HYBRID FUELING 101
Posted: Mon Mar 17, 2008 10:56 pm
Part 1: Using traditional hybrid fuel meters correctly
Question: How many of those who have constructed hybrid launchers have used either an online calculator to determine their fuel meter pressure, or have used 0.042*mix*ChamberVolume or similar to obtain the volume of fuel required for your hybrid's chamber?
Question 2: How many of those who have used the above method have had problems when using relatively strong burst disks and high mixes?
There seems to be a good deal of confusion surrounding the process of fueling hybrid launchers. Most users fail to understand the exact principles involved, and instead misconceive the integral components of fueling, thus resulting in problematic or sub maximum performance. This post will cover the most important components of proper hybrid fueling.
Let us first assume that we have a sealed chamber with a fixed volume of 100 cubic inches.
Common fuel meter pressure calculators assume that the pressure inside the chamber will NOT rise when the fuel is injected, meaning the incoming propane charge of 4.2 cubic inches will displace 4.2 cubic inches of air inside the chamber, giving a stoichiometric ratio of 95.8:4.2. This is probably quite accurate for standard combustion launchers, but hybrid launchers use sealed chambers, which requires a different approach.
Back to our chamber. Because is is sealed, the oxidizer volume is FIXED at 100 cubic inches. Assuming the chamber is at standard pressure, this 100 cubic inches is one portion of a 95.8% stoichiometric ratio. So, to obtain the proper quantity of propane, we must figure out what 100 cubic inches is 95.8% of (this means dividing by the fractional value of this, which is 0.958), then subtract the chamber volume. So, we have;
V<sub>P</sub> = 100/0.958-100
V<sub>P</sub> = 4.384 cubic inches (I am going to leave out significant figures for these calculations)
As you can see, this is not a significant difference at atmospheric pressure, but once you begin to increase the propane content of the fuel mixture, huge error margins can arise.
Now, because you will be adding an additional 100ci of air, this function will now scale linearly, and you can simply multiply the required volume of propane from the 1x calculation by a scale factor of 2.
V<sub>P</sub> = 4.384*2
V<sub>P</sub> = 8.77 cubic inches
Alternatively, you can simply multiply both volume terms by their corresponding mix numbers in the original calculation;
V<sub>P</sub> = (100*10)/0.958-(100*10)
You can now divide this figure by your fuel meter volume to obtain a meter pressure in atmospheres.
I am hoping you all understand this by now, but if any portion is boggling your mind, just ask about it.
Part 2: Using chamber pressure to achieve stoichiometry
DYI brought this up in another thread as he intends to use it on a hybrid launcher, so I figured the "chamber pressure meter" had a place in this thread.
As you could have guessed, the chamber pressure in a closed vessel will rise proportionally to the amount of gas added to it. This forms the basis for an accurate metering setup using only the chamber pressure as reference. Propane is injected until the chamber reaches a certain pressure, then the supply is cut off, and air is added.
To determine the chamber pressure required to obtain a specific volume of fuel, we must first calculate the volume of fuel required. Out comes the 100ci chamber, which will be using a 10x mix (lower mixes really aren't very practical with such a metering setup). The required fuel volume for stoichiometry is;
V<sub>P</sub> = (100*10)/0.958-(100*10)
V<sub>P</sub> = 44 cubic inches
Now, after adding the 44 cubic inches of propane to the 100 cubic inches of air, your chamber pressure will has risen. By how much exactly? To figure that out, you must first figure out what fraction of the chamber's volume you have added. In this case, it is 44/100 = 0.44. Because one atmosphere is equal to 14.7 psi, multiply the fraction by 14.7 to obtain your result.
P<sub>C</sub> = 0.44*14.7
P<sub>C</sub> = 6.44 psi
This pressure will be constant no matter the chamber volume, but will vary depending upon the fuel:volume ratio.
Hopefully this guide has been helpful, and will prevent people from inaccurately fueling hybrids in the future. As I said before, if there is something you don't understand, read it again until you do, or if that fails, ask about it.
