Rbick
February 20th, 2008, 10:36 AM
I decided to start my own thread for this, as the HE section seems kind of dead, and I don't want people to miss this as I think it is useful information. Hope you don't mind Mega.
I've noticed the lack of informtion on ETN on the internet, and no one is really sure how it measures up to other explosives other than it is "similar to PETN and OB positive". This is why I have gone through the trouble of making these calculations. Through the calculations, I will be posting the measurement of PETN for the given equation so we can compare the two. Incredibly, ETN exceeds PETN in a couple ways, although loses out in the end due to mols of gas produced and VoD. All PETN measurements were taking from "The Chemistry of Explosives" by J. Akhavan. If you don't feel like reading through all of it, the final calculations are listed at the bottom.
Purpose: To better understand the technical aspects of the explosive ETN and to be able to apply the knowledge/equations used to other explosives.
Information that follows contains: Heat of Formation, Heat of detonation, Heat of explosion, Gas output, Power index in relation to H Form and Gas output, and Temperature of explosion. I also will add a little formula for calculating approximate VoD at a given density.
Heat of explosion and detonation:
Heats of Formation:
ΔHf(ETN) = -483.2 KJ mol = ΔH1 (PETN = -538)
ΔHf g(CO2) = -393.7 KJ mol
ΔHf g(H2O) = -242 KJ mol
CO will not be mentioned since ETN is OB positive. Oxygen and Nitrogen gas are not counted. Heat of formation is the amount of heat energy released when a given chemical compound "forms" when they establish new bonds.
ΔH2 = ΣΔHf g = (4) -393.7 + (3)-242 = -2300.8 KJ mol
The numbers multipled by the Hf of the gases where determined by the number of mols of each gas released from detonation. These can be seen in the decomposition of ETN:
C4H6N4O12 --> 4CO2 + 3H2O + 22 + 1/2O2
ΔHd (detonation) = ΔH2 - ΔH1 = -2300.8 - -483.2 = -1817.6 KJ mol
Converted to KJ Kg = Q = -1817.6 (1000)/302 = -6018.54 KJ Kg
The 302 is grams per mol for ETN, which actually exceeds PETN in energy per gram, albeit slightly.
ΔHd is the heat of detonation. It is the amount of heat energy released by 1 mol of explosive. A negative number means an exothermic reaction, which is not suprising since we are dealing with explosives. ΔHd for PETN = -1831 KJ mol and Q for PETN = -5794 KJ Kg
Gas Output:
As we saw before from the decomposition of ETN, 9.5 mols of gas are liberated per mol of ETN. According to the Ideal Gas law, 1 mol of any type of gas will occupy the same amount of space, regardless of atomic weight. For instance, at STP (standard temp. and pressure), 1 mol of a given gas will occupy 22.4L of space. This factor is part of what gives an explosive its heaving force. Also involved is the amount of heat produced, which will effect how "hard" these gasses heave. To find the volume of gas produced, we take the number of mols in gas: 91/2 and multiply it by the amount of space each mol will occupy : 22.4L
91/2 (22.4) = 212.8L = volume of gas per mol of ETN
To convert this to mols of gas per gram, take 212.8/302 = .7046 L g
mols of gas per gram for PETN = .780 L g
Temperature of explosion:
This is were things get interesting, and some guess work must be done. Temp of explosion is exactly what it sounds like. Te is the maximum temp which the products can reach under adiabatic conditions. In the following equations, whether a number is negative or not is paid not attention as there is no addition or subtraction of negative numbers and we know that temp of explosion will be positive. For this problem we use equation:
Te = Q/ΣCv + Ti
Q = Heat of detonation that we figured out earlier, ΣCv = Summation of specific heat capacities of the products, Ti = Initial temperature, which we will say is 0*C = 273K
The equation is then rearranged to Q = ΣCv (Te - Ti)
This is done because heat capacities of the products change with temperature, making this equation difficult to work straight forward. So this is where we play the guessing game, more or less. We look at RDX frpm a different source and see that its Temp of explosion is 4186K. We can assume ETN will have a higher Te since our previously calculations on heat energy have been more than RDX. So we will guess 4500K and plug it into the equation. Note: This wasn't my first actual guess, it took me a few times...
