# 23 Practice Problems

Published on November 7, 2007

Author: Flemel

Source: authorstream.com

Forestry Problem Index (click on appropriate number below):  Forestry Problem Index (click on appropriate number below) Forestry #1 * Forestry #2 * Forestry #3 * Forestry #4 * Forestry #5 * Forestry #6 * Forestry #7 * Forestry #8 * Forestry #9 * Forestry #10 * Forestry #11 * More #1 * More #2 * More #3 * More #4 * More #5 * Go To Right-of-Way or Aquatic Problems Index * Forestry Problem Index (continued) Click on the appropriate number:  Forestry Problem Index (continued) Click on the appropriate number More #6 * More #7 * More #8 * More #9 * More #10 * And more #19 * And more #20 * And more #21 * And more #22 * And more #23 * Extra #a * Extra #b * ROW & Aquatic Problem Index (Click on desired problem):  ROW & Aquatic Problem Index (Click on desired problem) Right of way #1 * Right of way #2 * Right of way #3 * Right of way #4 * Right of way #5 * Right of way #6 * Right of way #7 * Right of way #8 * Right of way #a * Aquatics #1 * Aquatics #2 * Aquatics #3 * Aquatics #4 * Aquatics #5 * Aquatics #6 * Aquatics #7 * Go To Forestry Problem Index * Forestry problem #1:  Forestry problem #1 FORESTRY PROB #1:  FORESTRY PROB #1 How much Accord is needed? 3 gallons 3% solution FORESTRY PROB #1 1/2 GO TO FORESTRY PROBLEM INDEX:  GO TO FORESTRY PROBLEM INDEX 3 gallons x 0.03 = .09 gallon However it is easier to measure this as liquid ounces-- so… 0.09 gal x 128 liq oz / gal = 12 liq oz FORESTRY PROB #1 2/2 Forestry Problem #2:  Forestry Problem #2 FORESTRY PROB #2:  FORESTRY PROB #2 400 gallon tank 20 gal/ac rate 3 lb ai/ac 70% ai product FORESTRY PROB #2 1/4 FORESTRY PROB #2 2/4:  Treated acres per tank 400 gal / tank / 20 gal / ac = 20 ac / tank FORESTRY PROB #2 2/4 FORESTRY PROB #2 3/4:  FORESTRY PROB #2 3/4 Treated acres per tank 20 ac / tank Pounds active ingredient needed 20 ac x 3 lb ai / ac = 60 lb ai FORESTRY PROB #2 4/4:  Treated acres per tank 20 ac / tank Pounds active ingredient needed 60 lb ai Pounds of product needed 60 lb ai / 0 .7 lb ai / lb WP = 86 lb WP FORESTRY PROB #2 4/4 GO TO FORESTRY PROBLEM INDEX Forestry Problem #3:  Forestry Problem #3 FORESTRY PROB #3:  FORESTRY PROB #3 Calibration check Require 16 gpa 5% tolerance Sprayed 30’ x 690’ Expended 7.5 gals FORESTRY PROB #3 1/4 Slide14:  How much area sprayed? 30 ft x 690 ft / 43,560 sq. ft / ac = 0.475 ac FORESTRY PROB #3 2/4 Slide15:  How much area sprayed? .475 ac How many gpa? 7.5 gal / 0.475 ac = 15.78 gpa FORESTRY PROB #3 3/4 Slide16:  How much area sprayed? 0.475 ac How many gpa? 7.5 gal / 0.475 ac = 15.78 gpa Rate O.K.? 15.78 gpa / 16 gpa = 98.6% of desired Rate is O.K. (only 1.4% off) FORESTRY PROB #3 4/4 GO TO FORESTRY PROBLEM INDEX Forestry Problem #4:  Forestry Problem #4 FORESTRY PROB #4:  FORESTRY PROB #4 300 gal tank 30 gal mix per ac 1.5 gal Garlon 4 / ac 0.5% adjuvant 4 liq oz drift control / 100 gal mix FORESTRY PROB #4 1/5 Slide19:  Acres per tank? 300 gal / tank / 30 gal / ac = 10 ac / tank FORESTRY PROB #4 2/5 Slide20:  Acres per tank? 10 ac / tank Gallons of Garlon 4? 