Assistance on connections to Power Supply

[KeepItSimpleStupid]

I noticed that there is a mention of wall wart as alternative to the DC SSR setup.
Do correct me if I understand it wrongly,
1. A wall wart with volt control of 0 to 10v will be used.
2. The positive end of the wall wart is connected to the positive output of the DC power supply and the negative end of the wall wart is connected to the “PAR” input of the DC power supply (GPS-3030)
How do I connect the potentiometer and a 75watt wire wound resistor to this setup?

You really want to ignore this stuff. Initially I didn't realize that the controller provides a current limited voltage to turn on a LED or OPTO DC SSR, so let's stay simple.

The "wall wart" would have taken the place of the internal current limited supply (the 27 V) and/or provided power for the home made circuit that was decided would be abandoned. I have still been thinking about it. A nice linear very low current bi-polar +-15 V supply is the easy way out for the circuit that's gelling around in my head with no actual components attached.

"WE" never really told you how to connect an external voltage source to control the power supply nor have I tried to decipher the master/slave stuff. For now, it's unimportnant.

on the other hand
If I use a DC SSR, how do i connect it to the system with the PID and 75 watt wire wound resistor?

If you can connect a LED, you can connect the SSR. Take the 12 A version of the SSR here: https://www.omega.com/pptst/SSRDC100V.html. but you are probably on the other side of the pond than me. I think you mentioned your mains voltage of 220/240 or so. Farnell may be a source for the SSR.

One side of the SSR would connect like a LED. Note that it also has a built-in LED too. The specs of the SSR are incomplete too, but that company I've used before because it's a lot of different manufacturer's under one roof or a one-stop shop. They used to re-label their stuff. So a fluke thermometer would be come an Omega thermometer. So, observe the polarity when connecting the LED side. The datasheet is here: https://www.omega.com/temperature/pdf/SSRDC100V.pdf

Note that the input (LED side) and the output side have polarity markings. Earlier I told you how to connect the LED side.

So now think of the output side that also has polarity markings. It MAY be useful to use a fuse sized for the expected load, but it would just act like a SPST switch (observe polarity) using the output terminals of the power supply. So, everything goes in series. PS+Fuse+SSR+resistor in any order observing the polarities You would be able to set a lower voltage to get finer temperature control. The fuse, strictly isn't needed.

So, this way you would be applying brief power to the resistor at the front panel voltage setpoint of the power supply. The SSR will have some series resistance. If it gets hot, it needs to get heat sinked.

 

[Les Jones]

Hi zexclo,
It is quite possible that you have damaged the output of the PID controller by connecting your potential divider to it with a total resistance of 3 ohms. The manual says the minimum value should be 500 ohms.

The reason for the discussion about using a wall wart and opto coupler to control the power supply output is to avoid possible damage to the DC power supply when the mains to it is switched on and off fairly rapidly if an SSR was used to switch it's mains input. The mains to the DC power supply would be permanently on but its output would be switched between zero volts and the required voltage the heater. Switching the output on and off this way should not damage the DC power supply. The opto coupler and wall wart are required to provide isolation between the PID output and the control input of the DC power supply.

Edit. I've just read "KISS's" post 41. This type of SSR switching the DC output of the power supply is a much tidier solution than the one I have just described.

Les.

 

[KeepItSimpleStupid]

The tubes are small carrying liquid.

Earlier, the tube were carrying air!

 

[KeepItSimpleStupid]

Hi zexclo,
It is quite possible that you have damaged the output of the PID controller by connecting your potential divider to it with a total resistance of 3 ohms. The manual says the minimum value should be 500 ohms.

Your going goofy on me. If you use the built in resistor, 1 or 2 ohms to a few hundred means nothing.

The reason for the discussion about using a wall wart and opto coupler to control the power supply output is to avoid possible damage to the DC power supply when the mains to it is switched on and off fairly rapidly if an SSR was used to switch it's mains input. The mains to the DC power supply would be permanently on but its output would be switched between zero volts and the required voltage the heater. Switching the output on and off this way should not damage the DC power supply.

You need sleep. Switching the mains of the existing power supply never came up although I do remember a transformer discussion. You and I didn't realize that there was already a current limited source available. The SSR side provides the isolation.

The opto coupler and wall wart are required to provide isolation between the PID output and the control input of the DC power supply.
Les.

You need more sleep. The current iteration would not use the control input of the power supply. Isolation is provided by the SSR, not that it matters in the current implementation.

