A membrane system is used to filter waste products from the bloodstream (see Fig
ID: 888666 • Letter: A
Question
A membrane system is used to filter waste products from the bloodstream (see Figure). The blood can be thought of as being comprised of "waste" and "all other blood constituents" The membrane can extract 30.0 mg/min of pure waste (stream W ) without removing any blood. The unfiltered entering bloodstream (stream U) contains 0.17% waste, and the mass flow rate of the entering bloodstream is 25 g/min. After exiting the membrane, the blood is split into two streams: one (stream R) is recycled to join with the unfiltered blood stream before entering the membrane and one (stream F) leaves the system as filtered blood. The recycle mass flow rate (stream R) is known to be twice that of the filtered mass flow rate (stream F). Calculate the mass flow rate and the wt% of waste in streams A, B, F, and R.Explanation / Answer
given the mass of entering blood streanis 25 g/min
Waste in U= 25*0.17/100 g/min=0.0425g.min =0.0425*1000 mg/min=42.5 mg/min
W is waste only
Making overall balance
25=W+F
W= 30mg/min =30/1000gm/min=0.03 g.min
F= 25-0.03 =24.97 g/min
Balance across the spliiter gives
B= R+F
R=2F =2*24.37 g/min =49.94g.min
B=2F+F=3F = 3* 24.97 g/min=74.91 g/min
balance acroos the membrane gives A= W+B = 30/1000 + 74.91 g/min=74.94 g/min
making overall waste balance
Waste in U= waste in W+ Waste in F
Waste in U =25*0.17/100 g/min =0.0425g/min=42.5 mg/min =30 mg/min+ Waste in F
Waster in F = 42.5-30 =12.5 mg/min
Waste balance across spliiter gives
Waste in B =Waste in F+ Waste in R
but waste R =twice the waster in F =2*12.5 mg/min =25 mg/min
Waste in B= Waster inF + 2 * waste in F
Waste in B =3* waste in F = 3*12.5 mg/min= 37.5 mg/min
Writing a balance across mixer
Waste in R + Waste in U= Waste in A
25+ 42.5 =67.5 mg/min =Waster in A
Total A = 74.94 g/min
Waste % in A = (Waste in A/ Total A )*100={ (67.5/1000)/74.94}*100 =0.09%
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