-5
\$\begingroup\$

what electrical components do i need to heat up a small container of about 45mm in diameter and 65mm tall to a temperature of about 50C with 3 AA batterries

\$\endgroup\$
  • 2
    \$\begingroup\$ Don't confuse heat with temperature. You need a power source and a resistance wire. \$\endgroup\$ – Dampmaskin Jun 21 '17 at 13:26
  • 1
    \$\begingroup\$ If filled with fluid like coffee, you will need about 22V LiPo with 10A load lasting not very long with 2 ohms or 200W inside a thermos. Thus your question has a false assumption. \$\endgroup\$ – Sunnyskyguy EE75 Jun 21 '17 at 13:27
  • 1
    \$\begingroup\$ The amount of power required to maintain a temperature of 50°C could be anywhere from a tiny fraction of a watt to many watts depending on the insulation and the ambient temperature and other conditions. Even with no heat loss, the heat capacity and required warm-up time (and starting temperature) will impose a minimum power level. The heater is just a resistor. To control the temperature at a setpoint would require a few more components such as a sensor, a few resistors and an amplifier. The details depend strongly on the specifications. \$\endgroup\$ – Spehro Pefhany Jun 21 '17 at 13:32
  • \$\begingroup\$ @SpehroPefhany will some sort of a thermistor or a thermocouple help to regulate the temperature and also how many resistors (and what values since i will be using three 1.5V batteries) will i need? thank you \$\endgroup\$ – Rob Jun 21 '17 at 13:37
  • 1
    \$\begingroup\$ Update question with CLEAR Specs from now on please. Mass, thickness, material thermal resistance, insulation space, surface area, temp tolerance error. Otherwise you are overlooking something. and any answers will be incomplete. The best question has specs.in a small table. Then you wont waste as much of our time asking questions \$\endgroup\$ – Sunnyskyguy EE75 Jun 21 '17 at 15:02
1
\$\begingroup\$

Let's to walk through the math that is required to engineer a solution.

Firstly, as some commenters have pointed out, you didn't supply all of the necessary information.

  • When you say heat air to 50° C, you failed to specify the starting temperature. So let's assume it is 20° C for now.
  • You also didn't specify the insulating properties of the container. This is important because it affects the amount of heat required to raise the temperature as well as maintain it. So for this example, we will assume that the container is so well insulated, that once heated it requires no additional heat to maintain the temperature. This is not real world of course but it at least allows us to approximate the amount of heat required to do the initial heating task.
  • You also did not specify in what amount of time you would like to raise the temperature. You can imagine that the faster you want to heat it, the more heat it will take to do the job. For this example, we will assume 2 minutes.
  • You didn't specify what is inside of the container. For this example, we will assume it is air.

So the task is to raise the air temperature inside the cylinder by 30° C.

The volume of the air in a cylinder is given by r2hπ so your cylinder has a volume of 103378 mm3 or ~0.000103 m3.

The density of air at 20° C is 1.025 kg/m3 so the cylinder contains 0.000103m3*1.025 kg/m3 or 0.00106 kg (~1 g) of air.

The specific heat of air at 20° C is 1.005 kJ/(kg C). So to raise the volume of air in the cylinder by 30° C, it will take 0.00106 kg * 30° C * 1.005 kJ/(kg C) or ~0.032 kJ (32 J).

Power in Watts is equal to Joules/seconds. So if we wish to raise the temperature in 120 seconds, the source of heat would need to be 32 Joules / 120 seconds or 0.267 watts over a 120 second period.

Three alkaline cells would have a voltage of 4.5 volts when configured as a battery. We can calculate the resistor required since R = E2/P. So 4.52/0.267 yields ~75 ohms.

Let's check if this is a reasonable current for the battery to supply. Since I=E/R, we have 4.5 volts / 75 ohms or 60 mA. The battery can supply ~1150 mAh or 5000 J (see here). So this seems reasonable at least for the initial heating cycle.

This is what is called a first order analysis. You then have to consider other factors such as the air in the container is not being circulated so the heat will not evenly disburse on its own. You should come back to the question of insulation since we know it won't be perfect, some of the heat will be lost to the surrounding atmosphere. This suggests that a lower value of resistor will be required to make up for this heat loss. If you need to keep the air at that temperature, then additional heat and calculations will be required. You should also revisit any other assumptions or simplifications to determine their impact on the first order analysis.