PLAN: Keep the DC supply at some fixed value and apply that voltage modulated by PID to the resistor (Heater). Don't turn the power on and off to the controller and don;t play with the analog programming port in the current iteration.

There wasn't a plan A. The controller was likely found in a closet somewhere with a lot of dust on it, Plan A would have used a 0-10 process output to drive the DC supply, There could be caveats with this design too. i.e. Is the maximum of the PS too much? Does the controller have a "TAG" that limits Pout?

 

[zexclo]

You really want to ignore this stuff. Initially I didn't realize that the controller provides a current limited voltage to turn on a LED or OPTO DC SSR, so let's stay simple..

Ok, will not be following and do it.

Yeah, in my region we are using 220/240. I am probably going down to the electronics store to do a purchase for the DC SSR, as the shipping might take too long and I wish to complete this setup by end of this week, before my supervisor explodes.

Just in case, the shops do not carry this product, what specs of the DC SSR I have to look out for besides the 18-32V for input signal.

My bad, the surrounding is air, but the tubes is carry fluids

=====

Hi Les,

Is there any way for me to check if the output PID controller has been damaged?

Hi zexclo,

The reason for the discussion about using a wall wart and opto coupler to control the power supply output is to avoid possible damage to the DC power supply when the mains to it is switched on and off fairly rapidly if an SSR was used to switch it's mains input. The mains to the DC power supply would be permanently on but its output would be switched between zero volts and the required voltage the heater. Switching the output on and off this way should not damage the DC power supply. The opto coupler and wall wart are required to provide isolation between the PID output and the control input of the DC power supply.

Thanks for the explanations!

Ok, I shall follow the recommendations by Kiss’s” post #41.

 

[zexclo]

Your going goofy on me. If you use the built in resistor, 1 or 2 ohms to a few hundred means nothing.



You need sleep. Switching the mains of the existing power supply never came up although I do remember a transformer discussion. You and I didn't realize that there was already a current limited source available. The SSR side provides the isolation.



You need more sleep. The current iteration would not use the control input of the power supply. Isolation is provided by the SSR, not that it matters in the current implementation.

PLAN: Keep the DC supply at some fixed value and apply that voltage modulated by PID to the resistor (Heater). Don't turn the power on and off to the controller and don;t play with the analog programming port in the current iteration.

There wasn't a plan A. The controller was likely found in a closet somewhere with a lot of dust on it, Plan A would have used a 0-10 process output to drive the DC supply, There could be caveats with this design too. i.e. Is the maximum of the PS too much? Does the controller have a "TAG" that limits Pout?

LOL. the controller was indeed pass down from others with quite a bit of dust on it. PID Controllers are not cheap in my region and easy cost upwards of 100USD in my region. My controller should not be damaged. Glad to know that, haha

 

[zexclo]

One side of the SSR would connect like a LED. Note that it also has a built-in LED too. The specs of the SSR are incomplete too, but that company I've used before because it's a lot of different manufacturer's under one roof or a one-stop shop. They used to re-label their stuff. So a fluke thermometer would be come an Omega thermometer. So, observe the polarity when connecting the LED side. The datasheet is here: https://www.omega.com/temperature/pdf/SSRDC100V.pdf

Note that the input (LED side) and the output side have polarity markings. Earlier I told you how to connect the LED side.

So now think of the output side that also has polarity markings. It MAY be useful to use a fuse sized for the expected load, but it would just act like a SPST switch (observe polarity) using the output terminals of the power supply. So, everything goes in series. PS+Fuse+SSR+resistor in any order observing the polarities You would be able to set a lower voltage to get finer temperature control. The fuse, strictly isn't needed.

So, this way you would be applying brief power to the resistor at the front panel voltage setpoint of the power supply. The SSR will have some series resistance. If it gets hot, it needs to get heat sinked.

Hi Kiss, from the explanation
Can I sum it up as:

one side of the SSR would be connected to the PID.
while on the other side, one port will be connected to the DC power supply, while one end will be connected to the wire wound resistor.
Am wondering, the wire wound resistor will not be working since its only connected at one side?
And how do I connect the Power Supply to the SSR, do I set the DC supply to slave, and connect the SSR wire to "SER" of the DC power supply?

 

[Les Jones]

Hi zexclo,
The control input of the SSD is connected to the PID controller. (The polarity must be correct. Terminal 12 on the PID to + input on the SSR , Terminal 13 on the PID to - input on the SSR.) The power supply is configured as a stand alone power supply. (Nothing is connected to the rear panel terminals. The output will be on all of the time.) The positive output of the power supply will connect to the positive output terminal of SSR. The negative output terminal of the SSR will connect to one end of the heater resistor. The other end of the heater resistor will be connected to the negative output of the power supply.

