Relay modification for ign sw discussion

Topic: Relay modification for ign sw discussion (Read 10430 times) previous topic - next topic

Relay modification for ign sw discussion

Reply #90 – December 27, 2012, 01:08:37 AM

I knew i should not have discussed non linear loads.......... sorry........... ill try to avoid linear and non linear loads and I2R loss.

Relay modification for ign sw discussion

Reply #91 – December 27, 2012, 04:28:10 AM

Quote from: jcassity;405363

i think 30's are a little light,, considering low voltage, thats the conundrum, max size of the fuse before you change to a breaker. i was doing

if the gray wire can see about 20A, then 20A x 12.5v in run condition is 250w.
in a low voltage condition, lets say 10v, thats 250w / 10v is 25A.

to size the fuse / cb, that would be 25A x 1.2= 30A required ocp.

30A is on the cuff per relay... just my toughts,, would you do 30's?

i think the real issue is the voltage drop from the starter relay up to the yellow wires.

id be interested to see what my voltage is up to the gray Run wire now that im gonna feed it with a totally new feeder. Id like to see 13.5 to 14, that would get my current down.

that is correct scott, when battery source voltage is low, the current demand for an online device will increase.
what tom is saying is that he has changed the resistive value to adjust the wattage and that is not the correct condition we are in because we cant go around changing the internal resistance of our radio, EATC, our lighting ect at a whim so the wattage demand is a constant... thats why they rate things they way they do at the engineering level.

having said that, your math you posted was correct simulating a low battery condition. taking into account the power rating of the items are constant,, and also taking into account the actual voltage you measured at the ign sw was much lower than what you had at the beginning of the yellow wire,,,=ie= voltage drop.

so using 12.5v as the estblished source potential was prudent to guage the OCP you chose.

a quick google on battery state of heath and "coupe de fouet" as shown by a random white paper you found was a good resource.

as for the power comany example, it holds water for a bit but with power factor correcting transformers and capacitor banks, these little ajustments in the grid allow the power co to efficiently shed load in hard times or peak times and has very little to do with your homes actual current carrying consumption< this part becomes very tricky to think through.

if the other examples were true, then there would be no need for the varios voltages out there we can buy like 120v, 120/240, 120/208 1p, 120/208 3ph, 277/480v 3ph ect ect.

what was said above meant that taking i device that use to run on one leg of 120v has no gain to move it to 240v 2pole circuit because what is beind said is that if you do so, then your going to draw more current.

oh well, DC power is a specialized industry for a reason.

Relay modification for ign sw discussion

Reply #92 – December 27, 2012, 04:41:33 AM

oh yeah, the battey itself is infact dropping in current carrying capacity or its C/x rating as well as the voltage lowering. its storage capacity in both voltage potential and current storage DOES in fact go down but not exactly in a linear manner.
so yes, the current does go down (((***INSIDE***))) the battery

but......

this has nothing to do with the loads connected to the battery. the effect on the battery is invisible to the load ,, it does not care.
the load is going to draw what it can draw up until such point low voltage input for operation has been reached or something shorts or nothing shorts out and things just shut off. But, during the condiiton of a "LOAD" demanding "WATTS" which are established values that really cant be changed, we see now that in order to operated the load, since the voltage is lower, the "LOAD" will obviously demand slightly more current.

We are just trying to establish if the 80A cb is too much or not, since we dont want to allow the source 80cb to remain on if there is in fact the possibility of high current that could melt the wires thats all.

we're way off course i guess but does the last drawing look ok for sheading load off the ign sw in this manner?

again, im concerned about the 80A breaker being too big.

Relay modification for ign sw discussion

Reply #93 – December 27, 2012, 06:59:38 AM

Ok them why does my headlights get brighter when i start my car. Reason the voltage goes up and the headlights get brighter. Their fore the wattage goes up. SIMPLE

When the headlights are on with the car not running it is running on 12V when the car starts the headlights are running on 14V So the wattage goes up due to increased voltage

Voltage X Current = WATTS You cant change this JAY. When the voltage or current is increased the wattage goes up not down. I cant follow what you are saying. It makes no sense to me. If the current goes down when voltage goes up why then does the high speed portion of the blower circuit have a slave relay. Simple because the faster it goes the current rises. I will prove this this morning I think!!!

Relay modification for ign sw discussion

Reply #94 – December 27, 2012, 12:00:52 PM

Of course the lights get brighter with the engine running...been that way for a couple days now. Load...probably has not so much to do with it as simply a step up from from a flat 12v to somewhere around 13.8, add a load such as a heater fan, or twin e fans on the radiator....do you lights get brighter then? Mine sure as hell don't...they dim down for a second, but come back up.

