Volume of Oil in Cylindrical Tanks

by:

Scott Burrell

In this set we want to consider the problem context of the storage of fluids in cylindrical tanks that have been installed laying on their side. The common problem characteristic is that the DEPTH of the fluid is known, determine the volume.

First Situation

This problem was seriously proposed to one of my EMT 725 students by their Superintendent. At the Superintendent's home the furnace used fuel oil stored in an underground tank. It was known that the tank was installed level on its side and that it was 36 inches in diameter and 48 inches long. Using a stick dipped through the fill tube, the superintendent determined he had 10 inches of oil in the tank. He really did NOT want to know how to calculate the amount of oil. He knew from experience that it was February and he would need about 40 gallons of oil to finish the season. Would he have enough oil?

An estimate may suffice. For example, the surface of the oil is a rectangle 48 incles long and a little less than 18 incles wide. So the volume would be less than 48 X 36 X 10 cubic inches. This is the volume of rectangular parallelepiped. At 231 cubic incles per gallon, this given an overestimate of 75 gallons. In fact if we approximate the oil in the tank by a trianglular prism with altitude of 10 and base of 36 and lenght of 48 its volume is half of this parallelepiped, or 37.5 gallons estimated. That estimate is probably pretty close, but who would want to assure their boss that he has enough oil in the tank to finish out the winter on the basis of this estimate?

We may want to calculate the volume.

How many gallons of oil are in the gas tank?

The volume of a cylinder is the area of its base times the height. Since the tank is lying on its side, the base is the part of the circle where the 10 inches of oil are located. The drawing below is a diagram of the base.

The area of the base is the sector of the circle minus the area of the triangle. Since the radius of the circle is 18 and the height of the oil is 10, h=8. X can be found using the Pythagorean Theorem.

8^2 +x^2 =18^2
64+x^2=324
x^2=260
x=2(sqrt(65))

The area of the triangle is (1/2)bh. From the picture, we see that x= (1/2)b. The area of the triangle is xh.

Area of triangle = xh
= 2(sqrt(65))(8)
=16(sqrt(65)) in^2

To find the area of the sector, we must first find theta. Theta can be found using the arcos.

cos^-1 (theta) = 8/18

The area of the sector is (1/2) r^2 (theta). From the picture, the angle marked theta is actually half of the angle of the sector. Thus the area of the sector is r^2 (theta).

Area of sector = r^2 (theta)
=(18^2)(1.11)
=359.72 in^2

The area of the oil is the area of the sector minus the area of the triangle.

Area of oil = Area of sector - Area of triangle
=359.72-16(sqrt(65))
=230.72 in^2

The volume of oil is the area of the oil times the length of the tank.

Volume of oil = Area of oil * 48
=230.72 * 48
=11,074.67 in^3

However, the question asks for the number of gallons. The cubic inches must be converted to gallons. There are 7.481 gallons/cubic foot.

(11,074.67 in^3/1) * (1 ft^3/1,728 in^3) * (7.481 gallons/1 ft^3) = 47.94 gallons

There are approximately 48 gallons of oil in the tank.

Extension: This problem was extremely lengthly. I would not want to do this every time I wanted to know how much oil was in the tank. It would be more time efficient to calibrate the stick used to measure the depth of the oil so that the number of gallons could be read off of it.

To make a calibrated stick, I created an Excel spreadsheet that calculates the amount of oil in gallons for each 1/2 inch. The stick would be made by placing a mark at every 1/2 inch. The mark would be labeled with the number of gallons for that particular 1/2 mark.