10 ac x 1.5 gal / ac = 15 gal FORESTRY PROB #4 3/5 Slide21:  Acres per tank? 10 ac / tank Gallons of Garlon 4? 15 gal Adjuvant? 300 gal x 0.005 = 1.5 gal FORESTRY PROB #4 4/5 GO TO FORESTRY PROBLEM INDEX:  Acres per tank? 10 ac / tank Gallons of Garlon 4? 15 gal Adjuvant? 1.5 gal Drift control agent? 300 gal x 4 liq oz / 100 gal = 12 liq oz FORESTRY PROB #4 5/5 GO TO FORESTRY PROBLEM INDEX Forestry Problem #5:  Forestry Problem #5 FORESTRY PROB #5:  FORESTRY PROB #5 30 acres to be sprayed 20 gal spray mix / ac (gpa) 5 qt prod / ac 4 liq oz drift retardant / 100 gal mix FORESTRY PROB #5 1/5 Slide25:  Total mixture needed 30 ac x 20 gal /ac = 600 gal FORESTRY PROB #5 2/5 Slide26:  Total mixture needed 600 gal 2,4-DP needed 5 qt / ac x 30 ac = 150 qt (37.5 gal) FORESTRY PROB #5 3/5 Slide27:  Total mixture needed 600 gal 2,4-DP needed 37.5 gal Drift retardant needed 4 liq oz / 100 gal x 600 gal = 24 liq oz FORESTRY PROB #5 4/5 GO TO FORESTRY PROBLEM INDEX:  Final Solution in tank 600 gallons total mixture as 37.5 gal 2,4-DP 24 oz retardant 562.25 gal of water FORESTRY PROB #5 5/5 GO TO FORESTRY PROBLEM INDEX Forestry Problem #6:  Forestry Problem #6 FORESTRY PROB #6:  FORESTRY PROB #6 60 acres to be treated 3 qt Garlon 4 + 1 qt Accord / ac FORESTRY PROB #6 1/2 GO TO FORESTRY PROBLEM INDEX:  Gallons Garlon 4? 60 ac x 3 qt / ac = 180 qt / 4 qt / gal = 45 gallons Gallons Accord? 60 ac x 1 qt / ac = 60 qt / 4 qt / gal = 15 gal FORESTRY PROB #6 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem #7:  Forestry Problem #7 FORESTRY PROB #7:  FORESTRY PROB #7 40 gal / ac output at 4 mph new speed 8 mph FORESTRY PROB #7 1/2 GO TO FORESTRY PROBLEM INDEX:  Rule of thumb Double the speed = 1/2 the coverage 40 gal / ac at 4 mph --> 40/2 gal / ac or 20 gal / ac at 8 mph FORESTRY PROB #7 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem #8:  Forestry Problem #8 FORESTRY PROB #8:  FORESTRY PROB #8 Calibrated to deliver 25 gpa at 15 psi pressure increased to 60 psi FORESTRY PROB #8 1/3 Slide37:  New rate? Rule of thumb Rate increases as a function of the square root of the pressure increase 60 psi / 15 psi = 4 FORESTRY PROB #8 2/3 GO TO FORESTRY PROBLEM INDEX:  New rate? New pressure is 4 x old pressure 4 x pressure = 2 x the rate 25 gal x 2 = 50 gpa FORESTRY PROB #8 3/3 GO TO FORESTRY PROBLEM INDEX Forestry Problem #9:  Forestry Problem #9 FORESTRY PROB #9:  FORESTRY PROB #9 40 acres to be treated Accord used at 1:2 in water Inject 1 ml dilution / cut Injection on 2” centers 400 stems per acre Average dbh is 7” FORESTRY PROB #9 1/9 Slide41:  == Reorganize data == 40 acres to