Homework Assignment # 2 due Friday, September 1 st

Transcription

1 Homework Assignment # 2 due Friday, September 1 st 1. The electrostatic charge generation for hydrated lime powder during pneumatic transport can be described by the following relationship: q/m= Aw b u c where q/m is the ratio of charge/mass (C/g), i.e. coulombs/gram, w is the mass flow rate of the powder (g/s), and u (m/s) is the mean velocity of the powder in the tube. The following data [Liang et al., Ind. Eng. Chem. Res., 35, 2748 (1996)] have been obtained for charge buildup: q/m (C/g) w (g/s) u (m/s) 3.81E E E E E E E E E E E E E E E (a) What is the linearized form of the relationship between (q/m), w, and u? ln(q/m) = ln A + b ln w + c ln u (b) From the linearized form of the equation above and the data provided, use a spreadsheet program to calculate the best fit values for A, b, and c. Include your data and regression statistics.

2 ln q/m ln w ln u SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 15 ANOVA df SS MS F Significance F Regression E 30 Residual E E 06 Total Coefficients Standard Error t Stat P value Lower 95% 3.54E Intercept Upper 95% Lower 95.0% Upper 95.0%

3 X Variable X Variable E E c) What is the original equation with the correct parameters? q/m = 1.4 X 10-6 w u A mixture of propane and air containing 10 mass % propane flowing at a rate of 100 mol/hr is to be diluted with pure air to reduce the propane mass% to 6. (20 pts) a) What is the mole fraction of N 2 in the final stream? b) Would you expect an experiment of this nature? Why or why not? a).1/44+.9/29 = 1/MW MW = g/mol 100 mol/hr*30.02 g.mol = 3002 g 300 g of propane and g of air 300 g/.06 = 5000 = m 1.94 * 5000 = 4700 g of air = moles of air 300 g of propane/44 g/mole = 6.82 moles of propane total moles = 169 moles moles of N 2 =.79 * = 128 moles/169 moles =.76 mole fraction b) No, it s in the flammability limit

4 3. Using the problem with species A, B and C we did in class: a) write all three species balances and the overall balance b) show that these equations are not independent. c) solve for the unknowns if the mass fraction of A in the second input stream is 60 % and the mass flow rate of the bottom stream is 2000 kg/hr. (15 pts) a) A: x A (5300) = m 2 +.7(1200) B:.03m 1 + (1 x A )5300 = x B (1200) +.60m 3 C:.97m 1 =.40m 3 + (.3 x B )(1200 overall m 1 = m 2 + m b) If you add a,b,and c, they will equal the overall balance. c) A:.6*5300 = m 2 +.7*1200 m 2 = 2340 kg/hr overall: m = m 1 = 240 kg/hr B:.03* *5300 =.6* x B *1200 x B = 0.77

5 4. An artificial kidney is a device that removes water and waste metabolites from the blood. In one such device, the hollow fiber hemodialyzer, blood flows from an artery through the insides of a bundle of hollow cellulose acetate fibers, and dialyzing fluid, which consists of water and various dissolved salts, flows on the outside of the fibers. Water and waste metabolites - principally urea, creatinine, uric acid, and phosphate ions - pass through the fiber walls into the dialyzing fluid, and the purified blood is returned to a vein. Suppose that at some time during a dialysis the arterial and venous blood conditions are as follows. Dialyzing fluid Blood from an artery Purified blood to a vein Dialysate Arterial (entering) Blood Venous (exiting) Blood Flow Rate: 200 ml/min 195 ml/min Urea (H2NCONH2) Concentration: 1.90 mg/ml 1.75 mg/ml From these data, it is clear that water and urea are being removed from the blood at rates of 5 ml/min and mg urea/min, respectively. (a) (2%) Show the calculations for determining the rates of water and urea removal which are stated above? (b) (4%) If the dialyzing fluid enters at the rate of 1500 ml/min, calculate the concentration of urea in the exiting dialysate in units of mg urea/ml (neglect the urea volume)? (c) (4%) Suppose we want to reduce the patient s urea level from an initial value of 3.5 mg/ml to a final value of 0.8 mg/ml. If the total blood volume is 4 liters and the average rate of urea removal is that calculated in part a, how long in minutes must the patient be dialyzed (neglect the loss in total blood volume due to the removal of water in the dialyzer)?

6 Solution Arterial Blood 200 ml/min 1.9 mg urea/ml Venous Blood 195 ml/min 1.75 mg urea/ml Dialyzing Fluid 1500 ml/min Dialysate q (ml/min) c (mg urea/min) (a) Water Removal Rate = 200 ml/min ml/min = 5 ml/mon Urea Removal Rate = (1.9)(200) mg urea/min - (1.75)(195) mg urea/min = mg urea/min (b) q = = 1505 ml/min c = [38.75 mg urea/min]/[1505 ml/min] = mg urea/ml (c) [( ) mg removed/ml][1 min/38.75 mg removed][1000 ml/1 liter][4 liter] = 279 minutes (= 4.65 hours)