Transcript Accounting for Mass - Texas A&M University

Solution to
Ind. Ex. #1
Chicago
fuel
Detroit
fuel
Indianapolis
initial mass
final mass
= (60,000+8,000+6,000)
= 64,500 lbm
= 74,000 lbm
cargo = 3,500 lbm
final– initial = in – out = in - (fuel + cargo)
fuel = initial - final + in - cargo
= 74,000 lbm - 64,500 lbm + 0 –3,500 lbm
= 6000 lbm
Solution to Pairs Exercise #1
1. Describe the system
Inputs
Output
1.0 kg solution
2.3% sugar
X kg dry sugar
Mixer
Y kg final solution
18.0 % sugar
100% sugar
boundary
2. It is a batch process
3. Sketch the system and inputs and output
4. Label known amounts and compositions
5. Labeled unknowns X and Y
Solution to Pairs Exercise #1
(con’t)
6. Write mass balance equations
It starts empty and ends empty, so there is
no accumulation of mass in mixer, therefore
input = output
 Because we have 2 unknowns, we need 2
equations
 total amount: 1.0 kg + X kg = Y kg
 sugar amount: 0.023×1.0 + 1.00×X = 0.180×Y

• Note: we could have written the second equation
for water instead of sugar
0.977×1.0 + 0×X = 0.820×Y
• Which makes the solution easier?
Solution to Pairs Exercise #1
(con’t)
7. Solve for unknown variables
Y = 0.977/0.82 = 1.191 kg solution, and
X = Y - 1.0 = 0.191 kg dry sugar
8. Check: (Note: It is not always possible
to do an exact check on a mass balance
problem.)
Here we could substitute values for X and
Y in the “sugar” equation and verify the
answer.
 The value does seem reasonable.

Hint to Pairs Exercise #2
Inputs
10,000 lbm/h wet corn
25.0% water
Output
X lbm/h water
Drier
100% water
Y lbm/h dried corn
14.0% water
QUESTION: Why can’t we solve this problem “by
inspection”? We have 25% water in and 14% water out,
so water removed is 11%, or 1100 lbm/h.
No! No! No!
Water in is 25% of 10,000, but water out, in the corn, is
14% of Y, not of 10,000.
Solution to Pairs Exercise #2
In = Out
Corn: (1-0.25)10,000 lbm/h = (1-0.14) Y
Y  (1 0.25)10,000 lbm / h  8721 lbm / h
1- 0.14
Water: 0.25 (10,000 lbm/h) = 0.14 Y + 1.0 X
X  0.25(10,000 lbm / h) - 0.14Y
 0.25(10,000 lbm / h) - 0.14(8721lbm / h)
1279 lbm / h
Solution to Pairs Exercise #4
100 kg
0.0% fat
2.5% protein
X kg
2.0% fat
2.1% protein
Y kg
3.5% fat
1.9% protein
Z kg
1.6% fat
2.2% protein
Solution to Pairs Exercise #4 (con’t)
Total:
Fat::
Protein:
100 + X + Y = Z
(0) 100 + 0.020 X + 0.035 Y = 0.016 Z
0.025 (100) + 0.021 X + 0.019 Y = 0.022 Z
1.000 X +1.000 Y -1.000 Z = - 100
0.020 X +0.035 Y - 0.016 Z = 0
0.021 X + 0.019 Y - 0.022 Z = - 2.5
1.000 1.000 1.000  X 
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100
0.020 0.035  0.016 Y  0
0.021 0.019  0.022 Z  2.5
X = 129
Y = 57
Z = 286
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