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24V resistive load on 12V system?

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a.s.elliott(at)cox.net
Guest

Posted: Mon Feb 12, 2007 7:24 am Post subject: 24V resistive load on 12V system?

I want to make sure I understand this properly.  Please advise if I  have made some conceptual errors.

  

 I have a heated pitot tube designed for a 24V system.  It is supposed  to draw 8 amps at 24V.  This makes its fixed resistance, according to Ohms'  Law:

  

         R = V/I = 24/8 = 3 Ohms  

  

 So if I install this pitot tube in a 12V system, it should draw only 4  amps:

  

         I = V/R = 12/3 = 4 A

  

 But, it will only work at about 1/2 the heat level it had in the 24V  system.

  

 I can measure the stabilized temperature rise, but how hot should it have  to get to be usable in the 12V system?  I am thinking that since I don't  have any anti-ice on the plane, I never fly in the clouds when the OAT is below  freezing, so that if I can get a 25°C rise from the tube, I should be  OK.  Yes?

  

 I am *guessing* that the AN-spec 24V tube was designed heat from -56°C, the  nominal temp above 36K', and that 1/2 the rated power should work fine, while  also saving me some current draw.

  

 Thoughts?

  

 Andy Elliott, Mesa, AZ

N601GE (reserved)

601XL/TD/QB, Corvair,  building...

   [quote][b]

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nuckollsr(at)cox.net
Guest

Posted: Mon Feb 12, 2007 7:58 am Post subject: 24V resistive load on 12V system?

At 08:16 AM 2/12/2007 -0700, you wrote:



 	  Quote: 	
		  I want to make sure I understand this properly.  Please advise if I have 

made some conceptual errors.



I have a heated pitot tube designed for a 24V system.  It is supposed to 

draw 8 amps at 24V.  This makes its fixed resistance, according to Ohms' Law:



         R = V/I = 24/8 = 3 Ohms



So if I install this pitot tube in a 12V system, it should draw only 4 amps:



         I = V/R = 12/3 = 4 A



But, it will only work at about 1/2 the heat level it had in the 24V system.


	


   Actually, 1/4th the heat . . . or a bit more. What you need to do

   is hook your tube up to a 14V power supply and stir it in a bath

   of crushed ice and water. Measure current after it stabilizes -

   this may take a minute or so.



   Pitot tubes have a strong positive temperature coefficient of

   resistance. Here's some data I plotted off a batch of tubes

   a few years ago.



http://www.aeroelectric.com/Pictures/Curves/Pitot_Heater_R_Plot.pdf



   When you drop the voltage, not only does current go down by

   1/2, so does applied voltage . . . therefore the resulting power

   is 1/4 that of the 28v operation.



 	  Quote: 	
		  

I can measure the stabilized temperature rise, but how hot should it have 

to get to be usable in the 12V system?  I am thinking that since I don't 

have any anti-ice on the plane, I never fly in the clouds when the OAT is 

below freezing, so that if I can get a 25°C rise from the tube, I should 

be OK.  Yes?



I am *guessing* that the AN-spec 24V tube was designed heat from -56°C, 

the nominal temp above 36K', and that 1/2 the rated power should work 

fine, while also saving me some current draw.


	


   Keep in mind that the goal for a de-ice system is to

   melt ice. When super-cooled water hits your pitot tube,

   the energy in the tube has to overcome the heat-of-

   crystalization. Know that heated pitot tubes are a

   match in a big dark room when it comes to flying in

   an icing environment. Any heat is better than no

   heat . . . but running a 28v tube on 14v is waaayyyy down

   on the efficacy curve for "any heat".



   Bob . . .



        ----------------------------------------

        ( IF one aspires to be "world class",  )

        ( what ever you do must be exercised   )

        ( EVERY day . . .                      )

        (                  R. L. Nuckolls III  )

        ----------------------------------------

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frank.hinde(at)hp.com
Guest

Posted: Mon Feb 12, 2007 8:00 am Post subject: 24V resistive load on 12V system?

Heating effect is P=I*I*R

  

 Thus in your 28V system P= 8*8*3 = 192  Watts

  

 In your 12V system  P= 4*4*3 = 48  Watts

  

 Thus in a 12 system the same pitot only produces a quarter  of the 24V's heating effect. As Heating effect is proportional to the change in  temperature for two identical mass flow situations then the temp rise will only  be 25% of what it was previously.

  

 Now do you even need a heated Pitot?..Highly debatable even  in an IFR platform and I assume you intend your Zodiac as a VFR machine?...If so  then you certainly don't need a heated pitot.

  

 Frank

  

  



   From:  owner-aeroelectric-list-server(at)matronics.com  [mailto:owner-aeroelectric-list-server(at)matronics.com] On Behalf Of Dr.  Andrew Elliott

Sent: Monday, February 12, 2007 7:17 AM

To:  aeroelectric-list(at)matronics.com

Subject: 24V  resistive load on 12V system?

 

 I want to make sure I understand this properly.  Please advise if I  have made some conceptual errors.

  

 I have a heated pitot tube designed for a 24V system.  It is supposed  to draw 8 amps at 24V.  This makes its fixed resistance, according to Ohms'  Law:

  

         R = V/I = 24/8 = 3 Ohms  

  

 So if I install this pitot tube in a 12V system, it should draw only 4  amps:

  

         I = V/R = 12/3 = 4 A

  

 But, it will only work at about 1/2 the heat level it had in the 24V  system.

  

 I can measure the stabilized temperature rise, but how hot should it have  to get to be usable in the 12V system?  I am thinking that since I don't  have any anti-ice on the plane, I never fly in the clouds when the OAT is below  freezing, so that if I can get a 25°C rise from the tube, I should be  OK.  Yes?

  

 I am *guessing* that the AN-spec 24V tube was designed heat from -56°C, the  nominal temp above 36K', and that 1/2 the rated power should work fine, while  also saving me some current draw.

  

 Thoughts?

  

 Andy Elliott, Mesa, AZ

N601GE (reserved)

601XL/TD/QB, Corvair,  building...

[quote]



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[b]

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