Question 10.2.1: Mass Balance on a Water Storage Tank A 12.5-m³ tank is being......
Mass Balance on a Water Storage Tank
A 12.5-m³ tank is being filled with water at a rate of 0.050 m³ /s. At a moment when the tank contains 1.20 m³ of water, a leak develops in the bottom of the tank and gets progressively worse with time. The rate of leakage (m³ /s) can be approximated as 0.0025t, where t(s) is the time from the moment the leak begins.
1. Write a mass balance on the tank and use it to obtain an expression for dV/dt, where V is the volume of water in the tank at any time. Provide an initial condition for the differential equation.
2. Solve the balance equation to obtain an expression for V(t) and draw a plot of V versus t.
1. The total mass of the tank contents is M(kg) = ρ(kg/m³)V(m³), where ρ = 1000 kg/m³ is the density of liquid water. Then
accumulation (kg/s) = \frac{d(\rho V)}{dt} = \rho \frac{dV}{dt}
(The second step follows from the fact that the density of the liquid water in the tank is independent of time and so may be taken out of the derivative.)
input (kg/s) = ρ(kg/m³)(0.05 m³/s) = 0.05ρ
output (kg/s) = ρ(kg/m³)[0.0025t(m³/s)] = 0.0025ρt
Substituting these terms into the water balance equation (accumulation = input – output) and canceling ρ yields the differential equation
\boxed{\begin{matrix}\frac{dV}{dt} = 0.050 m^{3}/s – 0.0025 t \\ t=0 , V= 1.2 m^{3}\end{matrix} }
Verify that each term in the equation (including dV/dt) has units of m³/s.
2. To solve the equation, we separate variables (bring dt to the right-hand side) and integrate from the initial condition (t = 0, V = 1.2 m³) to an arbitrary time, t, and corresponding volume, V.
dV (m^{3}) = (0.050 – 0.0025 t)dt \Longrightarrow \int_{1.2 m^{3}}^{V}{dV} = \int_{0}^{t}{(0.050 – 0.0025 t)dt} \\ \Longrightarrow V \left. \Large{]} \right.^{V}_{1.2 m^{3}} = \left(0.050 t – 0.0025 \frac{t^{2}}{2} \right) \left. \Large{]} \right.^{t}_{0} \\ \Longrightarrow V(m^{3}) = 1.2 m^{3} +0.050 t – 0.00125t^{2}
Check 1: When t = 0, V = 1.2 m³ (confirming the given initial condition). \pmb{✓}
Check 2: dV/dt = 0.050 – 0.0025t [differentiating V(t) yields the original equation for dV/dt]. \pmb{✓}
A plot of the derived expression for V(t) is as follows:
Initially, the filling causes the volume of the tank contents to increase, but as the leak gets larger the tank begins to drain. The maximum volume is 1.7 m³ , well below the tank capacity of 12.5 m³ . At about t = 57 s the contents drain completely. The mathematical formula for V predicts negative volumes after this time, but physically the volume must remain at zero (the liquid discharges as fast as it is poured in). The actual solution of the balance equation is therefore
\boxed{\begin{matrix} V(m^{3}) =1.2 +0.050t – 0.00125t^{2} &0\leq t \leq 57 s\\=0 &t>57 s \end{matrix} }
The plot shown above should be changed in the range t > 57 s to a line coincident with the t axis.