\$\endgroup\$
1
\$\begingroup\$

I have done this before with heating Xtals HC49 cases with foam insulation, raising Xtal 50'C in 10 seconds due to low mass with 1/4W using only several 1206 SMD resistors, comparator and switch to drive resistors thermally coupled to DUT.

You can design based on your Rth of your case for heat rise of coil relative to container. If say 10'C/W then choose 1 Watt heater. if 20'C/W then reduce power <1 W. Choose 70'C as your safe wire max temp with margin to allow for poor contact or poor container thermal resistance and avoid stress to thin glass unless pyrex then higher thermal resistance demands higher surface area of heater thus lower Rth of heater to contact. < 10'C/W

In my case (pun intended) I used Polyamid or FPD or Kapton copper clad with SMD resistors and Thermistor using 2oz copper thus rigid and shaped to DUT inside 1cm of foam . You may use flat nichrome wire with an epoxy form with high metal content to make a case surrounded by 1cm min of flame resistance blue foam. THe resistor temp will be far greater than the contents due to poor Rth of your case unless you take care to minimize air gaps and encapsulate with low Rth compound ( not silicone).. Due to thermal time constant on off control is adequate with 0.1'C hysteresis or positive feedback ratio of output/input voltage swing/ temp swing. So it may be < 1% R ratio of feedback R to Req of Vin(+)

  • m= 15mg alum container + 5mg fluid
  • specific heat, c
  • fluid ~ 4.2 J/gm-°C
  • alum ~ 0.9 J/gm-°C

  • Thus energy needed = \$ E= cmΔT = (50-20)° ((15 * 0.9)+(5 * 4.2))~1kJ\$

E = P·t = V·I·t

  • Duracell AA: (2.85 Ah) x (1.5 V) x (3600 s) ~ 15 kJ is used at 20 h discharge rate
  • assuming you dont want to wait 20h
  • amp-hour rate of 2.85Ah reduced to 2h=discharge rate from Coppertop datasheet is 0.3W constant load with capacity reduced to 1.8Ah
  • 2h service life yields 750mA constant current to 0.8V dead. or an energy capacity of 1Vavg*0.75A*2h*3600s = 5.4kJ or about 1/3rd of the 20h capacity. Reducing the time even further degrades the Joule capacity even faster due to internal Pd *Rjc temp rise from 0.15 Ohm ESR

Expect to buy a lot of batteries or get a Lipo cell.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ thank you very much for your helpful input. once again i apologize for my vagueness. \$\endgroup\$ – Rob Jun 21 '17 at 16:17
  • \$\begingroup\$ Right LED will be dim but used as a Zener, Left LED is dim bugt with ultrabright 1~20 Cd 5mm LED is brightly visible at 1mA load. \$\endgroup\$ – Sunnyskyguy EE75 Jun 21 '17 at 16:52
0
\$\begingroup\$

You need a resistor. For heating purposes, wire with high resistivity, like nichrome, is often used.

\$\endgroup\$
  • \$\begingroup\$ gee that seems pretty obvious \$\endgroup\$ – Sunnyskyguy EE75 Jun 21 '17 at 15:46
  • \$\begingroup\$ @Tony: That was the point, although it answers the question as written. \$\endgroup\$ – Olin Lathrop Jun 21 '17 at 15:52
  • \$\begingroup\$ True yet barely worth even a comment . He's probably an IE not an EE \$\endgroup\$ – Sunnyskyguy EE75 Jun 21 '17 at 16:49
0
\$\begingroup\$

You can wrap a flex polyimide heater (or wrap some NiCr wire) around the inner part of your well-insulated housing. Imagine a thin metal tube with a heater wrapped around it. Somewhere in the tens of ohms, so about 1W of power.

1W is not a lot of power, you will need decent thermal design, which is outside the scope of this question.

The temperature control could be as simple as this:

enter image description here

Use your AA batteries (4.5V) rather than the 3.3V supply. Bypass with 1000uF/6.3V as well as the 2uF ceramic. Connect the output to a logic-level P-channel MOSFET, source to +4.5, drain to heater and ground the other side of he heater. The chip is available with a 45°C or 55°C setpoint (factory set). The sensor (shown as inside a uP) can be a 2N4403. It has a large 5°C hysteresis so the temperature will be "about" 50°C for some definition of "about", and since no specifications were rendered, this meets the specs.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.