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Hermes_Trismegistus
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| Quote: | Originally posted by rikkitikkitavi
actually wet and dry air has almost the same heat capacity, so using a fan shows no difference...
/rickard |
You're joking right!
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Organikum
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The thing used with HID lamps is no transformer but a so called "ballast".
The best solution si to use an bigger transformer (or for HID lamps an ballast of higher quality) as this in the end is cheaper and less hassle than
all this cooling with oil and timberbricks and fans. For the HID ballasts I would say some deeper investigation of this "electricity"
phenomen would bring up astonishing solutions, for example that there is no need for a separate ballast for every lamp, as a ballast only limits the
amperage and does this for one as for ten lamps, doesnt matter at all......
(you need actually two ballasts, one for starting one lamp after the other and one for running ALL lamps)
hehe.....
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rikkitikkitavi
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no, i am serious , wet and dry air has almost similar heat capacity.
so you wont see any difference when using a fan in dry or humid air for convective cooling.
that is what I meant, it could be interpreted otherwise i realise.
Technically, the convective heat transfer coefficient will be a bit different depending on humiditiy but not much...
/rickard
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axehandle
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| Quote: |
no, i am serious , wet and dry air has almost similar heat capacity.
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Interesting. Do you have the numbers? I'm a bit confused, since sitting in a sauna brings very different experiences based on the humidity of the
air.
I'm not taking sides here, I just want the numbers. It's quite interesting.
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Hermes_Trismegistus
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Sigh!
| Quote: | Originally posted by Organikum
The thing used with HID lamps is no transformer but a so called "ballast".
(you need actually two ballasts, one for starting one lamp after the other and one for running ALL lamps)
hehe..... |
Y'know Orgi, I'd hate to condradict you (cuz yer such a clever chemist)
but...
I called the ballast a transformer because I wanted people unfamiliar with HID's to know what I was talking about, and also because it is a
transformer, transforming the high amperage/ low voltage house hold AC current into the lower amperage / higher voltage DC current required by the HID
bulb.
The other portion of the HID light is commonly called the "starter".
What it really is, is a giant capacitor whose function is to initiate the arc through in the metal vapour inside bulb(simultaneously heating the metal
to the vapour point)
The other important parts that are so small (and usually attatched to the step up transformer) are usually overlooked but are just as important to the
function of the system...
they are..:
a full-wave rectifier connected to an AC power line to produce an unregulated pulsating DC output, A power factor correction (PFC) circuit. The PFC
circuit includes a first semiconductor electronic switch whose activation is controlled to bring the input current and voltage in time-phase with one
another, thereby imparting a high power factor raring to the ballast. The pulsating DC output of the PFC circuit is applied to the storage capacitor
circuit (whose function we discussed earlier),a second electronic switch whose activation is controlled to discharge the capacitor so as to cause the
capacitor circuit to yield in its output, a regulated DC which is fed to a power control (PC) circuit.
The PC circuit includes a third electronic switch whose activation is controlled to maintain the wattage of power supplied to the HID lamp at the
rated wattage of the lamp.
This is a necessary part of the sytem because as the bulb heats up the wattage requirements change, and in the long haul they also change as the bulb
ages.
----------------------------------------------------
I am not so well versed in electronics and electrical systems as I might seem in a general sense, but at one point the food on my table depended on
the proper functioning of those damn lights and I REALLY didn't feel comfortable letting repairmen into my house!
So I learned how to troubleshoot.
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Hermes_Trismegistus
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Jeeez...what are they teaching in school these days!
| Quote: | Originally posted by rikkitikkitavi
no, i am serious , wet and dry air has almost similar heat capacity.
/rickard |
Our air is composed of mostly Oxygen and Nitrogen and water vapour.
The specific heat capacity of Nitrogen is 1042 J/KgK
the specific heat capacity of Oxygen is less, approx 800-900 J/KgK
whereas the specific heat capacity of water vapour is extremely high ~2000 J/KgK
so as you can plainly see, varying the ratio of water to other gases varies the heat capacity quite significantly.
[Edited on 5-3-2004 by Hermes_Trismegistus]
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rikkitikkitavi
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no , i didnt learn it at school , but at university...
simulation with chemcad shows that air , saturated with water vapour at 20 C has a cp of 1,02kJ/kgK, increasing temperature to 40 C, but still
saturated air, cp increases to 1,047 kJ/kgK. a 2 % change...
it will different at higher temperatures , where the air can take higher water content before it saturates, but I was assuming you used a fan to cool
it with normal air that surrounds us , and you dont put your transformer in a sauna...
speaking of saunas, the reason for the high temperature is that water is condensing on
all surfaces ,especially a 35 C human body and then gives away its condensation heat
, about 2250 KJ/kg.
/rickard
[Edited on 6-3-2004 by rikkitikkitavi]
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Hermes_Trismegistus
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Yes...but air saturation levels are NOT constant!
| Quote: | Originally posted by rikkitikkitavi
simulation with chemcad shows that air , saturated with water vapour at 20 C has a cp of 1,02kJ/kgK, increasing temperature to 40 C, but still
saturated air, cp increases to 1,047 kJ/kgK. a 2 % change...
