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Calculating Heat Loss when there is a stirrer present

  1. Sep 17, 2011 #1
    Hi.

    I have a problem. I have been calculating the heat loss through the walls of a tank that is submerged in a larger tank acting as an ambient reservoir. To do this I filled the tank with relative hot fluid initially (28 degrees C to be precise, the ambient was at room temperature of 20.4 degrees C) and then monitored the decay of temperature within the tank itself using some T type thermocouples and LabView. To relate this to a U-value for the container, which was what I wanted to derive, I used Newton's law of cooling:

    T(t) = TA + (TH-TA) e-kt, where k=AU/mc, A the surface area of the inside of the tank, m the mass of fluid within the tank, and c the specific heat capacity of that fluid. The validity of using this model, and hence the lumped capacitance model, was ensured by stirring the fluid within the smaller tank using a magnetic stirrer.

    Unfortunately, due to the size of the stirrer it had to be placed inside the smaller tank itself and therefore adds some heat flux. The power rating of the is 7W from what I gather (it is a telemodul 7 immersible stirrer). However, I went about finding the appropriate U-value as follows, and to infer the heat flux added: I analysed the heat loss in stages, from 28-27, 27-26 and so on, which gave me U-values which dropped off, starting from approx. 4.5, 4, 3.5, 3 and 2. I assumed that the U-value associated with the largest heat difference (28-27, U-value=4.5) was the most accurate as this is where the small heat flux from the stirrer should have least impact. Where as the smaller temperature difference, 24-23 say, will result in less heat loss through the walls of the container and hence the heat flux from the stirrer will be more dominating. With this in mind I then take the highest U-value, 4.5, calculate what the heat loss at the 4 degree temperature difference should be using Q=U*A*DT. Then I subtract the experimentally found heat loss at this DT, using the same formula but with the smaller U-value found (in this case 2), from the heat loss just found with the higher U-value to infer an approximate heat flux of the stirrer. Once this has been found it can be used in the same relation Q=U*A*DT by adding it to the result (as the heat loss will be actually be, the heat loss found + the heat input by the stirrer) and then rearranged to obtain a new U-value. The whole process can then be iterated until it converges to an answer, which it does, but it gives me a U-value of 6.618 which seems reasonable for the tank in mind (all walls are composite construction, acrylic sheet (1.25cm) either side with a 1cm air gap sandwiched between the sheets) but it gives me a heat flux of the stirrer of 18W which can't surely be right given that the power of the module is just 7W. I would just like somebodies help as to where I am going wrong, or perhaps a step of this thinking which isn't logical.

    Many thanks and sorry for the long post but I feel it is best to be clear of the process.
     
  2. jcsd
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