be treated 400 stems per acre Average dbh is 7” Injection on 2” centers Inject 1 ml dilution / cut Accord used at 1:2 in water FORESTRY PROB #9 2/9 Slide42:  Number of stems to treat 40 ac x 400 stems / ac = 16,000 stems FORESTRY PROB #9 3/9 Slide43:  Number of stems to treat 16,000 stems Number of cuts per stem 7 in x 3.1416 = 21.98 in circumference 22 in/stem / 2 in/cut = 11 cuts per stem FORESTRY PROB #9 4/9 7in dbh x 3.1416 (-- “pi”) = 21.98 in circumference :  7in dbh x 3.1416 (-- “pi”) = 21.98 in circumference 7” FORESTRY PROB #9 5/9 Graphically this is: 22 in/stem / 2 in/cut = 11 cuts / stem :  22 in/stem / 2 in/cut = 11 cuts / stem x x x x x 2 in x x x x x x FORESTRY PROB #9 6/9 Slide46:  Number of stems to treat 16,000 stems Number of cuts per stem 11 cuts per stem Total number of cuts 16,000 stems x 11 cuts / stem = 176,000 cuts FORESTRY PROB #9 7/9 Slide47:  Total number of cuts 176,000 cuts Total ml of dilution needed 176,000 cuts x 1 ml / cut = 176,000 ml FORESTRY PROB #9 8/9 GO TO FORESTRY PROBLEM INDEX:  Total number of cuts 176,000 cuts Total ml of dilution needed 176,000 ml = 176 L Total Accord needed 176 L / 3.785 L/gal = 46.5 gal 46.5 gal / 3 [2 water :1 Accord] = 1.5 15.5 gals Accord FORESTRY PROB #9 9/9 GO TO FORESTRY PROBLEM INDEX Forestry Problem #10:  Forestry Problem #10 FORESTRY PROB #10:  FORESTRY PROB #10 Pond size - average 20’ wide - 5’ deep - 100’ long Spill 5 gal Accord -- Kill bluegill? Kill rainbow trout? FORESTRY PROB #10 1/9 Slide51:  Volume of pond? 20 ft x 5 ft x 100 ft = 10,000 cu ft FORESTRY PROB #10 2/9 Slide52:  Volume of pond? 10,000 cu ft Convert to metric 10,000 cu ft x 28.32 L / cu ft = 283,200 L FORESTRY PROB #10 3/9 Slide53:  Volume of pond in metric 283,200 L How much Accord in pond in metric? 8.3 lb/gal water x 1.23 (spec. grav. of Accord) = 10.2 lb/gal Accord 5 gal x 10.2 lb /gal x 453.6 gm / lb = 23,133.6 gm x 1,000 mg / gm = 23,133,600 mg FORESTRY PROB #10 4/9 Slide54:  Volume of pond in metric 283,200 L How much Accord in pond? 23,133,600 mg Average Accord in pond? 23,133,600 mg / 283,200 L = 81.7 mg / L (= ppm) FORESTRY PROB #10 5/9 Slide55:  Average Accord in pond? 81.7 ppm From MSDS -- LD50 s Bluegill = >1,000 mg/L Channel catfish = >1,000 mg/L FORESTRY PROB #10 6/9 Slide56:  Average Accord in pond? 81.7 ppm Neither bluegill nor catfish are expected to be seriously affected by this spill FORESTRY PROB #10 7/9 Slide57:  Had Roundup been spilled in the pond at the same 81.7 ppm From MSDS -- LD50 s are Bluegill = 5.8 mg/L Channel catfish = 16 mg/L FORESTRY PROB #10 8/9 GO TO FORESTRY PROBLEM INDEX:  Had Roundup been spilled in the pond at the same 81.7 ppm Both would be expected to be seriously affected bluegill more than catfish FORESTRY PROB #10 9/9 GO TO FORESTRY PROBLEM INDEX Forestry Problem #11:  Forestry Problem #11 FORESTRY PROB #11:  FORESTRY