Question: How many of those who have constructed hybrid launchers have used either an online calculator to determine their fuel meter pressure, or have used 0.042*mix*ChamberVolume or similar to obtain the volume of fuel required for your hybrid's chamber?
Question 2: How many of those who have used the above method have had problems when using relatively strong burst disks and high mixes?
There seems to be a good deal of confusion surrounding the process of fueling hybrid launchers. Most users fail to understand the exact principles involved, and instead misconceive the integral components of fueling, thus resulting in problematic or sub maximum performance. This post will cover the most important components of proper hybrid fueling.
Let us first assume that we have a sealed chamber with a fixed volume of 100 cubic inches.
Common fuel meter pressure calculators assume that the pressure inside the chamber will NOT rise when the fuel is injected, meaning the incoming propane charge of 4.2 cubic inches will displace 4.2 cubic inches of air inside the chamber, giving a stoichiometric ratio of 95.8:4.2. This is probably quite accurate for standard combustion launchers, but hybrid launchers use sealed chambers, which requires a different approach.
Back to our chamber. Because is is sealed, the oxidizer volume is FIXED at 100 cubic inches. Assuming the chamber is at standard pressure, this 100 cubic inches is one portion of a 95.8% stoichiometric ratio. So, to obtain the proper quantity of propane, we must figure out what 100 cubic inches is 95.8% of (this means dividing by the fractional value of this, which is 0.958), then subtract the chamber volume. So, we have;
V<sub>P</sub> = 100/0.958-100
V<sub>P</sub> = 4.384 cubic inches (I am going to leave out significant figures for these calculations)
As you can see, this is not a significant difference at atmospheric pressure, but once you begin to increase the propane content of the fuel mixture, huge error margins can arise.
Now, because you will be adding an additional 100ci of air, this function will now scale linearly, and you can simply multiply the required volume of propane from the 1x calculation by a scale factor of 2.
V<sub>P</sub> = 4.384*2
V<sub>P</sub> = 8.77 cubic inches
Alternatively, you can simply multiply both volume terms by their corresponding mix numbers in the original calculation;
V<sub>P</sub> = (100*10)/0.958-(100*10)
You can now divide this figure by your fuel meter volume to obtain a meter pressure in atmospheres.
I am hoping you all understand this by now, but if any portion is boggling your mind, just ask about it.
Part 2: Using chamber pressure to achieve stoichiometry
DYI brought this up in another thread as he intends to use it on a hybrid launcher, so I figured the "chamber pressure meter" had a place in this thread.
As you could have guessed, the chamber pressure in a closed vessel will rise proportionally to the amount of gas added to it. This forms the basis for an accurate metering setup using only the chamber pressure as reference. Propane is injected until the chamber reaches a certain pressure, then the supply is cut off, and air is added.
To determine the chamber pressure required to obtain a specific volume of fuel, we must first calculate the volume of fuel required. Out comes the 100ci chamber, which will be using a 10x mix (lower mixes really aren't very practical with such a metering setup). The required fuel volume for stoichiometry is;
V<sub>P</sub> = (100*10)/0.958-(100*10)
V<sub>P</sub> = 44 cubic inches
Now, after adding the 44 cubic inches of propane to the 100 cubic inches of air, your chamber pressure will has risen. By how much exactly? To figure that out, you must first figure out what fraction of the chamber's volume you have added. In this case, it is 44/100 = 0.44. Because one atmosphere is equal to 14.7 psi, multiply the fraction by 14.7 to obtain your result.
P<sub>C</sub> = 0.44*14.7
P<sub>C</sub> = 6.44 psi
This pressure will be constant no matter the chamber volume, but will vary depending upon the fuel:volume ratio.
Hopefully this guide has been helpful, and will prevent people from inaccurately fueling hybrids in the future. As I said before, if there is something you don't understand, read it again until you do, or if that fails, ask about it.
^^As clear as can be (I got a 404 Error).
