RA005-Using-Filter-Aids

Some of the fastest growing market segments in the dust collector field are those where wet scrubbers are being replaced and the application of dust collectors to general ventilation systems. Both of these areas involve filtering gas streams where conditions are near the dew point. There is danger that condensed droplets may plug the dust collector media.

While it is not necessary that each engineer have ability to make calculations to determine if dew points are reached by mixtures of hot and cold flows or through temperature drops in ductwork and through the collectors, it is very desirable for him to be able to follow a set of calculations made by the home office. The ability to understand the concepts usually can mean improving customer confidence.

The key to Psychometrics is to know how the tables are made. Once you understand the tables, the curves of a “psychometric chart” become clear. Attached is a condensed set of tables, which are adequate to handle most applications with sufficient accuracy for our requirements.

To begin the development of the chart, refer to Figure 1 and the attached “psych” tables. An analogy considers that we have a flexible bag and put in a pound of "bone dry" air. Next we raise and lower the temperature and measure the volume of the bag. We then place these volumes in column D next to the associated temperature. (As a point of interest, this volume can be calculated, since the air mixture obeys the gas laws.) You will note the volume for dry air for one pound of gas is 13.35 cu. ft. at 70oF and 16.63 cu. ft. at 200oF. This is "specific volume" for air.

Next we place a quantity of water in the bag. We can weight this water. We raise the temperature of the space around the bag and therefore heat the gas and the water in the bag. After a long period of time, the air will become saturated by evaporating water. We can determine the increased volume of the bag and the quantity of water evaporated. Repeat this procedure for each temperature. We now can record the total weight of water evaporated for each temperature and record this weight in column A.

At each temperature a precise amount of water will be evaporated until you reach 212oF and above. Now the bag will expand until all water is evaporated. Any mixture of vapor and air can be produced once the air temperature exceeds 212oF.

vapor

dry air

FIGURE 1

At each temperature below 212oF, there will be a precise new volume larger than the dry air volume. This will be entered in column F and is the volume of air plus water vapor that a pound of air can carry at a specific temperature. You will note the units are: Cubic feet of mixed air, and, vapor per pound of dry air.

Next is the most important concept: How we obtain column E, the volume of the water vapor per pound.

Refer to the table, see line 110oF:

From column D, the volume for one pound of dry air is 14.36 cu. ft.

From column A, the amount of water you can vaporize is 0.0593 pound vapor per pound of air.

From column F, the total volume of the mixture is 15.73 cu. ft. per pound of dry gas (air).

1) Now we will calculate column E:

15.73 cu. ft. mixture

-14.36 cu. ft. air

1.37 cu. ft of vapor / lb of dry air

2) This gives us cubic feet of water vapor per pound as follows:

1.37 cu. ft. / 0.0593 pounds of vapor = 23.02 cu. ft./lb vapor [column E]

3) Now we can determine the density of the saturated mix:

Wt. of dry air + wt. of vapor = 1lb + 0.0593 lb = 0.0673 lb/cu.ft. [column C]

Volume of mixture 15.73 cu. ft

Now we will assume we have 50% relative humidity and we will find the density. That means we have 50% of the maximum humidity the gas (air) can hold.

0.50 x 0.0593 (max humidity) = .02965 lb. vapor/pound of dry air.

4 (a) Find new volumes:

Air = 1 lb. x 14.36 cu. ft./lb = 14.36 cu.ft. [column D @ 110oF]

Mixture = 14.36 + (23.02 x 0.02965) = 15.042 cu.ft.

4 (b) Density = Wt. of dry air + wt. of vapor = 1 + .02965 = 0.0685 lb./cu.ft.

Volume of mixture 15.042

As you might expect, this is a higher density than we determined in (3).

5) Next we lower temperature on the mixture until condensation appears. If we go up column A until the number is less than 0.02965 lb, we will find the condensation point. It is between 85F and 90F (interpolating we get 88F). This is the Dew point.

Next we will consider column H. This is the heat, in btu /lb, needed to raise temperature of dry air from one value to another.

6) You will note that if interpolated the table considers 32F as 0 btu/lb. Values below 32F are negative. We could have chosen a table with 0F or minus 100F as 0 btu/lb. In heating and cooling calculations we consider differences in energy levels so this selection of zero value as a reference is of no concern.

