Question

Two tanks containing a liquid are placed in series so that the first discharges into the second and the second discharges into a waste outlet. Let \(q_{1}(t)\) and \(q_{2}(t)\) be the flow rates out of the two tanks respectively, and let the height of liquid in each of the tanks be \(h_{1}(t)\) and \(h_{2}(t)\). respectively. The two tanks are identical and each has a constant cross-sectional area \(A\). The outflow from each tank is proportional to the height of liquid in the tank. At \(t=0\) the height of liquid in the first tank is \(h_{0}\) and the second tank is empty. (a) Derive and solve the differential equation for \(h_{1}(t)\). (b) Hence find \(q_{1}(t)\). (c) Derive and solve the differential equation for \(h_{2}(t)\). (d) Hence find \(q_{2}(t)\).

Short answer

Question: Find the equations for the heights of the liquid (\(h_1(t)\) and \(h_2(t)\)) and the flow rates (\(q_1(t)\) and \(q_2(t)\)) in both tanks. Answer: The equations for the heights of the liquid are: 1. \(h_1(t) = h_0 \cdot e^{-\frac{k}{A}t}\) 2. \(h_2(t) = \frac{h_0 \cdot A}{k} \left(1 - e^{-\frac{k}{A}t}\right)\) The equations for the flow rates are: 1. \(q_1(t) = k \cdot h_0 \cdot e^{-\frac{k}{A}t}\) 2. \(q_2(t) = h_0 \cdot A \left(1 - e^{-\frac{k}{A}t}\right)\)

Step-by-step solution

Derive the differential equation for \(h_1(t)\)

Since the outflow of the first tank is proportional to the height (\(h_1\)) of the liquid in the tank, we can represent the flow rate \(q_1(t)\) as a constant proportionality (\(k\)) times the height of the liquid in the first tank. $$ q_1(t) = k \cdot h_1(t) $$ Taking the derivative of \(h_1(t)\) with respect to time \(t\): $$ \frac{dh_1(t)}{dt} = -\frac{q_1(t)}{A} $$ We replace \(q_1(t)\) with \(k \cdot h_1(t)\). $$ \frac{dh_1(t)}{dt} = -\frac{k}{A} \cdot h_1(t) $$

Solve the differential equation for \(h_1(t)\)

The differential equation for \(h_1(t)\) is a first-order linear homogeneous differential equation with constant coefficients. To solve it, we can use separation of variables: $$ \frac{dh_1(t)}{h_1(t)} = -\frac{k}{A} dt $$ Integrate both sides: $$ \int \frac{dh_1(t)}{h_1(t)} = -\frac{k}{A} \int dt $$ $$ \ln \left| h_1(t) \right| = -\frac{k}{A}t + C $$ Taking the exponential of both sides: $$ h_1(t) = e^{-\frac{k}{A}t + C} = e^{-\frac{k}{A}t} \cdot e^C $$ Let \(C = \ln{h_0}\), then we get: $$ h_1(t) = h_0 \cdot e^{-\frac{k}{A}t} $$

Find \(q_1(t)\)

By substituting the equation for \(h_1(t)\) back into the expression for \(q_1(t)\), we find: $$ q_1(t) = k \cdot h_1(t) = k \cdot h_0 \cdot e^{-\frac{k}{A}t} $$

Derive the differential equation for \(h_2(t)\)

The inflow to the second tank is equal to the outflow from the first tank, \(q_1(t)\). The outflow from the second tank is given by: $$ q_2(t) = k \cdot h_2(t) $$ Taking the derivative of \(h_2(t)\) with respect to time \(t\): $$ \frac{dh_2(t)}{dt} = \frac{q_1(t)}{A} - \frac{q_2(t)}{A} $$ Replacing \(q_1(t)\) and \(q_2(t)\) with expressions in terms of \(h_1(t)\) and \(h_2(t)\): $$ \frac{dh_2(t)}{dt} = \frac{k \cdot h_1(t)}{A} - \frac{k \cdot h_2(t)}{A} $$

Solve the differential equation for \(h_2(t)\)

