Problem 60 - Calculus Single Variable

Question

A volume \(V_{0}\) of gas is held at pressure \(p_{0}\) in a reservoir. The gas is discharged through a nozzle of opening area \(A\) into a region at lower pressure \(p\). Then the rate of discharge (in units of weight/time) is given by $$ A \sqrt{\frac{2 g \gamma}{\gamma-1} \frac{p_{0}}{V_{0}}\left(\left(\frac{p}{p_{0}}\right)^{2 / \gamma}-\left(\frac{p}{p_{0}}\right)^{(\gamma+1) / \gamma}\right)} $$ where \(\gamma\) is the adiabatic constant of the gas. What is \(p / p_{0}\) when the rate of discharge is greatest?

Step-by-Step Solution

Answer

The discharge rate is greatest when \( \frac{p}{p_{0}} = \left(\frac{2}{\gamma+1}\right)^{\gamma} \).

Step 1: Understanding the Problem

We need to find the ratio \( p / p_{0} \) that maximizes the rate of discharge of the gas formula provided. The formula for the discharge rate is a function of \( p / p_{0} \). We will focus on finding the \( p / p_{0} \) that yields the highest value of this function.

Step 2: Set Up the Function to Maximize

The function we need to maximize is: \[ f(x) = x^{2/\gamma} - x^{(\gamma+1)/\gamma} \] where \( x = \frac{p}{p_{0}} \). This is derived from the formula given for discharge rate.

Step 3: Find the Derivative

Calculate the derivative of the function \( f(x) \) with respect to \( x \). The derivative \( f'(x) \) is \( \frac{2}{\gamma}x^{2/\gamma-1} - \frac{(\gamma+1)}{\gamma}x^{(\gamma+1)/\gamma-1} \).

Step 4: Set Derivative to Zero

To find the critical points, set the derivative \( f'(x) = 0 \): \[ \frac{2}{\gamma}x^{2/\gamma-1} = \frac{(\gamma+1)}{\gamma}x^{(\gamma+1)/\gamma-1} \].

Step 5: Solve for Critical Points

Simplify the equation: \[ 2x^{2/\gamma} = (\gamma+1)x^{(\gamma+1)/\gamma} \]. Move terms to one side: \[ 2 = (\gamma+1)x^{1/\gamma} \]. Solving for \(x \) gives \[ x = \left(\frac{2}{\gamma+1}\right)^{\gamma} \].

Step 6: Determine Maximal Condition

To ensure this critical point is a maximum, examine the second derivative or analyze the behavior of \( f'(x) \) around this critical point. Based on the typical behavior for this type of function, the maximum discharge condition obtained is valid.

Key Concepts

Gas Dynamics

Gas dynamics is the study of gas flow and its interactions with different boundaries, such as nozzles or pipes. It focuses on understanding how gases behave under various conditions of pressure, temperature, and volume, especially when they move at high speeds.
In the exercise, we consider gas dynamics to determine how gas flows through a nozzle from a reservoir with higher pressure to a region of lower pressure. The discharge process involves calculating the rate at which gas is expelled based on the pressure differences and other factors like the opening area of the nozzle.

• Pressure plays a central role: the higher the difference in pressure, the faster the gas moves.
• The concept of discharge relates to the amount of gas (by weight) passing through the nozzle per unit of time.

The formula given in the exercise provides a mathematical representation to calculate this discharge rate, considering factors like the adiabatic constant \( \gamma \) and gravitational acceleration \( g \). Understanding the basics of gas dynamics is essential for solving the exercise, as it involves optimizing the gas flow rate through a nozzle.

Derivative Calculation

Derivative calculation is a fundamental concept in calculus used to find the rate at which a function changes at any point. It helps in understanding how the output of a function  in this case, the discharge rate  changes when there's a small change in input variables.
In our problem, the function \( f(x) \) needs to be differentiated to find the condition that maximizes the discharge rate. The derivative \( f'(x) \) is calculated as:
\[ f'(x) = \frac{2}{\gamma}x^{2/\gamma-1} - \frac{(\gamma+1)}{\gamma}x^{(\gamma+1)/\gamma-1} \]
This result shows how the discharge rate changes with respect to the variable \( x = \frac{p}{p_{0}} \).

• We use derivative rules on power functions to get these expressions.
• The derivative tells us where the function's slope is zero, i.e., potential points of maximum or minimum discharge.

Derivative calculations provide the critical information needed to identify crucial points for optimizing the discharge formula given.

Critical Points Analysis

Critical points in calculus are where the function's derivative is zero or undefined. These points help identify where the function may achieve its maximum or minimum values.
In this exercise, we set the derivative \( f'(x) = 0 \) to find critical points. Solving this equation helps determine the exact conditions \( x \) under which the discharge rate is optimized. Simplifying gives us:
\[ 2 = (\gamma+1)x^{1/\gamma} \], leading to the critical point \( x = \left(\frac{2}{\gamma+1}\right)^{\gamma} \).

• Critical points give potential locations for the highest or lowest values of functions.
• By solving for \( x \), we determine the optimal \( \frac{p}{p_{0}} \) ratio.

This analysis is crucial as it provides the specific condition for the gas discharge rate to be at its peak. Knowing how to identify and analyze these points is essential in optimization tasks.

Adiabatic Process

The adiabatic process is a type of thermodynamic process in which there is no heat exchange with the surroundings. This is essential when observing real-world systems, as it simplifies calculations by assuming that all changes in the system are due to pressure and volume changes alone.
In the context of our exercise, the adiabatic constant \( \gamma \) plays a significant role. It represents the ratio of specific heat capacities (i.e., heat capacity at constant pressure to that at constant volume) of the gas, influencing how pressure and volume affect the discharge rate.

• An adiabatic process assumes quick changes, such as the rapid discharge of gas, preventing heat transfer.
• The constant \( \gamma \) captures intrinsic properties of the gas, pivotal in defining how compressible or expansive it is under pressures.

Understanding the adiabatic process and the significance of \( \gamma \) is crucial. It provides insights into how to adjust and control conditions in systems dealing with gas flow and dynamic pressure changes, as shown in the exercise.