Turbulence - AP Physics 2

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Question

Turbulent flow is characterized by __________.

Answer

Turbulent flow is caused by sufficiently high flow rates, and is characterized by chaotic and irregular flow patterns. Streamlines represent the curves tangent to the point of the direction of flow; thus, if a flow is turbulent the streamlines will be chaotic as well.

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Question

Which of the following is an example of turbulent flow?

Answer

The wisps of smoke from a blown out candle start out a smooth, but eventually swirl and become chaotic and turbulent. The other examples can be intuitively thought of as smooth flow, and thus are laminar.

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Question

A baseball is thrown at a catcher with a high velocity and the ball passes right by the glove of the catcher. Which of the following scenarios will occur because of this?

Answer

As the baseball passes by the glove, the air surrounding the glove increases in velocity. The increase in air velocity will cause a decrease in pressure. The decrease in pressure causes the ball to be pulled towards the glove.

Therefore the correct answer is that the ball will be pulled toward the glove because of the decreased air pressure.

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Question

Methanol is traveling through a cylindrical tube with a diameter of . What is the maximum velocity of methanol that maintains laminar flow?

$\rho_m = 792\frac{kg}{m^3}$

$\mu_m = 0.611x10^{-3}Pa\cdot s$

Answer

We will use the expression for Reynold's number for this problem:

$Re = \frac{\rho_m vD_H}{\mu_m}$

Where is the hydraulic diameter, and for a cylindrical tube,

Rearranging for velocity:

$v = \frac{Re\mu_m}{\rho_m D}$

For laminar flow: $Re \leq 10$

Plugging in our values into the expression, we get:

$v = \frac{10(.611x10^{-3}Pa\cdot s)}{\left ( 792\frac{kg}{m^3}\right )(0.02m)}$

$v = 3.9x10^{-4}\frac{m}{s}$

Methanol has a relatively low viscosity, so it only has a very small range for laminar flow and quickly becomes turbulent

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Question

What is the Reynold's number of water flowing through a circular tube of diameter at a rate of $6\frac{m}{s}$ ?

Assume $\mu_w=1.002x10^{-3}Pa\cdot s$ and $\rho_w = 1,000\frac{kg}{m^3}$

Answer

We will use the expression for Reynold's number for this problem:

$Re = \frac{\rho_w vD_H}{\mu_w}$

Where is the hydraulic diameter, and for a cylindrical tube,

Plugging in our values, we get:

$Re = \frac{\left ( 1,000\frac{kg}{m^3}\right )\left ( 6\frac{m}{s}\right )(0.08m)}{\left ( 1.002x10^{-3}Pa\cdot s\right )}$

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Question

What is the Reynold's number of water flowing through a fully filled rectangle duct that is at a velocity $v = 2.5\frac{m}{s}$ ?

Assume $\mu_w=1.002x10^{-3}Pa\cdot s$ and $\rho_w = 1,000\frac{kg}{m^3}$

Answer

We will use the expression for Reynold's number for this problem:

$Re = \frac{\rho_w vD_H}{\mu_w}$

For a fully filled rectangular duct, the hydraulic diameter is:

$D_H = \frac{2wh}{w+h}$

Plugging in values:

$D_H = \frac{2(0.5m\cdot 0.3m)}{0.5m + 0.3m} = 0.375m$

Now we can plug our values into the original expression:

$Re = \frac{\left ( 1,000\frac{kg}{m^3}\right )(2.5\frac{m}{s})(0.375m)}{\left ( 1.002x10^{-3}Pa\cdot s\right )}$

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Question

Honey is flowing through a rectangular duct with width at a velocity of $0.8\frac{m}{s}$ . What is the depth of the honey if the Reynold's number is ?

$\rho_h = 1,420\frac{kg}{m^3}$

$\mu_h = 10Pa\cdot s$

Answer

We will use the expression for Reynold's number for this problem:

$Re = \frac{\rho_w vD_H}{\mu_w}$

Rearranging for hydraulic diameter, we get:

$D_H = \frac{Re\mu_w}{\rho_wv}$

For a partially filled rectangular duct, the expression for hydraulic diameter is:

$D_H = \frac{4hw}{2h+w}$

$\frac{4hw}{2h+w} = \frac{Re\mu_w}{\rho_wv}$

Rearranging for height:

$4hw = \left (\frac{Re\mu_w}{\rho_wv} \right )(2h+w)$

$h\left (4w -\frac{2Re\mu_w}{\rho_wv} \right )= \left (\frac{Re\mu_w}{\rho_wv} \right )w$

$h= \frac{\left (\frac{wRe\mu_w}{\rho_wv} \right )}{\left (4w -\frac{2Re\mu_w}{\rho_wv} \right )}$

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Question

By what factor does the Reynolds's number change for water flowing through a circular tube as the cross-sectional area of the tube gradually triples?

Answer

Let's begin with the expression for Reynold's number:

$Re_1 = \frac{\rho v_1D_{H_{1}}}{\mu}$

$Re_2= \frac{\rho v_2D_{H_{2}}}{\mu}$

Dividing the second expression by the first, we get:

$\frac{Re_2}{Re_1}= \frac{\left (\frac{\rho v_2D_{H_{2}}}{\mu} \right )}{\left (\frac{\rho v_1D_{H_{1}}}{\mu} \right )}$

$\frac{Re_2}{Re_1}= \frac{v_2D_{H_{2}}}{v_1D_{H_{1}}}$ (1)

From the problem statement, we are told:

$3\frac{\pi D_1^2}{4}=\frac{\pi D_2^2}{4}$

$D_2 = \sqrt{3}D_1$

Also, for a circular tube:

Therefore:

$D_{H_{2}}=\sqrt{3}D_{H_{1}}$ (2)

Now we need to determine the change in velocity. From the law of continuity, we know that:

$\dot{V_1}=\dot{V_2}$

Rearranging:

$v_2 = v_1\frac{A_1}{A_2}$

Where:

Thus:

$v_2=\frac{v_1}{3}$ (3)

Plugging in expressions (2) and (3) into (1), we get:

$\frac{Re_2}{Re_1}= \frac{\sqrt{3}v_1D_{H_{1}}}{3v_1D_{H_{1}}}$

$\frac{Re_2}{Re_1}=\frac{\sqrt{3}}{3}$

$\frac{Re_2}{Re_1}= 0.58$

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