Thermodynamics calculation help needed - Page 1
Forum section: electronics / general

author: MrOmnos
author_rank: Frequent Contributor
timestamp: June 07, 2018, 07:10:26 am
content: I am trying to build a set up to warm fluids up to a certain degree and keep it in an around the that temperature. I know I would require a heating element and few temperate sensors to make a closed loop system but I am not good at thermodynamics.Here's is the setup.Heater is going to be a off the market plate heater probably copper. So, what I need is to calculate the power of the heater that will be required. What are the parameters and formulas I would need to know to calculate the power required for a fixed flow rate and fixed time.For e.g. At a flow rate of 0.3L/mins how much power would it required to heat the liquid flowing in the tube from 20C to 30C in 10 seconds.

author: kridri
author_rank: Contributor
timestamp: June 07, 2018, 07:50:09 am
content: So in an ideal world we can assume that all the heat generated by the heating element will be adsorbed by the fluid.Thus we can state that the power equals the mass flow times change in temperature times specific heat of the fluid (P = m*cp*(Tout-Tin)).And the power for an restive electrical heater can be given by P = U*I.In a real word there will be some losses associated and these can be estimated more accurately (ex.http://www.thermopedia.com/content/946/).But by using the formulas of above and a coefficient of efficiency you can work out the basis.GreetzKris

author: Zero999
author_rank: Super Contributor
timestamp: June 07, 2018, 07:51:02 am
content: You need to know the specific heat capacity of the fluid, before this can be calculated.For example, water has a specific heat capacity of 4.1813J⋅g−1⋅K−1, meaning it takes 4.1813J of energy to raise the temperature of 1g (roughly 1ml or 1cm3) by 1oK or 1oC. Therefore to raise the temperature of 1ml of water by 1oC in 1 second, requires 4.1813W of power.In reality, there will be losses, which depend on the ambient temperature and difficult to estimate, so slightly more power will be required.Here's a list of specific heat capacities of different materials.https://en.wikipedia.org/wiki/Heat_capacity#Table_of_specific_heat_capacities

author: ElektroQuark
author_rank: Unknown
timestamp: June 07, 2018, 09:28:32 am
content: Feed the liquid for the lower part and take it from the upper one. This will avoid air retentions.

author: MrOmnos
author_rank: Frequent Contributor
timestamp: June 07, 2018, 10:55:48 am
content: Quote from: ElektroQuark on June 07, 2018, 09:28:32 amHmmm...nice idea!

author: apis
author_rank: Super Contributor
timestamp: June 07, 2018, 11:08:28 am
content: Quote from: kridri on June 07, 2018, 07:50:09 amMinor correction:This formula would give you theenergyneeded to heat a certain mass by a certain temperature:E=m*cp*(Tout-Tin)E=energym = masscp = heat capacity(Tout-Tin) = temperature differencePoweris energy per time, so you would have to divide E by the time (e.g. 10 seconds) to get the (minimum) required power. All units should beSI unitsof course.

author: JohnnyMalaria
author_rank: Super Contributor
timestamp: June 07, 2018, 01:00:03 pm
content: You have to take into account the thermal diffusivity of all the materials and every material is different. Certainly, you need to consider the choice of liquid, what you use to attach the tube to the heater, the thickness of the coil, flow rate, tube length, tube diameter etc. It seems a bit like determining what heat sink to use to dissipate heat from a power transistor.Based on your description, I don't think a plate heater is the way to go. I'd put your coil in a water bath and heat that instead. You'll get far better heat transfer.Frankly, to come up with the right answer using thermodynamics isn't going to be easy. You can estimate the ideal case (100% efficient, instant heat transfer) and use that as your minimum plate power requirement. Then, say, multiply that by four. Since you plan to use a feedback loop, just try it. I believe if you start in the right ballpark then experimental optimizing is the way to go.You could read up on geothermal heating to see if there are any simplified equations.

author: mbless
author_rank: Regular Contributor
timestamp: June 07, 2018, 05:30:01 pm
content: Quote from: apis on June 07, 2018, 11:08:28 amI would give them the benefit of the doubt and assume they couldn't type ṁ

author: apis
author_rank: Super Contributor
timestamp: June 07, 2018, 06:04:02 pm
content: Quote from: mbless on June 07, 2018, 05:30:01 pmI'm sure he knows that, probably better than me. But someone reading this might not and get confused. Doesn't matter who is right about every detail just that the end result is correct.

author: kridri
author_rank: Contributor
timestamp: June 07, 2018, 06:44:41 pm
content: @apis@mblessOff course I know that it need to be massflow (that is also the reason why I stated in my text "the power equals themass flowtimes change in temperature times specific heat of the fluid ".I was only unable to find the option to type ṁ in stead of m.Thank you for the correctionregardsKris