Hi All:
Either I'm having a problem with my calculus, but - I want to know how ro calculate the volume of beer in a given lenght of beer lines. Now, lets use an example:
Diameter of the beer line is 0.7 cm and the length (or h) is 12 cm Result:
3.142 X 72 X 12 = 1847.50 cm3
So, the volume of the line is 1847.50 cm3
Each cm3 holds 0.001 Liter of liquid, in our example it will be
1847.50 X 0.001 = 1.8475 Liter Is that correct?
Now, how is the oz/ft belew calculated?
The inner diameter of the tubing (ID) determines the liquid contents:
3/16” ID = 1/6 oz/ft
1/4” ID = 1/3 oz/ft
5/16” ID = 1/2 oz/ft
3/8” ID = 3/4 oz/ft
Also, at what point are we using 3/16"ID, 1/4"ID, 5/16"ID and 3/8" ID?????
I know is a lot of question, but I hope the our Micro Matic friends or anyone on the forum can help.
Cheers
Either I'm having a problem with my calculus, but - I want to know how ro calculate the volume of beer in a given lenght of beer lines. Now, lets use an example:
Diameter of the beer line is 0.7 cm and the length (or h) is 12 cm Result:
3.142 X 72 X 12 = 1847.50 cm3
So, the volume of the line is 1847.50 cm3
Each cm3 holds 0.001 Liter of liquid, in our example it will be
1847.50 X 0.001 = 1.8475 Liter Is that correct?
Now, how is the oz/ft belew calculated?
The inner diameter of the tubing (ID) determines the liquid contents:
3/16” ID = 1/6 oz/ft
1/4” ID = 1/3 oz/ft
5/16” ID = 1/2 oz/ft
3/8” ID = 3/4 oz/ft
Also, at what point are we using 3/16"ID, 1/4"ID, 5/16"ID and 3/8" ID?????
I know is a lot of question, but I hope the our Micro Matic friends or anyone on the forum can help.
Cheers
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