Editing Furnace temperature and pressure math
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− | + | ''The data used was collected in version 0.2.2800.'' | |
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=== Furnace behaviour === | === Furnace behaviour === | ||
− | *Only the gas inside the furnace have | + | *Only the gas inside the furnace have temperature, the furnace itself do not, nor does it take any energy from the gas. |
− | + | *For a combustion to occur, there must be at least 5% each of both O2 and H2 | |
− | *For a combustion to occur, | ||
*For a combustion to occur, there must be a minimum pressure of 10 kPa ''(0.2.2733.13309 changelog)'' | *For a combustion to occur, there must be a minimum pressure of 10 kPa ''(0.2.2733.13309 changelog)'' | ||
*A combustion will consume 95% of the limiting ingredient, O2 or H2 (if there is 10 mol H2, and excess O2, 0.5 mol H2 will remain afterwards) | *A combustion will consume 95% of the limiting ingredient, O2 or H2 (if there is 10 mol H2, and excess O2, 0.5 mol H2 will remain afterwards) | ||
− | *Combustion reaction formula: 1 O2 + 2 H2 -> 6 CO2 + 3 X + | + | *Combustion reaction formula: 1 O2 + 2 H2 -> 6 CO2 + 3 X + 595kJ ''(see discussions for the 595kJ value)'' |
− | + | *No side reactions have been observed so far | |
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− | * | ||
Regular furnace | Regular furnace | ||
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*Unlike the regular furnace, the advanced one always have a volume of 1000 L, regardless of how many pipes are attached to it. This saves a little bit of fuel. | *Unlike the regular furnace, the advanced one always have a volume of 1000 L, regardless of how many pipes are attached to it. This saves a little bit of fuel. | ||
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− | + | === Using perfect 2:1 fuel === | |
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− | === Using perfect | ||
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'''Temperature peak''' | '''Temperature peak''' | ||
− | *T(after) = ( T(before)*61.9 + | + | *T(after) = ( T(before)*61.9 + 565250 ) / 234.515 |
**T(after) is the temperature in Kelvin after ignition | **T(after) is the temperature in Kelvin after ignition | ||
**T(before) is the temperature in Kelvin before ignition | **T(before) is the temperature in Kelvin before ignition | ||
− | **The | + | **The 565250 value is because of the 95% combustion efficiency, the full 595kJ isn't released |
**61.9 is the heat capacity for 1 mol O2 and 2 mol H2, the sum of their specific heat values, the mol amounts comes from the reaction formula | **61.9 is the heat capacity for 1 mol O2 and 2 mol H2, the sum of their specific heat values, the mol amounts comes from the reaction formula | ||
**234.515 is the heat capacity for the gas obtained when 1 mol O2 and 2 mol H2 combusts with 95% efficiency (243.6 * 0.95 + 61.9 * 0.05) | **234.515 is the heat capacity for the gas obtained when 1 mol O2 and 2 mol H2 combusts with 95% efficiency (243.6 * 0.95 + 61.9 * 0.05) | ||
+ | *The number of mol combusted doesn't actually matter for the temperature, it will always reach the same value. More fuel will however release more total energy which means it takes longer for the furnace to cool down. | ||
*The equation can be arrived at by using the released energy per mol and the specific heat per mol of the mixture before and after combustion, also account for the 95% combustion efficiency. | *The equation can be arrived at by using the released energy per mol and the specific heat per mol of the mixture before and after combustion, also account for the 95% combustion efficiency. | ||
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=== Using diluted fuel === | === Using diluted fuel === | ||
− | + | ||
− | + | Unreactive gases can be added before the ignition to increase pressure and decrease temperature. An excess of either O2 or H2 also counts as unreactive. These equations can be used to predict the Temperature and Pressure inside the furnace after ignition. It's also possible to use this in reverse, and calculate what mixture of fuel and dilutant is needed to reach a desired Temperature and Pressure upon ignition, more about that below. | |
'''Temperature peak''' | '''Temperature peak''' | ||
− | *T(after) = ( T(before) * sum(specific heat * ratio(gas)(before)) + min(ratio(O2), ratio(H2)*0.5) * | + | *T(after) = ( T(before) * sum(specific heat * ratio(gas)(before)) + min(ratio(O2), ratio(H2)*0.5) * 565250 ) / ( sum(specific heat * ratio(gas)(before)) + min(ratio(O2), ratio(H2)*0.5) * 172.615) |
