Question

Tank A is conical in shape with a circular cross section. The diameter of tank A is 0.05 m at
its base. The diameter increases with height such that all sides of the tank slope out at a ratio
of 1:10 (1 horizontal:10 vertical).
(for example at a height of 2 m the tank has a diameter of 0.45 m)

Can some one please explain this, also what is the Formula used to solve this?

Answer 1

@satellite73 would you know?

Answer 2

@mathslover

Answer 3

you have not posted a question that asks to be solved ...

Answer 4

Amistre has a point :D

Answer 5

@amistre64 has a point, what is the formula for a conical cylinder? to find the volume.

Answer 6

volume of a conical cylinder is (1/3)*pi*(r^2)*height

Answer 7

what class is this from?

Answer 8

the volume of your cone will be the entire cone minus the small cone at the bottom. Using the numbers given, the radius of the tiny cone is
0.05/2 = 0.025
Its height will be 10 times bigger or 0.25

Answer 9

Problem Solving 4 - I have 3 subjects left till I graduate as an Engineer in Australia. Thanks, I could not get a straight answer from google.

Answer 10

Thanks phi great explanation, I understand now.

Answer 11

You have to be careful of the details. the height of the *entire cone* (including the tiny cone) is h+0.25
the radius at height h will be (h+0.25)/10

as a check, when h=2, r = 2.25/10= 0.225 and the diameter will be 2*0.225= 0.45 which matches with their
(for example at a height of 2 m the tank has a diameter of 0.45 m)

Answer 12

thanks I wish I had another medal.

Answer 13

post what you get for the volume of the truncated cone as a function of height h.

Answer 14

i am making a model in Simulink this is for two tanks which are joined which support the movement of a piston.

Answer 15

(1/3)*pi*(r^2)*height

rearranged for

height = Vol / (1/3)*pi*(r^2)

Answer 16

You could also use this formula for the volume of a truncated cone
http://keisan.casio.com/has10/SpecExec.cgi?id=system/2006/1223372110

Answer 17

that right phi?

Answer 18

yes, though I would write 1/(1/3) as 3, so
but we have to be careful, because we have a truncated cone

Answer 19

ok thanks.

Answer 20

if we let H be the height of the entire cone, and h be the height of the truncated cone, you can say
H= h+0.25

the volume of the entire cone is
\[ V_{\text{entire}} = \frac{\pi}{3} r^2 H \]
the volume of the truncated cone is the entire volume minus the little cone
\[ V_{\text{truncated}} = \frac{\pi}{3} r^2 H - \frac{\pi}{3} 0.025^2 \cdot 0.25 \]

using the fact that r= H/10, we can write

\[ V_{\text{truncated}} = \frac{\pi}{3} \frac{H^2}{10^2} H - \frac{\pi}{3} \frac{0.25^2}{10^2} \cdot 0.25 \\
V_{\text{truncated}} = \frac{\pi H^3}{300} - \frac{\pi 0.25^3}{300}
\]

Answer 21

you can solve for H, and then h= H-0.25