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

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%

answer: 43/157

this is the answer and if you try to prove me wrong your a meanie

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option