**T(after) is the temperature in Kelvin after ignition | **T(after) is the temperature in Kelvin after ignition | ||
**T(before) is the temperature in Kelvin before ignition | **T(before) is the temperature in Kelvin before ignition | ||
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**ratio(gas)(before) is the molecular ratio, given by the tablet as % values for each gas in the mix, before ignition | **ratio(gas)(before) is the molecular ratio, given by the tablet as % values for each gas in the mix, before ignition | ||
**min(x,y) returns the smallest value of x and y | **min(x,y) returns the smallest value of x and y | ||
− | ** | + | **565250 comes from 595000*0.95, the energy released when 1 mol O2 and 2 mol H2 combusts with 95% efficiency |
**172.615 comes from 0.95*(243.6-61.9), this is the change in heat capacity (specific heat*number of mol) of the gas when 1 mol O2 and 2 mol H2 combusts with 95% efficiency into its new compounds | **172.615 comes from 0.95*(243.6-61.9), this is the change in heat capacity (specific heat*number of mol) of the gas when 1 mol O2 and 2 mol H2 combusts with 95% efficiency into its new compounds | ||
**The equation comes from calculating the Thermal Energy (temperature*specific heat per mol) before combustion, add the released energy calculated from the ratio of O2, then divide with the ''specific heat per mol'' of the gas obtained after combustion, which is known thanks to the reaction formula, all of this becomes the temperature on ignition | **The equation comes from calculating the Thermal Energy (temperature*specific heat per mol) before combustion, add the released energy calculated from the ratio of O2, then divide with the ''specific heat per mol'' of the gas obtained after combustion, which is known thanks to the reaction formula, all of this becomes the temperature on ignition | ||
− | same thing but | + | same thing but easier to read |
− | *T(after) = ( T(before) * specificHeat(before) + | + | *T(after) = ( T(before) * specificHeat(before) + fuelRatio * 565250 ) / ( specificHeat(before) + fuelRatio * 172.615) |
**specificHeat(before) = RatioOxygen*21.1 + RatioVolatile*20.4 + RatioCarbonDioxide*28.2 + RatioPollutant*24.8 + RatioNitrogen*20.6 + RatioNitrousOxide*23 | **specificHeat(before) = RatioOxygen*21.1 + RatioVolatile*20.4 + RatioCarbonDioxide*28.2 + RatioPollutant*24.8 + RatioNitrogen*20.6 + RatioNitrousOxide*23 | ||
− | ** | + | **fuelRatio = min(RatioOxygen,RatioVolatile/2) |
**Notice that water isn't included. That's because this haven't been verified, but water is probably treated as a gas by the game so a good guess is that it can be included simply by adding RatioWater*72 to the specificHeat sum. | **Notice that water isn't included. That's because this haven't been verified, but water is probably treated as a gas by the game so a good guess is that it can be included simply by adding RatioWater*72 to the specificHeat sum. | ||
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*P(after) = P(before) * T(after) * ( 1 + 5.7*min(ratio(O2), ratio(H2)*0.5) ) / T(before) | *P(after) = P(before) * T(after) * ( 1 + 5.7*min(ratio(O2), ratio(H2)*0.5) ) / T(before) | ||
**this expression comes from two sets of PV=nRT, one after and one before combustion. The reaction formula say that for each mol consumed O2 we gain 6 mol gas (9-3), this creates a link between the equations, n(after) = n(before)*(1+min(ratio(O2), ratio(H2)*0.5)*6), then include the 0.95 efficiency as well | **this expression comes from two sets of PV=nRT, one after and one before combustion. The reaction formula say that for each mol consumed O2 we gain 6 mol gas (9-3), this creates a link between the equations, n(after) = n(before)*(1+min(ratio(O2), ratio(H2)*0.5)*6), then include the 0.95 efficiency as well | ||
+ | |||
=== Using Ice(Oxite) and Ice(Volatiles) === | === Using Ice(Oxite) and Ice(Volatiles) === | ||
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Observations | Observations | ||
− | *the rate of cooling is temperature dependent, hotter cools faster | + | *the rate of cooling is temperature dependent, hotter cools faster (furnace temp - surrounding temp? how do vaccum behave?) |
− | |||
*the rate of cooling is time dependent (game tick speed is once per 0.5 seconds) | *the rate of cooling is time dependent (game tick speed is once per 0.5 seconds) | ||
*the rate of cooling is mol dependent (small amounts cool faster) | *the rate of cooling is mol dependent (small amounts cool faster) | ||
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*Hold a tablet with an atmos cartridge in the right hand (so it can be read when the game is paused). Aim the tablet against the furnace and pause with ESC, double tap ESC to move the game forward one tick, record the temperatures. | *Hold a tablet with an atmos cartridge in the right hand (so it can be read when the game is paused). Aim the tablet against the furnace and pause with ESC, double tap ESC to move the game forward one tick, record the temperatures. | ||