Les.

 

[KeepItSimpleStupid]

Hi zexclo,
The control input of the SSD is connected to the PID controller. (The polarity must be correct. Terminal 12 on the PID to + input on the SSR , Terminal 13 on the PID to - input on the SSR.) The power supply is configured as a stand alone power supply. (Nothing is connected to the rear panel terminals. The output will be on all of the time.) The positive output of the power supply will connect to the positive output terminal of SSR. The negative output terminal of the SSR will connect to one end of the heater resistor. The other end of the heater resistor will be connected to the negative output of the power supply.

Les.

Good, you got some sleep.

 

[KeepItSimpleStupid]

Just in case, the shops do not carry this product, what specs of the DC SSR I have to look out for besides the 18-32V for input signal.

It may not be an "off the shelf item"unless you get lucky especially high current and DC. You might "dumb down" the specs a bit first and say a DC Solid State optoisolated relay with a rating greater than 3A. 3-32 VDC control is common. Why? There is at least a LED and a current limiter in there.

Not being able to obtain a power SSR, is a reason to re-visit the alternative control method, but the alternative might be harder.

This: https://uk.farnell.com/internationa...dp/1498213?ost=DC+ssr&categoryId=700000006239 "could work", but it's not as easy as the original component I selected. The voltage to your load would have to be < 20V AND you would have to make a small board adding more complexity and uncertainty.

Here's https://uk.farnell.com/schneider-el...-dd/ssr-din-rail-mount-50vdc-32vdc/dp/2103647 one. Now, the mounting system is different. DIN Rail. Note that the output is called MOSFET which is a more succinct technical term.

=====

Is there any way for me to check if the output PID controller has been damaged?

Connect a LED, short the thermocouple input. That makes the setpoint = room temperature. Now, you need some controller experience, but in a nutshell, you should be able to turn the LED ON and OFF using say values of 0 C and 100 C for the setpoints. (way below and way above). Knowing more about controllers, you could make the LED dim.

--

Controllers are complex beasts. When I used to test them, my choice was a 500 W incandescent light with a thermoucouple mounted on top of the bulb. I got great visual feedback with virtually no lag, but it was my "test bed". You could do similar. Similar, meaning with your lamp.

 

[zexclo]

Went down to the shops, they didnt carry the brands that was mentioned.

However they have this in stocks,
1. Crydom SSR d1d12
http://www.mouser.sg/Search/m_ProductDetail.aspx?Crydom/D1D12/&qs=KoN42VEC626efqu3OijruA==&_cdc=0

And
2. Fotek SSR-10DD-H
Cant seems to find the specs for that, but the specs for SSR-10DD is
http://www.photosensor.com.tw/Solid State Relay/Single Phase Solid State Relay (DC to DC SSR)/SSR-10DD.htm

3. Fotek SSR-25DD
Its label 5-50vdc and input label 3-32 vdc

personally would prefer this but am unsure if this fits my setup or is suited for the pid and power source.

Please kindly advice.
And as for the resistor, would the ohm for the wire wound resistor matters? An thinking of getting a 10ohm resistor

 

[KeepItSimpleStupid]

1 No, It's anAC relay
2. No, It's an AC relay - Specs: Higher voltage, AC Relay . http://www.fotek.com.hk/solid/SSR-3.htm
3. OK - http://www.amazon.com/Sunkee-Solid-...sim_328_5?ie=UTF8&refRID=127Y9B9355EDBXXW3J9Q
(Note the suggested heatsink)

==

Assuming a 3A, 30 V power supply;
P=I^R' --> (3*3)*10 or 90 W; then check the
voltage P = (V*2)/R =30*30/10 =90; see if a 30 V power supply would deliver 90 W
We actually have to use (30-1)*(30-1)/10 because of the <1 V drop across the relay. So 84.1 W; neglecting the tolerance of R.

