14.7 Viscosity and turbulence  (Page 6/14)

 Page 6 / 14

Summary

• Laminar flow is characterized by smooth flow of the fluid in layers that do not mix.
• Turbulence is characterized by eddies and swirls that mix layers of fluid together.
• Fluid viscosity $\eta$ is due to friction within a fluid.
• Flow is proportional to pressure difference and inversely proportional to resistance:
$Q=\frac{p-2{p}_{1}}{R}.$
• The pressure drop caused by flow and resistance is given by ${p}_{2}–{p}_{1}=RQ$ .
• The Reynolds number ${N}_{\text{R}}$ can reveal whether flow is laminar or turbulent. It is ${N}_{\text{R}}=\frac{2\rho vr}{\eta }$ .
• For ${N}_{\text{R}}$ below about 2000, flow is laminar. For ${N}_{\text{R}}$ above about 3000, flow is turbulent. For values of ${N}_{\text{R}}$ between 2000 and 3000, it may be either or both.

Key equations

 Density of a sample at constant density $\rho =\frac{m}{V}$ Pressure $p=\frac{F}{A}$ Pressure at a depth h in a fluid of constant density $p={p}_{0}+\rho gh$ Change of pressure with height in a constant-density fluid $\frac{dp}{dy}=\text{−}\rho g$ Absolute pressure ${p}_{\text{abs}}={p}_{\text{g}}+{p}_{\text{atm}}$ Pascal’s principle $\frac{{F}_{1}}{{A}_{1}}=\frac{{F}_{2}}{{A}_{2}}$ Volume flow rate $Q=\frac{dV}{dt}$ Continuity equation (constant density) ${A}_{1}{v}_{1}={A}_{2}{v}_{2}$ Continuity equation (general form) ${\rho }_{1}{A}_{1}{v}_{1}={\rho }_{2}{A}_{2}{v}_{2}$ Bernoulli’s equation $p+\frac{1}{2}\rho {v}^{2}+\rho gy=\text{constant}$ Viscosity $\eta =\frac{FL}{vA}$ Poiseuille’s law for resistance $R=\frac{8\eta l}{\pi {r}^{4}}$ Poiseuille’s law $Q=\frac{\left({p}_{2}-{p}_{1}\right)\pi {r}^{4}}{8\eta l}$

Conceptual questions

Explain why the viscosity of a liquid decreases with temperature, that is, how might an increase in temperature reduce the effects of cohesive forces in a liquid? Also explain why the viscosity of a gas increases with temperature, that is, how does increased gas temperature create more collisions between atoms and molecules?

When paddling a canoe upstream, it is wisest to travel as near to the shore as possible. When canoeing downstream, it is generally better to stay near the middle. Explain why.

The water in the center of the stream is moving faster than the water near the shore due to resistance between the water and the shore and between the layers of fluid. There is also probably more turbulence near the shore, which will also slow the water down. When paddling up stream, the water pushes against the canoe, so it is better to stay near the shore to minimize the force pushing against the canoe. When moving downstream, the water pushes the canoe, increasing its velocity, so it is better to stay in the middle of the stream to maximize this effect.

Plumbing usually includes air-filled tubes near water faucets (see the following figure). Explain why they are needed and how they work.

Doppler ultrasound can be used to measure the speed of blood in the body. If there is a partial constriction of an artery, where would you expect blood speed to be greatest: at or after the constriction? What are the two distinct causes of higher resistance in the constriction?

You would expect the speed to be slower after the obstruction. Resistance is increased due to the reduction in size of the opening, and turbulence will be created because of the obstruction, both of which will clause the fluid to slow down.

Sink drains often have a device such as that shown below to help speed the flow of water. How does this work?

Problems

(a) Calculate the retarding force due to the viscosity of the air layer between a cart and a level air track given the following information: air temperature is $20\phantom{\rule{0.2em}{0ex}}\text{°C}$ , the cart is moving at 0.400 m/s, its surface area is $2.50\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2},$ and the thickness of the air layer is $6.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\text{m}$ . (b) What is the ratio of this force to the weight of the 0.300-kg cart?

a. $3.02\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−}3}\phantom{\rule{0.2em}{0ex}}\text{N}$ ; b. $1.03\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−}3}$

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