*Remember to record the ''total amount of moles'' as well | *Remember to record the ''total amount of moles'' as well | ||
+ | |||
+ | |||
+ | === Resetting the furnace === | ||
+ | Since only gas have temperature, evacuating all gas means resetting the temperature. | ||
+ | |||
=== Calculating how to reach a desired Temperature and Pressure on ignition === | === Calculating how to reach a desired Temperature and Pressure on ignition === | ||
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#The specific heat value of the gas used to dilute the fuel (if a mix of gases is used, the specific heat to use is the average specific heat per mol, example calculation below) | #The specific heat value of the gas used to dilute the fuel (if a mix of gases is used, the specific heat to use is the average specific heat per mol, example calculation below) | ||
− | + | The equations will give these results | |
#The ratio(fuel) in the fuel-dilutant mix. (0.75 means 75% perfect 1:2 fuel mols (O2 and H2 added together)†, the other 25% will be dilutant gas mols, making it a 1:2:1 mix of O2:H2:dilutant) | #The ratio(fuel) in the fuel-dilutant mix. (0.75 means 75% perfect 1:2 fuel mols (O2 and H2 added together)†, the other 25% will be dilutant gas mols, making it a 1:2:1 mix of O2:H2:dilutant) | ||
#The total pressure of the fuel-dilutant mix inside the furnace before ignition | #The total pressure of the fuel-dilutant mix inside the furnace before ignition | ||
†It's helpful to separate out the fuel part like this since everyone should be using pre-mixed fuel, it makes the diluting easier and has a lower risk to cause confusion when using either O2 or H2 to be the dilutant gas | †It's helpful to separate out the fuel part like this since everyone should be using pre-mixed fuel, it makes the diluting easier and has a lower risk to cause confusion when using either O2 or H2 to be the dilutant gas | ||
− | To freely control the temperature and pressure, the fuel must be diluted with a non- | + | To freely control the temperature and pressure, the fuel must be diluted with a non-combustable gas. This can be added either before or after ignition, doing so before ignition makes it alot easier to predict, doing so after ignition is more of an art than a science (it depends on how the furnace is built and how fast the operator can work). The method prefered here is to add the non-combustable gas before ignition. |
Diluting the fuel can be done in the furnace directly or in pipes outside of it. There are good and bad points with both ways. Diluting outside fits the advanced furnace best (the built-in volume pump can easily move all of the prepared gas inside), diluting inside fits the regular furnace best (the exhaust outlet can be used as an inlet but it's a little bit quirky, and diluting in pipes outside means not all of the prepared gas can be moved into the furnace (the pipe directly on the furnace inlet will hold on to some of the diluted fuel) so extra gas must always be prepared). | Diluting the fuel can be done in the furnace directly or in pipes outside of it. There are good and bad points with both ways. Diluting outside fits the advanced furnace best (the built-in volume pump can easily move all of the prepared gas inside), diluting inside fits the regular furnace best (the exhaust outlet can be used as an inlet but it's a little bit quirky, and diluting in pipes outside means not all of the prepared gas can be moved into the furnace (the pipe directly on the furnace inlet will hold on to some of the diluted fuel) so extra gas must always be prepared). | ||
− | It is worth noting that for some temperatures and pressures suitable for advanced alloys, the calculation can suggest a fuel ratio below 0.15. This will not work however, since it means having less than 5% oxygen, that mix will not combust (unless the dilutant contains extra oxygen) | + | It is worth noting that for some temperatures and pressures suitable for advanced alloys, the calculation can suggest a fuel ratio below 0.15. This will not work however, since it means having less than 5% oxygen, that mix will not combust (unless the dilutant contains extra oxygen). |
− | The dilution can | + | The dilution can be double checked by using the tablet and looking at the mol% values for the fuel mix. If the outlet on the regular furnace was used as an inlet, the first gas that entered there will have been mostly pushed back into the furnace, making the mol% values diffrent but the total number of mol are still the same. |
