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# How Long Is Ice Cream Safe On Your Counter? - PowerPoint PPT Presentation

How Long Is Ice Cream Safe On Your Counter?. By: Felicia Marshall and Seth Winsor. {. D = 21.5 cm. {. T S = -16.1  C. L = 12.5cm. Measurements. T  = 18.2  C. Vanilla Bean. Assumptions.

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By: Felicia Marshall and Seth Winsor

D = 21.5 cm

{

TS = -16.1C

L = 12.5cm

Measurements

T = 18.2C

Vanilla Bean

• Thermophysical properties for ice cream found on internet (not specific to our brand or flavor) are close enough and constant throughout cooling process

• Ice cream (finite cylinder) can be modeled as a sphere

• Plastic container does not affect heat transfer

• Our experiment ranges up to but excluding melting (no phase change effects)

• Convective heat transfer coefficient (h) is constant and calculated from the properties at the initial film temperature

• Heat transfer by condensation is negligible

• Heat transfer from forced convection is negligible

TF = (TS + T)/2

= 1.05C  274K

air = 3.66E-3 K-1

air = 13.58E-6 m2/s

Prair = .714

air = 19.1E-6 m2/s

kair = .1466 W/mK

• Ice Cream

• Ts = -16.1C  274K

• cp, IC = 3500 J/kgK

•  IC = 600 kg/m3

• k IC = .3 W/mK

• IC = k IC /(cp, IC IC)

= 1.428E-7 m2/s

Properties

g = 9.81 m/s2

Ra = gair(Ts-T)L3/(airair)

= (9.81 m/s2)(3.66E-3 K-1)(34.3 K).125m3/[(13.58E-6 m2/s)(19.1E-6 m2/s)]

= 9,328,399

Nu = {.825+.387Ra1/6/[1+(.492/Pr)9/16]8/27}2

= {.825+.387(9,328,399)1/6/[1+(.492/(.714))9/16]8/27}2

= 30.625

h = Nukair/L = 30.625(.1466 W/mK)/.125m

= 35.917 W/m2K

*=n=1Cn exp(-nFo) 1/(nr*) to 4 terms*

*=(T-T)/(Ti-T)

Fo = t/r02 =.017794 (at t = 24 min)

r* = r/r0=1

Cn = 4[sin(n)-ncos(n)]/[2n-sin(2n)] ---> Solve for C1, C2, C3, C4

1-ncot(n) = Bi --- > Solve for 1, 2, 3, 4

Bi = hr/kIC = (35.917 W/m2K)(.1075m)/ .3 W/mK = 12.87

1, 2, 3, 4 = 2.9018, 5.8269, 8.7875, 11.7846

C1, C2, C3, C4 = 1.9513, -1.8229, 1.6529, -1.4762

* Approximate solution in the book is only accurate for Fo > .2

Finally,

T = T + (TS - T)[C1exp(-12 IC t/r2)sin(1)/ 1 +

C2exp(-22 IC t/r2)sin(2)/ 2 +

C3exp(-32 IC t/r2)sin(3)/ 3 +

C4exp(-42 IC t/r2)sin(4)/ 4]

• With an added offset the analytical result can match up to the experimental result fairly well

• Rate of temperature change starts out high and decreases over time.

• Transient heat equations may not be valid at the surface

• The thermocouple we used seemed to have a slow response time; it took a minute and a half to measure the initial temperature, temperature measurements could contribute to error

• Many assumptions and simplifications; propagation of error could be cause for big difference between experimental and analytical results.