We substitute the expression for \(h_1(t)\) into the differential equation for \(h_2(t)\) and solve it using an integrating factor: $$ \frac{dh_2(t)}{dt} = \frac{k \cdot h_0 \cdot e^{-\frac{k}{A}t}}{A} - \frac{k \cdot h_2(t)}{A} $$ Multiply both sides by the integrating factor \(e^{\frac{k}{A}t}\): $$ e^{\frac{k}{A}t} \frac{dh_2(t)}{dt} + \frac{k}{A} e^{\frac{k}{A}t} h_2(t) = \frac{k \cdot h_0}{A} e^{\frac{k}{A}t} $$ The left side is now the derivative of the product \(e^{\frac{k}{A}t} h_2(t)\). $$ \frac{d}{dt} \left( e^{\frac{k}{A}t} h_2(t) \right) = \frac{k \cdot h_0}{A} e^{\frac{k}{A}t} $$ Integrate both sides: $$ e^{\frac{k}{A}t} h_2(t) = \frac{h_0 \cdot A}{k} e^{\frac{k}{A}t} + C $$ $$ h_2(t) = \frac{h_0 \cdot A}{k} + Ce^{-\frac{k}{A}t} $$ Since the second tank is initially empty, \(h_2(0) = 0\): $$ 0 = \frac{h_0 \cdot A}{k} + C $$ $$ C = - \frac{h_0 \cdot A}{k} $$ So the final expression for \(h_2(t)\) is: $$ h_2(t) = \frac{h_0 \cdot A}{k} \left(1 - e^{-\frac{k}{A}t}\right) $$

Find \(q_2(t)\)

By substituting the equation for \(h_2(t)\) back into the expression for \(q_2(t)\), we find: $$ q_2(t) = k \cdot h_2(t) = k \cdot \frac{h_0 \cdot A}{k} \left(1 - e^{-\frac{k}{A}t}\right) $$ $$ q_2(t) = h_0 \cdot A \left(1 - e^{-\frac{k}{A}t}\right) $$

Key concepts

Flow Rate Calculation

Flow rate calculation plays a crucial role in understanding how liquids move through systems, such as tanks in series. When liquids are transferred from one tank to another, the flow rate determines how quickly the liquid travels.
In our tank system, the flow rate out of the tanks is proportional to the liquid height in each tank. This means if we know how high the liquid is in tank 1, we can easily calculate the flow rate using a constant of proportionality, often symbolized by the letter \( k \). The equation for flow rate \( q_1(t) \) out of the first tank is given by \( q_1(t) = k \cdot h_1(t) \). This simple relationship helps us track how much liquid is leaving tank 1 at any given time.
Knowing the flow rate is essential because it connects the physical state of the liquid in the tanks to the mathematical equations we use to model the system. It provides a link between theory and the actual observations of systems commonly seen not only in academic settings but in real-world applications like water treatment and chemical processing.

Liquid Tank Models

Liquid tank models are fascinating because they provide a way to understand how liquids behave in confined spaces. The model in this exercise involves two tanks in series. The first tank empties into the second tank, and the second tank eventually empties through a waste outlet.
Each tank has a cross-sectional area \( A \), which defines how wide or narrow the tank is. An interesting aspect of these models is that the flow out of each tank depends on the liquid's height inside them. The higher the liquid, the faster it can flow out.

• The height \( h_1(t) \) in the first tank starts at a known value \( h_0 \) when time \( t = 0 \).
• The second tank begins empty, making it initially have a height \( h_2(0) = 0 \).

These models not only help in understanding but also controlling liquid flow in industrial systems. By analyzing how each tank behaves, engineers can plan for effective designs and operations in processes involving liquids.

First-Order Linear Differential Equations

First-order linear differential equations are fundamental tools for modeling change over time. Their linearity makes them particularly easy to work with mathematically, offering solutions that are usually neat exponential functions.
In our tank problem, the equation governing the height of liquid in the first tank:\( \frac{dh_1(t)}{dt} = -\frac{k}{A} \cdot h_1(t) \). This is a classic example of a first-order linear differential equation. The main characteristic is that it can be written in the form \( \frac{dy}{dt} + Py = Q \), with \( P \) and \( Q \) being constants or functions of \( t \).
Solving these equations often involves finding an integrating factor or using separation of variables. For the problem given, by separating variables and integrating, we find a solution \( h_1(t) = h_0 \cdot e^{-\frac{k}{A}t} \). This result shows an exponentially decaying system, common for systems losing a proportional part of their quantities over time. Understanding these equations is key to predicting how systems evolve and adjust to changes, making them indispensable in fields like engineering, physics, and even economics.