PROB #11 Pond size 25’ wide x 4’ deep x 100’ long 1 gal Garlon 3A spill Bluegill and ducks are of concern FORESTRY PROB #11 1/5 Slide61:  Volume of the pond 25 ft x 4 ft x 100 ft = 10,000 cu. ft FORESTRY PROB #11 2/5 Slide62:  Volume of the pond 10,000 cu. ft 1 cu ft = 7.48 gals 10,000 cu ft x 7.48 gal/cu ft = 74,800 gal FORESTRY PROB #11 3/5 Slide63:  Volume of the pond 74,800 gal gal Garlon / gal pond x 1,000,000 = ppm Garlon in pond 1 / 74,800 x 1000000 = 13.4 ppm FORESTRY PROB #11 4/5 GO TO FORESTRY PROBLEM INDEX:  Concentration in the pond 13.4 ppm At this concentration there is little risk to either fish (bluegill = 471 ppm) or ducks (>10,000 ppm) from the spill FORESTRY PROB #11 5/5 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #1:  Forestry Problem More #1 FORESTRY More #1:  FORESTRY More #1 3 gal backpack 0.4 liq oz Arsenal / gal 2 liq oz Bullseye / gal 0.5% Cide-Kick FORESTRY PROB #More 1 1/2 GO TO FORESTRY PROBLEM INDEX:  0.4 liq oz Arsenal / gal x 3 gal = 1.2 liq oz Arsenal 2 liq oz Bullseye / gal x 3 gal = 6 liq oz Bullseye 0.005 Cide-Kick x 3 gal x 128 liq oz / gal = (1.92) = 2 liq oz Cide-Kick FORESTRY PROB #More 1 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #2:  Forestry Problem More #2 Forestry Problem More #2:  Forestry Problem More #2 FORESTRY More #2:  FORESTRY More #2 200 gal nurse tank Weighing 60 lb 5 liq oz Garlon 4 / gal water Spec grav of triclopyr = 1.08 Water = 8.3 lb / gal FORESTRY PROB #More 2 1/2 GO TO FORESTRY PROBLEM INDEX:  At a specific gravity of 1.08 the effect is negligible on the total mixture weight so use 8.3 for entire mix 1,000 lbs max weight – 60 lb tank = 940 lb allowable mixture / 8.3 lb/gal = 113.25 gal allowable mixture Probably wise to keep the load to 100 gal FORESTRY PROB #More 2 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #3:  Forestry Problem More #3 FORESTRY More #3:  FORESTRY More #3 30 gal tank 3 liq oz Garlon 4 / gal 0.4 liq oz Arsenal / gal 0.5% Cide-Kick FORESTRY PROB #More 3 1/2 GO TO FORESTRY PROBLEM INDEX:  3 liq oz Garlon 4 / gal x 30 gal = 90 liq oz Garlon 4 0.4 liq oz Arsenal / gal x 30 gal = 12 liq oz Arsenal 0.005 Cide-Kick x 30 gal x 128 liq oz / gal = 19.2 liq oz Cide-Kick FORESTRY PROB #More 3 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #4:  Forestry Problem More #4 FORESTRY More #4:  FORESTRY More #4 Four nozzle boom 8003 Spraying Systems flat fan nozzles 40 psi is being used through the boom FORESTRY PROB #More 4 1/2 GO TO FORESTRY PROBLEM INDEX:  8003 nozzle applies 0.3 gal/min at 40 psi at an 80o angle 4 nozzles x 0.3 gpm / nozzle = 1.2 gpm FORESTRY PROB #More 4 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #:  Forestry Problem More # FORESTRY More #5:  FORESTRY More #5 Spray covers an 8 foot wide swath Puts out 1.2 gpm Desire 5 gal / ac of the tank mix FORESTRY PROB #More 5 1/2 GO TO FORESTRY