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#1
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ml wrote:
well if one had to buy the jado it might be expensive, i can get one much cheeper than list and somone here posted an incorrect list price anyways , - then again if you add up the total cost of the competition you'd at worst case be in the ball park the goal was at first to power a hf/rig 100w tx for a few days on a backup power source looking at batteries/fuelcells i realized this is to much load so i figured my 2m righ on low pwr is a compramise @1week at 100w tx, i started to put together on paper what i need, a pretty big charger lots of very big heavy batteries , and they are VERY expensive, if the batteries went 'low' and the power wasn't restored say an extended blackout then the battery life goes down the tubes now i live in an apt, so too many batteries even if free arent practical code prevents me from using a generator in the building, so would the coop board, and the noise, smell vibration and mostly a generator running would attract 'undesiables' it would be kinda cheep and storing say desil wouldn't be a big deal if say i had a pvt house again thou getting extra fuel in a extended blackout would be an issue i had a demo of this fuel cell i was really impressed, is the cost and power output totally there no i guess not a few years from now prob but the cost of the fuel cell unit alone is comparable to the battery cost and i don't have to worry about the space, and other issues associated w/batteries i can get a small tank of hydrogen that would power the unit for longer than i was planning for it's cheep and i can get it very easy even if theyre was an extended blackout (details omitted) it's quite meets code, very small, very light dosn't pollute or stink seemed rather sexy the jado can also be powered from small cartridges making the unit selfcontained and very portable, however this isn't a requirement of mine but cool benni the only con for me, seems to be a)not totally going to give me 100w at full load b) i wish the unit was more than a puny 100w the larger units i would have to pay full list for and are therefore outta reach so solar /wind again might be nice in a pvt home but nothing i'd rely on it usually starts simple, i wanted somthing better than battery power much appreciation and mostly also because i like to tinker and learn seems like somthing cool to play with and experiment Ok, I have applied a simple power budget to what you stated your original desire was, as opposed to your cut down 2m 5W system. Using a rig I own, an FT-897, on HF. Assume 90% receive at 1 amp. Assume 10% transmit at 15 amps with 100 watts output, (measured as opposed to the 22 amps claimed in the docs) This gives 2.4 amps average consumption. (Check my math, the doctor did a lot during my pre-op today) 1 week is 168 hours which gives us 403 ampere hours. 100AH deep cycle batteries are around 100 each (or less), So that's about $400 for all the batteries you need for a week without sleep. A modern top of the line charger, such as produced by Schumacher, is less than $100, does automatic desulfation, will charge at 50 amps or greater, and weighs about 6 pounds. So now we are at under 500 dollars for an emergency power for any ham that needs to run an HF rig for a whole week. I just don't see why you want to use the fuel cell, it makes no sense. tom K0TAR |
#2
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Tom Ring wrote:
. . . 1 week is 168 hours which gives us 403 ampere hours. . . . which at 13.6 volts is about 5480 watt-hours. This is the energy storage requirement. The OP asked about using a capacitor. The energy stored in a capacitor is C * V^2 / 2. For capacitance in farads and voltage in volts, the result is joules, or watt-seconds -- you need 5480 * 3600 ~ 20,000,000 joules in round numbers. So suppose you wanted to store this same amount of energy in a capacitor, and you had a switching regulator which would handle 50 volts maximum input voltage. Solving for C, assuming it's charged to 50 volts, C = 20,000,000 * 2 / (50^2) = 16,000 farads. This assumes you can get all the energy out of the capacitor, which requires your regulator to work down to zero volts. But you'll get 3/4 of the energy out of it if your regulator cuts off at 25 volts, 7/8 if it cuts off at 12.5 volts, etc. How does that sound? Roy Lewallen, W7EL |
#3
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In article ,
Roy Lewallen wrote: Tom Ring wrote: . . . 1 week is 168 hours which gives us 403 ampere hours. . . . which at 13.6 volts is about 5480 watt-hours. This is the energy storage requirement. The OP asked about using a capacitor. The energy stored in a capacitor is C * V^2 / 2. For capacitance in farads and voltage in volts, the result is joules, or watt-seconds -- you need 5480 * 3600 ~ 20,000,000 joules in round numbers. So suppose you wanted to store this same amount of energy in a capacitor, and you had a switching regulator which would handle 50 volts maximum input voltage. Solving for C, assuming it's charged to 50 volts, C = 20,000,000 * 2 / (50^2) = 16,000 farads. This assumes you can get all the energy out of the capacitor, which requires your regulator to work down to zero volts. But you'll get 3/4 of the energy out of it if your regulator cuts off at 25 volts, 7/8 if it cuts off at 12.5 volts, etc. How does that sound? Roy Lewallen, W7EL well i guess it sounds like i wont have any 16,000 farad cap's stacked up in my myplace anytime soon, but i understand the math behind it now thanks perhaps i'll just get a small cap to handle a brief second keydown dip smooth things out a bit thanks |
