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[fluid mechanics] The velocity profile, \(v\), of a fluid is given by \(\begin{aligned} v=& 3 x^{2}-x^{3} \\ &(x \text { is between } 0 \text { and } 3) \end{aligned}\) where \(x\) represents distance. Plot the graph of \(v\) against \(x\), considering \(x=0,0.5,1,1.5,2.0,2.5\) and \(3.0\)

Short Answer

Expert verified
Calculate values of \( v \) for given \( x \), tabulate results, and plot graph.

Step by step solution

01

- Identify the function and the range

The velocity profile is given by \( v = 3x^2 - x^3 \). The distance \( x \) ranges from \( 0 \) to \( 3 \). Points to consider are: \( x = 0, 0.5, 1, 1.5, 2, 2.5, \text{ and } 3 \).
02

- Calculate velocity at each point

Substitute each value of \( x \) into the equation \( v = 3x^2 - x^3 \).1. For \( x = 0 \), \( v = 3(0)^2 - (0)^3 = 0 \).2. For \( x = 0.5 \), \( v = 3(0.5)^2 - (0.5)^3 = 0.75 - 0.125 = 0.625 \).3. For \( x = 1 \), \( v = 3(1)^2 - (1)^3 = 3 - 1 = 2 \).4. For \( x = 1.5 \), \( v = 3(1.5)^2 - (1.5)^3 = 6.75 - 3.375 = 3.375 \).5. For \( x = 2 \), \( v = 3(2)^2 - (2)^3 = 12 - 8 = 4 \).6. For \( x = 2.5 \), \( v = 3(2.5)^2 - (2.5)^3 = 18.75 - 15.625 = 3.125 \).7. For \( x = 3 \), \( v = 3(3)^2 - (3)^3 = 27 - 27 = 0 \).
03

- Tabulate the points

Create a table with values of \( x \) and corresponding \( v \).\[ \begin{array}{|c|c|} \hline x & v \ \hline 0 & 0 \ 0.5 & 0.625 \ 1 & 2 \ 1.5 & 3.375 \ 2 & 4 \ 2.5 & 3.125 \ 3 & 0 \ \hline \end{array} \]
04

- Plot the graph

Using the table, plot the points \( (x, v) \) on a graph with \( x \) on the horizontal axis and \( v \) on the vertical axis. Connect the points with a smooth curve.

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Key Concepts

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

Velocity Function Plotting
Plotting the velocity function involves a few important steps. First, identify the velocity profile, which in this case is given by the equation \( v = 3x^2 - x^3 \). This describes how the velocity, \(v\), of a fluid changes with distance, \(x\). The range of \(x\) goes from 0 to 3. To plot the function, substitute specific values of \(x\) into the equation to get the corresponding \(v\) values.

Once we have calculated the velocities for points like \( x = 0, 0.5, 1, 1.5, 2, 2.5, \) and \( 3 \), we can tabulate them for easy reference. For example, at \( x = 1 \), the velocity \( v \) equals 2. Doing this for all chosen points gives a series of (x, v) pairs like (0, 0), (0.5, 0.625), etc.

The next step is to plot these points on a graph. Place \(x\) values on the horizontal axis and \(v\) values on the vertical axis. Finally, connect the plotted points with a smooth curve to visualize how the fluid's velocity changes with distance. This graphical representation helps in understanding the fluid's behavior better.
Fluid Dynamics
Fluid dynamics studies the movement of liquids and gases. It involves understanding how forces, like pressure and gravity, interact with fluid particles. This branch of physics has practical applications in engineering, meteorology, medicine, and even computer graphics.

Consider our velocity profile equation \( v = 3x^2 - x^3 \). It reflects how fluid velocity changes over distance. Such an equation is derived from principles governing the motion of fluids.
This might include:
  • Continuity Equation: Ensures mass is conserved in the fluid flow.
  • Navier-Stokes Equations: Describe how viscous fluids move under different conditions of force and pressure.
Real-world applications of fluid dynamics include designing pipe systems, predicting weather patterns, and optimizing airflow over airplane wings. Understanding fluid dynamics allows engineers and scientists to predict and control how fluids will behave under various circumstances, making it a critical area of study.
Mathematical Modelling
Mathematical modeling involves using equations and computational techniques to simulate real-world systems and predict their behavior. The velocity profile \( v = 3x^2 - x^3 \) is an example of such a model in fluid mechanics.

By inputting values of \(x\) (distance) and calculating \(v\) (velocity), we create a mathematical model of how a fluid flows over a distance. This model helps predict the velocity at any given point within the specified range.

Mathematical models in fluid mechanics often use:
  • Differential Equations: To describe how fluid properties like velocity and pressure change over space and time.
  • Boundary Conditions: Specific conditions at the boundaries of the physical system, such as the walls of a pipe.
  • Numerical Methods: Computer algorithms that solve complex equations to simulate fluid flow.
These tools enable engineers to design systems like water distribution networks, oil pipelines, and air conditioning systems efficiently. Models can also help in research areas, such as understanding blood flow in arteries or predicting pollutant dispersion in the atmosphere. Accurate modeling leads to better design, control, and optimization of fluid systems in various industries.

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Most popular questions from this chapter

[fluid mechanics] The following equations are the velocity distributions \(v\) for a fluid: a \(v=2 x^{2}-x^{3}\) b \(v=4 x^{2}-2 x^{3}\) where \(x\) is the distance and is between 0 and 2 . Sketch the graphs of \(v\) against \(x\) on the same axes.

across a variable load resistor of resistance \(R\), is given by \(V=\frac{10 R}{5+R}\). Plot the graph of \(V\) versus \(R\) for \(0 \leq R \leq 10\).

[mechanics] The velocity, \(v\), of a particle is given by $$ v=48-3 t^{2} \quad \text { where } 0 \leq t \leq 4 $$ Sketch the graph of \(v\) against \(t\). What can you conclude about the acceleration of the particle?

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[electrical principles] The voltage, \(V\), across a variable load resistor of resistance \(R_{\mathrm{L}}\), is given by $$ V=\frac{E R_{\mathrm{L}}}{R+R_{\mathrm{L}}} $$ where \(E\) is the source e.m.f. and \(R\) is the source resistance. Plot the graphs (on different axes) of \(V\) versus \(R_{\mathrm{L}}\) for the corresponding values: a \(E=60\) volts, \(R=10 \Omega\) for \(0 \leq R_{\mathrm{L}} \leq 20\) b \(E=15\) volts, \(R=3 \times 10^{3} \Omega\) for \(0 \leq R_{\mathrm{L}} \leq 6 \times 10^{3}\) c \(E=10\) volts, \(R=15 \times 10^{3} \Omega\) for \(0 \leq R_{\mathrm{L}} \leq 30 \times 10^{3}\) From each of your graphs, determine the value of \(V\) at \(R_{\mathrm{L}}=R\). Do you notice any relationship between \(V\) and \(E\) at \(R_{\mathrm{L}}=R\) ? Show algebraically that if \(R_{\mathrm{L}}=R\) then \(V=\frac{E}{2}\).

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