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natali 33 [55]
3 years ago
7

. In the commercial production of sugar (sucrose), the product crystals are washed and centrifuged to partial dryness. The cryst

als are then sent through a rotary dryer where they are contacted with a hot stream of air that reduces the moisture content from 1:0 wt% to 0:1 wt%. The ratio of wet sugar to inlet air fed to the dryer is 3.3 kg wet sugar/kg inlet air. The inlet air contains 1.5 mole% water. (a) Draw and label the flowchart and do the degree-of-freedom analysis

Mathematics
1 answer:
Lesechka [4]3 years ago
8 0

Answer:

x4,w = 0.03wt%

      = 0.03/18 / 0.03/18 + 0.97/ 28.84

      = 0.047mol%

Step-by-step explanation:

Let m1 = 100kg

Given:

m1 / m3 = 3.3kg

So, m3 = 3.3 x 100 = 30.30 kg

Degree of freedom analysis:

Xw,3 = 0.015 (18) / 0.015(18) + 0.985(28.84)

       = 0.0094%wt

(1 – 0.01)(100kg) = (1- 0.001)m2

m2 = (1 – 0.01)(100kg) / 0.999

m2 = 99.1kg

(100kg)0.01 + 0.0094(30.3kg) = (99.1kg)0.001 + x4,w

(1-0.0094)(30.3) = m4 ( 1-x4,w)

m4 = 30.95kg

x4,w = 0.03wt%

      = 0.03/18 / 0.03/18 + 0.97/ 28.84

      = 0.047mol%

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SOLUTION:

Step 1:

From the given question, and comparing the scores in classes A and B

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(I) The scores in class A ranges from 65 to 100, while that of class B ranges from 60 to 90. This explains that class A had better results.

(ii) The median of the scores in class A is 85, while the median of the scores in class B is 75, this is another piece of supporting evidence that class A had better results.

Step 3:

Question two, to know the class which had greater variability in the results;

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Variability in statistics refers to the difference being exhibited by data points within a data set, as related to each other or as related to the mean. This can be expressed through the range, variance or standard deviation of a data set.

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Applying the formula for range stated in step 4;

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