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# e aCUOQ m tgnt Ot: L aK.U i l 'IUW W Hat ' -aU y vu &amp;quot;'&amp;quot;Y u vvu &amp;quot; w uy-- ~ - to-. - -J . W et air containing 4.0 m ole%...

This question was answered on Feb 16, 2011. View the Answer
Wet air containing 4.0 mole% water vapor is passed through a column of calcium chloride pellets.
The pellets adsorb 97.0% of the water and none of the other constituents of the air. The column
packing was initially dry and had a mass of 3.40 kg. Following 5.0 hours of operation, the pellets are
reweighed and found to have a mass of 3.54 kg.
(a) Calculate the molar flow rate (mol/h) of the feed gas and the mole fraction of water vapor in the
product gas. .
(b) The mole fraction of water in the product gas is monitored and found to have the value calculated
in part (a) for the first 10 hours of operation, but then it begins to increase. What is the
most likely cause of the increase? If the process continues to run, what will the mole fraction of
water in the product gas eventually be?

e aCUOQ m tgnt Ot:: L aK.t:U i l 'IUW W Hat ' -aU y vu &quot;'&quot;Y u vvu• &quot; w uy-- ~· - to·-. - -J • . . ..
W et air containing 4.0 m ole% water v apor is p assed through a column o f calcium chloride pellets.
The pellets adsorb 97.0% o f t he w ater and n one o f t he o ther c onstituents o f t he air. The column
packing was initially dry and h ad a mass o f 3.40 kg. Following 5.0 h ours o f o peration, the pellets a re
r eweighed and found t o h ave a mass o f 3.54 kg.
( a) C alculate the molar flow r ate ( mol/h) o f t he feed gas a nd t he mole fraction o f w ater vapor in the
product gas.
.
( b) T he m ole fraction o f w ater i n the product gas is monitored and found t o have the value calculated in p art (a) for the first 10 h ours o f o peration, b ut t hen it begins to increase. What is the
most likely cause o f t he increase? I f t he process continues t o r un, what will t he m ole fraction o f
w ater in the product gas eventually be? .t'robl~ms 163 &quot;' Q T he indicator dilution method is a technique used to determine flow rates o f fluids in channels for
which devices like rotameters and orifice meters cannot b e·used (e.g., rivers, blood vessels, i md largediameter pipelines). A stream o f a n easily measured substance (the tracer) is injected into the channel at a known r ate a nd the tracer concentration is measured a t a p oint far enough downstream o f
t he injection point for the tracer t o b e c ompletely mixed with the flowing fluid. The larger the flow
r ate o f t he fluid, the lower the tracer concentration a t t he measurement point.
A gas stream t hat c ontains 1.50 m ole% C 02 flows through a pipeline. Twenty (20.0) kilograms
o f C 02 p er m inute is i njected into the line. A sample o f t he gas is drawn from a point in the line 150
m eters downstream o f t he injection p oint a nd found t o c ontain 2.3 mole% C 0 2 .
( a) Estimate the gas flow r ate (kmoVrnin) upstream o f t he injection point.
( b) E ighteen seconds elapses from the instant the additional C 0 2 is first injected t o t he time the
C 0 2 c oncentration a t t he measurement point begins t o rise. Assuming t hat t he tracer travels a t
t he average velocity o f t he gas in the pipeline (i.e., neglecting diffusion o f C 02 ) , e stimate t he
r \ average velocity (m/s). I f the molar gas density is 0.123 kmoVm3 , what is the pipe diameter?
/ 4.25. 1 variation o f t he indicator dilution method (see preceding problem) is used t o m easure total blood
l~ olume. A known amount o f a t racer is injected into the bloodstream and disperses uniformly
throughout t he circulatory system. A blood sample is then withdrawn, the tracer concentration in
the sample is m easured, a nd t he measured concentration [which equals (tracer injected)/( total blood
volume) if n o t racer is lost through blood vessel walls] is used t o d etermine t he total blood volume.
I n o ne such experiment, 0.60 c m3 o f a s olution containing 5.00 mg/L o f a dye is injected into
