GENERAL

A plant loses about 3% of its production rate per 1000 feet of elevation due to the thinning of the air.

Plants are designed to remove 5% moisture. The higher the moisture, the lower the production rate.

To Calculate A Batch Plant’s Rate Of Production:

• Choose a weigh point on your aggregate scale (must be in a bin you are waiting on so that the pointer is rising slowly)
• With a stopwatch, determine the amount of time it takes the scales to return to this point – Time in seconds=(s)
• Divide (s) into the number of seconds per hour (3600)
• Multiply that answer by the average batch size (b)
• Assume the following: s=45seconds, b=5000# or 2.5 tons
• EXAMPLE: 3600 div. 45= 80 batches per hour
• 80x 2.5 = 200tph

To Calculate A Drum Plant’s Rate Of Production:

• Multiply the aggregate tons per hour (x) times 100% minus the moisture content (y)
• Add the oil tons per hour (z)
• Assume the following: x= 209, y= 4.3%, z= 11.76
• EXAMPLE: 209x .957= 200.01
• 200.01+ 11.76= 211.77tph

To Convert Horsepower To KW:

• Multiply horsepower (h) by .7457
• Assume the following: h= 350hp
• EXAMPLE: 350x .7457= 261 KW

To Calculate The Length Of A Slat Conveyor At A 45 Degree Angle:

• Measure height (h), multiply by 1.41416
• Assume the following: h= 65′
• EXAMPLE: 65x 1.41416= 91.9204 feet long slat at 45 degrees

To Calculate The Speed In Feet-Per-Minute Of A Slat Conveyor, The Following Information Is Needed:

• Motor RPM- (rpm)
• Motor sheave size- (ms)
• Driven sheave size- (ds)
• Gear box ratio- (gbr)
• Head sprocket size- (hs)
• Assume the following: Rpm= 1755, ms= 8″, ds= 14″, gbr= 25.13 to 1 and hs= 9″
• Calculate the sheave ratio, i.e.: 8″ drive, 14″ driven
• Multiply motor speed by the answer = input RPM
• Input RPM divided by gear box ratio x diameter of head pulley x .2618= slat speed in Feet Per Minute
• EXAMPLE: 8 divided by 14 = .571 x 1755 = 1002.86 input rpm div. 25.13 gbr = 39.91 head shaft rpm 39.91 x 9 = 359.19 x .2618 = 94.04 FPM

ASPHALT OIL

To Find Asphalt % By Total Mix:

• Add oil to aggregate then divide oil by total of the two
• Assume the following: 250 tons aggregate used, 15 tons oil used
• Example: 250agg + 15oil = 265 15 div 265 = .0566% oil by total

To Calculate Where You Should Be Running With A Known Aggregate Tph:

• TPH of dry aggregate x % of oil, div. 100-% of oil = oil tph
• Assume the following: 70tph aggregate (dry), 5.4% oil
• Example: 100% – 5.4% = 94.6 70 tph x 5.4% = 378 div 94.6 = 3.99 tph of oil

To Find The Number Of Gallons Delivered On A Particular BILL OF LADING:

• Net weight in pounds (x), divide 8.328 (pounds per gallon of water, which has a specific gravity of 1.0) = y
• Divide y by the specific gravity (z) listed on the delivery ticket for the oil you are using
• Assume the following: x= 66,920, z= 1.0273
• Example: 66,920 divide 8.328= 8,035.54 (y)
• 8,035.54 divide 1.0273 (z)= 7,822 gallons at 60 degrees Fahrenheit

To Find Asphalt Gallons Per Ton:

• Look up AC pounds per gallon at the oil’s temperature (x)
• Divide that number into 2000
• Assume the following: x = 7.81
• Example: 2000 div. 7.81 = 256.08 gallons AC per ton at the chosen temperature

To Convert Gallons Of Oil To Tons:

• Look up AC pounds per gallon at the oil’s temperature (x)
• Multiply that number by number of gallons
• Divide that number by 2000
• Assume the following: x = 7.81
• Example: 7.81 ppg x 25,000 gals = 195,250# div
• 2000 = 97.625 tons of oil at (x) temperature