PROBLEM INDEX:  43,560 sq ft / 8 ft = 5445 ft 5 gal / 1.2 gal / min = 4.17 min 5445 ft / 4.17 min = 1306.8 ft / min 1306.8 ft / min / 5280 ft / mi = .2475 mi / min x 60 min / hr = 14.85 mi / hr FORESTRY PROB #More 5 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #6:  Forestry Problem More #6 FORESTRY More #6:  FORESTRY More #6 Maximum safe speed is 5 mph From previous problem we know that we should be going 14.8 mph to get the desired output ( = 3 times the safe speed) FORESTRY PROB #More 6 1/2 GO TO FORESTRY PROBLEM INDEX:  If we reduce speed to 5 mph we triple the rate of application, so we must reduce flow of product Changing the spray nozzles from 8003 to 8001 nozzles will reduce the output per nozzle from 0.3 gpm to 0.1 gpm which reduces the output the desired amount (What you would not do is adjust pressure in the system) FORESTRY PROB #More 6 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #7:  Forestry Problem More #7 FORESTRY More #7:  FORESTRY More #7 5 pt / ac Velpar L Spot apply 2 ml per spot Determine appropriate spacing for the spots (spot grid application) FORESTRY PROB #More 7 1/2 GO TO FORESTRY PROBLEM INDEX:  5 pt / ac x 0.125 gal / pt x 3,785 ml / gal = 2366 ml / ac 2366 ml / ac / 2 ml / spot = 1183 spot / ac 43,560 sq ft / ac / 1183 spot / ac = 36.8 sq ft / spot 38.6 sq ft = 6.066 ft or a 6’ x 6’ grid FORESTRY PROB #More 7 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #8:  Forestry Problem More #8 FORESTRY More #8:  FORESTRY More #8 2600 stems / ac to be treated 135 ac 3 ml Garlon 4 per stem FORESTRY PROB #More 8 1/2 GO TO FORESTRY PROBLEM INDEX:  135 ac x 2,600 stems / ac x 3 ml / stem = 1,053,000 ml 1,053,000 ml = 1,053 Liters 1,053 L / 3.785 L / gal = (278.2 gal) or 280 gals FORESTRY PROB #More 8 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #9:  Forestry Problem More #9 FORESTRY More #9:  FORESTRY More #9 50 gal nurse tank Ratio of 1 : 5 (Garlon 4 : JLB Oil Plus) FORESTRY PROB #More 9 1/2 GO TO FORESTRY PROBLEM INDEX:  Garlon 4 1/ 6 x 50 gal = 8.3 gal JLB Plus 5/6 of mix x 50 gal = 41.7 gal FORESTRY PROB #More 9 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #10:  Forestry Problem More #10 FORESTRY More #10:  FORESTRY More #10 15 gal containers of JLB Oil Plus Packaged by 8 containers / pallet Need 280 gal of mix 5/6 of mix is JLB Oil Plus FORESTRY PROB #More 10 1/2 GO TO FORESTRY PROBLEM INDEX:  15 gal / jug x 8 jug / pallet = 120 gal / pallet 5/6 x 280 gal = 233.33 gal needed 233 gal / 120 gal / pallet = 2 pallet FORESTRY PROB #More 10 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #19:  Forestry Problem More #19 FORESTRY More #19:  FORESTRY More #19 15 gpa at 4 mph Reduce to 2 mph Half speed = double rate, or 2 x 15 gpa = 30 gpa FORESTRY PROB #More 19 1/1 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #20:  Forestry Problem More #20 FORESTRY