#4
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Roy Lewallen wrote in
: Tom Ring wrote: . . . 1 week is 168 hours which gives us 403 ampere hours. . . . which at 13.6 volts is about 5480 watt-hours. This is the energy storage requirement. The OP asked about using a capacitor. The energy stored in a capacitor is C * V^2 / 2. For capacitance in farads and voltage in volts, the result is joules, or watt-seconds -- you need 5480 * 3600 ~ 20,000,000 joules in round numbers. So suppose you wanted to store this same amount of energy in a capacitor, and you had a switching regulator which would handle 50 volts maximum input voltage. Solving for C, assuming it's charged to 50 volts, C = 20,000,000 * 2 / (50^2) = 16,000 farads. This assumes you can get all the energy out of the capacitor, which requires your regulator to work down to zero volts. But you'll get 3/4 of the energy out of it if your regulator cuts off at 25 volts, 7/8 if it cuts off at 12.5 volts, etc. How does that sound? Roy Lewallen, W7EL Send that to me in CW and I'll have a look at it. SC |
#5
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#6
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In article ,
Tom Ring wrote: ml wrote: omitted) it's quite meets code, very small, very light dosn't pollute or stink seemed rather sexy the jado can also be powered from small cartridges making the unit selfcontained and very portable, however this isn't a requirement of mine but cool benni the only con for me, seems to be a)not totally going to give me 100w at full load b) i wish the unit was more than a puny 100w the larger units i would have to pay full list for and are therefore outta reach so solath and experiment Ok, I have applied a simple power budget to what you stated your original desire was, as opposed to your cut down 2m 5W system. Using a rig I own, an FT-897, on HF. Assume 90% receive at 1 amp. Assume 10% transmit at 15 amps with 100 watts output, (measured as opposed to the 22 amps claimed in the docs) This gives 2.4 amps average consumption. (Check my math, the doctor did a lot during my pre-op today) 1 week is 168 hours which gives us 403 ampere hours. 100AH deep cycle batteries are around 100 each (or less), So that's about $400 for all the batteries you need for a week without sleep. A modern top of the line charger, such as produced by Schumacher, is less than $100, does automatic desulfation, will charge at 50 amps or greater, and weighs about 6 pounds. So now we are at under 500 dollars for an emergency power for any ham that needs to run an HF rig for a whole week. I just don't see why you want to use the fuel cell, it makes no sense. tom K0TAR tom, am i to understand you had some surgery today? if so, i hope all is well with you and taking the time to write is appreciated i am guessing roy's calculation of the total ampre hrs needed is closer to being correct, but it's 5am and i kinda don't know how to calculate that so i dunno as stated in my earlier post i don't have the formulas so i can't act to confirm but i nod my head and presume it's all correct based on roys math you are correct about what the rig is doing mine is exactly simular i would calculate based on 20amps re your last comments on why i would want a fuel cell, it should be very clear in all my many posts 5480 watt-hours of battery is more $$ and large than i have practical room for i can get the fuel cell for practically nothing and i just thought it might be able to offer some useful power to top off a battery to provide some b/u power why not? and also be a fun experiment it runs quiet , clean and i can get all the fuel i want free always, i would never be able to get fossel fuels in a blackout or other emerg so now i inch a drop closer knowing how many what hours my original abandoned goal would draw , however my current stated goal is to just power my 5w tx output(draws4amps) 2m mobile rig and try to calculate how much the battery would be drained in a 1min tx and then see if the 12v 7amps comming out of the jado would recharge it fast enough i believe that is what must be calculated for just not sure how do it, unless you still say a 7amp battery is still 'ok' my goal was to be able to calculate so a 7amp battery(or what ever is needed optically this seems like a too low a #)) w/a 1min 4amp load would be ??%drained then based on manuf spec's for the battery we could calculate how much power &time would be needed to top it back off then it would be simple enough to see if the output of the jado is sufficient i think thats all thats needed i think that's simple but i dunno how to calculate it thanks all i really wanted to do |
#7