To Find Out How Many Tons Of Oil Are In An AC Tank:

• Check oil temperature (t)
• Measure liquid, convert inches to gallons (g)
• Find temperature on compensation chart and get conversation factor (f)
• Read oil pounds-per-gallon* at 60 degrees (p)
• Assume: t= 325 degrees
• g= 12,834 gallons, f= .9105, p= 8.456
• Example: 8.456 x .9105= 7.699ppg at 325 degrees
• 7.699ppg x 12,834gals = 98,811.37 pounds
• 98,811.37 # divide 2,000 = 49.41tons of oil

To Figure Out How Much Mix You Can Make With A Given Quantity Of Oil:

• Tons oil divided by percent of oil in mix = yield
• Example: 28 tons div. .055%= 509.09 tons of hot mix produced

To Find Pounds-Per-Gallon Of Any Oil:

• Find specific gravity listed on ‘bill-of-lading’ (g)
• Multiply pounds-per-gallon of water (8.33)* by the specific gravity (g) of the oil
• Assume: g= 1.0273 {actual reading for Oregon PBA-2} w= 8.33
• Example: 8.33 x 1.0273= 8.56pounds-per-gallon for the oil at 60 degrees

* Actual ppg of water

To Find The US Gallons Capacity Of A Cylindrical Tank:

• Measure the diameter (d) and the length (l)
• Square the diameter, multiply by length, then multiply by .0034
• Assume: d= 95″, l= 337″
• Example: 95x 95x 337x .0034= 10,341 gallons

BAGHOUSE

It requires approximately 165 CFM of air to produce 1 ton of hot mix.

To Find The Correct Baghouse Size For A Particular Drum:

• Radius x radius x 3.142 (pi) x *1000= CFM required
• Example: Assume a 9 foot drum. 4.5 x 4.5 = 20.25 x 3.142 = 63.6255 x 1000 = 63,625.5 CFM
• This drum would require a 65,000 CFM baghouse

*1000 equals the air velocity through the dryer/mixer. This figure is usually down around 750 to 850fpm, but using 1000fpm give a little bit of oversizing which is a good thing since you don’t want the air flow through the baghouse to limit production rates.

To Find The Square Feet Of Cloth In A Particular Baghouse, You Need To Know The The Diameter Of Bag (D), Length Of Bags (L), And Number Of Bags (N):

• The formula is: D x 3.142 = A
• A div 12 = B
• B x L = SF (sq. ft. per bag)
• SF x N = total sq. ft. of cloth in baghouse

To Find Air To Cloth Ratio (Should Be 5.1 To 1 In Batch, And 5.5 To 1 In Drum Plants)

• Determine square footage of an individual bag (X)
• Multiply that by the number of bags (Y)
• Divide that answer by baghouse CFM (Z). X times Y = total square feet of cloth (sq/ft). Z divide sq/ft = air to cloth ratio

Most manufacturers figure drum air velocity at between 900 & 1000 FPM.

CONVEYORS

To Find The Speed Of The Conveyor:

• Calculate the sheave ratio, i.e.: 10″ drive, 18″ driven
• 10 divided by 18 = .555 Multiply motor speed by the answer = input RPM
• Input RPM divided by gear box ratio x diameter of head pulley x .2618= belt speed in FPM
• Example: Assume an 1800 RPM motor, a 25-1 gear box and an 28″ head pulley: 1800 x .555= 999 divide 25 = 39.96 x 28= 1118.88 x .2618= 292.92 FPM

To Zero Ramsey 10-201 Belt Scales (Total Tons In Lifetime)

• Press: Set-up – 76 – Enter
• 1853 – Enter – Enter

PUMPS

• HL225 .01898 gallons per revolution
• K225 .1055 gallons per revolution
• L225 .23 gallons per revolution
• 2″Q34 .23 gallons per revolution
• 3″Q34 .61 gallons per revolution