More #20:  FORESTRY More #20 15 gpa at 20 psi Increase pressure to deliver 30 gpa (30 gpa / 15 gpa)2 x 20 psi = 4 x 20 psi = 80 psi FORESTRY PROB #More 20 1/1 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #21:  Forestry Problem More #21 FORESTRY More #21:  FORESTRY More #21 Prescription: 8 gal herbicide / 200 gal mix Require only 50 gal of mix 50 gal / 200 gal x 8 gal = 2 gal FORESTRY PROB #More 21 1/1 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #22:  Forestry Problem More #22 FORESTRY More #22:  FORESTRY More #22 1.5 gal / ac As a 25 gal mix in water 30 ac FORESTRY PROB #More 22 1/2 GO TO FORESTRY PROBLEM INDEX:  1.5 gal / ac x 30 ac = 45 gals herbicide 30 ac x 25 gal / ac = 750 gal total 750 gal – 45 gal = 705 gal water FORESTRY PROB #More 22 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem More #23:  Forestry Problem More #23 FORESTRY More #23:  FORESTRY More #23 3 gal / min 200 ft / 2 min 40 ft wide swath FORESTRY PROB #More 23 1/2 GO TO FORESTRY PROBLEM INDEX:  200 ft / 2 min x 40 ft = 4,000 sq ft / min 43,560 sq ft / ac / 4,000 sq ft / min = 10.89 min / ac 10.89 min / ac x 3 gal / min = 32.67 gpa FORESTRY PROB #More 23 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem #a:  Forestry Problem #a Forestry Problem #a 40 ac 3000 stem / ac 2 ml / stem:  Forestry Problem #a 40 ac 3000 stem / ac 2 ml / stem FORESTRY PROB #More a 1/2 40 ac x 3000 stem / ac x 2 ml / stem = 240,000 ml 240,000 ml / 1,000 ml / L = 240 L = approximately 60 gallons :  40 ac x 3000 stem / ac x 2 ml / stem = 240,000 ml 240,000 ml / 1,000 ml / L = 240 L = approximately 60 gallons FORESTRY PROB #More a 2/2 GO TO FORESTRY PROBLEM INDEX Forestry Problem #b:  Forestry Problem #b Forestry Problem #b:  Forestry Problem #b 40 ac 200 stem / ac stem average 5” dbh injected on 3” centers 1 ml solution / cut FORESTRY PROB #More b 1/4 5 in dbh x 3.1416 (-- “pi”) = 15.7 in circumference :  5 in dbh x 3.1416 (-- “pi”) = 15.7 in circumference 5” FORESTRY PROB #More b 2/4 15.7 in/stem / 3 in/cut = 5 cuts/ stem :  15.7 in/stem / 3 in/cut = 5 cuts/ stem x x x x x 3 inches FORESTRY PROB #More b 3/4 5 cut / stem x 1 ml / cut x 200 stem / ac x 40 ac = approximately 40,000 ml = 40 L or 10 gallons __________ Note that this number is a little low since the average stem required 5.2 cuts (= 10.5 gal solution):  5 cut / stem x 1 ml / cut x 200 stem / ac x 40 ac = approximately 40,000 ml = 40 L or 10 gallons __________ Note that this number is a little low since the average stem required 5.2 cuts (= 10.5 gal solution) FORESTRY PROB #More b 4/4 GO TO FORESTRY PROBLEM INDEX R-O-W Problem #1:  R-O-W Problem #1 RIGHT-OF-WAY PROB #1:  RIGHT-OF-WAY PROB #1 10 mi of R-O-W 15’ each side of road R-O-W PROB # 1 1/3 Slide118:  How many sq ft 15 ft x 2 x 5280 ft / mi x 10 mi = 1,584,000 sq ft R-O-W PROB # 1 2/3 GO TO R-O-W & AQUATIC PROBLEM INDEX:  How many sq ft 1,584,000 sq ft How many acres? 