So, you can expect to get about 84 W; neglecting the resistor tolerance. If you had a 10% resistor, redo the calcs above with (10-0.1) and (10+0.1) or 10+-10%)

Aside::
You MIGHT find that series or parallel resistors might work better for you. The parallel formula is: 1/Rt=1/R1+1/R2+...1/Rn; . Power dissipations would add. e.g. 8 ohm 1 W in parallel with 8 ohm 1 W would b a 4 ohm 2 W. Just remembering an 8ohm 100 or 200 W (typical audio load that I can relate to) resistor will be pretty big, but it's not as wierd as a 1 Meg 200 W resistor which I had to replace, Check the physical size and see how it integrates with your setup. Two identical resistors in parallel is a special case. Rt=R/2
More resistors give you more surface area. Resistors in series add.

Aside #1: Wiring losses could play a small part.

 

[zexclo]

Thanks KISS, will be getting number 3, will be getting the heat sink too. I do not want my SSR to fail on me halfway when I am running the whole setup!

====

Awesome, was kinda lost earlier wondering how am I going to start with the resistor calculation as the guy over at the counter was asking me about the current and voltage going through the resistor, as the too high a current will cause the resistor to fail. So grateful that you helped with the guide on the calculations and choosing the resistors!

Yeah, the person showed me some resistor, the physical size gets bigger as the resistance and watts increase. interesting learning and eye widening experience!
the 1 meg 200w resistor gotta be huge!

 

[Tony Stewart]

Is air always 22'C intake? Or is some recirculated ?

You will probably find some temperature variation along the tube, depending on water temp, insulation and number of chicks. So test with multiple sensors. Use whatever power source is avail. The one you specified needs 0-10V input.

 

[KeepItSimpleStupid]

Use whatever power source is avail.

Agreed and suggested earlier - basically to see if it works thermally.

The one you specified needs 0-10V input.

or use the manual Front Panel controls for 0-30 V , 3A output. The controller was found in a closet and wasn't "selected" for the project so the easiest thing to do was the switch the output with an opto SSR designed for DC.

In the 80's I used a proprietary controller and AC SCR unit. Later, the SCR unit and the controller used selectable standard process signals like 4-20 mA. Later some of our processes used a controller and DC supply like this should have had. Some issues would still have to be checked.

 

[zexclo]

Is air always 22'C intake? Or is some recirculated ?

You will probably find some temperature variation along the tube, depending on water temp, insulation and number of chicks. So test with multiple sensors. Use whatever power source is avail. The one you specified needs 0-10V input.

The surrounding air, outside the box is re-circulated. The area is cooled by air conditioning. Will definitely test with multiple sensors, thanks for the advice.

The manual Front Panel controls for 0-30 V , 3A output. The controller was found in a closet and wasn't "selected" for the project so the easiest thing to do was the switch the output with an opto SSR designed for DC.

In the 80's I used a proprietary controller and AC SCR unit. Later, the SCR unit and the controller used selectable standard process signals like 4-20 mA. Later some of our processes used a controller and DC supply like this should have had. Some issues would still have to be checked.

Yeah, most of the parts are found in the closet or recycled from previous seniors' set ups.
Reduce, reuse and recycle.. lol.
I havent had the chance to get back to my lab and test the LED, out shopping the whole day, and I only manage to the the LEDs. The counter staff wasnt sure if the LEDs were 20mA, they just told me its 3V standard LED. Nonetheless, just bought a few, will be back in lab earliest on Thursday, hopefully with all the components bought and ready to test.

 

[Tony Stewart]

Have checked, the PID is working and pulsing at appr. 27 VDC from pin #12 and #13 when connected to a multi-meter.

In response to post on #33, I aim to keep the temperature tolerance to app 1 degrees. Do feel free to ask me questions on my setup and tolerances/information that is needed

Hey Tony,


In response to the post on #30


1. The mass to be heat is air measuring 120cm x 40cm x 40cm.


2. Temperature rise should be from 22 to 25 degree Celsius. The set temperature will be 25 degree and heating will occur when temperature drops below 25 degree.


3. The enclosure (the box containing the 120cm x 40cm x 40cm) is made of acrylic thickness of 0.5cm.


4. By allowed temperature error for regulation, I believe you meant by time for the temperature to reach set temperature of 25 degree Celsius. For now, as I am quite new to this circuitry system, I aim to keep it under 30minutes and hopefully later under 10minutes when I am better.


5. By max response time, I inferred that the response of the PID upon detection of temperature difference and action is taken?


6. Sensor will be located at the other side, top end of the box. The heater near one end where the fan will be blowing out the heat, and the sensor will be at the other end, top corner.

----

In response to #36

b. am currently using a thermocouple attached to the PID as the temperature sensor


Do let me know if you need other design information and also feel free to voice out your recommendations. Cheers.

