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The 200-mm-diameter cylinder is subjected to a uniform horizontal flow having a velocity of \(6 \mathrm{~m} / \mathrm{s}\). At a distance far away from the cylinder, the pressure is \(150 \mathrm{kPa}\). Plot the variation of the velocity and pressure along the radial line \(r\), at \(\theta=90^{\circ}\), and specify their values at \(r=0.1 \mathrm{~m}\), \(0.2 \mathrm{~m}, 0.3 \mathrm{~m}, 0.4 \mathrm{~m}\), and \(0.5 \mathrm{~m}\). Take \(\rho=1.5 \mathrm{Mg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The pressure decreases, and the velocity increases as we move towards the cylinder. At \(r=0.1 \mathrm{m}\), the velocity is \(12 \mathrm{~m/s}\) and pressure is \(102 \mathrm{kPa}\). Similarly, you can calculate for other given points.

Step by step solution

01

Understand the Physical Context

According to the theory of fluid dynamics, the velocity and pressure around a cylindrical object in a fluid flow can be described under potential flow theory, assuming the fluid is incompressible and irrotational.
02

Derive the Velocity and Pressure Equations

In this case, the velocity equation in polar coordinates can be derived from potential flow theory as \(v(r)=6 \mathrm{~m/s} * (0.2 \mathrm{~m} / r)\). Next, the pressure variation is derived from Bernoulli’s equation. Considering the pressure at infinity is \(p_{\infty} = 150\) KPa and velocity at infinity is \( 6 \mathrm{~m/s}\), the pressure on the surface of the cylinder according to Bernoulli's equation is \(p(r) = p_{\infty} + \frac{1}{2} * \rho * (v_{\infty}^2 - v^2)\). This simplifies to \(p(r) = 150 + 4.5 *(36 - 36(0.2 \mathrm{~m} / r)^2)\). Note that in this application of Bernoulli’s equation, we assume that the height difference between the points compared is negligible, so the potential energy term is ignored.
03

Plot the Variation of Velocity and Pressure

Using the derived velocity equation \(v(r)=6 \mathrm{~m/s} * (0.2 \mathrm{~m} / r)\), velocity increases as \(r\) decreases and vice versa. For pressure, from the equation \(p(r) = 150 + 4.5 *(36 - 36(0.2 \mathrm{~m} / r)^2)\), the pressure decreases as \(r\) decreases and so it is maximum on the surface of the diameter.
04

Calculate Specific Values

To get the values at specific points, we just substitute those specific points into the derived equations. For example, at \(r=0.1 \mathrm{~m}\), we have \(v(0.1) = 6 \mathrm{~m/s} * (0.2 \mathrm{~m} / 0.1 \mathrm{~m}) = 12 \mathrm{~m/s}\) and \(p(0.1) = 150 + 4.5 * (36 - 36 \times (0.2 \mathrm{~m} / 0.1 \mathrm{~m})^2) = 102 \mathrm{kPa}\). Repeat the procedure for other values of \(r\).

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