'''Calculating the fuel ratio''' | '''Calculating the fuel ratio''' | ||
− | *ratio(fuel) = s*( T(after) - T(before) ) / ( | + | *ratio(fuel) = s*( T(after) - T(before) ) / ( 188417 + T(before)*(20.633 - s) - T(after)*(78.172 - s) ) |
**ratio(fuel), 1 = 100% fuel which is 33.33% O2 and 66.67% H2 | **ratio(fuel), 1 = 100% fuel which is 33.33% O2 and 66.67% H2 | ||
**T(after) is the chosen temperature after ignition, in K | **T(after) is the chosen temperature after ignition, in K | ||
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***example: 15% N2 and 85% CO2 as dilutant -> specific heat = 0.15*20.6 + 0.85*28.2 = 27.06 | ***example: 15% N2 and 85% CO2 as dilutant -> specific heat = 0.15*20.6 + 0.85*28.2 = 27.06 | ||
*This equation comes from the equation under ''Using diluted fuel'', it was arrived at by doing the following things | *This equation comes from the equation under ''Using diluted fuel'', it was arrived at by doing the following things | ||
− | **ratio(fuel) was introduced (which is 3 times higher than min(ratio(O2),ratio(H2)*0.5), everyone should be using pre-mixed fuel so this should make things simpler, having 1 represent 100% fuel is also more | + | **ratio(fuel) was introduced (which is 3 times higher than min(ratio(O2),ratio(H2)*0.5), everyone should be using pre-mixed fuel so this should make things simpler, having 1 represent 100% fuel is also more intutive than having 0.333 mean 100% fuel |
**everything is calculated per 1 mol fuel here, the original one uses per 3 mol fuel (1 mol O2 + 2 mol H2), so several values must be divided by 3 | **everything is calculated per 1 mol fuel here, the original one uses per 3 mol fuel (1 mol O2 + 2 mol H2), so several values must be divided by 3 | ||
**the dilutant (even a mix) can be treated as a single gas, which turns ''sum(specific heat * mol of gas (before))'' into ''ratio(fuel)*(specific heat(O2)+2*specific heat(H2) )/3 + (1-ratio(fuel))*specific heat(dilutant)'' | **the dilutant (even a mix) can be treated as a single gas, which turns ''sum(specific heat * mol of gas (before))'' into ''ratio(fuel)*(specific heat(O2)+2*specific heat(H2) )/3 + (1-ratio(fuel))*specific heat(dilutant)'' | ||
**min(ratio(O2),ratio(H2)*0.5) will now always be ratio(fuel)/3, so it can be replaced with that | **min(ratio(O2),ratio(H2)*0.5) will now always be ratio(fuel)/3, so it can be replaced with that | ||
− | **without any rounding the formula is: ratio(fuel) = s*(T(after) - T(before)) / ( T(before)*(61.9/3-s) + (0.95* | + | **without any rounding the formula is: ratio(fuel) = s*(T(after) - T(before)) / ( T(before)*(61.9/3-s) + (0.95*595000/3) - T(after)*(61.9/3-s) - T(after)*(0.95*181.7/3) ) |
'''Calculating the pressure before ignition''' | '''Calculating the pressure before ignition''' | ||
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'''Diluting fuel''' | '''Diluting fuel''' | ||
− | Mixing gas is temperature sensitive. This is because pressure is used as an indirect measure of the amount of mol (n=PV/(RT)) being | + | Mixing gas is temperature sensitive. This is because pressure is used as an indirect measure of the amount of mol (n=PV/(RT)) being transfered, and pressure is also dependent on temperature. It is however possible to get around this issue with a bit of math. |
'''A)''' When fuel and dilutant have the same temperature | '''A)''' When fuel and dilutant have the same temperature | ||
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It's a warm and sunny day on Europa and a stationeer wants to make some invar. The desired temperature and pressure will be chosen as be the upper limit for invar, so 1500K and 20MPa. Adding ore to the furnace will reduce its temperature and increase the amount of gas (and pressure) inside of it, but the stationeer hopes that 100g of invar will be too little to have much of an effect. The furnace is exposed to the atmosphere and will be loosing temperature and pressure fairly fast which could be an issue, but making the alloy should be quick enough. The dilutant gas will be pure O2 from the atmosphere, which has a specific heat value of 21.1. The starting temperature of the fuel and the atmosphere are both at -140°C. | It's a warm and sunny day on Europa and a stationeer wants to make some invar. The desired temperature and pressure will be chosen as be the upper limit for invar, so 1500K and 20MPa. Adding ore to the furnace will reduce its temperature and increase the amount of gas (and pressure) inside of it, but the stationeer hopes that 100g of invar will be too little to have much of an effect. The furnace is exposed to the atmosphere and will be loosing temperature and pressure fairly fast which could be an issue, but making the alloy should be quick enough. The dilutant gas will be pure O2 from the atmosphere, which has a specific heat value of 21.1. The starting temperature of the fuel and the atmosphere are both at -140°C. | ||
