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.

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

[show] 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?

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

[show] Solution

showStep 1 Let S - sugar and W - water Let the entering stream be ${\dot {m}}_{1}$, ${\begin{aligned}{\dot {m}}_{1}&={\dot {V}}\cdot \rho \\&={\dot {V}}\cdot \rho _{ref}\cdot SG\\&=75L/min\cdot 1kg/L\cdot 1.035\\&=77.625kg/min\\\end{aligned}}$ showStep 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?

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

[show] Solution

showStep 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 $m_{5}$ ${\begin{aligned}{\dot {m}}_{5}&={\frac {{\dot {w}}_{1,s}\cdot {\dot {m}}_{1}}{{\dot {w}}_{5,s}}}\\&={\frac {77.625kg/min\cdot 0.012}{0.67}}\\&=1.39kg/min\end{aligned}}$ showStep 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, ${\begin{aligned}{\dot {m}}_{2}&=0.7\cdot ({\dot {m}}_{2}+{\dot {m}}_{4})\\&=0.7\cdot 76.235kg/min\\&=53.36kg/min\end{aligned}}$