Question

A water storage tank initially contains \(400 \mathrm{~m}^{3}\) of water. The average daily usage is \(40 \mathrm{~m}^{3}\). If water is added to the tank at an average rate of \(20[\exp (-t / 20)] \mathrm{m}^{3}\) per day, where \(t\) is time in days, for how many days will the tank contain water?

Short answer

The tank will contain water for approximately 10 days.

Step-by-step solution

- Understand the problem

The tank starts with 400 cubic meters (m³) of water. Water is used at a constant rate of 40 m³ per day, while it is replenished at a rate of 20[exp(-t/20)] m³ per day. We need to find the time in days when the water in the tank will be depleted.

- Set up the equation for water in the tank

Let the amount of water in the tank at time t be W(t). Initially, W(0) = 400. The differential equation to represent the water in the tank is: \( \frac{dW}{dt} = -40 + 20 \text{exp}(-t/20) \)

- Solve the differential equation

Rewrite the differential equation: \( \frac{dW}{dt} = -40 + 20 \text{exp}(-t/20) \)Separate variables and integrate both sides: \( \frac{dW}{dt} = -40 + 20 \text{exp}(-t/20) \)Integrate: \( \frac{dW}{dt} = -40t - 400 + 400e^{\frac{-t}{20}} \)

- Determine the integration constant

When t = 0, W(0) = 400, solving for C:\( 400 = -40 \times 0 - 400 + 400e^{0}+C\)\( C = 400\)

- Solve for when the tank is empty

Set W(t) = 0 and solve for t: \( 0 = -40t - 400 + 400e^{\frac{-t}{20}}+400 \)\( 0 = -40t + 400e^{\frac{-t}{20}} \)Rearrange and solve: \( 40t = 400-400e^{\frac{-t}{20}} \)\( t = 10(10-e^{\frac{-t}{20}}) \)

Key concepts

Differential Equations in Thermodynamics

Differential equations are crucial in thermodynamics for modeling systems and predicting their behavior over time. In the context of our water storage tank problem, we use a differential equation to represent changes in the water volume.

The differential equation set up in this problem is: \(\frac{dW}{dt} = -40 + 20 \text{exp}(-t/20)\). Here, \(\frac{dW}{dt}\) represents the rate of change of water in the tank. The term \(-40\) reflects the constant daily water usage, while the term \(20 \text{exp}(-t/20)\) models the replenishing water flow that decays exponentially with time.

Solving this differential equation helps us understand how the water volume decreases and when the tank will be empty. Knowing how to solve these equations is essential for engineers to predict system behavior and make necessary adjustments over time.

Water Storage Tank Problems

Water storage tank problems, like the one we’re exploring, often involve balancing input (water addition) and output (water usage). Such systems are modeled to predict the time at which specific conditions (like an empty tank) are met.

In this situation, we start with an initial amount of 400 cubic meters of water in the tank. The daily usage is a constant 40 cubic meters, while the water is added at a decaying rate given by \(20 \text{exp}(-t/20)\, \text{m}^3 \text{per day}\).

To solve these problems:

• First, set up the differential equation based on given rates.
• Next, integrate to find the function of water volume over time.
• Then, apply initial conditions to find the integration constant.
• Lastly, solve for when the water volume (W(t)) becomes zero.

Understanding the inflow and outflow rates helps design and manage systems like water tanks efficiently.

Exponential Decay in Engineering

Exponential decay is a fundamental concept in engineering used to describe processes that diminish over time. In our tank problem, the refill rate \(20 \text{exp}(-t/20)\, \text{m}^3 \text{per day}\) follows an exponential decay pattern.

This is typical in many engineering scenarios where a quantity decreases at a rate proportional to its current value.

• Exponential decay is characterized by the formula: \(Q(t) = Q_0 \text{exp}(-kt)\), where \(Q(t)\) is the quantity at time t, \(Q_0\) is the initial quantity, and k is the decay constant.
• In our scenario, 20 cubic meters is the initial replenishment rate and 1/20 is the decay constant.

Through understanding exponential decay, engineers can predict how quickly resources are depleted, make informed decisions, and optimize resource use efficiently. This is vital in fields ranging from chemical engineering to environmental management.