In a small box, Air temperature control due to turbulence , convection and forced air with high exchange rates is a highly non-linear problem and not easy to solve with a simple point source heater with directional radiant heat. Uniformity within 1 deg in a small box may be difficult to maintain only 3 deg above ambient. Distribution of heat and absoption of heat may be difficult to make homogenious.

Volume : 192 k cu.cm. minus objects inside. Or 6.7 cu.ft.

A typical CPU fan using 5W has an "unloaded" volume rate of ~30 cfm or 30/6.7=>4 air exchanges per minute.
Thus your response time will be rapid . it will be more than 0.25 minute because air circulates and some air may flow thru very fast while surface air moves very little. Heat conduction by light bulb radiance is relatively fast as well but dependant on direction and absorption of the surface.

Thus you may want to significantly lower the air exchange time to 5-10 minutes and have more slow RPM circulation inside to make the temperature more uniform by your design. Then exhaust air speed may affect your system response and PID optimal gain parameters significantly. If you start with more stable open loop design, with stable air temp, then the regulation will be far easier to manage. I would test your choice of fans, locations, RPM and heat sources and record temperature using at least 4 thermal sensors. We call this thermal profiling. You can use low cost thermistors that you calibrate with separate pots then fixed R's to a stable 2.50V from a 5.00 V ref at 25'C 0r all give the same divider voltage and then mount in 4 zones and begin testing ... before you try to get fancy with PID's and SSR's.

one low coas solution might use an old small CPU heatsink and 12 V fan at low RPM (6~8V) with a TO-220 transistor mounted as a cheap heat source and spreader, and power supply solution . Then transistor Vbe bias can be your heat control input by either PWM or linear derived from Thermistor bridge mixer Op Amp using 4 sensors to inverting input.

Then you should consider heat loss outside, air turbulence inside, max acoustic noise , max linear air speed inside, and any other significant factors.

As a comparison, for heating reptiles in a zoo, they just use heat lamps at a distance with free air, perhaps to 35'C. So reconsider, warm air flow version vs gentle radiant heat conduction methods and effects.

 

[KeepItSimpleStupid]

Voltage Programming of a Single Supply explained a bit.

OK, SER means series. Since your more likely to put batteries in series, think of SER as Voltage
PAR means parallel and PARallel means you increase the current, so lets mentally relabel PAR as Current

So, we can mentally re-lable the switch to Setpoint with the positions INTERNAL, Voltage and Current
Now it's easy:
If you want the knobs to control the power supply use INTERNAL
If you want an external 0-10V signal to control the voltage use Voltage(SER)
If you want an external 0-10V signal to control the Current use Current(PAR) position,

then there are terminals with the heading of SLAVE INPUT and SER and PAR
We can mentally re-label these as the Programming Input terminals(Slave Input) with Voltage programming(SER)and Current programming(PAR)

Figure 5-4 tells, that we need to connect the negative of our programming input (0-10V) to the positive output of the power supply (seems stupid that that terminal is not there and it's probably due to space and the _ side of the 0-10V source to either the current or voltage programming terminal.

So, here I reduced the seemingly complex nomenclature to cover programming a single supply.

We should be told the input impedance of the programming terminals, but I could not find that specification.

==

IDENTICAL power supplies can be set up to double the voltage or double the current or "track". An example of "tracking" would be when you wanted identical + and - V control with one knob.

We're not using any of these modes including voltage programming. I took the simplest case of using an analog voltage to program a single power supply primarily by relabeling the terminals.

 

[zexclo]

Hi zexclo,
The control input of the SSD is connected to the PID controller. (The polarity must be correct. Terminal 12 on the PID to + input on the SSR , Terminal 13 on the PID to - input on the SSR.) The power supply is configured as a stand alone power supply. (Nothing is connected to the rear panel terminals. The output will be on all of the time.) The positive output of the power supply will connect to the positive output terminal of SSR. The negative output terminal of the SSR will connect to one end of the heater resistor. The other end of the heater resistor will be connected to the negative output of the power supply.

Les.

Hey Les,

Finally manage to get the components, fix up and trial today. Test the circuits and initial test run is alright, Will be doing more tests on the temperature setting, the rate of temperature raise etc on monday.

