Team H problems (Calinisan, Figueroa, Justiniani)

1. A 250 mL can of Red Bull, with unknown composition of caffeine, and other ingredients, is

mixed with a stream containing 20mL CH2Cl2 and 5g Na2CO3 (density=2.54g/mL). Physical

properties make it possible to separate the mixture of Na2CO3 with the other ingredients and

water from the mixture of caffeine with CH2Cl2. To isolate caffeine (density = 1.23 g/mL), the

latter mixture is heated to the boiling point of CH2Cl2 (39.6 C). However, for this problem, 6 mL

of the water in the feed stream is left with the mixture of caffeine with CH2Cl2. So to fully isolate

caffeine with CH2Cl2 before evaporation, 0.94mL of Na2SO4 is mixed into the stream. Determine

the amount of caffeine present in the energy drink (in mg) and the composition in each stream.

Note: The volume of the stream containing the mixture of Na2CO3 with the other ingredients and

water is 9.434 times that of the volume of the stream containing the mixture of caffeine with

CH2Cl2.

Solution:

Block Flow Diagram of the problem:

*By isolating the first mixing process in the system, we can solve for the stream

compositions and volume (assuming that no chemical reactions take place, only physical).

5g Na2CO3*(mL/2.54g) = 1.97 mL Na2CO3

General Equation: IN = OUT

n_f = 20 mL CH2Cl2 n_e = 1.97 mL Na2CO3

stream1 = 9.434*stream2

OMB: 250 mL + 21.97 mL = stream1 + stream2

stream1 = 245.90 mL stream2 = 26.07 mL

CMB: others n_b = n_d = stream1 - n_e - 220 = 23.93 mL others

caffeine n_a = n_g = stream2 - n_f - 6 = 0.066 mL caffeine

From the solved values for the first mixing process, the rest of the unknowns can be solved.

stream3 = 6 + 0.94 = 6.94 mL

n_h = 6 mL H2O n_i = 0.94 mL Na2SO4

stream4 = n_f + n_g = 20.066 mL

stream5 = n_j = 20 mL CH2Cl2 stream6 = n_k = 0.066 mL caffeine

After solving all compositions and isolating caffeine, it can then be converted to milligrams.

0.066 mL caffeine*(1.23g/mL)*(1000mg/g) = 81.18 mg of caffeine

Remark: The actual Red Bull energy drink contains 80 mg of caffeine for every 250 mL serving.

This amount is approximately the same as to that of a serving of coffee.

2. Aluminum Appendix

a. Find a trendline for the specific heat of aluminum.

b. Red Bull recycles aluminum cans to save up to 95% energy consumption compared

to using new aluminum cans. Suppose that per year, the company is able to

collect an average of 1M cans for recycling by melting. Given that each empty

can has a mass of 2oz (56.7 g), aluminum has the latent heat of 321kJ/kg and a

melting point of 660.3 C, determine how much energy is consumed in recycling

the old cans if they are melted from a temperature 37 C.

c. If the company doesn't recycle their cans, how much will be the actual yearly energy

consumption of Red Bull?

Solution:

a. A trend line can be manually solved for or it can be determined by using a spreadsheet.

For this problem, it will be solved for by spreadsheet since the long method was not

discussed in class.

Given the following data above, its graph and trend line equation is as follows:

(Cp) = ((T) - 0.6964)/0.0006

b. Assuming that there is no change in potential and kinetic energies, and no shaft work is

affecting the system, the overall equation to be used will be (delta)H = Q where Energy =

mQ is the is the value being solved for.

Block Flow Diagram of the problem:

(delta)H = Cp(delta)T + latent heat = ((660.3 - 37) - 0.6964)/0.0006 + 321 = 1,037,993.667 kJ/kg

Energy per can = 0.0567kg*1037993.667kJ/kg = 58,854.24 kJ

Energy for 1M cans = 5.89*10^10 kJ

c. Energy saved per 1M cans= (Actual energy used w/o recycling) - (5.89*10^10kJ)

= 95%(Actual energy used w/o recycling)

Actual energy used w/o recycling = 1.18*10^12 kJ

3. Taurine is produced by reacting aziridine and sulfurous acid:

C2H5N + H2SO3 → C2H7NO3S

The product stream contains 47.2% of C2H5N and 18.1% H2SO3. The feed to the reactor

contains only aziridine and sulfurous acid. Calculate the following:

a. fractional conversion of the limiting reactant

b. percentage where the other reactant is in excess

c. if the molar flow rate of the feed stream is 150 mol/s, determine the extent of reaction.

Solution:

Block Flow Diagram of the problem:

a. Extent of Reaction Balance

C2H5N: 47.2 = n_a - E

H2SO3: 18.1 = n_b - E

C2H7NO3S: 34.7 = E

n_a=81.9 mol C2H5N n_b=52.8 mol H2SO3 n_feed = 134.7 mol enters feed (60.8% C2H5N, 39.2% H2SO3)

n_a/n_b = 1.55 v_a/v_b = 1 (limiting reactant is H2SO3)

Fractional Conversion (Limiting Reactant) = (n_in - n_out) / n_in

= (52.8 - 18.1) / 52.8 = 0.66

b. Percent Excess = (n_in - n_stoich) / n_stoich = (81.9 - 52.8) / 52.8 = 0.55 = 55%

c. n_feed= 150 mol (basis:1second)

n_a=0.608*n_feed = 91.2 mol C2H5N

n_b=0.392*n_feed = 58.8 mol H2SO3

n_H2SO3(product) = n_b - n_b*0.66 = 19.99 mol H2SO3

Extent of Reaction Balance

C2H5N: n_C2H5N(product) = n_a - E

H2SO3: 19.99 = n_b - E

E=38.81 n_C2H5N(product) = 52.39 mol C2H5N