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#1
02-10-2011, 06:13 PM
 Unregistered Guest Posts: n/a
Herbicide/ lbs to gal to oz

The herbicide is a granule, it is 71.25% active ingredient by weight, the container holds 4 lbs.

I need to spray 1 oz per acre. The mixing tank holds 40 gallons. The main tank holds the 2,000 gallons. The computer can control the injector pump from the mixing tank from .o1 to 999 oz per acre. The computer uses the speed of the truck to adjust the injector pump speed.

How much of the herbicide do I need to add to the mixing tank, and what does the injector pump need to be set at to spray 1 oz per acre?
#2
02-11-2011, 03:10 AM
 JohnS Moderator Join Date: Dec 2007 Location: SE Michigan, USA Posts: 9,445 Rep Power: 19
Re: Herbicide/ lbs to gal to oz

Quote:
 Originally Posted by Unregistered The herbicide is a granule, it is 71.25% active ingredient by weight, the container holds 4 lbs. I need to spray 1 oz per acre. The mixing tank holds 40 gallons. The main tank holds the 2,000 gallons. The computer can control the injector pump from the mixing tank from .o1 to 999 oz per acre. The computer uses the speed of the truck to adjust the injector pump speed. How much of the herbicide do I need to add to the mixing tank, and what does the injector pump need to be set at to spray 1 oz per acre?
There are several issues here, and I think I need more info:

1) There are two kinds of ounces, weight and volume. The weight ounce is 1/16 of a pound, so the 4 lb container is 64 ounces. The fluid ounce is 128 fl oz per gallon, so the mixing tank is 5120 fl oz.

2) Is the application rate per acre 1 oz of the granules or 1 oz of active ingredient. If the latter, at 71.25% purity, 1.4 oz of granules per acre would be required to get 1 oz/acre active ingredient. The 4 lb container is enough for either 64 acre or 45.6 acre.

3) Do you know the solubility of the granules? Just to get them to dissolve, you may need a certain amount of water is the mixing tank. Do the directions recommend a dilution for application, or just give the 1 oz/acre figure.

4) I understand that you can vary the injection rate from the mixing tank. At what rate is water pumped from the main tank? How many gallons of fully diluted mixture are being applied per acre, how many acres will a full main tank cover?

5) How many acres are you trying to treat?
#3
02-11-2011, 05:04 AM
 Unregistered Guest Posts: n/a
Re: Herbicide/ lbs to gal to oz

Quote:
 Originally Posted by JohnS There are several issues here, and I think I need more info: 1) There are two kinds of ounces, weight and volume. The weight ounce is 1/16 of a pound, so the 4 lb container is 64 ounces. The fluid ounce is 128 fl oz per gallon, so the mixing tank is 5120 fl oz. 2) Is the application rate per acre 1 oz of the granules or 1 oz of active ingredient. If the latter, at 71.25% purity, 1.4 oz of granules per acre would be required to get 1 oz/acre active ingredient. The 4 lb container is enough for either 64 acre or 45.6 acre. 3) Do you know the solubility of the granules? Just to get them to dissolve, you may need a certain amount of water is the mixing tank. Do the directions recommend a dilution for application, or just give the 1 oz/acre figure. 4) I understand that you can vary the injection rate from the mixing tank. At what rate is water pumped from the main tank? How many gallons of fully diluted mixture are being applied per acre, how many acres will a full main tank cover? 5) How many acres are you trying to treat?
(1) correct

(2) 1 oz of active ingredient.

(3) Sorry about not being clear on this. The directions say fill your mixing tank half way with water. Add the herbicide, and finish filling with water. The directions does not recommend a dilution rate. It only says to apply no more than 3 oz per acre per year. We will be spraying 3 times this year.

(4)What we are spraying is the sides of the road. The truck has six spray nozles. We will adjust the nozles and water pressure from the main tank to get a even coverage of twenty feet. Not shure how many acres the main tank will cover. The computer control system knows the width that each nozle covers. And when the nozles are on or off. Then the computer takes the truck speed and the width being sprayed to get the acres being sprayed.
#4
02-11-2011, 03:19 PM
 JohnS Moderator Join Date: Dec 2007 Location: SE Michigan, USA Posts: 9,445 Rep Power: 19
Re: Herbicide/ lbs to gal to oz

OK, if it is 1 oz active ingredient per acre, the 1.4 oz granules per acre must be applied and the 66 oz container covers 45.6 acres (a strip 20' wide by 18.8 miles, or half that if doing a 20' strip on each side).

If you mix the 64 oz in the 40 gal mixing tank, you have 5120 fl oz in the mixing tank to cover 45.6 acres, so you want to apply 112.3 fl oz per acre. Mix with sufficent water from the main tank for wetting and coverage.

I don't know how realistic it is that you want to cover 18.8 miles by a 20 ft strip. To cover less, you have to weigh out part of container, and recalculate the application rate. How much to weigh out depends on the area to be covered, 1.4 oz per acre (to get 1 oz active ingredient)

In the mixing tank, 40 gallons will yield 5129 fl oz of mixture, divide that by the acres from step 1 to get the application rate,
#5
02-11-2011, 04:57 PM
 Unregistered Guest Posts: n/a
Re: Herbicide/ lbs to gal to oz

Quote:
 Originally Posted by JohnS OK, if it is 1 oz active ingredient per acre, the 1.4 oz granules per acre must be applied and the 66 oz container covers 45.6 acres (a strip 20' wide by 18.8 miles, or half that if doing a 20' strip on each side). If you mix the 64 oz in the 40 gal mixing tank, you have 5120 fl oz in the mixing tank to cover 45.6 acres, so you want to apply 112.3 fl oz per acre. Mix with sufficent water from the main tank for wetting and coverage. I don't know how realistic it is that you want to cover 18.8 miles by a 20 ft strip. To cover less, you have to weigh out part of container, and recalculate the application rate. How much to weigh out depends on the area to be covered, 1.4 oz per acre (to get 1 oz active ingredient) In the mixing tank, 40 gallons will yield 5129 fl oz of mixture, divide that by the acres from step 1 to get the application rate,
Ok, this is the information that I needed. I see some of the things that I was doing wrong. First, I was trying to add the 64 oz from the container to the 5120 oz of water in the mixing tank to have a total of 5184 oz. When I was trying to figure out how much of the 5184 oz in the mixing tank I needed to spray per acre to get the 100% of active ingredient.

I now see that the first thing to figure out is how many acres a container will cover. Then you can divide the amount in the mixing tank by the number of cover acres to get the application rate.

As far as recalculating the application if spraying less than 20' ft, the computer reduces the application rate when you start turning nozles off.

Thank you.

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