First I will list the specific heat capacities of the products at 4500K:
CO2 = 50.43 J mol, H2 = 42.3 J mol, N2 = 27.154 J mol, 1/2 O2 = 21.056 J mol
To find ΣCv, Multiply the given specific heats by the number of mols of gases from detonation. This will give you 403.984
Now subtract the Ti which we decided would be 273K from our guessed Te, 4500K.
4500 - 273 = 4227
Now we have the eqaution all set up:
Q4500K = 403.984(4227) = 1707640
This is in J mol, to convert it, multiply it by 1000 and get 1707.64 KJ mol. This is how we check ourselves. Earlier, we calculated that the heat of detonation for ETN was -1817.6 KJ mol, not 1707.64. This is how we know our guess is too low. So now we guess again.
This time we will try 5000K. Specific heats for products at 5000K are:
CO2 = 50.949 J mol, H2 = 43.137 J mol, N2 = 27.397 J mol, 1/2 O2 = 21.248 J mol
ΣCv = 409.244
Te - Ti = 5000 - 273 = 4727
Q5000K = 409.244 (4727) = 1934520 J mol, or 1934.52 KJ mol
Now this guess is too high! This is where your graph drawing skillz come in to play. On your Y axis, you label Heat Liberated in KJ and on the X you have your Temp of explosion. Plot the calculations for 4500K and 5000K and draw a straight line through them. Then, plot on the Y axis the actual heat of detonation we calculated earlier, or 1817.6 KJ mol. Now draw a straight line from that point through the line you drew for the other two points. Where they intersect is your accurate Temp of explosion for ETN. Excel could prove helpful, but I'm not fluent in excel, so I had to draw it out.
I put it at approx. 4721.25K = Te. I havn't calculated the Temp of explosion for PETN yet, but RDX, as mentioned early is 4186K, exceeded greatly by ETN. Remember that these measurements are approximate, as not all enviornmental factors are being accounted for.
Power index and explosive power
Now lastly, in a comparison to other explosive, we can produce a power index and explosive power. Explosive power is the heat of explosion multiplied by gas output in mols per gram of explosive and then divided by 10.
Explosive power = 6018.54KJ Kg-1 (.7046) / 10 = 424.06
Explosive index is this value multiplied by the explosive power of a standard. In this case, we will use TNT as the standard, which has the explosive power at 314.3.
So we take 424.06/314.3 (100) = 134.9% for ETN compared to TNT. Some other explosives are listed below. This is where ETN loses out, as it doesn't have a very impressive gas output. There are other ways to compare explosives power that include VoD, but this is a simple way to do it.
PETN = 143.7%, RDX = 145%, NG = 145.8%
VoD Calculations:
One misconception I see commonly is the thought that an explosives VoD is constant regardless of diameter or density. This is Very false. I am not aware of the equations for the effects of diameter, but I can tell you that all homogenous explosives have a critical diameter and the smaller it gets, the lower the VoD. The following equations are based on density alone and assume perfect diameter.
Vp1 = Vp2 +3500 (p1 - p2)
V 1 and 2 are the VoDs for densities p 1 and 2 respectively. 3500 is a constant.
Lets try PETN for example. VoD of 8400 m/sec at max density of 1.77 g/cm3. A density of this amount could probably only be reached through professional means (eg hydrolic press). So lets see what happens when the home chemist presses his to 1.2 g/cm3
8400 = x +3500 (1.77 - 1.2)
8400 = x + 1995
x = 8400 -1995 = 6405 m/sec
:o Thats a big difference! Not exactly disappointing, but definitely makes a substantial drop in VoD. This can be applied to ETN, which is rumored to have a VoD of approx. 8000 m/sec. The following equation can be used to find approx. VoD, but I can't seem to figure it out. Let me know if you can!
Vpx = 430 (nTd)1/2 + 3500(px - 1)
Here, V is the VoD of a given density p, n is number of mols per gram of gas produced and T is temp in kelvin at which the detonation occurs. Good luck!
