My Project

Started by NePaSmoKer, January 01, 2011, 02:47:43 PM

Previous topic - Next topic

0 Members and 1 Guest are viewing this topic.

NePaSmoKer

Ok gimmies some input.  I really dont want to use the original element cuz it draws 15 amps  :'(

The heater. There is a plate that site over the heater.


The left knob is the temp, right is the timer which is not working.



Bottom of electronics. There is a 20 cfm blower in there to.



This is looking across the heater at the hole in back, the blower pushes air thru this and up the back wall channel (not shown) and thru louvers in the channel to distribute heat to each tray per louver.



There is a thermocouple under this plate.



The cabinet is double door and totally insulated.


Temp bulb on top door.





Bottom with electric slide out removed.


Dimensions.







I'm thinking 2 brinkman 1500 watt elements and just hard wiring the blower?

Tenpoint5

I'm thinking my 70 cfm blower might be a bit much for mine.
Bacon is the Crack Cocaine of the Food World.

Be careful about calling yourself and EXPERT! An ex is a has-been, and a spurt is a drip under pressure!

Friar_Tuck

That is one nice cabinet there!
Won't two 1500 watt elements take 25 amps?  Or are they 220 elements?
I have had to tame the fan down on my beast a bit, but really like the fan running.
It seems to cut down on my total time in smoker.
Keep us posted.
Jim

BuyLowSellHigh

#3
If I remember correctly that is a 120 VAC cabinet.

If the heating element alone draws 15 amps, then it should be ~ 1800 watts.  If the total draw is 15 amps then a low side estimate (not knowing how much the fan draws) might be 1500 watts.

I am NO expert, but I think both the current element and the Brinkman element are simple resistance heaters.  If my thinking cap is on right then I believe the way to determine the expected current draw is to use Ohm's Law.

A 1500 watt element at 120 VAC should have a resistance of ~ 9.6 ohms an draw 12.5 amps.  If you wired two in parallel you would have a net resistance of 4.8 ohms giving 3000 watts and the current draw should be 25 amps.  But, if you wired them in series  the resistance would be 19.2 ohms giving 750 watts of power drawing 6.25 ohms amps.

If it less power, meaning less heat, that you want I think it would be cheaper and easier to just install one lower power element. or two much lower power elements in parallel  - like the double element setup many have used on the Bradley to get 1000 watts.
I like animals, they taste good!

Visit the Recipe site here

NePaSmoKer

Quote from: BuyLowSellHigh on January 01, 2011, 11:56:31 PM
If I remember correctly that is a 120 VAC cabinet.

If the heating element alone draws 15 amps, then it should be ~ 1800 watts.  If the total draw is 15 amps then a low side estimate (not knowing how much the fan draws) might be 1500 watts.

I am NO expert, but I think both the current element and the Brinkman element are simple resistance heaters.  If my thinking cap is on right then I believe the way to determine the expected current draw is to use Ohm's Law.

A 1500 watt element at 120 VAC should have a resistance of ~ 9.6 ohms an draw 12.5 amps.  If you wired two in parallel you would have a net resistance of 4.8 ohms giving 3000 watts and the current draw should be 25 amps.  But, if you wired them in series  the resistance would be 19.2 ohms giving 750 watts of power drawing 6.25 ohms.

If it less power, meaning less heat, that you want I think it would be cheaper and easier to just install one lower power element. or two much lower power elements in parallel  - like the double element setup many have used on the Bradley to get 1000 watts.



Ummmmm...................yeah    :D

SL2010

Quote from: NePaSmoKer on January 02, 2011, 05:35:19 AM
Quote from: BuyLowSellHigh on January 01, 2011, 11:56:31 PM
If I remember correctly that is a 120 VAC cabinet.

If the heating element alone draws 15 amps, then it should be ~ 1800 watts.  If the total draw is 15 amps then a low side estimate (not knowing how much the fan draws) might be 1500 watts.

I am NO expert, but I think both the current element and the Brinkman element are simple resistance heaters.  If my thinking cap is on right then I believe the way to determine the expected current draw is to use Ohm's Law.

A 1500 watt element at 120 VAC should have a resistance of ~ 9.6 ohms an draw 12.5 amps.  If you wired two in parallel you would have a net resistance of 4.8 ohms giving 3000 watts and the current draw should be 25 amps.  But, if you wired them in series  the resistance would be 19.2 ohms giving 750 watts of power drawing 6.25 ohms.

If it less power, meaning less heat, that you want I think it would be cheaper and easier to just install one lower power element. or two much lower power elements in parallel  - like the double element setup many have used on the Bradley to get 1000 watts.



Ummmmm...................yeah    :D



Did you get all that  ??? ??? ??? ??? ??? ??? ??? ??? ??? ???

NePaSmoKer

Quote from: SL2010 on January 02, 2011, 05:41:47 AM
Quote from: NePaSmoKer on January 02, 2011, 05:35:19 AM
Quote from: BuyLowSellHigh on January 01, 2011, 11:56:31 PM
If I remember correctly that is a 120 VAC cabinet.

If the heating element alone draws 15 amps, then it should be ~ 1800 watts.  If the total draw is 15 amps then a low side estimate (not knowing how much the fan draws) might be 1500 watts.

