introduction to air pollution - workbook

CONVERSION OF UNITS

Problem :

An SO2 concentration is given as 830mg/m3 at 25°C and 1 atm. Express this concentration in parts per

million (ppm).

Solution :

Step 1

Concentration of SO2 is 830 mg/m3

AT STP conditions (25°C and 1 atm), one mole of gas occupies 24.5 L

(V=nRT/P).Molecular Weight of SO

2 is 64g/mol

Step 2

ppm =

concentration(10 mg/m3) x 24.5 L/mol

molecular weight(g/mol) x 1000 (mL/m3*g)

Step 3

ppm =

830 (mg/m3) x 24.5 L/mol

64 (g/1 mol SO2) x 1000 (mL/m3*g)

= 0.32 ppm

Exercise: Carbon Monoxide concentration at 90°C and 6 atm is 90 mg/m3. Express this concentration in ppm.(0.016 ppm)

Problem :The exhaust from a 1981 Honda contains 1.5% by volume of carbon monoxide. Compute the concentration

of CO in milligrams/m3 at 25°C and 1 atm of pressure.

Solution :

Step 1

1 percent by volume = 104 ppm.

Introduction to Air Pollution - Workbook http://www.eng.utoledo.edu/~akumar/IAP1/aprg_remains/workbook/cha...

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1.5 percent by volume = 1.5*104 ppm.Molecular Weight of CO is 28 g/mol

Step 2

concentration (mg/m3) =

ppm x molecular weight(g/mol) x 1000 (mg*L/m3*g)

24.5 (L/mol)

Step 3

concentration (mg/m3) =

1.5 x 104 ppm x 28(g/mol) x 1000 (mg*L/m3*g)

24.5 (L/mol)

= 1.71 x 107 ug/m3 = 1.71 x 104 mg/m3

Exercise: An exhaust gas containing 3.2 percent by volume SO2 is released at 25°C and 1 atm. Compute the

concentration in mg/m3.

(8.36 x 104 mg/m3)

Weight Fraction, Mole Fraction, and Average Molecular Weight

BASIC EQUATIONS :

Density : mass

volume

Specific gravity : density of a given substance

density of water =

density of a given substance

1 g/cm3

For example Specific gravity of methanol is given as 0.91. Therefore density of methanol is 0.91 multiplied

by density of water = 1 g/cm3.Supposing a mixture has A, B, and C as its components. Then,

Weight fraction of A = ( Wt. of A ) / ( Total Wt. )Mole fraction of A = ( Moles of A ) / ( Total moles )Number of moles of A = ( Wt. of A ) / ( Mol. Wt of A )Total Wt = Wt(A) + Wt(B) + Wt(C).Total moles = moles

A + moles

B + moles

C.

Problem :Determine the mole fraction, weight fraction and average molecular weight of all the components in a


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mixture, if contains 20 lbs of O2, 30 lbs of NO2 and 10 lbs of SO2.

Solution :Step 1Find the molecular weight(mol.wt.) of each gas.mol.wt. of O2 = 32g/mole.

mol.wt. of SO2 = 64g/mole.

mol.wt. of NO2 = 46g/mole.

Step 2Determine the number of moles of each gas.

moles of O2 = 20 lbs / (32g * 10-3kg/g * 2.2 lb/kg) = 284.1

moles. moles of NO2 = 30 lbs / (46g * 10-3kg/g * 2.2 lb/kg) = 296.4 moles.

moles of SO2 = 10 lbs / (64g * 10-3kg/g * 2.2 lb/kg) = 71 moles.

Total moles = 284.1 + 296.4 + 71 = 651.1.Step 3Determine mole fractions.mole fraction O

2 = 284.1 / 651.5. = 0.436.

mole fraction NO2 = 296.4 / 651.5. = 0.455.

mole fraction SO2 = 71.0 / 651.5 = 0.109.

Step 4Determine the weight fractions.Wt. fraction. O

2 = 20/(20+30+10) = 2/6.

Wt. fraction. NO2 = 30/60. = 1/2

Wt. fraction. SO2 = 10/60 = 1/6

Step5Determine the average mol.wt. of mixture.Average mol. wt. = (mol.wt of A * mole fraction of A) + (mol.wt of B * mole fraction of B) + (mol.wt of C *mole fraction of C).Avg. mol.wt. = 32*0.436 + 46*0.455+64 *0.109 = 41.858 g/mole.

