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8/15/2019 Chapter 1 (Part 2) Process Variables
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Chapter 1(Part 2)
Processes and Process
Variables
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Evaluation of performance of processoperation requires the knowledge of theamounts composition conditions of
materials that enter and leave eachprocess units!
"his chapter presents methods ofcalculating variables that characteri#ethe operation of processes andindividual
process units!
Processes and Process Variables
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Process
Process$ an% operation that cause a ph%sical orchemical change in a substance! Can consist of
several process unit!
Process streams connecting process units andform the process &ow sheet!
Chemical engineer is responsible to design andoperate the process!
Process
'nitnput*eed +utputProduct
Process
'nit +utputProduct
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,
Process Variables???
- "he quantities used to describe a processand these must be measured andcomputed
PROCESS .Feed/
0P'" .Products/
+'"P'"
- "o design or anal%#e a process we need to knowthe amounts compositions and condition ofmaterials entering leaving and within theprocess!
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Processes and
Process Variables
Density and
Specific Gravity
Flow rate
Chemical
Composition
Pressure
Temperature
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ensit% 3peci4c Volume
ensit% (ρ) mass per unit volume of a substanceensit% of a substance can be used as a conversion
factor to relate the mass and the volume of thesubstances!
'nit5 gcm67 kgm67 lbmft6!
3peci4c Volume volume per unit mass of a substance inverse of densit% 'nit5 cm6g7 m6kg7 ft6lbm
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"he densit% of CCl, is 1!898 gcm67 what isa) :ass of 2; cm6 of CCl,
b) Volume of
20 cm3 1.595 g= 31.9 g
cm3
6.20 lbm 454 g cm3
= 1760 cm3
1 lbm 1.595 g
Try This…
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Specifc Gravity3peci4c =ravit% (3=)>atio of the densit% (ρ) of a substance to the densit%
of a reference (ρref ) substance at a speci4c condition5
ensit% of water at ,?C is used as a referencedensit%7ρref =ρ@2+(l) (,?C) A 1!;;; gcm6
A 1;;; kgm6
A
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B liquid has a 3= of ;!8;! *inda) ensit% in gcm6
b) ensit% in lbmft6
c) :ass of 6 cm6 of this liquidd) Volume occupied b% 1 g of this liquid
Try This…
8/15/2019 Chapter 1 (Part 2) Process Variables
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3olution
a) ρ =0.5 1 g
= 0.5 g/cm3cm3
b) ρ =0.5 62.43 lbm
= 31.215 lbm
/ft3
ft3
c) 3 cm3 0.5 1 g
= 1.5 gcm3
d) 18 g cm3
= 36 cm30.5 g
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Flow Rate
Continuous process involve movement ofmaterials from one point to another withcertain rate!
*low rate$ the amount of material thatmoves into or out of a process unit perunit time
*low rate can be eDpressed as 5◦ :ass &ow rate (masstime) A◦ Volumetric &ow rate (volumetime) A◦ :olar &ow rate5 (molestime )
mQ
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12
Relat!o" # $ea%!e%t to mea%&'e) to
(f eloc!t*+ v !% mea%&'ed+ # t,e" calc&lated a%
m
Qm ρ =
AQ ν =
Flow RateFlow Rate
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-,e ma%% flow'ate of ",ea"e $ρ=0.659g/cm3) !" a !e !% 6.59 g/%. ete'm!"e t,eol&met'!c flow'ate of t,e ",ea"e
Try This…
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*low meter is a device mounted in aprocess line that provides a continuousreading of the &ow rate in the line!
"wo commonl% used &ow meter arerotameter and ori4ce meter!
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Flow meterFlow meter
Rotamete'
'!f!ce mete'$ba%ed o" 'e%%&'e
d'o)
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ChemicalComposition
Moles and
Molecular Weiht
Mass and
Mole Fractions
!verae
Molecular Weiht
Concentration
Parts per Million "ppm#
$ Part per %illion "ppb#
Chemical CompositionChemical Composition
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ole% olec&la' e!g,t
Atomic weight - mass of an atom based on carbonisotope 12C.
Molecular weight - Sum of the atomic weights ofatoms that represent a molecule of the compound.
