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Old 01-17-2013, 08:20 AM   #11
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Quote:
Originally Posted by herk7769 View Post

Your sewer pipes should contain no water since the valves are supposed to be closed.
Unless he's got 2 foot of pipe or more between tank and shut-off valve.

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Old 01-17-2013, 08:48 AM   #12
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Originally Posted by f1100turbo View Post
Unless he's got 2 foot of pipe or more between tank and shut-off valve.

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I can see where that would be a problem, but in this case a pipe heater would have been installed. I suspect his system is not working (or not being used) correctly.

The pads need to be turned on well in advance of the actual cold temperatures. Once stuff freezes there is just not enough heat from the pads to defrost a frozen anything.

In the aircraft we have two types of ice control systems; one "de-ices" and an the other "anti-ices."

A "de-icer" allows ice to form, then it is removed while frozen. Like wing "boots" that let with wing coat with ice and the boot is inflated breaking the ice off or a prop blade deicer that cracks off ice on the blade using prop rotation forces to sling it away.

An "anti-ice" system is turned on BEFORE you go near ice. It warms the part preventing ice from forming (YOU HOPE). Used on a wing, hot engine air is blown into a cavity behind the leading edge keeping the wing well above outside air temperature so ice does not form. When used on props it is used in conjunction with the de-ice on the blades, but only on the hub where rotation forces are insufficient to sling the ice off if it allowed to build. They "modulate on and off" to maintain a temperature range at all times.

Our tank heaters are ANTI-FREEZING systems and as such MUST be on before temps drop to where you might need them. Waiting until the tanks are cold enough to freeze then turning them on will not work as the system will be overwhelmed. Just not enough BTU to melt something that has already frozen.
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Old 01-17-2013, 09:01 AM   #13
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Makes sense to me .


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Old 01-17-2013, 10:14 AM   #14
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Quote:
Originally Posted by herk7769 View Post
... Lower voltages increase amps to maintain wattage.
I believe these pads are purely resistive so amps would decrease at lower voltages. They would have to be fed by a voltage regulator circuit (and operated at a lower voltage than supply) in order to do what you suggest.
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Old 01-17-2013, 11:42 AM   #15
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Barry, let me look into this. I remember doing this exercise during a discussion on low power line voltage.
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Old 01-17-2013, 11:56 AM   #16
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Barry here is the result of my review:

Quote:
Originally Posted by BarryD0706 View Post
I believe these pads are purely resistive so amps would decrease at lower voltages. They would have to be fed by a voltage regulator circuit (and operated at a lower voltage than supply) in order to do what you suggest.
Some definitions are in order.

Resistive Load - An electrical load which is characteristic of not having any significant inrush current. When a resistive load is energised, the current rises instantly to it's steady-state value, without first rising to a higher value. An electrical load in which voltage and current are converted to energy in the form of heat; i.e., an electrical heater, incandescent bulb.

Inductive Load - An electrical load which pulls a large amount of current (an inrush current) when first energized. After a few cycles or seconds the current "settles down" to the full-load running current. The time required for the current to "settle down" depends on the frequency or/and the inductance value of the Inductive load

Being purely resistive, a 1000 watt heater would draw 1000 watts.

My logic using a constant power requirement went like this.
Power = Volts x Amps

Power is measured in watts so a pad that draws 120 watts of power when "on" will draw 10 amp at 12 volts.

That same pad at 13 volts would be 120/13 or 9.2 amps.

For your assertion to be correct, the power requirement would have to vary and not stay fixed. As I understand it, an inductive load varies by applied voltage (like your air conditioner). As the voltage goes down the amount of amps required to turn over the compressor and fan motor goes up A LOT until everything comes up to speed.

Other threads on similar topics:

Low voltage and air conditioning

Re-thinking Hardwire Surge Guard...Opinions?
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Old 01-17-2013, 12:27 PM   #17
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I found my error.
Those who like seeing me wrong, please bookmark this thread.

When I realized my simultaneous equations were not giving me the results I expected, I threw out all assumptions and went "back to the drawing board".

Here is why you are correct and a valuable lesson is here for everyone who uses a Resistance heater in low power conditions.

My example below assumed the power required stayed constant. It does not.
Power AVAILABLE is voltage dependent. That was my mistaken assumption.

So back to my simultaneous equation.

V=IR or Voltage = amps times circuit resistance in Ohms.

Assuming the resistance of the heating wire in our heating pad (using the ACTUAL numbers from the manual) stays constant for both incoming voltages and that heating coil is RATED at 78 Watts at 13.5 volts.

The instantaneous current required for that 78 Watts to be available is 5.8 amps at 13.5 volts. So, if we decrease the incoming voltage to 12 DCV, then what happens.

We need to solve 2 equations with 2 unknowns R of the heater and I2 of the lower voltage.
First equation:
V1 = I1 x R or 13.5 volts = 5.8 amps x R
Second equation:
V2=I2xR or 12.0 volts = I2 (UNKNOWN amps) x R

Solve the second equation for R

R = 12 volts / I2 (unknown amps)

Substituting the second equation for R in the first equation

13.5 volts = 5.8 amps x (12 volts/I2)

Solve for I2 = (5.8 x 112)/13.5

I2 = 5.15 amps

So the Available power of our "heater" at 12 volts and 5.15 amps is 61.8 Watts. This is only 79% of the "heat" available at 13.5 watts or 21 % LESS efficient.
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Old 01-17-2013, 12:30 PM   #18
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Quote:
Originally Posted by herk7769 View Post
...
My logic using a constant power requirement went like this.
Lou,
there is nothing in a resistive load that indicates that it will have a constant power requirement. A lightbulb, or these pads, will have reduced power usage/output at reduced voltage.

I thought perhaps you had actually measured increased amps at your TT input, and thinking about that made me realize that we are both right.

If you're looking at the operation of these pads as your TT battery voltage drops, then what I said is true. If you're looking at current into your TT from the line, the converter is keeping constant voltage on the battery so the TT input current goes up when the TT input voltage input drops! That makes everything that's 12v in your trailer look inductive (or at least constant-power).
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Old 01-17-2013, 12:33 PM   #19
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I think I did not learn anything (I am old gimmeee a break!).

My PREVIOUS error on this board was this very same problem! We were talking about running the water heater on electricity or propane as I recall. The result of THAT thread was "if you want hot water in the summer; use propane" for the exact same reason here.

here is is! Elect water heater

And only September too. Twice in the same year on the same problem...
VERY
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Old 01-17-2013, 01:14 PM   #20
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I'm old too, so we all need a break. To simplify matters in DC power we have to understand the known from the un-known. If the heaters are normally 12 v and we know that they are 120 watts then (P=V*A) (120= 12* A) (A=10 amps) Now we need to know the resistance of the heat pad. (E=I*R) (12=10*R) (R= 1.2 ohms) Now the resistance is the only thing that cannot change in this case it's fixed , therefore if the voltage drops so does the power. If the volts are now 10 V and R is 1.2 power is (E=I*R) (10=I*1.2) (I=8.333) Now power or watts will be (P=V*A) (P=10*8.33) P=83.33 watts less heat. The numbers used are for illustrative purposes only!! You're pretty good for a flyboy I too spent 36 years in this mess I mean Aviation. Have a good day!
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