/rickard
[Edited on 6-3-2004 by rikkitikkitavi]
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You're figures aren't showing us the effect of Humidity change, Because your figures don't use constant Temperature.
You aren't even effectively showing us the effect of temperature change because you are varying the level of humidity(saturation level varies
dramatically with T, warm air absorbs far more water than cold air.)
Your math may be correct, but the basis of your calculations is fundamentally flawed.
So please, input those numbers again with absolutely dry air at 20 deg C and saturated air at 20 deg C (or dry air at 40 deg C and sat air at 40 deg
C)
We await your figures.
edit: (courtesy)
[Edited on 6-3-2004 by Hermes_Trismegistus]
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Organikum
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Might you please cool down a little bit?
It might help to clearly define the QUESTION again as it seems to me that here is the main problem.
Actually it is by no way clear to me WHAT exactly you are arguing about and I have the impression it isnt clear to the gladiators in this arena also.
Or at least it seems clear to every single gladiator but the definitions dont match.
regards
ORG
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rikkitikkitavi
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I dont like to be called fool, whatever way you put it. So I suggest you tag it down and read what I write carefully. I dont try to owerrun anyone
with data, as I believe you would have the capacity to grasp my posts.
The values for cp that I have calculated are for air saturated with water at respective temperature, so the amount of water , measured in kg/kg dry
air IS different.
If you dont trust the numbers and believe I am bullshitting you, please look up a air-humidity chart and se the numbers yourself.
Dry air 20 C ,0 kg H2O/kg dry air, cp 1,00 kJ/kg K
saturated air 20 C, 0.015 kg H2O/kg dry air , cp 1,02 kJ/kg K
saturated air 40 C, 0,05 kgH2O/kg dry air, cp = 1,04 kJ/kg K
dry air 40 C , 0 kg H2O/kg dry air , cp = 1,00 kJ/kg K
Now I would like to see you show us why cp should variate so much with a megre 5 % change in composition (0,05 kg H2O/kg dry air) you can assume cp
for each component to be constant over such small temperature range.
I use chemcad and similar programs almost daily as tools to design equipment like cooling towers, heat exchangers e t c,To my experience a 80000 Us$
program is quite correct, but if you prefer I can calculatet it by hand to.
Engineering is applied science, actually.
/rickard
[Edited on 6-3-2004 by rikkitikkitavi]
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Hermes_Trismegistus
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| Quote: | Originally posted by rikkitikkitavi
Dry air 20 C ,0 kg H2O/kg dry air, cp 1,00 kJ/kg K
saturated air 20 C, 0.015 kg H2O/kg dry air , cp 1,02 kJ/kg K
saturated air 40 C, 0,05 kgH2O/kg dry air, cp = 1,04 kJ/kg K
dry air 40 C , 0 kg H2O/kg dry air , cp = 1,00 kJ/kg K
[Edited on 6-3-2004 by rikkitikkitavi] |
I didn't call anyone a fool, but obliterated any reference to the word, to ensure not misunderstandings occur.
(this is, after all, just an academic disagreement)
for clarification and confirmation, are your figures indicating that a Kg of air, at 20 C is saturated with water when it contains 15 grams of water
and a kg of air at 40 degrees C is saturated when it contains 50 grams of water?
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rikkitikkitavi
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yes.
/rickard
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IgnorantlyIntelligent
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Does anyone know how one might aquire one of those huge transformers used by the city(the ones on the telephone poles) Does anyone know the ratio or
w/e for those? Like 1:30? inpututput
I'm not against illegal activities but it would seem very difficult to steal a massive transformer...
[Edited on 11-3-2004 by IgnorantlyIntelligent]
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Hermes_Trismegistus
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OK rikki.....I checked out your figures in my spare time.
You're right, I was wrong. Humidity levels have suprisingly little effect on the heat capacity of air.
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rikkitikkitavi
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never mind, errare humane est
or maybe alea jacta est ?
/rickard
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chemoleo
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I happen to have an extremely big transformer with a 6 kW output, input 220V, output 42 V, 120 A.
I have used this transformer on occasion, the foremost success was to make Calcium carbide.
Anyway, the reason why I couldnt use the transformer much was because it tripped the fuse of the house electricity system, but *not* because of
drawing too much power out of it, but rather, once the damn thing is switched off! Even if the transformer is not hooked to something, i.e. it is
running but not drawing power!
I guess this is some inductive effect, where, when the magnetic fields collapse, backinduce a current into the system.
It is NOT because the transformer is earthed, and leaking current (simply because it's not earthed).
Now, what's the solution to avoiding this?
I guess one needs a capacitor or two in the right places, but unfortunately I don't know enough about electrical engineering to know what uF /V
specifications the capacitor should have, and also as I don't know how to connect the capacitors.
Anyone got some ideas?
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Mr. Wizard
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I think you are right about the field collapsing and causing a spike of voltage into your primary winding. perhaps you could hook up a high wattage
load across the primary just before you disconnect and have it absorb the spike? A capacitor might work too, but I don't know how to calculate
its size. Instead of cutting off the power to the transformer suddenly, you could add more resistance to the input until you had the current at a
safe level, and then disconnect it. You could use old heating elements from electric stoves. How many amps does it draw without a load on the
secondary? How many square centimeters of iron in the cross section of the core?
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