1,584,000 sq ft / 43,560 sq ft / ac = 36.4 ac R-O-W PROB # 1 3/3 GO TO R-O-W & AQUATIC PROBLEM INDEX R-O-W Problem #2:  R-O-W Problem #2 RIGHT-OF-WAY PROB #2:  RIGHT-OF-WAY PROB #2 15 feet on both sides 5 mi R-O-W 2 gal / ac R-O-W PROB # 2 1/2 GO TO R-O-W & AQUATIC PROBLEM INDEX:  15 ft x 2 x 5 mi x 5280 ft / mi = 792,000 sq ft 792,000 sq ft / 43560 sq ft / ac = 18.2 ac 18.2 ac x 2 gal / ac = 36.4 gal R-O-W PROB # 2 2/2 GO TO R-O-W & AQUATIC PROBLEM INDEX R-O-W Problem #3:  R-O-W Problem #3 RIGHT-OF-WAY PROB #3:  RIGHT-OF-WAY PROB #3 25 gpa at 8 mph Speed reduced to 4 mph 25 gpa x 8 mph / 4 mph = 50 gpa R-O-W PROB # 3 1/1 GO TO R-O-W & AQUATIC PROBLEM INDEX R-O-W Problem #4:  R-O-W Problem #4 RIGHT-OF-WAY PROB #4:  RIGHT-OF-WAY PROB #4 25 gpa at 6 mph Pressure reduced from 50 psi to 12.5 psi 20 gpa / 50 psi / 12.5 psi = 10 gpa R-O-W PROB # 4 1/1 GO TO R-O-W & AQUATIC PROBLEM INDEX R-O-W Problem #5:  R-O-W Problem #5 RIGHT-OF-WAY PROB #5:  RIGHT-OF-WAY PROB #5 400 gal tank 20 gal / ac 1.25 gal 2,4-D / ac 3 liq oz drift control agent / 100 gal 1% adjuvant R-O-W PROB # 5 1/2 GO TO R-O-W & AQUATIC PROBLEM INDEX:  400 gal / tank / 20 gal / ac = 20 ac / tank 20 ac x 1.25 gal / ac = 25 gal 2,4-D 400 gal x 3 liq oz / 100 gal = 12 liq oz drift control agent 400 gal x 0.01 = 4 gal adjuvant R-O-W PROB # 5 2/2 GO TO R-O-W & AQUATIC PROBLEM INDEX R-O-W Problem #6:  R-O-W Problem #6 RIGHT-OF-WAY PROB #6:  RIGHT-OF-WAY PROB #6 25 gal / ac required 6 gal expended on 20’ x 545’ area +/- 5% tolerance R-O-W PROB # 6 1/2 GO TO R-O-W & AQUATIC PROBLEM INDEX:  20 ft x 545 ft / 43560 sq ft / ac = 0.25 ac 6 gal / 0.25 ac = 24 gpa 24 gpa / 25 gpa = .96 96% rate is acceptable (range is 95 – 105%) R-O-W PROB # 6 2/2 GO TO R-O-W & AQUATIC PROBLEM INDEX R-O-W Problem #7:  R-O-W Problem #7 RIGHT-OF-WAY PROB #7:  RIGHT-OF-WAY PROB #7 25 gpa 20 mi 15 ft both sides 1.5 gal 2,4-D / ac 3 liq oz Poly Control / 100 gal mix R-O-W PROB # 7 1/2 GO TO R-O-W & AQUATIC PROBLEM INDEX:  20 mi x 5,280 ft / mi x 2 x 15 ft / 43,560 sq ft / ac = 72.7 ac 72.7 ac x 1.5 gal / ac = 109 gal 2,4-D 72.7 ac x 25 gal / ac = 1817.5 gal of mix – 109 gal 2,4-D = 1,708.5 gal water 1817.5 gal x 3 liq oz / 100 gal = 54 liq oz Poly Control R-O-W PROB # 7 2/2 GO TO R-O-W & AQUATIC PROBLEM INDEX R-O-W Problem #8:  R-O-W Problem #8 RIGHT-OF-WAY PROB #8:  RIGHT-OF-WAY PROB #8 5 gal Garlon 4 spilled into a 200’ x 320’ pond with average depth of 6’ Bluegill 96 hr LC50 is 0.87 mg/L R-O-W PROB # 8 1/2 GO TO R-O-W & AQUATIC PROBLEM INDEX:  200 ft x 320 ft x 6 ft x 28.32 L /cu ft = 10,874,880 L 8.33 lb / gal (water) x 1.08 (spec grav of G4) x 5 gal (G4) x 453.6 gm / lb x 1000 mg / gm = 20,389,300 mg in 5 gal G4 20,389,200 mg / 10,874,880 L = 1.88 mg (G4) / L (pond water) 1.88 mg/L > 0.87 mg/L – fish kill expected R-O-W PROB # 8 2/2 GO TO R-O-W & AQUATIC PROBLEM INDEX R-O-W Problem #a:  R-O-W Problem #a RIGHT-OF-WAY PROB #a:  RIGHT-OF-WAY PROB #a 6 mi of R-O-W 12’ each side of road 2 gal Krenite / ac R-O-W PROB # a 1/4 Slide141:  How many sq. ft 12’ x 2 x 5280’ / mi x 6 mi = 760,320 sq ft R-O-W PROB # a 2/4 Slide142:  How many sq ft 760,320 sq ft How many acres? 