Math/Physics/Science question? Its pretty tough.?

Question

Suppose you have a perfectly spherical water tank with an inside diameter of 8.6 metres. If the drain at the bottom of the tank can't handle a hydrostatic pressure of more than 50 kilopascals, what is the maximum volume of water, in litres, that can be contained in the tank? Assume that gravitational acceleration is exactly 9.81 m/s2. Please round to the nearest 10 litre increment, and please submit only a number for your answer. (For example, if you calculate the answer to be 16277 litres, submit 16280 as your answer)

Thanks for your help.

Answer 1

The hydrostatic pressure at the base of a column of water is equal to:

P = D*g*h

where D is the density of water (1000 kg/m^3)

g is the acceleration of gravity

h is the height of the column.

A pressure of 50 kPa therefore corresponds to a column of water of height:

h = (50*10^3 Pa)/(9.81 m/s^2 *1000 kg/m^3)

h = 5.097 m

The spherical tank has a total interior volume of:

V = (4/3)*pi*(4.3 m)^3

V = 333,038.1 liters

The maximum volume of water that the tank can hold without failure of the drain is equal to the total interior volume minus the unfilled volume. The unfilled volume has the shape of a spherical cap of height (8.6 m - 5.097 m) = 3.503 m. The volume of a spherical cap, in terms of the radius of the sphere (R) and cap height (H) is given by (see source):

V_cap = (1/3) * pi * H^2 * (3*R - H)

The empty part of the sphere then has a volume of:

V_cap = (1/3) * pi * (3.503 m)^2 * (3*4.3 m - 3.503 m)

V_cap = 120.75 m^3 = 120,762.0 liters

The volume of water in the sphere is then:

V_water = V - V_cap = (333,038.1 - 120,762.0) liters

V_water = 212,276.1 liters

Rounded to the nearest 10 liters, this is 212,280 liters