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., Vtot = Vwater + Vethanol

The following data may or may not be useful:

 Chemical Specific Gravity Molecular Weight (g/mol) ethane --- 30.07 ethanol 0.789 46.07 methanol 0.792 32.04 water 1.000 18.016

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:

 Component Volume (ml) Density (g/ml) Mass (g) wt % (g/g total) Ethanol 25.00 0.789 19.725 0.208 Water 75.00 1.000 75.000 0.792

Total mass = 19.725 + 75.000 = 94.725 g

Recall that rfluid = (S.G.)rwater

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

 Component Mass (g) Molecular Weight (g/mol) Moles (mol) Ethanol 20.800 46.070 0.451 Water 79.200 18.016 4.390

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 P2 is 90 percent of P1? The following data may or may not be useful:

 Chemical Specific Gravity Molecular Weight (g/mol) ethanol 0.789 46.07 mercury 13.546 200.61 water 1.000 18.016

I have added two more lengths to the diagram so I can write:

Pleft = Pright

1 atm + rgas g z1 + rgas g h + rm g z2 = P2 + rgas g z1 + rm g h + rm g z2

We see that the rm g z2 cancels from both sides to give:

1 atm + rgas g z1 + rgas g h + = P2 + rgas g z1 + rm g h

And, the rgas g z1 term cancels from both sides as well to give:

1 atm + rgas g h + = P2 + rm g h

we can rearrange and plug in some other information to get:

1 atm - 0.9 atm = (rm - rgas) g h

but, rgas << rm so we can just ignore it to get:

0.1 atm = 1.01325*104 N/m2 = rm g h

Rearrange and plug in to get:

h = 10132.5 N/m2 * (1 kg /9.8 N) * (1 m3 / 13540 kg) = 0.0763 m = 76.3 mm

b) Qualitatively, what will happen to h if P2 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 30-foot high storage tank that ruptured. Twenty-one 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 lbm? And, what was the pressure at the bottom of the tank in pounds-force 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/ft3) * (1 ft3/7.408 gal) = 2.687*107 lbm

For the second part, we can use our old familiar equation of:

Pbottom = Ptop + rgh

In this case, Ptop = 1 atm, h = 30 ft, r = 1400 kg/m3 to give:

P = 14.696 psia + 1.4 * 62.43 lbm/ft3 * (1 lbf/1 lbm) * 30 ft * (1 ft2/122 in2)

P = 14.696 psia + 18.208 lbf/in2 = 32.9 psia

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