Exercise: For a mixture equal weights of O2 and SO2, determine the mole fractions and weight fraction of

each component and the average molecular weight of the mixture.(O2: 0.67 mol frac., 0.5 weigh frac.;

SO2: 0.33 mol frac., 0.5 weigh frac.;

average molecular weight: 42.6 g/mol)

ATMOSPHERIC PRESSUREDefinition


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Atmospheric pressure is the pressure exerted by the atmosphere on the liquid surface. It

is equal to 101.3 KPa absolute (760mm of mercury or 2116 lbf/ft2).

Absolute pressure :Gage pressure + Atmospheric pressure.

Problem :Determine the pressure, both absolute and gauge, exerted at the bottom of the column of liquid, with 1mheight and with a density of 1200 Kg/m3.Solution :Pgage = Density*accel. due to gravity *height of liquid column. = 1200 Kg/m3 * 9.8 m/s2 * 1 m = 11.760 N/m2

Pabsolute = 101.3 + 11.760 = 113.06.KPa.

Exercise: Determine the absolute and gauge pressure exerted at the bottom of the column of liquid, with 11

meters high and with a density of 1 Kg/m3.(P

gauge = 0.107 kPa, P

absolute = 101.407 kPa)

GAS LAWFormula :

P V = n R Twhere,

P: Absolute PressureV : Volume occupied.T : Absolute temperature.n : Number of molesR : Gas law constant. = 1.986 cal/gmole-K.

Absolute temperatures ( Rankine or Kelvin) are used in the ideal gas law.Temperature in F + 460 = Temperature in Rankine.Temperature in C + 273 = Temperature in Kelvin.Rewriting ideal gas law in terms of density,Density = mass/volume = n * mol.wt. of gas / Volume.where,

n = number of moles = (P*V ) / ( R*T).Density = P*mol.wt. / ( R * T)By combining Boyle's law and Charles law, we get

{ q * P / T } a = { q * P / T} swhere,


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q :Volumetric flow rateP : Absolute pressure.T :Absolute temperature.s : Standard conditionsa : Actual conditions.

Problem :Calculate the density of a gas whose molecular weight is 29 at 1 atm,absolute and 80°F.Solution :80 F = 80 + 460 = 560 Rdensity = P * mol.wt/RTR = 0.73 atm-ft3 /lb mol-R.density = 1*29/0.73*540 = 0.0735 lb/ft3.

Exercise: Calculate the density of the same gas at 3 atm and 212°F

(0.1773 lb/ft3)

Problem :Determine the actual volumetric flow rate in acfm assuming that pressure is constant, when the actualtemperature is 700 F. The standard conditions are 70 F and 3000 cfm.Solution :Temperaturestd = 70 F = 530 R.

Temperatureact

= 700 F = 1160 R.

qact = qstd

*(Tempact /

Tempstd

).

= 3000*(1160 / 530). = 6566.03 acfm.

Exercise: Determine the actual volumetric flow rate of the above furnace's releases at 1 atm and 65°F.(3268.86 L/hr)

Problem :

For an SO2 emission of 22000 kg/day and an exhaust gas flow rate of 5.0 million m3/hr (after the scrubber)

measured at 150°C and 1 atm of pressure, calculate the concentration (ppm) of SO2 in the exhaust gases.

Solution:Step 1 :

n = 22000kg

dayx 1000

g

kgx

g-mol

64 g

= 343750g-mol

day

Step 2:


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VSO2 = nRT

P =

343750 (g*mol/day) x 0.08206 (L*atm/gmol*°K) x (150+273)°K

1 atm

= 11932037 L/day

Step 3:

Vexhaust gases

= (5.0 x 106 m3/hour x 1000L/m3 x 24 hours/1 day)

= 1.2 x 1011 L/day

Step 4:For 1 day,

total volume = 1.2 x 1011 L

volume of SO2 = 1.1932037 x 107 L

Concentration SO2 =

VSO2

Vexhaust gases

= 1.1932037 x 107 L

1.2 x 1011 Lx 106 ppm

= 99 ppm

Exercise: Calculate the concentration of SO2 if the above measurements were made at 400°C and 2.5 atm.

(63 ppm)

STACK DISCHARGE VELOCITYDefinition :Stack discharge velocity is the velocity at which the exhaust gas from the process is discharged to theatmosphere.Problem :Determine the stack discharge velocity at 320°F if the standard conditions for the exhaust are 2000 scfm,60°F and 1 atm. The diameter of the stack is found to be 1.2 ft.Solution:Actual volumetric flow rates are always used to calculate stack discharge velocity.Pstd/Pact is usually equal to 1 as the gas is discharged into the atmosphere.

Step 1 :Calculate the qactual

Formula :qact = qstd * {Tact*Tstd} at constant pressure.Tact = 320 + 460 = 780 RTstd = 60 + 460 = 520 Rqstd = 2000 scfm.qact = 2000 * 780/520. = 3000 acfm.

