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Calculating time for object to reach ambient temperature in an oven

  1. Sep 19, 2011 #1
    Hi all,

    I am currently doing an engineering internship (still an undergrad), my field is electrical engineering so thermodynamics just confuses me!

    My job has been to come up with a good Environmental Stress Screening porgram (ESS) plan, while doing this I have realised that the industrial oven that this company has is, well, crap. I am not the only one to realise this and the wheels are moving on getting a new oven, although I need to highlight how bad the current oven is for the head dudes to give the green light.

    Right, to the actual problem.

    I am trying to figure how long it should theoretically take for the rack of components in the oven to reach the desired temperature. There will be practical tests for this but I could do with the theory to back me up.

    The plan is to set the oven to 80C (and -20C) and see how long it takes for the equipment to reach its steady state, now all the equipment is going to be running so getting it to it's hot steady state shouldn't be as hard as getting it to it's cold steady state.

    From what I can gather from the internet I am basically looking at estimating the thermal time constant of the system. Then according to the paper I am reading ( which is attached, highlights all the steps I have skipped to get to the thermal time constant), 4 time constants should take it to around 98% to it's max temp which is close enough for me.

    It gives the thermal time constant as:

    Tau = (M X Cp)/ (h X A)

    M=mass , Cp = specific heat capacity, h = convective heat transfer coefficient , A = surface area

    well mass is easy enough to get (120kg), specific heat capacity isnt that easy because of the wealth of different materials in the components (I can attach a photo of the components tomorrow to show the set up, I am assuming a Cp of 0.162 Btu/lb F),I realise this is imperial units btw, I don't mind taking a rough approximate for this though. The surface area I am not so sure about because of what surface to take as the 'area' ( the dimensions being 765mmX600mmX390mm), and the heat transfer coefficient I a have no idea how to calculate.

    I am assuming that the heat transfer coefficient is different for every oven? The power of the current oven is 3kw and the new oven is 51kw. I am guessing the power of the oven has to impact the transfer coefficient in some way because if it didn't then the time constant would be the same for both the garbagety oven and the new (hence no point getting a new oven). I have seen a few sources that give ranges for the coefficient of between 10-200(W/m2K) but that is quite a big range.

    I would just like to know if I am on the right track with estimating how long it would take for this thing to get to its steady state temp? Is there more information needed?

    Any help is greatly appreciated. I am very shaky with my thermodynamics but I am going to keep looking into this
     
  2. jcsd
  3. Sep 19, 2011 #2
    I am not sure how the attachment works on this forum so I have copy and pasted the relevant part of the pdf, sorry for the bad formatting

    "
    Thermal Cycling Frequency
    A thermal cycle has an associated ramp rate and dwell time. During thermal ramp, the
    mismatch in coefficients of thermal expansion for the constituent materials on a printed circuit board causes stresses to develop. During thermal dwell, the stress energy gets relieved by converting to strain or relative motion. For example, if solder joints are marginal, permanent deformation or damage occurs during the thermal dwell. If the solder joints are good, no permanent deformation or damage occurs. The point is the thermal dwell time must be sufficiently long for this stress-relaxation process to occur. If the thermal dwell time is not sufficiently long, the full benefit of HASS is not realized.

    Our experience indicates that ramp rates of 8oC/minute to 20oC/minute and dwell times of 15
    minutes to 20 minutes are sufficient to precipitate most manufacturing and component defects.
    Of the two, proper thermal dwell time is the more important parameter. This opinion is also
    supported by the work of numerous researchers. An excellent example of such work is the
    technical paper by Zhai, Sidharth and Blish II.

    In order to ensure that optimal thermal dwell times are achieved, the thermal time constant of the product must be estimated. The UUTs inside a HASS chamber are heated and cooled
    through forced-air convective heat transfer. The air is forced down along the side walls of the chamber and back up through the center ventilating fan. The general equation for convective heat transfer is as follows:

    q = h x A x ( Ta – T )
    Where:
    q = Heat flow (BTU/hr)
    h = Convective heat transfer coefficient (BTU/hr - ft2 - oF)
    A = Surface area normal to the direction of heat flow (ft2)
    T = Temperature of the UUT surface being heated or cooled (oF)
    Ta = Temperature of the chamber air environment (oF)


    The general equation for fundamental heat capacity is as follows:
    Q = M x cP x DT = M x cP x ( Tf – Ti )
    Where:
    Q = Amount of heat added or removed (BTU)
    M = Mass of the UUT being heated or cooled (lbs)
    cP = Specific heat of the UUT (BTU/lb - oF)
    DT = Temperature change of the UUT surface (oF)
    7
    Tf = Final temperature of the UUT surface being heated or cooled (oF)
    Ti = Initial temperature of the UUT surface being heated or cooled (oF)

    The preceding equation for heat capacity can be written in differential form as follows:
    DQ/dt = M x cP x dT/dt = q

    The first order derivatives for Q from the preceding 2 equations can be equated as follows:
    q = h x A x ( Ta – T ) = M x cP x dT/dt

    Consider what occurs during the temperature transition of a thermal cycle. Assume that the
    UUT and chamber air are both at their low temperature limit. Identify this initial UUT
    temperature set point as Ti. The chamber is then commanded to go to the high temperature
    limit. Identify the final UUT temperature set point as Tf. During the transition, the temperature set point for chamber air is made slightly higher than the UUT temperature set point in order to provide a more rapid thermal ramp rate for the UUT temperature. It takes time for the chamber air to reach its set point, but the time constant for chamber air is much less than the time constant for the UUT. This is because the mass of chamber air is much less than the mass of the UUT. Therefore, for the purposes of this discussion, the time it takes for the chamber air to change temperature will be neglected. If this approximation were not made, it would obfuscate the simple relationships that exist between the UUT time constant and physical parameters. Using this approximation, Ta becomes Tf.

