

Test #1  Problem #1 There are other alternative solutions, but here is mine A liquid stream contains 25 vol % ethanol and the rest is water. What would the molar flowrate (mol/hr) be if the mass flowrate is 100 g/hour? In this problem, assume that volumes are additive; i.e., V_{tot} = V_{water} + V_{ethanol} The following data may or may not be useful:
Solution: Choose a basis by volume because we are given volume percents. So, my basis will be 100 ml. Make a table of volumes and masses like we did in class:
Total mass = 19.725 + 75.000 = 94.725 g Recall that r_{fluid} = (S.G.)r_{water} Now we need to use the total mass flowrate to find out how much we have really. It is time to get rid of our basis assumption and use the real information now. Let's make another table and wrap this problem up: where: mass / MW [=] moles
Total = 0.451+ 4.390 = 4.84 moles total Test #1  Problem #2 a) A pipe with a gas in it is connected to a mercury manometer. What is h (in mm) if P_{2} is 90 percent of P_{1}? The following data may or may not be useful:
I have added two more lengths to the diagram so I can write: P_{left} = P_{right} 1 atm + r_{gas }g z1 + r_{gas} g h + r_{m} g z2 = P_{2} + r_{gas} g z1 + r_{m} g h + r_{m} g z2 We see that the r_{m} g z2 cancels from both sides to give: 1 atm + r_{gas }g z1 + r_{gas} g h + = P_{2} + r_{gas} g z1 + r_{m} g h And, the r_{gas }g z1 term cancels from both sides as well to give: 1 atm + r_{gas} g h + = P_{2} + r_{m} g h we can rearrange and plug in some other information to get: 1 atm  0.9 atm = (r_{m}  r_{gas}) g h but, r_{gas} << r_{m} so we can just ignore it to get: 0.1 atm = 1.01325*10^{4} N/m^{2} = r_{m} g h Rearrange and plug in to get: h = 10132.5 N/m^{2} * (1 kg /9.8 N) * (1 m^{3} / 13540 kg) = 0.0763 m = 76.3 mm b) Qualitatively, what will happen to h if P_{2} decreases further? We could plug in P2 = 0.8 atm to see what happens. Or, we can look at the physical picture and think about how reducing the pressure or force on the right side would affect the height. We can think about lifting a weight off the right side as an example. This will cause the fluid to slide to the right side farther and will increase h. Test #1  Problem #3 The great Boston molasses flood occurred on January 15, 1919. In it, 2.3 million gallons of molasses flowed from a 30foot high storage tank that ruptured. Twentyone people were killed and 150 more were injured. The estimated specific gravity of molasses is 1.4. What was the mass of molasses in the tank in lb_{m}? And, what was the pressure at the bottom of the tank in poundsforce per square inch? Assume the tank was open to the atmosphere at the top. The mass of the molasses is: m = V * r = 2,300,000 gal * 1.4 * (62.43 lbm/ft^{3}) * (1 ft^{3}/7.408 gal) = 2.687*10^{7} lbm For the second part, we can use our old familiar equation of: P_{bottom} = P_{top} + rgh In this case, P_{top} = 1 atm, h = 30 ft, r = 1400 kg/m^{3} to give: P = 14.696 psia + 1.4 * 62.43 lbm/ft^{3} * (1 lbf/1 lbm) * 30 ft * (1 ft^{2}/12^{2} in^{2}) P = 14.696 psia + 18.208 lbf/in^{2} = 32.9 psia 
© 2007 Arizona Board of Regents for The University of Arizona 