Next refer to column I. This is the amount of heat necessary to vaporize a pound of water at the given temperature. Below 32F it is theoretical since the heat to melt ice is not included.

7) It is relatively easy to develop column G which is the heat contained in a saturated mixture of one pound of air and its associated water vapor. The heat in the air is added to the heat in the vapor.

As a typical problem consider the following conditions pertinent to a foundry shakeout.

Exhaust volume 10,000 CFM

Sand generated 400# /lb of airflow at 8% moisture (relative humidity)

After shakeout sand has 5% moisture

Ambient temperature 60F

Dew point is 50F

Following is required: Will exhaust stream reach dew point?

8) From the chart ambient moisture is 0.0111 lb. vapor/lb dry gas {column A}

Volume of vapor at 60F is 21.05 cu. ft./ lb {column E}

Volume of air at 60F is 13.10 cu.ft./lb {column D}

8a) Volume of mixture per pound of air

13.10 cu. ft/lb + (0.0111 lb vapor./lb dry air. X 21.05 cu.ft.) = 13.33 cu.ft./lb dry gas {column F}

9) 10,000 CFM / 13.33 cu.ft./lb = 750 lb/min of dry gas

Therefore 750 X 0.0077 {column A @ 50F dew point} = 5.75 lb/min dry air

9 (a) Find total moisture in gas stream:

400 lb./min X (8% - 5%) = 12lb./min. moisture liberated from sand into gas stream.

9 (b) Total moisture:

5.75 lb./min + 12 lb./min = 17.75 lb/min

9 (c) Ratio of vapor to air:

17.75 lb.per min / 750 lb. per min = 0.02367 lb vapor/lb. Air

At 60F, the gas can only carry 0.0111 lb. vapor/lb. of air {column A}. This means the air stream will have condensed moisture in it, so it will plug a fabric bag house.

Going down column H until we reach a temperature where the gas stream could carry that much water vapor. This temperature is 85oF. The gas stream must be raised to 85o F to prevent condensation.

10) At 85F the gas can carry 0.0264 lb./vapor/lb. dry air: {column A}

It is assumed that the moisture liberated is in the form of vapor, and, we do not need to vaporize liquid droplets. To accomplish this, the gas stream is heated to 85F by mixing it with a side stream at 250F raised from ambient by a duct heater. The following heat balance is made:

Heat lost to gas stream = Heat gained by gas stream.

Where Wm = lb./min. of main stream

Ws = lb./min. of side stream

Ti = Initial temperature 60F

Tm = Fnal mix temperature 85F

Cp = Heating coefficient for air 0.246 btu/lb deg F

H1 = Heat content of vapor at 60F, 1098.3 btu/lb. {column I}

H2 = Heat content of vapor at 85F, 1087.5 btu/lb {column I}

Total moisture in air stream = 17.75 lb/min {from above}

The moisture content in the side stream can be ignored in this calculation.

11) (Wm X Cp X (Tm-Ti)) + (17.75 lb. X (H2-H1)) = Ws X Cp X (250-Tm)

(750 X 0.246 X 25) + (17.75 X 10.8) = Ws X 0.246 X 165 so Ws= 115 lb/min

12) Volume through side stream at ambient:

115 lb./min. x 13.33 cu. ft. {column F @ 60oF} = 1533 CFM

Heater size: Q = Ws X Cp X (250- T1)

Q = 115 lb./min. X 0.246 X (250-60) = 5,375 btu/min. = 322,500 BTU/HR

Find vapor ratio in mixture:

13 (a) Total moisture: 17.75 lb./min + (0.0076 lb./lb.{column A @ 50oF} X 115 lb/min.) = 18.63 lb./min.

13 (b) Total air: 750 lb./min + 115 lb./min = 865 lb./min

13 (c) Vapor/Air ratio: 18.63 divided by 865 = 0.0215 lb vapor./lb air.

Going down column A, we see that the dew point is less than 80F. By interpolating we can determine it is 79oF.

13 (d) Total air flow at 85oF:

Column D; Air displaces 13.73 cu. ft./lb

Column E; Vapor displaces 22.04 cu.ft./lb.

