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Unformatted text preview: PROBLEM SET 1&2 ‐ SOLUTION Question 1: a) Define the terms: mass flow rate, volumetric flow rate, density, mass fraction Mass flow rate: rate at which material enters/leaves a system expressed as mass/time Volumetric flow rate: rate at which material enters/leaves a system expressed as volume/time b) What is conserved in a chemical reaction? Total mass, mass of each individual element c) Define closed, open and isolated systems. Closed system exchanges heat but no mass with surroundings Open system exchanges both heat and mass with surroundings Isolated system exchanges no heat or mass with surroundings d) What is the material balance equation for an individual chemical species “A” in an open system? Which terms become zero if you have: a closed system, no reaction, or steady‐state? Mass A in – Mass A out + Mass A generated – Mass A consumed = Accumulation of A Closed system: Mass A in = Mass A out = 0 No reaction: Mass A generated = Mass A consumed = 0 Steady‐state: Accumulation of A = 0 Question 2: The kidneys carry out a separation process where blood enters at a rate of 1200 mL/min and is split into two streams: filtered blood and urine. We are specifically interested in the details of the separation of urea and a drug circulating in the patient’s blood by the patient. What is known is given below: Filtered
Blood Blood
VB = 1200 mL/min
Curea, B = 10 mg/dL
Cdrug, B = 100 uM Kidney VF = ?
Curea, F = ?
Cdrug, F = 90 uM
Urine
VU = 1.5 mL/min
Curea, U = 500 mg/dL
Cdrug, U = ? Find the unknown quantities in the filtered blood and urine streams. Assume density of all streams is the same as water = 1 kg/L. Answer: Use mass conservation equation for non‐reacting, steady‐state system. Overall balance: mB – mF ‐ mU = 0 Using densities: ρVB – ρVF ‐ ρVU = 0 divide by ρ: VB – VF ‐ VU = 0 VF = VB ‐ VU = 1200 mL/min – 1.5 mL/min = 1198 mL/min Balance on urea: murea,B – murea,F – murea,U = 0 Using concentrations: Curea,B VB – Curea,F VF – Curea,U VU = 0 Curea,F = (Curea,B VB – Curea,U VU )/VF = (10 mg/dL ∙ 1200 mL/min ∙ 0.01 dL/mL ‐ 500 mg/dL ∙ 1.5 mL/min ∙ 0.01 dL/mL) 1198 mL/min Curea,F = 0.094 mg/mL. Note that if you do not use the conversion factor 0.01 dL/mL, you get the answer 9.4 mg/dL, which is also correct. Balance on drug: Cdrug,B VB – Cdrug,F VF – Cdrug,U VU = 0 Cdrug,U = (Cdrug,B VB – Cdrug,F VF )/VU = (100 umol/L ∙ 1200 mL/min ‐ 90 umol/L ∙ 1198 mL/min) 1.5 mL/min Cdrug,U = 8120 umol/L or 8.12 mmol/L. Question 3: You are preparing a mixture of methanol (MeOH) by combining two solutions of MeOH, solution A and solution B, with mass fractions on 0.4 and 0.7 respectively. What is the mass fraction MeOH in the final mixture C (see figure)? A B 200 g
xM, A = 0.400 150 g
xM, B = 0.700
C Answer: Mass balance on MeOH: mM, A + mM, B = mM, C Using mass fractions: 200 g ∙ 0.400 + 150 g ∙ 0.700 = 185 g Total mass balance: mA + mB = mC 200 g + 150 g = 350 g Mass fraction of MeOH in C: xM, C = 185 g / 350 g = 0.529 Question 4: You have designed a centrifuge system to separate cells from a solution. A mixture enters the centrifuge system at a rate 1000 L/h and contains 500 mg cells/L. You can assume that the density is that of water (1 g/cm3). The inlet stream or feed (F), is separated to a cell‐free supernatant (S), and a pellet (P) consisting of a solution containing water and cells. Using the data in the figure below, calculate the mass flow rates mP (g/hr) and mS (g/hr) Supernatant (S)
Ccells, S = 0
Feed (F)
VF = 1000 L/h
ρF = 1.0 g/cm3
Ccells, F = 500 mg cells/L Centrifuge
Pellet (P)
Xcells, P = 0.5
Xwater, P = 1‐ Xcells, P = 0.5 Question 5: Corn‐steep liquor contains 2.5 wt% dextrose (i.e. the mass fraction of dextrose is 2.5/100 = 0.025) and 50 wt% water; the rest of the liquor stream is solids. Beet molasses contain 50 wt% sucrose, 1.0 wt% dextrose, and 18 wt% water; the rest of the molasses stream contains solids. Beet molasses is mixed with corn‐steep liquor and water in a mixing tank to produce a dilute sugar mixture. The exit stream contains 2.0 wt% dextrose and 12.6 wt% sucrose, and is ready to be fed into a fermentation unit. a) Draw a diagram of the system and label entering and exiting streams. b) What are the mass fractions of dextrose, sucrose, solids, and water in the exit stream? c) What is the ratio of the mass flow rate of the water stream to the mass flow rate of the corn‐
