question 11(1 point) Hollow-fiber membrane devices are used in a number of appli
ID: 701243 • Letter: Q
Question
question 11(1 point) Hollow-fiber membrane devices are used in a number of applications in bioengineering. A typical unit consists of thousands of small tubes packed in a tubular device (Figure). Components within the tubular fibers can be isolated from components outside the fibers based on solubility and molecular weight restrictions. Small molecules can easily diffuse across the membranes into the outer annular space. In one applications, hollow fiber device was used to concentrate a 1% (wt basis) bacterial suspension at a rate of 35 L/min (density is 1.028 kg/L). An aqueous buffer solution (density -1 kg L) is fed to the annular space at SL min Bacteria in the cell puspensicn are too farge to pass through the mem brane but due to oamotic pressure water hows into the buffet. In ca, preparatio the outlet bacterial suspension is 6% (wt basis), Then the mass flow rate ofoudet cell suspension is kg min and mass flow rate of buffer exiting the dericei_kg min 38 and 6 6 and 38 32 and 12 12 and 32 b View hint for Question 11Explanation / Answer
Given that the cell suspension has a density of 1.028 Kg/Land bacteria concentration of 1% by weight.
Volumetric flow rate of cell suspension = 35 L/min
Mass flow rate of cell suspension = Volumetric flow rate * density = 35 * 1.028 = 35.98 Kg/min
Given that 1% by weight is bacteria . So the flow rate of bacteria alone in the suspension = 1% of 35.98 = 0.01 * 35.98 = 0.3598 Kg/min
It is said that the bacteria is retained with the suspension while the water flows through the membrane. So the bacteria mass flow rate in the exit of suspension is same as at the inlet = 0.3598 Kg/min
But the exit cell suspension has bacteria by 6% by weight, which means that 6% of the total mass flow rate = 0.3598 Kg/min. Then,
Total mass flow rate = 0.3598 / 0.06 = 5.9966 Kg/min ( which is almost equal to 6 Kg/min.)
Now we have that at the inlet, 35.98 Kg/min of Cell suspension is entering and 5.9966 Kg/min is leaving. This means means the difference is the amount of water that has crossed the membrane.
Mass flow rate of water crossing the membrane and getting into buffer solution = 35.98 - 5.9966 = 29.9834 Kg/min
Volumetric flow rate of buffer = 8 L/min
Mass flow rate of buffer = Volumetric flow rate * density = 8 * 1 = 8 Kg/min
Total exit mass flow rate of buffer stream = Mass flow rate of buffer stream + Mass flow rate of water entering into the buffer stream = 8 + 29.9834 = 37.9834 Kg/min
The answers are 5.9966 ( 6 Kg/min) and 37.9834 ( 38 Kg/min)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.