Question

MODULE 5: ASSIGNMENT ENVM 590: AIR POLLUTION CONTROL. Module 5: Particulate Control 1. For a given dust loading initially and two identical collectors in series, why is the overall collection efficiency for the second usually less than that of the first collector? 2. An industrial furnace burns No. 4 residual oil with a heating value of 135,000 Btu/gal. Based on the emission factors given in Table 5-4, is a particulate collector required to meet a state emission standard of 0.10 lb/106 Btu? 3. Two particulate collectors are in series. The fractional efficiency for size dp in the upstream device is 80 percent, and for the downstream device the efficiency is 60 percent. Find the overall removal efficiency for size dp. 4. What is the required minimum efficiency of an electrostatic precipitator used as the second collector, when it is preceded by a cyclone separator having an efficiency of 80% if the desired overall efficiency is 98 %? 5. A spreader stoker is used to burn coal containing 8 per cent ash. The flue gas from burning 1 lb of coal is 170 ft3. What is the maximum dust loading of PM in combustion gas in grains per cubic foot?

Answer 1

1)

we know that

the dust collection is mainly through the force of inertia

now

higher the mass , higher the inertia

because of this the heavy particles remain in the first collector

so

this reduces the efficiency of the second dust collector

2)

now

according to the Table 5-4

for No.4 oil

PM value is 7 lb / 1000 gal oil burned

given that

heating value is 135000 Btu / gal

so

in this case

( 7 / 1000) / 135000 = 5.185 x 10-8

= 0.05185 lb / 10^6 Btu

now

the given standard is 0.1 lb / 10^6 Btu

so

the value obtained is within the standard

so

a particulate collector is not required

3)

we know that

for a series

overall removal efficiency = 1 - [ (1-n1) (1-n2) (1-n3) ---- (1-nz) ]

in this case

given that

n1 = 0.8

n2 = 0.6

so

using those values

we get

overall removal efficiency = ( 1 - [ (1-0.8) (1-0.6) ] )

overall removal efficiency= ( 1- 0.08)

overall removal efficiency = 0.92

so

the overall removal efficiency is 92 %

4)

now

overall efficiency = 1 - [ (1-n1) (1-n2) (1-n3) ---- (1-nz) ]

in this case

overll efficiency = 1 - [(1-n1) (1-n2) ]

given

overall efficiency = 0.98

n1 = 0.8

so

0.98 = 1 - [ ( 1- 0.8) ( 1-n2) ]

0.98 = 1 - [ 0.2 ( 1-n2) ]

0.2 x (1-n2) = 1 - 0.98

0.2 x ( 1-n2) = 0.02

1 - n2 = 0.1

n2 = 1- 0.1

n2 = 0.9

so

the required minimum efficiency is 90 %

5)

given a spreader stoker

in general

emission factor of a spreader stoker = 66 lb / ton coal

now

we know that

1 ton = 2000 lb

so

emission factor of a spreader stoker = 66 lb / 2000 lb coal

emission factor of a spreader stoker = 0.033 lb / 1 lb coal

now

given that

170 ft3 of flue gas is obtained by burning 1 lb of coal

also

emission factor = 0.033 lb / 1 lb coal

so

amount of PM = 1 x 0.033

amount of PM = 0.033 lb

we know that

1 lb = 7000 grains

so

amount of PM = 0.033 x 7000

amount of PM = 231 grains

now

maximum dust = amount of PM / amount of flue gas

maximum dust = 231 / 170

maxiumum dust = 1.3588 grains / ft3