Documentation:CHBE Exam Wiki/Midterm Exam 1 2016W/MT1 Question 1

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Maple syrup is produced by concentrating the sap from maple trees. An initial stream of sap containing only sugar and water is fed into a reverse osmosis unit to remove pure water and concentrate the sugar. Following that the sugar solution is further concentrated in an evaporator which again removes pure water.

Question 1a [15 points]

Draw a flowchart for this process showing the process units and labelling what components are contained in each stream.

Hint
How many unit operations are there in this process? Think about the input(s) and outputs(s) of each unit operation

Solution

Question 1b [10 points]

The initial stream of sap enters the process at a total flow rate of 75.0 L/min and has a specific gravity of 1.035, it contains 12000 ppm (by weight) of sugar, what is the flow rate of water into the process in kg/min?

Hint
Step 1 Determine the mass flow rate of the initial stream Step 2 Find the mass fraction of sugar in the stream. Next, determine the mass flow rate of water in the initial stream

Solution

Step 1 Let S - sugar and W - water Let the entering stream be ${\dot {m}}_{1}$, ${\begin{aligned}{\dot {m}}_{1}&={\dot {V}}\cdot \rho \\&={\dot {V}}\cdot \rho _{ref}\cdot SG\\&=75L/min\cdot 1kg/L\cdot 1.035\\&=77.625kg/min\\\end{aligned}}$ Step 2 The mass fraction of sugar, $w_{1,s}$ is 12 000 ppmw, ${\begin{aligned}{\dot {w}}_{1,s}&=12000ppmw\cdot 10^{-6}\\&=0.012\end{aligned}}$ Hence, the mass fraction of water $w_{1,w}$ is $0.988$ The mass flow of water in the entering stream, ${\dot {m}}_{1,w}$ ${\begin{aligned}{\dot {m}}_{1,w}&={\dot {w}}_{1,s}\cdot {\dot {m}}_{1}\\&=0.988\cdot 77.625kg/min\\&=76.69kg/min\end{aligned}}$

Question 1c [15 points]

The entire process produces a final solution containing 67.0 weight percent sugar. Of all the water removed in the process, 70.0% is removed by the reverse osmosis unit and the other 30.0% removed by the evaporator. How much water is removed in the reverse osmosis unit in kg/min?

Hint
Step 1 Using the overall mass balance of sugar, determine the mass flow rate of the final stream Step 2 Determine the amount of water removed from the process using the overall mass balance for water

Solution

Step 1 Using the overall sugar balance: ${\dot {w}}_{1,s}\cdot {\dot {m}}_{1}={\dot {w}}_{5,s}\cdot {\dot {m}}_{5}$ determine the flow of the exit stream $m_{5}$ ${\begin{aligned}{\dot {m}}_{5}&={\frac {{\dot {w}}_{1,s}\cdot {\dot {m}}_{1}}{{\dot {w}}_{5,s}}}\\&={\frac {77.625kg/min\cdot 0.012}{0.67}}\\&=1.39kg/min\end{aligned}}$ Step 2 Using the overall water balance, determine the amount of water removed from stream 2 and stream 4. ${\dot {w}}_{1,s}\cdot {\dot {m}}_{1}={\dot {w}}_{5,s}\cdot {\dot {m}}_{5}+{\dot {m}}_{2}+{\dot {m}}_{4}$ Rearranging the equation, ${\begin{aligned}{\dot {m}}_{2}+{\dot {m}}_{4}&={\dot {w}}_{1,s}\cdot {\dot {m}}_{1}-{\dot {w}}_{5,s}\cdot {\dot {m}}_{5}\\&=77.625kg/min\cdot 0.988-1.39kg/min\cdot 0.33\\&=76.235kg/min\end{aligned}}$ Since the 70% of the total water is removed by the osmosis unit, ${\begin{aligned}{\dot {m}}_{2}&=0.7\cdot ({\dot {m}}_{2}+{\dot {m}}_{4})\\&=0.7\cdot 76.235kg/min\\&=53.36kg/min\end{aligned}}$