A Tootsie Pop is made by heating sugar to a fluid state that it can be formed in

Question

A Tootsie Pop is made by heating sugar to a fluid state that it can be formed into a spherical shape (radius, r= 2 cm). It is important to allow this candy shape to cool before further processing/packaging so it retains its shape. Cooling is done by blowing ambient temperature air by the hot candy, which causes heat transfer from the candy to the air. It takes 20 minutes for the Tootsie pops to go from J&$8, 70 degree F, under these conditions. (See data below) The marketing would like to produce a new product, Super Tootsie pops, which have a 3 cm radius. These are made of the same material, only larger. Rather than do another set of experiments (never can tell when the marketing folks will change the size again!), you decide to develop an analytical model for this cooling process based on an energy balance. Using this model, how long will this larger product take to cool down below 70.1 degree F assuming all the conditions/properties are the same? Provide a plot this model for 0

Explanation / Answer

The general expression can be derived by keeping the ambient temperature and radius of Tootsie Pop as variable parameters, so that by putting different values of temperature and radius, required values can be calculated.

First we have to define the various parameter signs as follows:

= density of tootsie pop

r = radius of tootsie pop

cp = specific heat of tootsie pop

k = conductivity of tootsie pop

h = ambient heat transfer coefficient

Ta = ambient temperature

Ts = surface temperature of tootsie pop

A = surface area

For sphere, V/A = r/3 , where V = volume of sphere

Now in order to find the surface temperature, Lumped parameter analysis is to be used as follows:

Rate of dcrease of heat capacity = Rate of heat transfe from body

-mcpdT/dt = hA(T-Ta)

Ti to TdT/(T-Ta) = 0 to t-hA/(Vcp)

After integrating within the mentioned limits we wil get,

(T - Ta)/(Ti - Ta) = e[-hAt/cpV]    ---------------------- equation 1

Now putting the given values in the above equation:

(T - 70)/(250 - 70) = e[-3h1200/0.02cp]

T = 70 + 180e[-3h1200/0.02cp]

Now in order to get the temperature of tootsie as near as ambient temperature, we will take the difference between its surface and environment as 1 degree.

i.e. T = 70.1 degree Farenheight

70.1 = 70 + 180e[-3h1200/0.02cp]

0.1/180 = e[-3h1200/0.02cp]

Taking log both sides:

7.49 = 180000h/cp

cp/h = 34682.08 ---------------------------equation 3

So the general equation becomes:

T = Ta + 180e(-0.0000865t/r)

Now for larger diameter product, given parameters are:

r = 0.03 m

T = 70.1

Ta = 70

Ti = 250

Find t

So,

70.1 = 70 + 180e(-0.0000865t/.03)

from here,

t = 43.32 minutes

Now Plotting the t vs T for this model :

Put the value of t after converting it into seconds

Yes the model does make sense comparing to the plot from Part A. As the pattern of temperature distribution is similar.

The main benefit of developing analytical model is that this model is flexible for different sizes of product.

t T 0 250 1 221.43 2 197.40 3 177.18 4 160.17 50 70.03

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