1 No, It's anAC relay
2. No, It's an AC relay - Specs: Higher voltage, AC Relay . http://www.fotek.com.hk/solid/SSR-3.htm
3. OK - http://www.amazon.com/Sunkee-Solid-...sim_328_5?ie=UTF8&refRID=127Y9B9355EDBXXW3J9Q
(Note the suggested heatsink)

==

Assuming a 3A, 30 V power supply;
P=I^R' --> (3*3)*10 or 90 W; then check the
voltage P = (V*2)/R =30*30/10 =90; see if a 30 V power supply would deliver 90 W
We actually have to use (30-1)*(30-1)/10 because of the <1 V drop across the relay. So 84.1 W; neglecting the tolerance of R.

So, you can expect to get about 84 W; neglecting the resistor tolerance. If you had a 10% resistor, redo the calcs above with (10-0.1) and (10+0.1) or 10+-10%)

Aside::
You MIGHT find that series or parallel resistors might work better for you. The parallel formula is: 1/Rt=1/R1+1/R2+...1/Rn; . Power dissipations would add. e.g. 8 ohm 1 W in parallel with 8 ohm 1 W would b a 4 ohm 2 W. Just remembering an 8ohm 100 or 200 W (typical audio load that I can relate to) resistor will be pretty big, but it's not as wierd as a 1 Meg 200 W resistor which I had to replace, Check the physical size and see how it integrates with your setup. Two identical resistors in parallel is a special case. Rt=R/2
More resistors give you more surface area. Resistors in series add.

Aside #1: Wiring losses could play a small part.

Hi Kiss
I bought the Fotek SSR-25DD, and am using a 50W, 4.7ohm resistor.
Using the calculations guide, have modified my settings a to 3A and 10V from the power supply.

In a small box, Air temperature control due to turbulence , convection and forced air with high exchange rates is a highly non-linear problem and not easy to solve with a simple point source heater with directional radiant heat. Uniformity within 1 deg in a small box may be difficult to maintain only 3 deg above ambient. Distribution of heat and absoption of heat may be difficult to make homogenious.

Volume : 192 k cu.cm. minus objects inside. Or 6.7 cu.ft.

A typical CPU fan using 5W has an "unloaded" volume rate of ~30 cfm or 30/6.7=>4 air exchanges per minute.
Thus your response time will be rapid . it will be more than 0.25 minute because air circulates and some air may flow thru very fast while surface air moves very little. Heat conduction by light bulb radiance is relatively fast as well but dependant on direction and absorption of the surface.

Thus you may want to significantly lower the air exchange time to 5-10 minutes and have more slow RPM circulation inside to make the temperature more uniform by your design. Then exhaust air speed may affect your system response and PID optimal gain parameters significantly. If you start with more stable open loop design, with stable air temp, then the regulation will be far easier to manage. I would test your choice of fans, locations, RPM and heat sources and record temperature using at least 4 thermal sensors. We call this thermal profiling. You can use low cost thermistors that you calibrate with separate pots then fixed R's to a stable 2.50V from a 5.00 V ref at 25'C 0r all give the same divider voltage and then mount in 4 zones and begin testing ... before you try to get fancy with PID's and SSR's.

one low coas solution might use an old small CPU heatsink and 12 V fan at low RPM (6~8V) with a TO-220 transistor mounted as a cheap heat source and spreader, and power supply solution . Then transistor Vbe bias can be your heat control input by either PWM or linear derived from Thermistor bridge mixer Op Amp using 4 sensors to inverting input.

Then you should consider heat loss outside, air turbulence inside, max acoustic noise , max linear air speed inside, and any other significant factors.

As a comparison, for heating reptiles in a zoo, they just use heat lamps at a distance with free air, perhaps to 35'C. So reconsider, warm air flow version vs gentle radiant heat conduction methods and effects.

Thanks Tony, for the pointers and advice. Will probably install/insert more fans around for the air flow to be more homogeneous, with lower RPM of the fans.

Wow, the consideration of heat loss outside, air turbulence inside, max acoustic noise etc seems abit too much for me at the moment. I am unsure, as this is more of a design similar to an incubator, just maintaining the temperature in the box to be around 25 degree Celsius, with tube carrying fluids through the box.
Will definitely look into the areas, once the set up is done! Cheers

 

[zexclo]

Hey, guys!

Been busy connecting and testing the circuit this week. Thanks its currently working fine.
Managed to get the temperature controller working!

Am using a DC to DC SSR and a 50W resistor for the setup.

Although, my supervisor wants to improve on the system like reducing the "noise" and maybe adding a capacitor and improve on the temperature sensitivity(currently its able to maintain at 24.8 although we want to get 25 degrees Celsius)

So glad that you guys are so fore coming and kind to guide me in this mini project!Cheers