Eg.: !"gen atomic weight #$ % 1&.' g(g-mol thus) the 2 M* % 1&.' + 1&.' % ,2.' g(g-
mol .
nit: g(mol) g(mol) and lbm(lbmole
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ole% olec&la' e!g,tco"t.
t 1 mol of a %&b%ta"ce+ !t% ma%% !" g'am !% e&al to !t% molec&la'we!g,t.
amle: ; ,a% molec&la' we!g,t of 28g/mole.
-,e'efo'e
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19
amle :
34 g of >?3 !% e&!ale"t to ole%
$ >?3 = 17.0 g/mol)
" >?3 = 34 g/ $17.0 g/mol) = 2 mol
&'( )bmol
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a%% a"d ole F'act!o"
Process input or output streams can contain mi!tures ofli/uids or gases) solutions of one or more solutes in asol0ent.
need mass fraction and mole fraction to define thecompositions:
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a%% a"d ole F'act!o"
:ass fraction
nit: g A(g total g A(g total lbm A(lbm total
:ole fraction
'nit5 kmol Bkmol total7 lb$moles Blb$moletotal
masstotal
Aof mass= A x
molestotal
Aof moles= A y
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B miDture of gases has the following masscomposition5
+2 1
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, steps to convert from mass fractions to molesfractions5
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3olution
@a%!%: 100g of m!t&'e
;omo"e"ta%%
F'act!o" a%% ole% ole F'act!o"
! ! m! ! "! *!
2 0.16 16 32 0.500 0.152
; 0.04 4 28 0.143 0.044
;2 0.17 17 44 0.386 0.118
>2 0.63 63 28 2.250 0.686
-otal 1.00 100 3.279 1.000
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Bverage :olecular Geight
e'age olec&la' e!g,t
Mean molecular weight of a mi!ture #g(mol) lbm(lbmole$.
f " i is the mole fraction of the component i of the mi!ture
and Mi is the molecular weight:
f !i is the mass fraction of the component i of the mi!ture
and Mi is the molecular weight:
∑=++= com0o"e"tall!!2211 *.....**
∑=++=como"e"tall !
!
2
2
1
1
4
/.....
4
/
4
/
4
1
4
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2<
E!ample:
3etermine the a0erage molecular weight of refiner" waste gas has the followinganal"sis b" 0olume CH4- 78.0%, C2H6-10.0%, C3H8- 8.0%, C4H10-4.0%
• Strategy:
4 5ASS: 1'' mol refiner" gas
4 nit: 6g(6mole) g(mole
Vol.% kmol MW kg
CH4
C2H6
C3H
C4H!"
#$"
!"$"
$"
4$"
#$"
!"$"
$"
4$"
!6$"
3"$!
44$!
%$!
!24
3"!
3%2
232
!""$"" !""$"" 2!33
∴Mavg = 2133/100= 21.33
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2F
ConcentrationsConcentrations
%
mixtureof Volume
Aof masscA =
mixtureof Volume
Aof molesC
A =
Liter inmixtureof Volume
Aof molesMolarity =
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Iou have ;!;2 molar solution of 0a+@ (;!;2mol 0a+@ 1 J of solution)
@ow man% moles in 8 J of solutionsH
;!;2 mol 0a+@ D 8 J A ;!1 mol 0a+@ 1 J
Try This…
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Aa't% e' !ll!o" $m)Aa't% e' @!ll!o" $b)
-o e'e%% t,e co"ce"t'at!o"% of t'ace %ec!e% !" m!t&'e%of ga%e% o' l!&!d%.
a* 'efe' to ma%% 'at!o% $&%&al fo' l!&!d%) o' mole 'at!o%
$&%&al fo' ga%e%). ?ow ma"* a't% $!" g'am o' mole%) of t,e %ec!e% a'e
'e%e"t e' m!ll!o" o' b!ll!o" a't% of t,e m!t&'e.
m!= *! 106
b! = *! 109
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Pressure
'e%%&'e !% t,e 'at!o of a fo'ce to t,e a'ea o" w,!c,t,e fo'ce act% $A= F/).