Step 2 :


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Calculate crossectional area of the stack.

A = PI* D2/4.

=Pi * 1.22/4. = 1.130 ft2

Step 3 :Calculate discharge velocity.Discharge vel. = q

act / A

= 3000 / 1.13 = 2654.86 ft / min. = 44.2 ft / sec.

Note : The velocity calculated in the above problem is average velocity.

Exercise: Determine the stack discharge velocity at 125°F for a 2.4 ft diameter stack with the same standardexhaust conditions.(8.3 ft/sec)

REYNOLDS NUMBER (Re)Definition :Reynolds Number is the ratio of the inertial forces to viscous forces acting on a fluid. It provides informationabout the flow behavior of the fluid, i.e. whether the flow is laminar or transitional. It is defined:

Re = D*V*r

m=

D*V

u

D = diameter of duct

V = average velocity of fluid

m = viscosity of fluid

r = density of fluid

u = kinematic viscosity of liquid

In general,

laminar flow Re < 2000

transition region 2000 < Re < 4000

turbulent flow Re > 4000

although this varies from case to case

Problem :Determine the Reynolds number and the flow behavior for a gas flowing through a 1.7 ft diameter duct at a

velocity of 25 ft/sec. The density of the gas is 0.075 lb/ft3 and the viscosity is 1.16 x 10-5 lb/ft-s.Solution:Step 1:


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Re = D*V*r

m=

1.7 ft x 25 ft/sec x 0.075 lb/ft3

1.16 x 10-5 lb/ft*sec

= 2.75 x 105

Step 2:

2.75 x 105 > 4000 therefore flow is turbulent

Exercise: At what velocity would the above gas need to achieve laminar flow through the duct?(0.18 ft/sec)

REQUIRED HEAT RATEThe amount of heat required to a fluid which undergoes a temperature change can be defined as follows:

heat transferred = mCpDT

where, m = mass of fluid

Cp = heat capacity of fluid

DT = change in temperature

Furthermore, the rate of heat transfer, Q, can be defined:

Q = mC

pDT

t

where, t = time

Problem :Determine the rate of heat transfer if the mass flow rate of the gas stream is 1400 lb/min, when thetemperature of the gas is raised from 200°F to 1000°F. Cp is given as 0.26 Btu/lb-°F.

Solution:

Q = 1400 lb x 0.26 Btu/lb-°F x (1000°F-200°F)

1 min

= 2.912 x 105 Btu/min

Exercise: What would be the heat transferred over a period of one hour and 15 minutes?

(2.184 x 107 Btu)

GROSS HEATING VALUE (HVG)

Definition :Gross heating value of a mixture is given the sum of the heating value of each component multiplied by thatcomponents mole fraction, or

HVG = S(xi)(HVi)

where, xi = mole fraction of ith component

HVi = heat value of ith component


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Problem :Determine the HVG of a mixture which comprises of N2, CH4 C2H6, C3H6, and C4H10 with respective mole

fractions of 0.0515, 0.811, 0.0967, 0.0351, and 0.0056. Use the following data

Gas HVG (Btu/s-ft3)

N2 0

CH4 1013

C2H6 1700

C3H

6 1920

C4H

10 3218

Solution:HV

G = S(x

i)(HV

i)

= (0.0515 x 0)+(0.811 x 1013)+(0.0967 x 1700)+(0.0351 x 1920)+(0.0056 x 3218)

= 1071.35 (Btu/s-ft3)

Exercise: Find the HVG for the mixture if there were no nitrogen present.

(1129.55 (BTU/s-ft3))

HENRY'S LAWDefinition :Henry's law states that the partial pressure of the solute in equilibrium in solution is proportional to molefraction

pg = H*x

g

where, pg = partial pressure of gas (atm.)

H = Henry's las constant (atm/mole fraction)

xg = mole fraction of gas in solution

(Note: H is a function of temperature)

For and ideal gas,

pg = yg*P

where, Pg = total pressure

yg = mole fraction of gas in gaseous mixture

Problem :Determine the maximum mole fraction of SO2 that can be dissolved in a solution to exert a partial pressure of

0.01 atm at 80°F. Henry's law constant = 510 atm/mole fraction @ 1 atm, 80°FSolution:

xg = P

g

H =

0.01 atm

510 (atm/mole fraction)


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= 1.96 x 10-5

Exercise: Find the pressure exerted by the solution when the mole fraction of SO2 is 1.0.

(510 atm)

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Drop in your comments and suggestions to

Ashok Kumar, [email protected] Sud, [email protected]


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introduction to air pollution - workbook

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