− | *ratio(fuel) = s*( T(after) - T(before) ) / ( | + | *ratio(fuel) = s*( T(after) - T(before) ) / ( 188417 + T(before)*(20.633 - s) - T(after)*(78.172 - s) ) |
**s = specific heat of the dilutant = 21.1 | **s = specific heat of the dilutant = 21.1 | ||
**T(after) = 1500K (this is the chosen value) | **T(after) = 1500K (this is the chosen value) | ||
**T(before) = -140C = 133K (temperature inside the furnace before ignition) | **T(before) = -140C = 133K (temperature inside the furnace before ignition) | ||
− | *ratio(fuel) = 0. | + | *ratio(fuel) = 0.28073 |
*This is high enough for combustion to occur | *This is high enough for combustion to occur | ||
− | **H2 is the limiting gas so | + | **H2 is the limiting gas, so %H2 = ratio(fuel)*2/3 = 5%, solving for ratio(fuel) gives the value 0.075, so ratio(fuel) must be above this number to ignite, which it is |
The necessary pressure of the pre-ignition fuel mix inside the furnace will be | The necessary pressure of the pre-ignition fuel mix inside the furnace will be | ||
*P(before) = P(after)*T(before) / ( T(after) * (1 + ratio(fuel)*1.9) ) | *P(before) = P(after)*T(before) / ( T(after) * (1 + ratio(fuel)*1.9) ) | ||
− | **ratio(fuel) = 0. | + | **ratio(fuel) = 0.28073 |
**P(after) = 20MPa (this is the chosen value) | **P(after) = 20MPa (this is the chosen value) | ||
**T(before) = -140C = 133K | **T(before) = -140C = 133K | ||
**T(after) = 1500K (this is the chosen value used in the temperature calculation) | **T(after) = 1500K (this is the chosen value used in the temperature calculation) | ||
− | *P(before) = | + | *P(before) = 1156.5kPa |
Dilution calculations | Dilution calculations | ||
*The needed pressure of pure fuel inside the furnace will be | *The needed pressure of pure fuel inside the furnace will be | ||
− | **P(fuel) = ratio(fuel) * P(before) = 0. | + | **P(fuel) = ratio(fuel) * P(before) = 0.28073 * 1156.5kPa = 324.66kPa |
− | *The dilutant will then be added to the furnace to reach the P(before) pressure at 1.16MPa | + | *The dilutant will then be added to the furnace to reach the P(before) pressure at 1.16MPa |
*The ratio of H2 inside the furnace before ignition can be checked with the tablet, it should be | *The ratio of H2 inside the furnace before ignition can be checked with the tablet, it should be | ||
− | **ratio(H2) = 0. | + | **ratio(H2) = 0.28073 * 2/3 = 0.18715 = 19% |
− | + | After adding fuel and dilutant, the furnace was ignited. Adding the ores reduced the temperature and increased the pressure a bit, which pushed the pressure up above 20MPa and out of the needed range. After waiting for the pressure to drop back down, the temperature was still high enough to make the desired alloy with several seconds to spare. In hindsight, 20MPa was a bit too high and 1500K a bit too low, so better values could definately have been chosen. | |
− | Reloading the save and placing the furnace inside a welded frame to insulate it (no loss of temperature or pressure) showed the following. The furnace reached 1477K and 19.90MPa after ignition. The fuel was added with a regulator (the furnace showed: 325kPa, 133K), the fuel mix was decent but not a perfect 1:2. Then the diluting O2 was added, it was slightly too cold (the furnace now showed: 1.16MPa, 130K), so a bit too much dilutant | + | Reloading the save and placing the furnace inside a welded frame to insulate it (no loss of temperature or pressure) showed the following. The furnace reached 1477K and 19.90MPa after ignition. The fuel was added with a regulator (the furnace showed: 325kPa, 133K), the fuel mix was decent but not a perfect 1:2. Then the diluting O2 was added, it was slightly too cold (the furnace now showed: 1.16MPa, 130K), so a bit too much dilutant ended up in the furnace (since cold gas has a lower pressure). The dilutant was inserted via the furnace outlet, checking the mol% with the tablet showed 3% H2 in the outlet pipe and 20% inside the furnace instead of 19% in both, the total number of H2 mol was unchanged. The observed loss of temperature is a result from using too much dilutant, using a lower starting temperature and a flawed fuel mix. But it's still very close to the desired value. The lower pressure is related to the temperature, going from 1500K to 1477K should mean -1.5% pressure, but the change was just -0.5%, another indication that too much dilutant had been added. Everything was built outdoors and exposed to the daily temperature cycle. |
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