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Chapter 4: Problem 54

A gas flows along the \(x\) axis with a speed of \(V=5 x \mathrm{m} / \mathrm{s}\) and eressure of \(p=10 x^{2} \mathrm{N} / \mathrm{m}^{2},\) where \(x\) is in meters. (a) Determine the time rate of change of pressure at the fixed location \(x=1\) (b) Determine the time rate of change of pressure for a fluid particle flowing past \(x=1 .\) (c) Explain without using any equations why the answers to parts (a) and (b) are different.

Short Answer

Expert verified
The time rate of change of pressure at the fixed location \(x=1\) is \(0 \mathrm{N/m^2\,s}\). The time rate of change of pressure for a fluid particle flowing past \(x=1\) is \(100 \mathrm{N/m^2\,s}\). The difference is due to the movement of the fluid particle in the pressure gradient.

Step by step solution

01

Determine the rate of change of pressure at x=1

It is given that the pressure vary with position \(x\) as \(p = 10x^2\). The change in pressure with time at a fixed location is given by the partial derivative of pressure with respect to time. Since the pressure does not directly depend on time, the rate of change of pressure at the fixed location \(x=1\) will be zero.
02

Determine the rate of change of pressure for a fluid particle flowing past x=1

The change in pressure along the fluid particle is given by the convective derivative of the pressure. It can be computed using the formula:\[\frac{dp}{dt} = \frac{鈭俻}{鈭倀} + V \frac{dp}{dx}\]As mentioned before, \(\frac{鈭俻}{鈭倀}\) is zero, while \(\frac{dp}{dx}\) is the rate of change of pressure with respect to position. Given \(p = 10x^2\), \(\frac{dp}{dx}\) is \(20x\). At \(x =1\) and given \(V = 5\), by subbing these into the formula:\[\frac{dp}{dt} = 0 + 5 \times 20 \times 1 = 100 \mathrm{N/m^2\,s}\]
03

Explain the difference between answers to parts (a) and (b)

The rate of change of pressure at a fixed point is zero because there's no movement to cause a variation in pressure. On the other hand, for a fluid particle flowing along the x axis, it experiences changes in pressure due to its movement in the pressure gradient. Hence, the rate of change of pressure along a flowing fluid particle is non-zero.

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Key Concepts

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

Pressure Variation
Pressure variation in fluid dynamics is about how the pressure changes within a fluid flow related to space. In our exercise, the pressure of the gas flowing along the x-axis is given by the equation:
  • \( p = 10x^2 \)
This simple expression indicates that the pressure is not uniform across different points along the x-axis. In other words, pressure changes as the position \( x \) changes.
If we want to find how fast the pressure is changing as we move along the x-axis, we calculate the derivative with respect to \( x \). This gives us the rate of change of pressure with respect to space, \( \frac{dp}{dx} = 20x \), indicating a linear increase in pressure with distance from the origin.
It鈥檚 essential to understand that such spatial variation in pressure occurs independently of time, as long as you're looking at a fixed location on the axis.
Convective Derivative
The convective derivative is a vital concept in fluid dynamics that accounts for changes experienced by a moving fluid particle.
When a particle flows past different points, it undergoes changes not only due to its location but also owing to its motion.
  • The convective derivative formula is given by:
\[\frac{dp}{dt} = \frac{鈭俻}{鈭倀} + V \frac{dp}{dx}\]
  • \( \frac{鈭俻}{鈭倀} \) represents the rate of pressure change over time at a fixed position, zero in our case since pressure only varies with position.
  • \( V \frac{dp}{dx} \) signifies the pressure change a particle experiences due to movement through a pressure gradient.
In the exercise, with \( V = 5 \) m/s and \( \frac{dp}{dx} = 20x \), for \( x = 1 \), the equation becomes:
  • \( \frac{dp}{dt} = 5 \cdot 20 \cdot 1 = 100 \text{ N/m}^2 \text{/s} \)
This calculation shows how the moving particle experiences a pressure change of 100 N/m虏/s.
Rate of Change of Pressure
The rate of change of pressure is a measure of how quickly the pressure is increasing or decreasing. In our given exercise, the confusion might arise from comparing two different scenarios:
  • A fixed point along the x-axis (part a).
  • A fluid particle moving along the axis (part b).
When considering a fixed location at \( x = 1 \), the rate of change of pressure with time is zero because the pressure expression \( p = 10x^2 \) only changes with \( x \). Therefore, it does not change with time if \( x \) is held constant.
However, a fluid particle moving through the x-axis experiences changes in pressure because it flows through varying pressure levels along its path due to its velocity \( V = 5 \) m/s. From this perspective, the pressure does change as the particle moves.
This time-varying pressure for a moving particle is quantified by the convective derivative, which combines the movement's effect and is not present when considering a static point.

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