Last notes
One more point I'd like to make: For those of you having troubles with ETN yields, my advice is temp control is everything! Last night I had a dream of mixing 50g ammonium nitrate with 100mL sulfuric acid, never exceeding 20* C. The reason for this can be seen in [list]
this article here (http://www.roguesci.org/theforum/attachment.php?attachmentid=992&d=1190786433"), which is posted at RS. It may not be available as RS is going through some reconstruction, but it will be in the future I'm sure. In short, past a certain temperature, sulfuric acid stops reacting with ammonium nitrate to produce nitric acid and NOx fumes and other unwanted products are produced instead. The temp listed is 30* if I remember correctly. So anyway, I allowed the mix to cool to 10*C and then slowly added my 16g of erythritol at about 1-2g per minute. The temp never went above 12* and after adding all the erythritol, I let it nitrate for an additional 20 minutes.
The mixture at this point is thick like honey with ETN crystals and is dumped into distilled water, filtered, washed with bicarb, rinsed with distilled water, washed again in bicarb, and again rinsed. pH is tested with litmus and washing continues until it is at 7 pH. I havn't measured my dry product yet, but I'm guessing it will be around 24g if not more.
I recall in my first trial runs with ETN, I got shitty yields. I attribute this to being impatient with my nitration and letting the temperature get too hot during both the nitrate addition and the erythritol addition. Lack of nitric acid and excessive heat will MURDER your yield. I would get pissed because I would get such crappy yields. After a certain temperature, probably 20*C, the mix will no longer nitrate the erythritol, but will dehydrate and destroy it instead. I see many other people making this same mistake. So point being: Be patient, use proper cooling equpment (eg ice/salt bath), stay in the correct temperature range, and be safe. Oh yeah, I also took a picture in my dream of the yield, which I will post later.
So I hope this information is helpful in comparing ETN with other explosives. If you have any criticism, ideas, comments, please feel free. To sum it all up:
ETN info:
Heat of detonation: -1817.6 KJ mol
Heat of explosion: -6018.54 KJ Kg
Volume of gas produced: .7046 mol g
Temperature of explosion (approx): 4721.25K
Power index in comparison to TNT: 134.7%
I've noticed the lack of informtion on ETN on the internet, and no one is really sure how it measures up to other explosives other than it is "similar to PETN and OB positive". This is why I have gone through the trouble of making these calculations. Through the calculations, I will be posting the measurement of PETN for the given equation so we can compare the two. Incredibly, ETN exceeds PETN in a couple ways, although loses out in the end due to mols of gas produced and VoD. All PETN measurements were taking from "The Chemistry of Explosives" by J. Akhavan. If you don't feel like reading through all of it, the final calculations are listed at the bottom.
Purpose: To better understand the technical aspects of the explosive ETN and to be able to apply the knowledge/equations used to other explosives.
Information that follows contains: Heat of Formation, Heat of detonation, Heat of explosion, Gas output, Power index in relation to H Form and Gas output, and Temperature of explosion. I also will add a little formula for calculating approximate VoD at a given density.
Heat of explosion and detonation:
Heats of Formation:
ΔHf(ETN) = -483.2 KJ mol = ΔH1 (PETN = -538)
ΔHf g(CO2) = -393.7 KJ mol
ΔHf g(H2O) = -242 KJ mol
CO will not be mentioned since ETN is OB positive. Oxygen and Nitrogen gas are not counted. Heat of formation is the amount of heat energy released when a given chemical compound "forms" when they establish new bonds.
ΔH2 = ΣΔHf g = (4) -393.7 + (3)-242 = -2300.8 KJ mol
The numbers multipled by the Hf of the gases where determined by the number of mols of each gas released from detonation. These can be seen in the decomposition of ETN:
C4H6N4O12 --> 4CO2 + 3H2O + 22 + 1/2O2
ΔHd (detonation) = ΔH2 - ΔH1 = -2300.8 - -483.2 = -1817.6 KJ mol
Converted to KJ Kg = Q = -1817.6 (1000)/302 = -6018.54 KJ Kg
The 302 is grams per mol for ETN, which actually exceeds PETN in energy per gram, albeit slightly.
