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7/28/2019 Process and Process Variables
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Upon completion of this presentation, you should be ableto:
Calculate the mass/or volume given the density andvolume/or mass.
Correlate mass and volumetric flow rates.
Identify some flow meters. Calculate mass fractions from mol fractions and vice
versa.
Calculate the average molecular weight of a mixture.
Define pressure, temperature and their units.
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Mass: is a measure of the amount of a matter in the body
Mass and volume are correlated via density according to the
equation:
Density r = mass/volume (m/v)
Specific volume = 1/density
Specific gravity (SG) is the ratio of the density of a substance to
a reference density; usually taken the density of water at 4 0C,
which equals to 1g/cm3, thus
SG (dimensionless) r / r of water at 4 0C.
Mass and volumetric flow rates (mass or volume per time) are
correlated via density.
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1 mole of A or 1 gmol of A is the mass (g) of 6.02 x10-23 molecules of that species.
The molecular weight of a compound is the sum ofthe atomic weights of the atoms that constitute a
molecule of the compound. If the molecular weight of a substance is M, then
there are M kg/kmol, M g/mol, and M lbm /lb-mol ofthis substance.
The same factors used to convert masses from oneunit to another may be used to convert theequivalent molar units: there is 454 g/ lbm forexample, and therefore there is 454 mol/ lbmol.
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Molecular weight of CO2:
M.W. = 1* atomic weight of C + 2*atomic weight of O
= 1*12+2*16=44 g/molMol of CO2: = 100 g CO2* mol CO2/44 g CO2= 2.273 mol
lb-mol of CO2 : = 2.273 mol CO2*1 lb-mol/454 mol
= 0.005 lb-mol CO2
Mole of C: = 2.273 mol CO2*(1 mol C/mol CO2}
= 2.273 moles C
Example: Consider 100 g of
CO2
, Calculate:
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Mole of C = 2.273 mol CO2*(1 mol C/mol CO2}= 2.273 mol C
Mole of O = 2.273 mol CO2*(2 mol O/mol CO2}= 4.546 mol O
Mass of O2: = 2.273 mol CO2*(1 mol O2/mol CO2}*
(32 g O2/mol O2) = 72.74 g O2.
Practice: Try to find the mass of O using the moles ofO2.
Example: Consider 100 g of
CO2
, Calculate:
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Mole Fraction, Mass Fraction and Average
Molecular Weight
The mass fraction of a component in a mixture is the ratio of themass of this component to the total mass of the mixture; xA =
mass of A/total mass of mixture
The mole fraction of a component in a mixture is the ratio of the
moles of this component to the total moles of the mixture; yA =moles of A/total moles of mixture
The average molecular weight of the mixture can be calculated
using:
Mw
x
wM
1
orMwywM
i i
i
i
ii
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Example: Consider a mixture of gaseswith the following composition:
Gas mass %
O2 16
N2 63
CO2 17
CO 4
What is the molar composition and the average molecular weight ofthe gas mixture?
Solution: The steps to solve such problems are
PICK a Basis:100 g of the gas mixture.NOTE that solution will not
depend on the basis you pick!!! Calculate the mass of each gas in the mixture.
Calculate the number of moles of each gas in the mixture.
Calculate the total number of moles of the gas mixture.
Calculate the mol fraction of each component.
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Lets now start solving the problem
The mass of each gas = total mass*massfraction
For example: Mass of O2 = 100*0.16=16 g
Number of moles = mass /molecular weight
For example Moles of O2 = 16 g O2 *(1 mol O2 /32g O2) = 0.5 mol O2
Following same procedure for other gases, theresults are summarized in the table below
Total number of moles = moles of all gases.xi= number of moles of i/total number of
moles.
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Table 1: Solution of example 2
Total 100 % 100 g 3.279 mol 1
Average molecular weight = yi(Mwi) or = 1/(xi/(Mwi))
= 0.15*(32)+0.69*(28)+0.12*(44)+0.04*(28) =30.5 g/mol
=1/[(0.16/32)+(0.63/28)+(0.17/44)+(0.04/28)] = 30.5 g/mol
Gas Mass
%
Mass
(g)
Moles Mole
fraction
O2 16 16 0.5 0.15
N2 63 63 2.25 0.69
CO2 17 17 0.386 0.12
CO 4 4 0.143 0.04
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Pressure
Pressure is the ratio of force to area on which the force isacting.
Units of pressure: N/m2 (Pascal), or lb/in2 (psi).
Hydrostatic (static) pressure Po (atmospheric pressure)
Consider a vessel of area A that Area Acontains a liquid density ofr at a
height of h.
Recall from physics that the force
at the bottom of the tank is given byFbottom = Ftop + Fwt. of fluid , divide by A
(Fbottom /A) = (Ftop/A) + (Fwt. of fluid/A), or
(Fbottom /A) = (Ftop/A) + (Ahrg/gcA), then
Pbottom = Po + (hrg/gc)
column
cf fluid
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Absolute pressure: actual pressure
Gauge pressure: pressure relative to atmospheric pressure.
Pabsolute= Pgauge+ Patmospheric
If absolute pressure = zero perfect vacuum!
If gauge pressure = zero, pressure = atmospheric
pressure.
Manometers: are used to measure the pressure up to 3 atm.
There are three types of manometers:Open-end
manometers, Differential manometers, and Sealed-end
manometers. These manometers are shown in Figure
below.
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Figure: Types of manometers
P2
=Patm
Manometer fluid
P1
(a) Open-end
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P1 P2
(b) Differential
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P2
=0
P1
(c) Sealed-end
Figure 3.4-4 Manometers
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General Manometer Equation
Consider a manometer that has a fluid of densityrf and isused to measure P1 and P2, as shown in the figure below.
P1 P2
Manometer fluid
Density f
Fluid 2Density 2
(a) (b)
Fluid 1
Density 1
d1
h
d2
Figure 3.4-5 Manometer variables
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Notice that at same horizontal levels, the pressure will bethe same no matter how it is measured. Then
P1 + r1 (g/gc)d1 =
P2 + r2 (g/gc)d2 + rf(g/gc)h
Special Cases:
Ifr1 = r2 =r(differential
Manometer), then
P1
P2 = (rf-r) (g/gc)hIf fluids 1 & 2 are gases, then
P1P2 = rf (g/gc)h
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Temperature: is a measure of the average kinetic energy
possessed by the substance molecules.
Temperature can be measured using the centigrade scale(Celsius); oC, or the Fahrenheit scale; oF.
The centigrade scale is based on the fact that at 1 atm, the
boiling point of water is 100 oC, and its freezing point is 0
oC, while according to the Fahrenheit scale, they equal to212 and 32 oF, respectively.
Kelvin (K) scale is the same as the centigrade scale except
for the reference of zero.
Rankine (R) scale is the same as the Fahrenheit scaleexcept for the reference of zero.
Temperature can be converted from a scale to another
scale using the following correlations:
T(K) T( C) 273 15
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T(K) = T(oC) + 273.15
T(R) = T(oF) + 459.67
T(R) = 1.8 T(K)
T(o
F) = 1.8 T(o
C) + 32 (Can you derive this relation?!)Note the difference between the temperature and temperature
difference!!!!
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