91Ó°ÊÓ

The following pressures for the air flow in Problem 4.24 were measured: $$\begin{array}{cccc} & x=0 & x=10 \mathrm{m} & x=20 \mathrm{m} \\ \hline t=0 \mathrm{s} & p=101 \mathrm{kPa} & p=101 \mathrm{kPa} & p=101 \mathrm{kPa} \\ t=1.0 \mathrm{s} & p=121 \mathrm{kPa} & p=116 \mathrm{kPa} & p=111 \mathrm{kPa} \\ t=2.0 \mathrm{s} & p=141 \mathrm{kPa} & p=131 \mathrm{kPa} & p=121 \mathrm{kPa} \\ t=3.0 \mathrm{s} & p=171 \mathrm{kPa} & p=151 \mathrm{kPa} & p=131 \mathrm{kPa} . \end{array}$$ Find the local rate of change of pressure \(\partial p / \partial t\) and the convective rate of change of pressure \(V\) on \(/ \partial x\) at \(t=2.0 \mathrm{s}\) and \(x=10 \mathrm{m}.\)

Step by step solution

01

Calculate the local rate of change of pressure (\(\partial p / \partial t\))

The local rate of change of pressure at \(t=2.0 \mathrm{s}\) and \(x=10 \mathrm{m}\) can be calculated by taking the average of the change in pressure over the change in time between \(t=1.0 \mathrm{s}\) and \(t=3.0 \mathrm{s}\) at \(x=10 \mathrm{m}\). Accordingly, \(\partial p / \partial t = \frac{p(t=3.0 \mathrm{s}) - p(t=1.0 \mathrm{s})}{3.0 \mathrm{s} - 1.0 \mathrm{s}} = \frac{151 \mathrm{kPa} - 116 \mathrm{kPa}}{3.0 \mathrm{s} - 1.0 \mathrm{s}} = 17.5 \mathrm{kPa/s}\)

02

Calculate the convective rate of change of pressure (\(\partial p / \partial x\))

The convective rate of change of pressure at \(t=2.0 \mathrm{s}\) and \(x=10 \mathrm{m}\) can be calculated by taking the average of the change in pressure over the change in distance between \(x=0 \mathrm{m}\) and \(x=20 \mathrm{m}\) at \(t=2.0 \mathrm{s}\). Accordingly, \(\partial p / \partial x = \frac{p(x=20 \mathrm{m}) - p(x=0 \mathrm{m})}{20 \mathrm{m} - 0 \mathrm{m}} = \frac{121 \mathrm{kPa} - 141 \mathrm{kPa}}{20 \mathrm{m} - 0 \mathrm{m}} = -1 \mathrm{kPa/m}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Gradient

The pressure gradient is a crucial concept in fluid mechanics that describes how pressure changes across a distance. Imagine a field where you're gently sloping downward—this slope is similar to a pressure gradient, dictating how pressure varies in a given space. In the context of air flow, the pressure gradient gives insight into how pressure differences drive fluid movement. It is often represented by \(\frac{\partial p}{\partial x}\), where \(p\) is pressure and \(x\) is distance.

• A positive gradient indicates increasing pressure in the direction of flow.
• A negative gradient suggests decreasing pressure.
• In our exercise, this was calculated at \(t=2.0\) seconds to be \(-1 \ \text{kPa/m}\), indicating a decrease in pressure as you move through the air flow.

Understanding the pressure gradient helps in predicting fluid behavior and designing systems like pipelines.

Local Rate of Change

The local rate of change of pressure refers to how pressure varies with time at a specific position. Think of it as watching a balloon inflate—measuring how quickly the pressure changes at one point. This is particularly important for understanding dynamic environments where conditions aren't static.

• Mathematically, it's represented as \(\frac{\partial p}{\partial t}\), where \(p\) is pressure and \(t\) is time.
• In the example, the rate was found to be \(17.5 \ \text{kPa/s}\) at \(x=10 \ \text{m}\) and \(t=2.0\) seconds.
• This shows how rapidly the pressure is changing at that spot, highlighting areas of rapid inflation or deflation.

This kind of analysis is vital in weather forecasting and other fluid dynamics applications.

Convective Rate of Change

The convective rate of change combines both spatial and temporal aspects of pressure variation. In simpler terms, it accounts for changes due to movement such as wind carrying pressure differences along a path. Imagine the pressure as being transported with the air currents—this notion captures the essence of convection in fluids.

• The formula for convective change is given by \(V \frac{\partial p}{\partial x}\), where \(V\) is the velocity of the fluid.
• In our problem, it's evaluated at \(t=2.0\) seconds and \(x=10 \ \text{m}\), relating to the spatial derivative already calculated.
• Understanding convective change allows engineers to design ventilation systems and predict pollutant dispersion in the atmosphere.

By comprehending this rate, one can better anticipate how fluid systems behave under dynamic conditions.