ΔHd is the heat of detonation. It is the amount of heat energy released by 1 mol of explosive. A negative number means an exothermic reaction, which is not suprising since we are dealing with explosives. ΔHd for PETN = -1831 KJ mol and Q for PETN = -5794 KJ Kg
Gas Output:
As we saw before from the decomposition of ETN, 9.5 mols of gas are liberated per mol of ETN. According to the Ideal Gas law, 1 mol of any type of gas will occupy the same amount of space, regardless of atomic weight. For instance, at STP (standard temp. and pressure), 1 mol of a given gas will occupy 22.4L of space. This factor is part of what gives an explosive its heaving force. Also involved is the amount of heat produced, which will effect how "hard" these gasses heave. To find the volume of gas produced, we take the number of mols in gas: 91/2 and multiply it by the amount of space each mol will occupy : 22.4L
91/2 (22.4) = 212.8L = volume of gas per mol of ETN
To convert this to mols of gas per gram, take 212.8/302 = .7046 L g
mols of gas per gram for PETN = .780 L g
Temperature of explosion:
This is were things get interesting, and some guess work must be done. Temp of explosion is exactly what it sounds like. Te is the maximum temp which the products can reach under adiabatic conditions. In the following equations, whether a number is negative or not is paid not attention as there is no addition or subtraction of negative numbers and we know that temp of explosion will be positive. For this problem we use equation:
Te = Q/ΣCv + Ti
Q = Heat of detonation that we figured out earlier, ΣCv = Summation of specific heat capacities of the products, Ti = Initial temperature, which we will say is 0*C = 273K
The equation is then rearranged to Q = ΣCv (Te - Ti)
This is done because heat capacities of the products change with temperature, making this equation difficult to work straight forward. So this is where we play the guessing game, more or less. We look at RDX frpm a different source and see that its Temp of explosion is 4186K. We can assume ETN will have a higher Te since our previously calculations on heat energy have been more than RDX. So we will guess 4500K and plug it into the equation. Note: This wasn't my first actual guess, it took me a few times...
First I will list the specific heat capacities of the products at 4500K:
CO2 = 50.43 J mol, H2 = 42.3 J mol, N2 = 27.154 J mol, 1/2 O2 = 21.056 J mol
To find ΣCv, Multiply the given specific heats by the number of mols of gases from detonation. This will give you 403.984
Now subtract the Ti which we decided would be 273K from our guessed Te, 4500K.
4500 - 273 = 4227
Now we have the eqaution all set up:
Q4500K = 403.984(4227) = 1707640
This is in J mol, to convert it, multiply it by 1000 and get 1707.64 KJ mol. This is how we check ourselves. Earlier, we calculated that the heat of detonation for ETN was -1817.6 KJ mol, not 1707.64. This is how we know our guess is too low. So now we guess again.
This time we will try 5000K. Specific heats for products at 5000K are:
CO2 = 50.949 J mol, H2 = 43.137 J mol, N2 = 27.397 J mol, 1/2 O2 = 21.248 J mol
ΣCv = 409.244
Te - Ti = 5000 - 273 = 4727
Q5000K = 409.244 (4727) = 1934520 J mol, or 1934.52 KJ mol
Now this guess is too high! This is where your graph drawing skillz come in to play. On your Y axis, you label Heat Liberated in KJ and on the X you have your Temp of explosion. Plot the calculations for 4500K and 5000K and draw a straight line through them. Then, plot on the Y axis the actual heat of detonation we calculated earlier, or 1817.6 KJ mol. Now draw a straight line from that point through the line you drew for the other two points. Where they intersect is your accurate Temp of explosion for ETN. Excel could prove helpful, but I'm not fluent in excel, so I had to draw it out.
I put it at approx. 4721.25K = Te. I havn't calculated the Temp of explosion for PETN yet, but RDX, as mentioned early is 4186K, exceeded greatly by ETN. Remember that these measurements are approximate, as not all enviornmental factors are being accounted for.
Power index and explosive power
Now lastly, in a comparison to other explosive, we can produce a power index and explosive power. Explosive power is the heat of explosion multiplied by gas output in mols per gram of explosive and then divided by 10.
