Process and Process Variables

7/28/2019 Process and Process Variables

1/19


2/19

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.


3/19

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.
http://www.animationfactory.com/animations/web_text_e_g/graph_down/da503/http://www.animationfactory.com/animations/people_m_z/miscellaneous/100270/


4/19

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.


5/19

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:


6/19

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:


7/19

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


8/19

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.


9/19

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.


10/19

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


11/19

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


12/19

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.


13/19

Figure: Types of manometers

P2

=Patm

Manometer fluid

P1

(a) Open-end


14/19

P1 P2

(b) Differential


15/19

P2

=0

P1

(c) Sealed-end

Figure 3.4-4 Manometers


16/19

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


17/19

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


18/19

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


19/19

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!!!!

Documents

Process and Process Variables