an artery o f a grown man. A bout 10 minutes later, after the tracer has h ad time t o d istribute
itself uniformly throughout the bloodstream, a blood sample is withdrawn and placed in the
sample chamber o f a s pectrophotometer. A beam o f light passes through the chamber, and
the spectrophotometer measures the intensity o f t he transmitted beam a nd displays the value
o f t he solution absorbance (a quantity t hat increases with the amount o f light absorbed by the sample). The value displayed is 0.18. A c alibration curve o f a bsorbance A versus tracer concentration
C (micrograms dye/liter blood) is a straight line through the origin and the point (A = 0.9, C =
3 J.IWL). E stimate t he p atient's total blood volume from these data. (7 A liquid mixture containing 30.0 m ole% benzene (B), 25.0% t oluene (T), and the balance xylene
(X) is fed t o a distillation column. The bottoms product contains 98.0 m ole% X and n oB, a nd 96.0%
o f t he X in the feed is recovered in this stream. T he o verhead product is fed t o a s econd column.
The overhead product from the second column contains 97 .0% o f t he B in t he feed t o this column.
The composition o f this stream is 94.0 m ole% B and the balance T.
(a) D raw a nd label a flowchart o f this process and do the degree-of-freedom analysis t o p rove t hat
f or an assumed basis o f calculation, molar flow rates and compositions o f all process streams
can be calculated from the given information. Write in o rder t he equations you would solve
to calculate unknown process variables. I n e ach equation ( or p air o f simultaneous equations),
circle the variable(s) for which you would solve. D o n ot d o the calculations.
(b) Calculate (i) the percentage o f t he benzene in the process feed (i.e., the feed t o t he first column)
t hat emerges in the overhead product from the second column and (ii) the percentage o f t oluene
in the process feed t hat emerges i n t he bottom product from the second column. ··· --- ---, - ~0 J Fresh orange juice contains 12.0 w t% solids a nd t he balance water, and concentrated orange juice
.4.~ &quot;:::::::!' contains 42.0 w t% solids. Initially a single evaporation process was used for the concentration, but
\ volatile constituents o f t he j uice escaped with t he water, leaving the concentrate with.a fiat taste. T he
c urrent process overcomes this problem by bypassing the evaporator with a fraction o f t he fresh
juice. T he juice t hat e nters the evaporator is c oncentrated to 58 w t% solids, a nd t he evaporator
product stream is mixed with the bypassed fresh juiCe to achieve the desired final concentration.
( a) D raw a nd label a flowchart o f this process, neglecting the vaporization o f everything in the juice
b ut water. First prove t hat t he subsystem containing the point where the bypass stream splits
off from the evaporator feed has one degree o f freedom. ( If you think i t has zero degrees, try
determining the unknown variables associated with this system.) Then perform the degree-offreedom analysis for the overall system, the evaporator, and t he b ypass-evaporator product
mixing point, and write i n o rder t he equations you would solve t o d etermine all unknown stream
variables. In each equation, circle the variable for which you would solve, b ut d on't d o any
calculations.
.
( b) C alculate the amount o f p roduct (42% c oncentrate) produced p er 100 kg fresh juice fed t o t he
process and t he fraction o f th~leed t hat bypasses the evaporator.
(c) Most o f t he volatile ingredients t hat p rovide the taste o f t he concentrate are contained i n t he
fresh juice t hat bypasses the evaporator. You could get more o f these ingredients in the final
product by evaporating t o (say) 90% solids instead o f 5 8%; you could then bypass a greater
fraction o f t he fresh juice and thereby obtain an even b etter tasting product. Suggest possible
drawbacks t o this proposal.

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## This question was asked on Feb 15, 2011 and answered on Feb 16, 2011.

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