A'e%%&'e &"!t%: >/m2+ d*"e%/cm2+ lbf /!"2+ %!+ Aa.
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Hydrostatic PressureHydrostatic Pressure
- n term of @eight h(m) of column of &uid and has a
cross sectional area A (m2
) !- "he &uid has a densit% of K (kgm6) that a pressure
P;(0m2) is eDerted on the upper surface of the
column!
- "he pressure P at the base of the column is b%de4nition the force eDerted on the base divided b%the area A that force is the weight of the columnplus an% force acting on the top!
gh A F P
g Ah g m F
mg ma F
ρ
ρ
=∆=∆
=∆=∆
==
/
)()(
ghPP 0 ρ+=
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62
Pressure HeadPressure Head
- Ghen pressure is eDpressed in terms of aheight of &uid it is called &uid LheadM
- 'suall% water or mercur% is used
- @ead units are mostl% used for ver% lowpressures and eDpressed as Lmm @gL or
Lin @2+N- Converting between forcearea and head
units )( fluid head of gP area force
P h fluid →=
ρ
h i l ! "
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Atmospheric, Asolute ! "au#ePressure
"he atmospheric pressure can be thought ofas the pressure at the base of a column of&uid (air) located at the point measurement(e!g! at sea level)!
B t%pical value of the atmospheric pressure atsea level F
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Bbsolute pressures$ absolute value of the force perunit area eDerted on a surface b% a &uid!n a perfect vacuum the absolute pressure is#ero!
=auge Pressure $the pressure relative to the ambientmeasuring or atmospheric pressure at measurementpoint!
>elationship between absolute pressure and gaugepressure is5
Atmospheric, Asolute ! "au#eAtmospheric, Asolute ! "au#e
Pressure$Pressure$
atmopheric gaugeabsolute P P P +=
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Fluid Pressure Measurement
Common pressure$ measurement devices$Oourdon gauge and monometer!
:onometer$ '$shaped tube partiall% 4lled
with &uid of known densit%!
:anometer gives the measurement ofpressure in pressure dierence ∆P (P1$P2)!
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Pressure test equipment
:anometer Pressure gauge
2-36
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-eme'at&'e of a %&b%ta"ce !" a a't!c&la' %tate ofagg'egat!o" $%ol!d+ l!&!d+ o' ga%) !% a mea%&'e of t,eae'age !"et!c e"e'g* o%%e%%ed b* t,e %&b%ta"cemolec&le%.
Bome teme'at&'e mea%&'!"g de!ce%: 'e%!%ta"cet,e'momete'+ t,e'moco&le+ *'omete' a"dt,e'momete'.
TemperatureTemperature
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-,e follow!"g 'elat!o"%,! ma* be &%ed toco"e't a teme'at&'e e'e%%ed !" o"edef!"ed %cale &"!t to !t% e&!ale"t !" a"ot,e'<
Temperature%cont$Temperature%cont$
( ) ( )
( ) ( )
( ) ( )
( ) ( ) 32!1
!1
"#!$%&
1%!2#3
+=
=
+=
+=
C T F T
K T RT
F T RT
C T K T
…
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69
…
Temperature inter&alTemperature inter&al
the conversion actor !se" are1.# o$ 1.# oR 1 o$ 1oC
1oC 1 % 1 oR 1%
CAD :
1) F!"d t,e "&mbe' of celc!&% deg'ee% betwee" 32 oF a"d 212 oF
∆- $o;) = $212 E 32) 1o; = 100o;1.8 oF
2. F!"d t,e teme'at&'e of 32oF !" o;
-$oF) = 1.8 -$o;) 32
32 = 1.8 -$o;) 32 = 0 o;
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Consider the interval from 2;Q* to ;Q*
a) Calculate the equivalent temperature inQC and the interval between them!
;33.36.7)$26.6--G-
;26.6;1.8
3280F)$80-
;6.7;1.8
3220F)$20-
1.8
32F)-$;)-$
12
2
1
°=−−=−=
°=°
−=°
°−=°
−=°
−°=°
Try This…
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What ha&e you learn 'romWhat ha&e you learn 'rom
this Chapter(this Chapter(