I am NO expert, but I think both the current element and the Brinkman element are simple resistance heaters.  If my thinking cap is on right then I believe the way to determine the expected current draw is to use Ohm's Law.

A 1500 watt element at 120 VAC should have a resistance of ~ 9.6 ohms an draw 12.5 amps.  If you wired two in parallel you would have a net resistance of 4.8 ohms giving 3000 watts and the current draw should be 25 amps.  But, if you wired them in series  the resistance would be 19.2 ohms giving 750 watts of power drawing 6.25 ohms.

If it less power, meaning less heat, that you want I think it would be cheaper and easier to just install one lower power element. or two much lower power elements in parallel  - like the double element setup many have used on the Bradley to get 1000 watts.



Ummmmm...................yeah    :D



Did you get all that  ??? ??? ??? ??? ??? ??? ??? ??? ??? ???

Well i know how to wire from a switch panel, plugs, lights, switches but i am pretty electronically challenged when it comes to what BLSH said  :D

beefmann

one thing i would consider is to reverse the  flow of the  air and bring it from the top down and across the heater since heat naturally rises.

as for  more hear, id convert it to 220 and run 2 - 1500 watt 120 volt heaters  one on each leg and each heater controlled by a pid ,  with one pid being set 10 degrees lower then the other. , run the fan full time   

BuyLowSellHigh

Let me try this ... in a circuit the heating elements behave basically as resistors.

There are two basic ways to wire a  resistor in a circuit, as shown in the image below.  The squiggly lines labeled R1 and R2 represent the heating elements.



The top one is parallel wiring, the bottom one is series wiring.

In parallel wiring one end of each element (R1 and R2) connects to the same + supply, and the other end of each element connects to the same - supply (this is how the second heating element addition on the Bradley is commonly done).

In series wiring one end of one element (R1) connects to +, one end of the second element (R2) connects to -, and the two elements are connected two each other in between.

In parallel wiring the power of each element is preserved and additive 1500 watts + 1500 watts = 3000 watts, which at 120 VAC would draw ~ 25 amps

In series wiring the total power is reduced as the resistacne for current flow is the sum of the resistance for each element, which reduces total current flow and resulting net power.  In this case using two 1500 watt elements would give you not 3000 watts but 750 watts of heating power, which would draw ~ 6.25 amps (I mistakenly wrote ohms in the original post, now fixed).

Rick, a few questions -- how do you plan to use this beast, how much will you be putting in it and how hot do want it to be able to get with a full load?
I like animals, they taste good!

Visit the Recipe site here

lumpy

Wow......this is heavy conversation.
I'm going back to the sausage thread :D :D............something I understand

Good luck Nepas ;)

NePaSmoKer

Quote from: lumpy on January 02, 2011, 08:46:25 AM
Wow......this is heavy conversation.
I'm going back to the sausage thread :D :D............something I understand

Good luck Nepas ;)

Yeah i think i need to drink a few woodchucks, sit and stare at the cabinet and figur it out  :D

pensrock

I see no benefit of wiring the heating elements in series. He could just use a smaller heating element and get the same results.
I guess the question is..... What are you trying to do? Double the heat? Keep the heat the same? Or reduce the amount of heat?
Do you want to use 220VAC or 110VAC?
Maybe I missed something in the first few posts?  ???

BuyLowSellHigh

Quote from: pensrock on January 02, 2011, 03:56:44 PM

I see no benefit of wiring the heating elements in series. He could just use a smaller heating element and get the same results.
I guess the question is..... What are you trying to do? Double the heat? Keep the heat the same? Or reduce the amount of heat?



Prexactly what I said,  then asked.

Bottom line - to reduce current draw means less power and less heat. 

Increasing heat means more power and more current.  If that is desired than a change to 240 V would be the way to go. 

If less power/heat is desired, then either a smaller element or possibly a rheostat are options.
I like animals, they taste good!

Visit the Recipe site here

NePaSmoKer

110v

Doing sticks, SS and maybe jerky in it. I know there is going to be a Bradley smoke gen on it. The element just needs to get to 180*

BuyLowSellHigh

Rick, I don't have a good idea of how low you can go on heating power and still get to 180 °F with a meat load.  If the current 15 amp current draw is your concern then I believe that is the question.  MY complete WAG is that it is a combination of cabinet volume and meat load (lbs).

That cabinet is pretty good size, nearly 20 cu ft interior volume by my calculation looking at your measurements in the pictures you posted.  Currently a reasonable estmate of the heating element power would be ~ 1600 W.  That gives you about 80 W/cu ft.

Here are some comparisons:

Bradley six rack: 625 W with the SG, 3.15 cu ft (measured interior on mine) = 198 W/cu ft

Sausage Maker : from their smokehouse specifications interior measurements and stated power:

   20 lb model = 358 W/cu ft
   50 lb model = 198 W/cu ft
  100 lb model = 203 W/cu ft

Smokin' Tex 1500 = ~ 240 W/cu ft

Cookshack SMO66AmeriQue = ~229 W/cu ft

The common number seems to be about 200 W/cu ft, your current cabinet has ~ 80.  I think you can get it hot without a load in it, but as soon as you start loading meat to smoke and cook it's going to become a mass limited issue.
I like animals, they taste good!

Visit the Recipe site here