760,320 sq ft / 43,560 sq ft / ac = 17.45 ac R-O-W PROB # a 3/4 GO TO R-O-W & AQUATIC PROBLEM INDEX:  How many acres? 760,320 sq ft How many acres? 17.45 ac How much Krenite? 17.5 ac x 2 gal / ac = 35 gal R-O-W PROB # a 4/4 GO TO R-O-W & AQUATIC PROBLEM INDEX Aquatics Problem #1:  Aquatics Problem #1 AQUATICS PROB #1:  AQUATICS PROB #1 20 surface acres Average 2’ deep 20 ac x 2 ft = 40 ac ft AQUATIC PROB # 1 1/1 GO TO R-O-W & AQUATIC PROBLEM INDEX Aquatics Problem #2:  Aquatics Problem #2 AQUATICS PROB #2:  AQUATICS PROB #2 Pond: 100’ long x 500’ wide x 4’ average depth 100 ft x 500 ft x 4 ft / 43560 cu ft / ac ft = 4.6 ac ft AQUATIC PROB # 2 1/1 GO TO R-O-W & AQUATIC PROBLEM INDEX Aquatics Problem #3:  Aquatics Problem #3 AQUATICS PROB #3:  AQUATICS PROB #3 30 lb chemical / ac ft Using a 50% granule 300’ long x 10’ wide with average depth of 10’ AQUATIC PROB # 3 1/2 GO TO R-O-W & AQUATIC PROBLEM INDEX:  300 ft x 10 ft x 10 ft / 43,560 cu ft / ac ft = 0.689 ac ft 0.689 ac ft x 30 lb / ac ft x 2 (or divide by 0.5) = (41.34 lbs) = 42 lbs AQUATIC PROB # 3 2/2 GO TO R-O-W & AQUATIC PROBLEM INDEX Aquatics Problem #4:  Aquatics Problem #4 AQUATICS PROB #4:  AQUATICS PROB #4 1 gal / 2 ac ft 420’ long x 105’ wide by average 4’ deep 420 ft x 105 ft x 4 ft / 43,560 cu ft / ac ft x 1 gal / 2 ac ft = 2 gal AQUATIC PROB # 4 1/1 GO TO R-O-W & AQUATIC PROBLEM INDEX Aquatics Problem #5:  Aquatics Problem #5 AQUATICS PROB #5:  AQUATICS PROB #5 30 ft spray boom 50 ft / min 2 gal / min 30 ft x 50 ft / min / 43,560 sq ft / ac = 0.034 ac / min 2 gal / min / 0.034 ac / min = 58 gpa AQUATIC PROB # 5 1/1 GO TO R-O-W & AQUATIC PROBLEM INDEX Aquatics Problem #6:  Aquatics Problem #6 AQUATICS PROB #6:  AQUATICS PROB #6 10 surface ac x 4 ft Desired rate is 0.4 ppm From the label – 4 ft deep requires 4.8 gal / surface ac 4.8 gal / ac x 10 ac = 48 gal AQUATIC PROB # 6 1/1 GO TO R-O-W & AQUATIC PROBLEM INDEX Aquatics Problem #7:  Aquatics Problem #7 Aquatics Problem #7:  Aquatics Problem #7 AQUATICS PROB #7:  AQUATICS PROB #7 10 gal spill 2 lb a.i. / gal – toxicity based entirely on the a.i. 2 ac x 4’ deep 1 ppm may result in fish kill AQUATIC PROB # 7 1/2 GO TO R-O-W & AQUATIC PROBLEM INDEX:  2 ac x 4 ft x 43,560 sq ft / ac x 62.4 lbs / cu ft = 21,745,152 lb of water in lake 2 lb (ai) x 1,000,000 /21,745,152 lb (water) = 0.92 ppm 0.92 ppm < 1 ppm There should not be a significant fish kill AQUATIC PROB # 7 1/2 GO TO R-O-W & AQUATIC PROBLEM INDEX

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