Explosive power = 6018.54KJ Kg-1 (.7046) / 10 = 424.06
Explosive index is this value multiplied by the explosive power of a standard. In this case, we will use TNT as the standard, which has the explosive power at 314.3.
So we take 424.06/314.3 (100) = 134.9% for ETN compared to TNT. Some other explosives are listed below. This is where ETN loses out, as it doesn't have a very impressive gas output. There are other ways to compare explosives power that include VoD, but this is a simple way to do it.
PETN = 143.7%, RDX = 145%, NG = 145.8%
VoD Calculations:
One misconception I see commonly is the thought that an explosives VoD is constant regardless of diameter or density. This is Very false. I am not aware of the equations for the effects of diameter, but I can tell you that all homogenous explosives have a critical diameter and the smaller it gets, the lower the VoD. The following equations are based on density alone and assume perfect diameter.
Vp1 = Vp2 +3500 (p1 - p2)
V 1 and 2 are the VoDs for densities p 1 and 2 respectively. 3500 is a constant.
Lets try PETN for example. VoD of 8400 m/sec at max density of 1.77 g/cm3. A density of this amount could probably only be reached through professional means (eg hydrolic press). So lets see what happens when the home chemist presses his to 1.2 g/cm3
8400 = x +3500 (1.77 - 1.2)
8400 = x + 1995
x = 8400 -1995 = 6405 m/sec
:o Thats a big difference! Not exactly disappointing, but definitely makes a substantial drop in VoD. This can be applied to ETN, which is rumored to have a VoD of approx. 8000 m/sec. The following equation can be used to find approx. VoD, but I can't seem to figure it out. Let me know if you can!
Vpx = 430 (nTd)1/2 + 3500(px - 1)
Here, V is the VoD of a given density p, n is number of mols per gram of gas produced and T is temp in kelvin at which the detonation occurs. Good luck!
Last notes
One more point I'd like to make: For those of you having troubles with ETN yields, my advice is temp control is everything! Last night I had a dream of mixing 50g ammonium nitrate with 100mL sulfuric acid, never exceeding 20* C. The reason for this can be seen in [list]
this article here (http://www.roguesci.org/theforum/attachment.php?attachmentid=992&d=1190786433"), which is posted at RS. It may not be available as RS is going through some reconstruction, but it will be in the future I'm sure. In short, past a certain temperature, sulfuric acid stops reacting with ammonium nitrate to produce nitric acid and NOx fumes and other unwanted products are produced instead. The temp listed is 30* if I remember correctly. So anyway, I allowed the mix to cool to 10*C and then slowly added my 16g of erythritol at about 1-2g per minute. The temp never went above 12* and after adding all the erythritol, I let it nitrate for an additional 20 minutes.
The mixture at this point is thick like honey with ETN crystals and is dumped into distilled water, filtered, washed with bicarb, rinsed with distilled water, washed again in bicarb, and again rinsed. pH is tested with litmus and washing continues until it is at 7 pH. I havn't measured my dry product yet, but I'm guessing it will be around 24g if not more.
I recall in my first trial runs with ETN, I got shitty yields. I attribute this to being impatient with my nitration and letting the temperature get too hot during both the nitrate addition and the erythritol addition. Lack of nitric acid and excessive heat will MURDER your yield. I would get pissed because I would get such crappy yields. After a certain temperature, probably 20*C, the mix will no longer nitrate the erythritol, but will dehydrate and destroy it instead. I see many other people making this same mistake. So point being: Be patient, use proper cooling equpment (eg ice/salt bath), stay in the correct temperature range, and be safe. Oh yeah, I also took a picture in my dream of the yield, which I will post later.
So I hope this information is helpful in comparing ETN with other explosives. If you have any criticism, ideas, comments, please feel free. To sum it all up:
ETN info:
Heat of detonation: -1817.6 KJ mol
Heat of explosion: -6018.54 KJ Kg
Volume of gas produced: .7046 mol g
Temperature of explosion (approx): 4721.25K
Power index in comparison to TNT: 134.7%