    Writing the preceding equation in standard form, it falls into the category of a first-order linear homogeneous differential equation with constant coefficients:
    dT/ dt + { ( h x A ) / ( M x cP ) } x T - { ( h x A ) / ( M x cP ) } x Tf = 0

    Define thermal time constant, ז, as follows:

    ז = Thermal time constant = {( M x cP ) / ( h x A )} (hours)

    Substituting for the UUT thermal time constant, ז, the differential equation may be simplified as follows:
    dT/ dt + 1/ ז x T - 1/ ז x Tf = 0

    Since this is a linear homogeneous differential equation, the complete solution is the
    homogeneous solution.

    The solution is as follows:
    T ( t ) = { ( Tf – Ti ) x ( 1 - exp{ -t / ז } ) } + Ti
    DT( t ) = ( Tf – Ti ) x ( 1 - exp{ -t / ( { ז
    %DT( t ) = ( 1 - exp{ -t / ז } ) x 100
    Where:
    T = Surface temperature of the UUT being heated or cooled (oF)
    Tf = Final surface temperature of the UUT being heated or cooled (oF)
    Ti = Initial surface temperature of the UUT being heated or cooled (oF)
    DT = Temperature change of the UUT surface (oF)
    ז = Thermal time constant = {( M x cP ) / ( h x A )} (hours)
    M = Mass of the UUT being heated or cooled (lbs)
    cP = Specific heat of the UUT (BTU/lb - oF)
    h = Convective heat transfer coefficient (BTU/hr - ft2 - oF)
    A = Surface area normal to the direction of heat flow (ft2)


    Since the UUT temperature varies exponentially, the UUT will be considered to have reached its final temperature when it is within 3oC of the final UUT set point temperature, Tf. Since most HASS temperature swings are 50oC to 70oC, this 3oC means that the UUT must achieve at least approximately 96% of the full temperature swing. 96% of full temperature swing equates to approximately 4 thermal time constants as follows.

    %DT( t=4 1 ) = ( ז - exp{ -4 ז / ז } ) x 100 = ( 1 - exp{ -4 } ) x 100 ~ 98%

    After 4 thermal time constants, the UUT temperature is considered to have stabilized. Then the dwell time at temperature Tf may begin. As mentioned previously, the optimal dwell time is from 15 minutes to 20 minutes. Therefore, the total dwell time consists of a transition time of 4 thermal time constants plus a soak time of 15 minutes to 20 minutes.
    A complete thermal cycle consists of a dwell at hot temperature and an equal dwell at cold
    temperature. This defines the time period of the thermal cycle. An optimal thermal cycling
    frequency can then be defined which is the inverse of the thermal period. The transition time
    depends on the specific heat capacity of the UUT. Since the UUT is comprised of a number of materials, the specific heat capacity will be a composite based on the percent contribution of the constituent materials of the UUT. A utility has been developed to assist in calculating all the quantities needed to determine the optimal thermal cycling frequency."
     
  4. Sep 22, 2011 #3
    To get an exact answer it is more complicated. Do some research on "thermal circuits".
    Basically a thermal mass acts like a capacitor, and the heat generator a current source.
    Each mass (the UUT, the oven walls, oven racks etc is a separate capacitor). Connecting them
    are different thermal resistances with components from conduction, radiation and convection. The
    power of the oven determines the value of the drive current, so your 51kw oven can heat up 7x the drive current of your 13kw oven but the mass of its walls,
    racks etc is also probably much larger.

    The heating rate of the oven with a given mass inside may be one of it's specs.
    Assuming the UUTs are quite thermally conductive it may be all that you need.
     
  5. Sep 23, 2011 #4
    Well the best way to get an accurate answer for this is to test. If you cant test, then estimate! I dont have any idea what your 'rack of components' looks like, but try to make a guess as to the surface area. Then guess the heat transfer coefficient. Does the 'oven' have a fan to blow the air around? Does the rack have a fan? If not, then it is 'natural convection' and you can assume h=2 Btu/hr-ft2-F (I'm happy with 'imperial' units). Or you can guess h=4. It will change your answer but so what, it's just an estimate (guess)!

    As far as the rating of the oven, once you get your time constant, try to estimate how much power it takes to heat your rack in that time (m Cp * change in temp) = Energy (Btu); divide by time (get Btu/hour) & then convert units (divide by 3.4) to get watts. Is the oven powerful enough to do it?
     
  6. Sep 25, 2011 #5
    Thank you very much for the reply guys!

    Unfortunately I am off work for the few days but when I get back I'll try and do what you recommended.

    At the end of the day it will be backed up by a practical test so the theory is just for added confidence.

    Thanks again,
    Graeme
     
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