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![]() ml wrote: In article , Tom Ring wrote: ml wrote: omitted) it's quite meets code, very small, very light dosn't pollute or stink seemed rather sexy the jado can also be powered from small cartridges making the unit selfcontained and very portable, however this isn't a requirement of mine but cool benni the only con for me, seems to be a)not totally going to give me 100w at full load b) i wish the unit was more than a puny 100w the larger units i would have to pay full list for and are therefore outta reach so solath and experiment Ok, I have applied a simple power budget to what you stated your original desire was, as opposed to your cut down 2m 5W system. Using a rig I own, an FT-897, on HF. Assume 90% receive at 1 amp. Assume 10% transmit at 15 amps with 100 watts output, (measured as opposed to the 22 amps claimed in the docs) This gives 2.4 amps average consumption. (Check my math, the doctor did a lot during my pre-op today) 1 week is 168 hours which gives us 403 ampere hours. 100AH deep cycle batteries are around 100 each (or less), So that's about $400 for all the batteries you need for a week without sleep. A modern top of the line charger, such as produced by Schumacher, is less than $100, does automatic desulfation, will charge at 50 amps or greater, and weighs about 6 pounds. So now we are at under 500 dollars for an emergency power for any ham that needs to run an HF rig for a whole week. I just don't see why you want to use the fuel cell, it makes no sense. tom K0TAR tom, am i to understand you had some surgery today? if so, i hope all is well with you and taking the time to write is appreciated i am guessing roy's calculation of the total ampre hrs needed is closer to being correct, but it's 5am and i kinda don't know how to calculate that so i dunno as stated in my earlier post i don't have the formulas so i can't act to confirm but i nod my head and presume it's all correct based on roys math you are correct about what the rig is doing mine is exactly simular i would calculate based on 20amps re your last comments on why i would want a fuel cell, it should be very clear in all my many posts 5480 watt-hours of battery is more $$ and large than i have practical room for i can get the fuel cell for practically nothing and i just thought it might be able to offer some useful power to top off a battery to provide some b/u power why not? and also be a fun experiment it runs quiet , clean and i can get all the fuel i want free always, i would never be able to get fossel fuels in a blackout or other emerg so now i inch a drop closer knowing how many what hours my original abandoned goal would draw , however my current stated goal is to just power my 5w tx output(draws4amps) 2m mobile rig and try to calculate how much the battery would be drained in a 1min tx and then see if the 12v 7amps comming out of the jado would recharge it fast enough i believe that is what must be calculated for just not sure how do it, unless you still say a 7amp battery is still 'ok' my goal was to be able to calculate so a 7amp battery(or what ever is needed optically this seems like a too low a #)) w/a 1min 4amp load would be ??%drained then based on manuf spec's for the battery we could calculate how much power &time would be needed to top it back off then it would be simple enough to see if the output of the jado is sufficient i think thats all thats needed i think that's simple but i dunno how to calculate it thanks all i really wanted to do If your rig draws 4 amps (I assume that's on transmit; it would be an awful lot of power for a modern receiver), and the fuel cell can put out 7 amps continuously albeit at slightly reduced voltage, and I assume it can put out 4 amps at less of a reduction, why are you worried about it? Unless you are VERY long-winded with your transmissions, a small 7amp-hour sealed lead acid battery would give you a couple hours operating time even if it wasn't being continuously recharged. I'd actually opt instead for some voltage stabilization at the fuel cell output: a buck-boost regulator would do it. If you trust the fuel cell to do its job, then trust it. If you don't, then provide a backup. So what I read between the lines you've written is that you're going to get this fuel cell, but you don't really trust it. You don't even trust it to come close to what are apparently its published ratings. Pretty much all the formulas you need to figure all this stuff out: power = voltage times current (watts = volts * amps) energy = power times time (joules = watts * seconds) -- total energy needed = sum of (each current times the length of time that current is drawn) amp-hours = amps * hours = amps * seconds / 3600sec/hr battery energy storage in watt-hours = amp-hour rating * volts (assuming full charge) battery energy storage in joules = amp-hour rating * volts * 3600 sec/hour capacitor energy storage = (initial charge voltage)^2 * capacitance / 2 joules = watt-seconds = volts^2 * farads / 2 watt-hours = joules/3600 = volts^2 * farads / 7200 Where is it that you can get a license without knowing at least this much? Why so much trouble with the concepts of power and energy calculated from current, voltage and time? (Why is this posted to an antennas newsgroup?) Cheers, Tom |
#8
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K7ITM wrote:
-- total energy needed = sum of (each current times the length of time that current is drawn) You probably meant to say current times time is total charge. 73 de AC6XG |
#9
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![]() Jim Kelley wrote: K7ITM wrote: -- total energy needed = sum of (each current times the length of time that current is drawn) You probably meant to say current times time is total charge. 73 de AC6XG Yeah, probably something like that. ;-) Actually, what I meant to write was more like: -- total energy needed = voltage * sum of (each current times the length of time that current is drawn. Lesseee...unit analysis...volts * amps * time. OK, that looks better. Thanks, Jim. Cheers, Tom |
