i live on a small farm and recently bought an old gravity gas tank.the tank sits horizontal and is 38in in diameter by 48in long. i think i know how many gallons it holds[295?] but my question is after filling i have no idea how much fuel is inside? when putting a yard stick in till touching bottom how much fuel is left in tank at every inch starting at 37 down to one?
Moved - Gas in a horizontal tank
- Thread starter lade1
- Start date
Hello, lade1!
This is the end view of the tank.
It is turned 90o clockwise.
The equation of the circle is: \(\displaystyle x^2 + y^2 \,=\,361\)
The upper semicircle is: \(\displaystyle y \:=\:\sqrt{361-x^2}\)
The area from -19 to a is: .\(\displaystyle \displaystyle A \;=\;2\int^a_{\text{-}19}\sqrt{361-x^2}\,dx\)
Let \(\displaystyle x \:=\:19\sin\theta \quad\Rightarrow\quad dx \:=\:19\cos\theta\,d\theta\)
Substitute: .\(\displaystyle \displaystyle A \;=\;2\int 19\cos\theta(19\cos\theta\,d\theta) \;=\;722\int\cos^2\theta\,d\theta \)
. . . \(\displaystyle \displaystyle =\;722\int \left(\frac{1+\cos2\theta}{2}\right) d\theta \;=\;361\int(1 + \cos2\theta)d\theta\)
. . . \(\displaystyle \displaystyle =\;361\left(\theta + \tfrac{1}{2}\sin2\theta\right) + C \;=\;361(\theta + \sin\theta\cos\theta) + C\)
Back-substitute: .\(\displaystyle A \;=\;361\left(\arcsin\dfrac{x}{19} + \dfrac{x\sqrt{361-x^2}}{19}\right) + C\)
Evaluate: .\(\displaystyle A \;=\;361\left(\arcsin\dfrac{x}{19} + \dfrac{x\sqrt{361-x^2}}{19}\right)\,\bigg]^a_{\text{-}19}\)
This gives us the area of the shaded region.
Divide by \(\displaystyle 361\pi\) and get the percent of the tank that is filled.
This is the end view of the tank.
It is turned 90o clockwise.
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|The upper semicircle is: \(\displaystyle y \:=\:\sqrt{361-x^2}\)
The area from -19 to a is: .\(\displaystyle \displaystyle A \;=\;2\int^a_{\text{-}19}\sqrt{361-x^2}\,dx\)
Let \(\displaystyle x \:=\:19\sin\theta \quad\Rightarrow\quad dx \:=\:19\cos\theta\,d\theta\)
Substitute: .\(\displaystyle \displaystyle A \;=\;2\int 19\cos\theta(19\cos\theta\,d\theta) \;=\;722\int\cos^2\theta\,d\theta \)
. . . \(\displaystyle \displaystyle =\;722\int \left(\frac{1+\cos2\theta}{2}\right) d\theta \;=\;361\int(1 + \cos2\theta)d\theta\)
. . . \(\displaystyle \displaystyle =\;361\left(\theta + \tfrac{1}{2}\sin2\theta\right) + C \;=\;361(\theta + \sin\theta\cos\theta) + C\)
Back-substitute: .\(\displaystyle A \;=\;361\left(\arcsin\dfrac{x}{19} + \dfrac{x\sqrt{361-x^2}}{19}\right) + C\)
Evaluate: .\(\displaystyle A \;=\;361\left(\arcsin\dfrac{x}{19} + \dfrac{x\sqrt{361-x^2}}{19}\right)\,\bigg]^a_{\text{-}19}\)
This gives us the area of the shaded region.
Divide by \(\displaystyle 361\pi\) and get the percent of the tank that is filled.
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,760
Draw lines from the center of the tank to the top of the oil and draw a line from the center to the top of the oil. If the top of the oil is "d" inches from the bottom of the tank, that gives two right triangles with hypotenuse of length (1/2)(38)= 19 inches and one leg of length d- 19. The angle at the center, for each right triangle is \(\displaystyle arccos((d- 19)/19)\) so the entire angle, from one radius to the other, is \(\displaystyle 2arccos((d- 19)/19)\) radians. So that the area of that sector, from the center of the circle out to the circle itself, is \(\displaystyle \frac{2arccos((d-19)/19)}{2}\pi(19^2)\).
The length of the other leg of each right triangle, by the Pythagorean theorem, is \(\displaystyle \sqrt{38d- 19^2}\). So the length of the line showing the top of the oil is \(\displaystyle 2\sqrt{38d- 19^2}\) and the area of the triangle is \(\displaystyle (1/2)(19- d)\sqrt{38d- 19^2}\).
The area of the portion of the sector above the oil, then, is the difference between the two: \(\displaystyle \frac{2arccos((d- 19)/19}{2}\pi(19^2)- (1/2)(19- d)\sqrt{38d- 19^2}\). And, of course, the volume is that area multiplied by the length of the tank, \(\displaystyle 48\left(\frac{2arccos((d- 19)/19}{2}\pi(19^2)- (1/2)(19- d)\sqrt{38d- 19^2}\right)\)
Because you measured the depth of the oil in inches, that is in cubic inches. There are 231 cubic inches to a (U.S.) liquid gallon so divide that by 231 to convert to gallons.