steep liquor stream? a) Corn Steep Liquor 2.5 wt% dextrose 50 wt% water Rest is solids Beet Molasses 1.0 wt% dextrose 50 wt% sucrose 18 wt% water Rest is solids MIX Exit 2.0 wt% dextrose 12.6 wt% sucrose ? wt% water ? wt% solids 100% Water b) Steady‐state. No reaction. Overall balance: W + C + B = E Dextrose: 0.025 C + 0.01 B = 0.02 E Sucrose: 0.50 B = 0.126 E ‐> B = 0.252 E Substitute in dextrose balance: 0.025 C + 0.01(0.252 E) = 0.02 E ‐> C = 0.699 E Substitute results for B and C into overall balance: W + 0.699 E + 0.252 E = E ‐> W = 0.049 E Water: 0.5 C + 0.18 B + W = fW E fW = mass fraction of water in exit stream Substitute prior results for C, B, and W: 0.5 (0.699 E) + 0.18 (0.252 E) + 0.049 E = fW E Find fW = 0.444 or 44.4% Therefore mass fraction of solids in exit stream is fS = 1‐ 0.02 – 0.126 – 0.444 = 0.410 or 41.0% c) We want W/C. Divide overall balance by C: W/C + 1 + B/C = E/C ‐> W/C = E/C – B/C – 1 But E/C = 1/0.699 = 1.431 Since B = 0.252 E, B/C = 0.252 E/C = 0.252 * 1.431 = 0.361 Therefore W/C = 1.431 – 0.361 – 1 = 0.07 Question 6: An adult takes about 12 breaths per minute, inhaling roughly 500 mL of air with each breath. The molar composition of the inspired and expired gases are as follows: Species Inspired gas (%) Expired (%) O2 20.6 15.1 CO2 0.0 3.7 N2 77.4 75.0 2.0 6.2 H2O Assume that the temperature of the air in and air out are the same. Also, note that nitrogen is not transported into or out of the blood in the lungs, so the amount of N2in = amount of N2out a) Draw a diagram of the lung system and the streams in and out of it. b) Calculate the volumes of O2, CO2, and H2O transferred from the pulmonary gases to the blood or vice versa (specify which) per minute. a) Volume flow rate of inspired gas: 12 breaths/min * 500 ml/breath = 6000 mL/min Inspired gas 6000 mL/min 20.6% O2 0.0% CO2 77.4% N2 2.0% H2O MIX Expired gas ? mL/min 15.1% O2 3.7% CO2 75.0% N2 6.2% H2O Blood ? mL/min ?% O2 ?% CO2 0.0% N2 ?% H2O b) Balance on N2: 6000 mL/min inhaled * 0.774 = ? mL/min exhaled * 0.75 ‐> 6192 mL/min exhaled Balance on O2: 6000 mL/min inhaled * 0.206 = 6192 mL/min exhaled * 0.151 + O2 absorbed ‐> Oxygen absorbed = 6000*0.206 – 6192*0.151 = 301 mL/min Balance on CO2: 0 = 6192 mL/min exhaled * 0.037 + CO2 absorbed ‐> CO2 absorbed = ‐6192*0.037 = –229 mL/min or +229 mL/min released Balance on H2O: 6000 mL/min inhaled * 0.02 = 6192 mL/min exhaled * 0.062 + H2O absorbed ‐> Water absorbed = 6000*0.02 – 6192*0.062 = ‐264 mL/min or +264 mL/min released Question 7: 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 A
Mixer
to join with the unfiltered blood stream before entering the membrane and one (stream F) leaves the system as filtered blood. The recycle W
mass flow rate (stream R) is known to be twice Membrane R
that of the filtered mass flow rate (stream F). Unit
Calculate the mass flow rate and the wt% of waste in streams A, B, F, and R. Hint: the splitter B
does not change the composition of the stream Splitter
that it separates. A
U
Mixer 25 g/min
0.17% waste W
30 mg/min
100% waste Membrane Unit R B Splitter R = 2F F Overall balance around everything: 25 = 0.03 + F ‐> F = 24.97 g/min Waste balance around everything: 0.0017 * 25 = 0.03 + wF F ‐> wF F = 0.0125 ‐> wF = 0.0125/24.97 = 0.000501 or 0.0501% Overall balance around splitter: B = R + F since R = 2F as stated in problem, B = 3F ‐> B = 3*24.97 = 74.91 g/min and R = 2*24.97 = 49.94 g/min Splitter does not alter composition, just separates flow into two streams of same composition, therefore: wB = wR = wF = 0.0501% Overall balance around membrane: A = B + 0.03 ‐> A = 74.91 + 0.03 = 74.94 g/min U F Waste balance around membrane: wA A = wB B + 0.03 ‐> 74.94 wA = 74.91 wB + 0.03 Since wB = 0.0501%, use equation above to can find wA = 0.000901 or 0.0901% ...
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 Fall '10
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