#10
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![]() ml wrote: In article , Tom Ring wrote: ml wrote: omitted) it's quite meets code, very small, very light dosn't pollute or stink seemed rather sexy the jado can also be powered from small cartridges making the unit selfcontained and very portable, however this isn't a requirement of mine but cool benni the only con for me, seems to be a)not totally going to give me 100w at full load b) i wish the unit was more than a puny 100w the larger units i would have to pay full list for and are therefore outta reach so solath and experiment Ok, I have applied a simple power budget to what you stated your original desire was, as opposed to your cut down 2m 5W system. Using a rig I own, an FT-897, on HF. Assume 90% receive at 1 amp. Assume 10% transmit at 15 amps with 100 watts output, (measured as opposed to the 22 amps claimed in the docs) This gives 2.4 amps average consumption. (Check my math, the doctor did a lot during my pre-op today) 1 week is 168 hours which gives us 403 ampere hours. 100AH deep cycle batteries are around 100 each (or less), So that's about $400 for all the batteries you need for a week without sleep. A modern top of the line charger, such as produced by Schumacher, is less than $100, does automatic desulfation, will charge at 50 amps or greater, and weighs about 6 pounds. So now we are at under 500 dollars for an emergency power for any ham that needs to run an HF rig for a whole week. I just don't see why you want to use the fuel cell, it makes no sense. tom K0TAR tom, am i to understand you had some surgery today? if so, i hope all is well with you and taking the time to write is appreciated i am guessing roy's calculation of the total ampre hrs needed is closer to being correct, but it's 5am and i kinda don't know how to calculate that so i dunno as stated in my earlier post i don't have the formulas so i can't act to confirm but i nod my head and presume it's all correct based on roys math you are correct about what the rig is doing mine is exactly simular i would calculate based on 20amps re your last comments on why i would want a fuel cell, it should be very clear in all my many posts 5480 watt-hours of battery is more $$ and large than i have practical room for i can get the fuel cell for practically nothing and i just thought it might be able to offer some useful power to top off a battery to provide some b/u power why not? and also be a fun experiment it runs quiet , clean and i can get all the fuel i want free always, i would never be able to get fossel fuels in a blackout or other emerg so now i inch a drop closer knowing how many what hours my original abandoned goal would draw , however my current stated goal is to just power my 5w tx output(draws4amps) 2m mobile rig and try to calculate how much the battery would be drained in a 1min tx and then see if the 12v 7amps comming out of the jado would recharge it fast enough i believe that is what must be calculated for just not sure how do it, unless you still say a 7amp battery is still 'ok' my goal was to be able to calculate so a 7amp battery(or what ever is needed optically this seems like a too low a #)) w/a 1min 4amp load would be ??%drained then based on manuf spec's for the battery we could calculate how much power &time would be needed to top it back off then it would be simple enough to see if the output of the jado is sufficient i think thats all thats needed i think that's simple but i dunno how to calculate it thanks all i really wanted to do If your rig draws 4 amps (I assume that's on transmit; it would be an awful lot of power for a modern receiver), and the fuel cell can put out 7 amps continuously albeit at slightly reduced voltage, and I assume it can put out 4 amps at less of a reduction, why are you worried about it? Unless you are VERY long-winded with your transmissions, a small 7amp-hour sealed lead acid battery would give you a couple hours operating time even if it wasn't being continuously recharged. I'd actually opt instead for some voltage stabilization at the fuel cell output: a buck-boost regulator would do it. If you trust the fuel cell to do its job, then trust it. If you don't, then provide a backup. So what I read between the lines you've written is that you're going to get this fuel cell, but you don't really trust it. You don't even trust it to come close to what are apparently its published ratings. Pretty much all the formulas you need to figure all this stuff out: power = voltage times current (watts = volts * amps) energy = power times time (joules = watts * seconds) -- total energy needed = sum of (each current times the length of time that current is drawn) amp-hours = amps * hours = amps * seconds / 3600sec/hr battery energy storage in watt-hours = amp-hour rating * volts (assuming full charge) battery energy storage in joules = amp-hour rating * volts * 3600 sec/hour capacitor energy storage = (initial charge voltage)^2 * capacitance / 2 joules = watt-seconds = volts^2 * farads / 2 watt-hours = joules/3600 = volts^2 * farads / 7200 Where is it that you can get a license without knowing at least this much? Why so much trouble with the concepts of power and energy calculated from current, voltage and time? (Why is this posted to an antennas newsgroup?) Cheers, Tom |
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