Relay modification for ign sw discussion

Reply #95 – December 27, 2012, 12:22:46 PM

LOL,, its a good converstation to have either way,
im waiting for your current readings , its a neat thing to just be aware of thats all.

by the way, with your examples your posting, then why would a screwdriver self distruct if you put it across the batt posts....

seems odd that a load like a screwdriver would break the Law~~!!!

the screw driver is that load,, and its not a motor so whats the load demanding? its demanding voltage and current but the voltage goes down as the current goes up.

in your example, we should be able to lay a screwdriver __or a load bank__ across the battery posts and see the voltage drop with current until the battery is dead.
if thats the case then there would have never been any white papers drafted on coupe de fout or on thermal run away.

also,

when the voltage is higher, you dont need as much current to support the same amount of watts.,, plain and simple.

do it like this....
250kw gen running 120/208 3phase... thats 250000 / 208 / 1.73= approx 690A

250kw gen running 277/480v 3phase ... thats 250000 / 480 / 1.73= approx 300A

now it appears the 208 3ph service above gets you more amps and it does but the cost of running 208 3phase is very different.
Actually to get the most efficiencey and lower amperage, you want 480v 3ph so all your devices will draw less current.

same for batteries in that during its C/x discharge rating (capacity divided by time), as the voltage drops, the current will go up ***and what is meant is that the devices hooked up to teh battery will demand more current to compensate.

i thought for sure posting a random coup de foute chart would make you think for a moment but to your example, your headlamps are getting brigher because of your alternator maxing out voltage rating and current rating.

sorry but no law broke here, take a few minutes or days and ask around, you will be very hard pressed to find anyone who knows what i am saying because these are things people dont really think about or even know , not to thier fault either because its **WATTS.

trust me, you take everyting back to watts and your math will always be correct.,, thats the only way i can properly size hvac (btu), generators, rectifiers, inverters, converters or battery plants.

again, this is just a shootin the breeze converstation but i hope i got you curious, as smart as you are, your probably going to take this and stew on it then realize whats happening more than ever when loads are dispursed across a very small circular millimeter of copper,, and then see how voltage drop really is messing with you.

the whole reason for this side track is because im saying ((( I think the 80 ocp is too large))) and would like that to be the topic based on things i know but im willing to bend to others thoughts as well. For this little piece though, current **DEMAND** on a battery goes up as its battery voltage goes down, and thats not my words,,
check out C&D, Deka, Marathon, GNB and all the big battery makers out there... then i would expect you to come back and say,, hey j, this is purdy interesting stuff becausue you are actually right.

Relay modification for ign sw discussion

Reply #96 – December 27, 2012, 12:50:32 PM

another thought, and i forgot to push your button on this one but i was kinda hoping you'd ask why 6awg was my call out to feed the 80A breaker.
actually to be correct per nec , we exceed table 310.17 free air at 90degC because 8awg is sufficient.

to keep the 80A ocp, the wire is going to be exposed to heat off the engine, and its likely the wire feeding the 80A ocp is going to be in a wire loom (defined as raceway or conduits ect per the nec) so we must use table 310.16 instead.
The max ampacity of a 6awg conductor at 90degC from table 310.16 is 75A for copper if the insulation meets the qualifiers for this column.

there are two things to consider.....
you size your wire to the nec or you size your wire to the voltage drop method.

the formula to calculate the wire requried for a specified ocp using the voltage drop method is

11.1 x LL x amps / VD

11.1= a constant for copper
LL= total circuit loop distance for the hot and ground leg
amps= yeah thats amps
VD= desired or "allowable" voltage drop you are going to control

so in my case, im saying that im powering an 80A ocp and to do so, my wire footage for the hot and ground i will assume to be approx 10feet total (this allows for turns and bends ect), i want to allow only 1v loss across the circuit which also keeps the math simple.

11.1 x 10 x 80A / 1v= 8880 circular milimeters of copper required

so there are two things,,
does the circuit meet the voltage drop method more or does it meet the nec table 310.16? which one has the most strengent standard or requirement?

with the voltage drop method we require 8880cmil of copper and this number can be converted to wire size by refering to the NEC chapter 9, table 8.
8880cmil = a min #10 wire.

with the ampacity method, using the nec table 310.16, we find that to properly bond an 80A ocp to a wire,said wire must be equal to or greater than #4awg.
so....even my illustration is lean and is the reason i asked if the ocp can come down based on real loads and the almost word for word event Thunderchicken described by doing both windows down at once along with power trunk and blower ect.. unlikely events.

so if the ocp comes down to like 40A, it seems more appropriate but its more like 60 for safety.

my point again is i believe putting in the 80A ocp is going to "allow" the wire to burn before the breaker to open hence why even though i am lean on the 6awg feeder, its still helping but if i can get this ocp down some then things start making more sense.

