Cheesygodxx
27.02.2020 •
Chemistry
Brown sugar is a mixture of white sugar and molasses. Unrefined brown sugar contains a large portion of molasses. After one round of crystallization of this unrefined sugar, the brown sugar product contains 11.00 wt% molasses and the remainder white sugar. A second round of crystalization reduces the molasses content to 1.20 wt% molasses. A plant wishes to produce a brown sugar ideal for baking that contains 5.00 wt% molasses. In order to acheive this concentration of molasses, a plant feeds once-crystallized brown sugar into a recrystalizer. Some of this feed is diverted into a bypass stream that rejoins the sugar that leaves the crystalizer. Pure molasses leaves the crystalizer as a bypass stream. This process is described by the flow chart below Write the material balance equations for the species around the elements listed below. In the answer blank to the left of the equal sign, enter in an expression for the mass of the species entering the given element. In the answer blank to the right of the equal sign, enter in an expression for the mass of the species leaving the given element. Put your responses in terms of numeric mass fractions and the mass flow rate variables. Do not include units in your response. Open the hint panel for an identical flow chart you can view while filling in the responses below.
1 If the plant wishes to produce 450.0 g/min of brown sugar today, what is the feed rate, m1?
2.What is the flow rate of pure molasses from the crystallizer, m4?
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Ответ:
explanation:
the transformation T(a, b) can be change the pre image to the final image by following the rule as we explain below
if A(x, y) is a pre image, then A ' ( x', y') is its image, and x' = a+x, y' = b + y, where because of T(a, b) transformation
in our case Quadrilateral GHJK has vertices G(2, 3), so the coordinates of G”
can be found with
T(-4, -5) applied to G(2, 3) ⇒ G' (-4 +2, -5+3)=G' (-2, -2)
T(-4, -5) applied to G' (-2, -2) ⇒ G"(-4-2, -5-2) =G"(-6, -7)
the coordinates of G” are (-6, -7)