin Kerman
Link to comment
Share on other sites

This thread is quite old. Please consider starting a new thread rather than reviving this one.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...