how many galons of water can a drip pan of aevaporator hold if its dimensions are 8' long 8' wide and 9" deep?

Answer 1

8 feet = 96 inches
Therefore pan volume = 96 x 96 x 9 = 82944 cubic inches
(which is also 1359208 cubic cm).

1 milliliter of water, at "room temperature", occupies 1 cubic cm.

This means there are 1359208 ml = 1359.208 liters

1 US gallon = 3.785 liters ... 1 Imperial gallon = 4.546 liters.

Therefore your pan will hold (1359.208 / 3.785) = 359.1 US gallons, or (1359.208 / 4.546) = 299.0 Imperial gallons.

Answer 2

1st convert ft to inches. 8ft.= 96 inches.
96 " long X 96" wide X 9 " deep = 82,944 cu. in.

There is 1728 cu. in. in a cu. ft. 12 x 12 x 12 = 1728.

Therefore 82,944 cu. in. / 1728 cu. in. = 48 cu. ft.

1ft^3 Liquid = 7.48051948060182 Gallons US
7.48051948060182 Gallons x 48 cu. ft. = 359.06 Gallons US Liquid Pure Water.

If you are looking at dry capacity it is 308.568558 Gallons US Dry.

I've rechecked my resources (3 of them) and my math too. I don't know where the others are getting their info from.
Still, I could be wrong.

Answer 3

There are 7.49 U.S.gallons in one cubic foot.
8' x 8' x (9/12)' x 7.49 g/ft^3 = 359.52 U.S.gallons
That is some big refrigerator.

Answer 4

(8)(8)(.75)(1728)/231 =359 gallons