Relay modification for ign sw discussion

Reply #97 – December 27, 2012, 01:41:48 PM

Quote from: TOM Renzo;405495

Ok them why does my headlights get brighter when i start my car. Reason the voltage goes up and the headlights get brighter. Their fore the wattage goes up. SIMPLE

When the headlights are on with the car not running it is running on 12V when the car starts the headlights are running on 14V So the wattage goes up due to increased voltage

Voltage X Current = WATTS You cant change this JAY. When the voltage or current is increased the wattage goes up not down. I cant follow what you are saying. It makes no sense to me. If the current goes down when voltage goes up why then does the high speed portion of the blower circuit have a slave relay. Simple because the faster it goes the current rises. I will prove this this morning I think!!!

Tested on vehicle. 9007 on high: 4.8A at 12.2v for 58.56W, 5.25A at 14.4v for 75.6W. Tom, I yield to you. I was thinking in terms of wattage as a constant which, clearly, it is not.

Relay modification for ign sw discussion

Reply #98 – December 27, 2012, 01:50:49 PM

In your screwdriver across the battery example, the current goes up because the resistance (load) goes down to almost zero.
I=E/R Current = 12.5 ÷ 0 = infinity, or in reality the max the battery can supply.

Relay modification for ign sw discussion

Reply #99 – December 27, 2012, 02:04:14 PM

The notation you circled on the battery back-up diagram does not agree with the diagram. The diagram shows the voltage and current to the load dropping together until the ac comes back on.
You believe the notation.
I believe the diagram.

If a battery is the sole power source in a circuit, the internal battery current is equal to the load current.
For every electron that leaves the negative battery post one enters the positive battery post.

Relay modification for ign sw discussion

Reply #100 – December 27, 2012, 03:06:46 PM

OK BOYS AND GIRLS HERE WE GO!!!

H3 Lamp full 12 volts 5 Amps

H3 lamp 8 Volts 3 Amps

Chevy silverado blower motor

Full 12 volts 6 Amps Full speed

Full 10 volts 5 Amps Medium speed

Full 8 Volts 4 Amps Low Speed

Locked rotor blew a 20A fuse.

So as i thought and knew when the voltage goes down the current does also.

Now i did this this morning with a fully charged battery and on my Silverado

Relay modification for ign sw discussion

Reply #101 – December 27, 2012, 03:38:20 PM

Jay if i may you always revert back to AC on your calculations. DC ids different. From memory DC uses the entire circular mill on the wire capacity. AC uses only the outer portion of the wire. So basically speaking AC does not use the entire CM only the outer portion of said wire. Where as DC uses the entire CM for capacity and amp loads. Just saying.

Relay modification for ign sw discussion

Reply #102 – December 27, 2012, 03:48:58 PM

Jay Cars are DC, So wire can be thinner. Just a fact!!!

DC does not experience 'skin effect' that is inherent in AC (in AC, the current runs predominantly on the outside - or 'skin' - of the wire and not in the interior), meaning that for the same sized wire and same voltage, DC loses less energy per distance than AC does.

Stranded wire has more CM than solid further making it a smaller wire to carry bigger loads. Just saying!!!

Relay modification for ign sw discussion

Reply #103 – December 27, 2012, 03:58:48 PM

AWG AC DC

22    5a    5a
20    7.5a    8a
18    10a    10a
16    13a    20a
14    17a    40a
12    23a    60a
10    33a    100a
8    46a    150a

Relay modification for ign sw discussion

Reply #104 – December 27, 2012, 04:31:43 PM

Quote from: TOM Renzo;405517

Locked rotor blew a 20A fuse.

I think I can explain why it blows the fuse with a locked rotor.
When voltage is first applied to a coil it resists current flow. This is called inductive reactance. The reactance lasts until the magnetic flux field finishes building up around the coil. Then it goes away.
When a DC commutator motor is running it is constantly energizing different coils as it moves over the commutator segments.
When the motor stalls, the inductive reactance goes away once it finishes building the flux field for that segment.
Then all you have left is the resistance of the coil wire.