1 / 10

How Long Is Ice Cream Safe On Your Counter?

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.

york
Télécharger la présentation

How Long Is Ice Cream Safe On Your Counter?

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. How Long Is Ice Cream Safe On Your Counter? By: Felicia Marshall and Seth Winsor

  2. { D = 21.5 cm { TS = -16.1C L = 12.5cm Measurements T = 18.2C Vanilla Bean

  3. Assumptions • 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 • Radiation is negligible • Heat transfer by condensation is negligible • Heat transfer from forced convection is negligible

  4. Air 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

  5. Finding h 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

  6. Transient Convection *=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

  7. Transient Convection 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]

  8. Experimental Results

  9. Analytical/Experimental Comparison

  10. Conclusions • 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.

More Related