Volume = (865 lb./min X 13.73 cu. ft./lb) + (18.63 lb./min. X 22.04 cu.ft./lb) = 12,287 CFM

Density = 865 + 18.63 = 0.0700 lb./cu. ft.

12,287

14) Final conditions:

12,287 CFM

Dry bulb 85F

Dew point 79F

Heater size 322,500 btu/hr. or 94.5 Kilowatts

Now we will look at another case (see figure 2)

Air is cooled

500 cfm

dew point ?

90 deg at 70%

relative humidity

Figure 2

Determine if we get condensation

15) First determine maximum humidity in 90oF air. This is in column A or 0.0312 lb. vapor/lb. air:

70% x 0.0311 lb./lb. air = 0.0218 lb. vapor/lb air

From chart column A, we note dew point is below 80F or 79F by interpolation.

15 (a) The volume per pound of air is calculated as follows:

= (1.0 lb air X 13.85 cu. ft./lb {column D @90F}) + (0.0218 X 22.21 cu.ft./lb {column E @90F})

= 14.33 cu.ft./lb. of air

1 TON or 12,000 BTU/HR

AIR CONDITIONER

15 (b) air content = 500 ACFM divide by 14.33 cu.ft./lb. dry air = 34.89 lb./min

vapor content = 0.0218 lb. vapor./lb. dry air X 34.89 = 0.76 lb./min

16) Heat Content:

(34.89 lb./min X 13.94 btu/lb. {column H @90F) + (0.76 lb./min X 1101 btu./lb.{column I @ 90F)

= 1323 btu./min

Find heat removed and heat left:

12,000 btu/hr. divide 60 min./hr. = 200 btu/min.

17 (a) 1323 btu/min. - 200 btu/min = 1123 btu./min

18) Find btu/lb. for cooled gas:

1123 btu./min divide by 34.89lb./min = 32.19 btu./lb

If we go to column G to locate saturated temperature that has this heat content, we will find it is higher than 75F and less than 80F. Interpolating we get 77oF. This is below previously determined dew point of 79oF. This means we have condensation in the mixture.

Once you understand the columns, solving problems is a matter of simple algebra and no worse than a tax return.

AVAILABLE EQUIPMENT AND SERVICES

Free engineering consulting with the Dust Cleaning Institute Staff

Gary Berwick, P.Eng., Director QAM Dust Control Institute

Phone (800) 267-5585, Fax (866) 899-8954

Address: 240 Camille Cres, Waterloo, Ontario, N2K3B7

e-mail : garyb@qamanage.com

Lesson 6 Review

1) The Psychometric tables contain data about the following

( ) Properties of gas and water vapor at various temperatures

( ) Heat content of air and water vapor at various temperatures

( ) Are properties for atmospheric pressures at seal level

( ) All of the above

2) The Psychometric tables can help determine following:

( ) Dew point temperatures

( ) Water vapor content at different dew points

( ) Both of the above

3) Below 212 degrees temperature column A lists the weight of moisture at saturation that a pound of dry air can hold without condensation forming:

( ) True

( ) False

4) Column H lists the heat content of the water vapor in BTU per pound of steam or vapor. The heat needed to form the steam from liquid is included in this number:

( ) True

( ) False

If we mix 0.0158 pounds of water vapor at 70 degrees and 1 lb of dry air:

The volume of vapor is:

( ) 0.0158 lb / lb air x 21.45 cu. Ft. / lb =0 .34 cu. Ft. / lb air

( ) 0.0158 cu. Ft. x 13.35 cu. Ft./ lb = 0.211 cu. ft.

6) In the conditions described in (5) what is the total gas volume when we have one pound of air and 0.0 158 lb of water vapor.

( ) 13.35 cu. ft. + 0.34 cu. ft. = 13.69 cu. ft.

( ) 13.35 cu. ft.

7) Air density in examples (5) and (6) is:

( ) 13.69 cu. ft divided by (1.00 + 0.0158)

( ) From table under saturated density in #/cu.ft.

( ) 0.0742 #/cubic foot

( ) All of the above:

8) If the collector contains 0.318 pounds of vapor and 20 lbs of dry gas at 90 degrees and the collector is placed outdoors at 32 degrees. What approximate weight will condense inside the collector overnight:

( ) 0.300 lbs or 5 ounces.

( ) None

( ) Less than 2 ounces