Separate the formal model from inconsistent pump directions

The cylindrical tank in a research laboratory has a cross-sectional area of 4 m². To cool the tank, water is pumped in and out of the tank simultaneously. The volume and height of the water in the tank at any time tt minutes is given by VV (m³) and hh (metres) respectively. Clean water is pumped into the tank at a rate that is proportional to h2h^2 and the water is pumped out from the tank at a rate that is proportional to hh.

(i) Assume that the water does not overflow and that there is no change to the height of the water when h=10h=10, show that dhdt=kh(h−10)4\frac{dh}{dt}=\frac{kh(h-10)}4 where kk is a real constant. marks

The tank was initially filled with clean water to a height of 2 metres. When the height of the water is 5 metres, the volume of water is increasing at a rate of 5.5 m³ per minute.

(ii) Find the exact value of kk. Hence find hh in terms of tt. marks

(iii) Sketch a graph of hh against tt. Hence write down the minimum height of the cylindrical tank that will not result in the overflow of the water. marks

[2022/ASRJC/Prelim/I/11]

Total marks: 12

Solution

(i)

Rate balance

Show that dhdt=kh(h−10)4\dfrac{dh}{dt}=\dfrac{kh(h-10)}4 for a tank of cross-section 4 m², given inflow ∝h2\propto h^2, outflow ∝h\propto h and no change at h=10h=10.

Hint

  1. dVdt=in−out=Ah2−Bh\frac{dV}{dt}=\text{in}-\text{out}=Ah^2-Bh.

  2. V=4hV=4h.

Answer · (i)

h′=kh(h−10)/4h\prime=kh(h-10)/4.

(ii)

Solving

With h=2h=2 initially and dVdt=5.5\frac{dV}{dt}=5.5 at h=5h=5, find kk and h(t)h(t).

Explore this idea

Hint

  1. dVdt=4dhdt\frac{dV}{dt}=4\frac{dh}{dt}.

  2. Separate with partial fractions.

Answer · (ii)

Formal answer: k=−11/50k=-11/50, h=10/(1+4e−11t/20)h=10/(1+4e^{-11t/20}).

Incompatible with positive printed inflow∝h² and outflow∝h; no positive physical constants satisfy all data.

Explore the idea

The tank height

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of h at t = 02
Read the solution at t4
Try an example
h at t = 4
6.929
Rate dh/dt there
1.17
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

The height rises in an S-shape towards 10 m.

Ideas to try
  1. Start above 10: the height falls to 10.
  2. Read hh at t=4t=4 minutes.
(iii)

Interpreting the graph

Sketch hh against tt and give the minimum tank height that avoids overflow.

Hint

  1. As t→∞t\to\infty, e−11t/20→0e^{-11t/20}\to0.

Formal tank height
Formal tank heightScroll across to see the full diagram.

Answer · (iii)

Formal minimum tank height 10 m; height rises sigmoidally from 2 toward 10.

Conditional algebraic model, not a physically consistent solution of the printed pump description.