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Assignment 2 Essay

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Name: Veronica Malong

ID#: 0896981 Date: Sunday, March 22, 2015

Course #: FOOD 2010

Course Name: Principles of Food Science

Instructor: Massimo Marcone Assignment #: 2

Assignment Name: Assignment 2

# of Pages (including this one): 8

(For instructor’s use)

Grade:

2. Draw a flow diagram of the unit operations involved in sugar refining and at each step write a brief sentence explaining what is occurring. Use the information from SKIL site to develop your flow chart. Since the information does not start from the beginning of the refining process, please include an extraction step (extraction is explained in your textbook). Either milling or diffusion is …show more content…

The steam enters the heat exchanger as 50% quality steam and exits as water condensate at 85°C. Calculate the mass of steam required to heat 150 kg of cider. (For the cider, assume the Cp=3.651 kJ/kg°C, and latent heat=280.18 kJ/kg.)

Cider: T1 = 4  C T2= 65C Cp= 3.651 kJ/kg C Mass = 150 kg
Water: T1­ = 100 C T2 = 85C Latent Heat = 2257.06

Quality=MCpT

Quality of Cider:

=(150 kg)(3.651 kJ/kgC)(65-4C)

=33406.65 kJ

Quality of Water:

=(4.18 )(100-85C
=62.7 KJ/Kg of water

Quality of 50% Steam:
= (0.5)(2257.06 kJ) + 62.7
=1128.53+62.7
=1191.23 KJ

mass of steam = 33406.65/1191.23=28.04382865

Thus the mass of steam needed would be 28.044 kg

5.
a) Graph the following data on semi-log OR regular graph paper. Determine the D-value using the graph. Show your work.

D = (t₂ - t₁)/(log a – log b)

=(250-100)/[log(270000)-log(30)]

=150/3.9542

=37.9 min.

b) How long would it take to obtain a 3-D reduction in microbial load?

3-D reduction in microbial load = 3 x (37.934) = 113.80 min.

Therefore 113.80 minutes are required to obtain a 3-D reduction in microbial load.

6.
a) Graph the following data on regular OR semi-log graph paper (graph both thermal death time curves on the same sheet of paper). Use the graph to determine the z-value for each organism. Show your work.

Z = T2-T1/log a – log b

A:
Za=150-110/[log(700)-log(0.04)]
=40/4.243
=9.43 F

B:
Zb = 180-100/[log(500)-log(4.5)]
=80/2.0458
=39.11

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