Question

What is an appropriate differential equation, and what is the answer for V?

A cylindrical water tank is 3 meters in diameter and 6 meters tall. Water is allowed to drain out of a hole at the bottom. The flow rate is proportional to `sqrtd`, the square root of the depth of the water in the tank, and is therefore also proportional to `sqrtV` , the square root of the volume of the water in the tank.
a) Write an appropriate differential equation. Integrate to solve the equation for V.
b) `V_0=40m^3`, and at first `(dV)/(dt)= -1.6(m^3/min)`. Find the particular solution.
c) If water flowed consistently at `1.6m^3/min`, it would require 25 minutes for the tank to empty. Explain why this is not the case.

Answer 1

part a

The flow rate of part b shows us the form of the differential so that we may write a proportion:

`(dV)/(dt) prop sqrtV`

Replace the proportion with an equal sign and a proportionality constant, `k`:

`(dV)/(dt) = ksqrtV" [1]"`

Equation [1] is the appropriate differential equation.

The equation is separable:

`1/sqrtVdV = kdt`

Integrate both sides:

`int1/sqrtVdV = intkdt`

`2sqrtV = kt+C`

`sqrtV = 1/2kt+C`

`V = (1/2kt+C)^2" [2]"`

Equation [2] is a form of the solution (you can square the right side to obtain other forms).

part b

Equation [1] allows us to find the value of `k`:

`-1.6" "m^3/min = ksqrt(40m^3)`

`k ~~ -6.32" "m^(3/2)/min`

Equation [2] allows us to find the value of C:

`40m^3 = (1/2k(0)+C)^2`

`C = 2sqrt10m^(3/2)`

Substitute the values for `k` and `C` into equation [2]:

`V = (1/2(-6.32" "m^(3/2)/min)t+2sqrt10m^(3/2))^2" [2.1]"`

Equation [2.1] is the particular solution.

part c

The given flow rate multiplied by the time gives the following volume:

`(1.6m^3/min)(25min) = 40m^3`

The volume of the tank is:

`pi(1.5m)^2(6m) ~~ 42.4m^3`

If the initial condition of the tank is full, there will be `2.4m^3` remaining in the tank after 25 mins.

Answer 2

See below

Explanation:

(a)

On the basic of continuity, the flow rate , which is in units of `m^3 \ min^(-1)`, is also `(dV)/(dt)`, the change in volume of fluid in the vessel.

Therefore:

• `(dV)/(dt) = -k sqrtV, qquad k >0`

[That minus sign is there because the volume of fluid is decreasing through time.]

This separates:

`int (dV)/sqrtV = -k int \ dt`

And solves:

`2sqrtV = - kt + C`

You can re-arrange that if you like into: `V = V(t)`

(b)

Using the IV's:

• `2sqrt(40) = - k(0) + C implies C = 4sqrt(10)`

• `((dV)/(dt))_o = -k sqrt(V_o) implies - 1.6 = - k sqrt(40) implies k = (2 sqrt2)/(5sqrt5)`

I.e.:

`2sqrtV = - (2 sqrt2)/(5sqrt5)t + 4sqrt10`

`implies V(t) = ( 2sqrt10 - ( sqrt2)/(5sqrt5)t )^2`

(c)

`V(tau) = 0 implies tau = 50 " mins"`

In practise, it takes twice as long for the tank to drain. This is predictable. The more fluid in the vessel, the higher the pressure at the hole at the bottom. Gravity is at the centre of this dynamic. It is pulling the volume of fluid down toward to the hole such that the pressure/force on water particles at the hole is according to the mass of fluid still in the vessel.

[This is actually Toricelli's Law in disguise, which is a result of Bernoulli's Theorem: `P + rho g h + 1/2 rho v^2 = C`. ]