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Author Topic: Installing Heaters  (Read 2727 times)
Talon
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« Reply #15 on: February 26, 2006, 10:25:27 AM »

merwin - you just made me dizzy trying to understand that :P
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merwin10
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« Reply #16 on: February 26, 2006, 10:44:47 AM »

When you want heat at anchor here is one of the possible solutions! There is a marine heater company specializing in marine heaters for Artic usage called Espar. They make several solutions from hot air to hot water depending on your use! They are pricey but made for the marine evironment. Require a small diesel tank 5 gallons would last all season, they use 0.07 gallons per hour on high heat. Does require exhaust system complete with muffler and a piping or vent system to get the water or air to the cabin. Water system has a demand circulator pump both work off of a remote themostat. Here is the information on the unit!

http://www.espar.com/pdfs/marine.pdf

Mike -  :idea:
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« Reply #17 on: February 26, 2006, 10:47:32 AM »

Hey, Just trying to pass the info along. Don't want anyone to try something and spend money only to find out it won't work as they wanted!

Mike -  :wink:
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« Reply #18 on: February 26, 2006, 02:58:30 PM »

I understand what you mean, That is why I have not run it off the inverter. Luckily when it has been cold enough to need the heater we have been hooked up to shorepower all night. Didn't think it would last long off the inverter. Man, i wish i was good at math and figure stuff out like you can merwin!!!
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« Reply #19 on: February 27, 2006, 03:12:51 PM »

Hi Mervin,

Just a quick question about your calcs on amp hours, I understand how you got the current @ 120V from P=VI but how did you get the current @ 12v to = approx 90A?

I'd like to go through everything I have on my boat and work out total amp hour requirements...just out of interest  :wink:

Matt
« Last Edit: December 31, 1969, 08:00:00 PM by mg2006 » Logged

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merwin10
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Ok
« Reply #20 on: February 27, 2006, 06:53:01 PM »

mg2006,

Here is the secret which you have some of the information already. As I stated the number was not real accurate, more of a guess-timate. Ok you know if you have watts and you divide by volts you get amps.

1000/120=8.3 amps or 1000/12=83.3 so where is the 6.7 amps

That was easy! So where is the 6.7 amps. That comes in several forms, line loss, efficiency factor, planning on inductive resistive loading. I didn't or better couldn't really figure the real numbers because I did not have the actual specs of distance between the battery and inverter, inverter efficientcy, size wire, temperature, type of inductive loading at startup, and if the heater unit would be pure resistive load or combination of resitive and inductive load.

So the 6.7 amps was a guess based on other installations.

So why are these losses important. Well a small difference in voltage equals a large difference in current. Lets say we have 11 volts DC at the Inverter, due line loss, not to crazy!

1000/11=90.9 amps.

Now since we all know inverters are not 100% efficent, more like 90%, so to produce 1000 watts at 120 volts you need to put in 1100 watts at DC volts.

1100/11=100 amps

You are getting the idea! of course this is all theoretical due to the lack of ture specifications. But here I have only touch two of the losses, line loss and effieciency

Look at my post under electrical and electronics forum shore power question. The same thing happens with your shore power, line loss.

When figuring out your power requirements there is also a duty cycle to consider. In the case of the heater I used 100%. Let's think of a 1000 watt microwave. You run the microwave 4 times a day for a total of 10 minutes.

Using the same process! 1100w/11v=100a however for only 10 minutes.
So 10 minutes is one sixth of an hour, hence you would use 16.6 amp minutes or .6 amp hours per day. Big difference! So a 75 amp hour battery would handle that just find.

The thing you need to do is to make a list of the options and duty cycle. Looks like this working on a 24 hour day.


Loads.........A standby....StandbyHours......High A....High A Hours....AHR Day
Instruments......0.............0......................0.5..............10................5.0
GPS.................0.............0......................0.5..............10................5.0
DGPS...............0.............0......................0.5..............10................5.0
Radar.............2.3............4......................3.9..............2.0..............16.8
VHF................0.8............8......................5.0..............0.2...............7.2
Autopilot.........0.1............8......................5.0..............1.5...............8.3
Engines.............0............0......................30.0............2.0...............60.0
Cabin lights.......0............0.......................5.0.............3.0...............15.0
Navigation.........0............0.......................1.0.............3.0................3.0
Anchor lights......0............0.......................1.0.............8.0................8.0
Batt monitor.....0.1..........24........................0................0.................2.4
Bilge pumps.......0............0.........................0...............0....................0
Waste water.......0............0.........................0...............0....................0
Windless............0............0.......................70.0............0.1...............2.4
Blowers..............0............0.......................4.0.............4.0...............12.0
CO2 detector......0.1........24........................0...............0..................2.4
Washdown Pump.0............0........................12..............0...................0
Freshwater Pump0............0........................12.............1.0...............12.0

...................................................................Total Amp Hours.......164.5


Hope this helps!

Mike -  :wink:
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« Reply #21 on: February 27, 2006, 07:28:53 PM »

This thread may get moved to merwins domain in the electric/electronics forum. Great stuff! Makes much more sense now.
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« Reply #22 on: February 27, 2006, 07:45:52 PM »

Talon,

I would have moved it but I don't have privs. So move it!

Mike -  :wink:
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« Reply #23 on: February 28, 2006, 10:50:50 AM »

Time for another lesson you all have read the discussion here and I have probably written some words that you may not have considered before.
So I thought I would explain the whys and where fors!

One of the failures of inverters is ripple voltage on the dc leads from the battery to the inverter. This ripple voltage causes premature failure of the filter capacitors in the inverter.

So where is this ripple coming from? Well in is induces into the battery cables by the battery cables themselves or by other cables they run parallel too.

Well how does this happen and how can you reduce the effect of this.

First you need to understand how this inductance is produced. When an electrical current (flow) passes thru a wire a magnetic field is created. You all remember your six grade science. Take a normal nail wrap it with some wire attach a battery WOLLA you have a magnet. Well as this magnetic field builds it induces a voltage in any conductor nearby. Thus there is one of the problems. However, the building of the magnetic field also induces a voltage in the original conductor in the opposite direction of the current that produced it in the first place, called self inductance. How much self-induced reverse current is proportional to the size of the loop? This is how transformers work! The larger the loop or number of loops the great the inductance.

So your battery cables aren't making any loops so there is not a problem, wrong! Your battery cables positive and negative makes up one big loop, thus how they are positioned or arranged makes a difference as to how much self induced voltage there will be. The more the distance the more the self induced current will be. Interesting it the magnetic fields in the positive and negative conductors are opposite of each other. So by putting the positive and negative wires together and tapping them every four inches or so the self induced magnetic fields tend to cancel each other out.

This effect can be measured in electronics the unit of measurement is called Henry like resistance is called Ohm. At a foot of separation you will double the number of henries of that of taped battery leads. Of course in reality we measure this in micro-henries, so when taped it measures about 3uH and at a foot 6uH.

Since the induced voltage varies as a rated of change of the current in a conductor, the voltage could be three or four times the original voltage. The induced current works against the applied current which in turn causes loss of inverter performance and reduced efficiency.

All this adds up to needing to apply more current to run the inverter and shortens the battery life and run time. So to sum this all up! Tape your battery leads together right after they come off the battery terminals and keep the runs as short as possible, use large over size cable, avoid running the cables parallel to other wires and keep the battery up to charge.

Hope this helps - I tried to explain this in the simplest form I could!

Mike -  :wink:
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« Reply #24 on: March 10, 2006, 09:12:08 AM »

Oh ... shoot .......



I thought you said ... HEADERS !

 wink
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