where {jatex options:inline}p= pressure, \; \rho = density, \; v= fluid \; speed, \; g=9.81m/s^2, \; h=height {/jatex}

If the fluid is flowing horizontally we can write {jatex options:inline}p+ \frac{1}{2} \rho v^2 = constant{/jatex}

The Bernoulli Effect explains why having a strong heart that pumps efficiently will generally result in a lower blood pressure. The heart is a pump. If it can pump more strongly, the blood will flow faster and blood pressure will fall as a result of the Bernoulli effect.

Blood pressure is important because high blood pressure can cause kidney failure and heart problems. The muscles of the heart can be strengthened by exercise, as with any muscles.]]>

The thinner the glass tube, the higher the water will crawl.

Consider the middle tube. At a heightthe water exerts aand to get to this height there must be a force between the water and glass of magnitude whereis the surface tension between the water and glass. Setting these two equal gives

This equation explains the diagram – the water climbs higher with decreasing radius.

Nature uses capillary action in a great many places. Sap uses capillary action to circulate through trees and plants.

]]>If a group of another sort of molecules enter the liquid at one location, they also experience this random motion. As time progresses, the molecules move or diffuse away from the point at which they enter the liquid into the surrounding liquid. Diffusion is the slow net movement due to random collisions of atoms and molecules away from areas in which they are concentrated into areas in which they are less concentrated.

Diffusion occurs in all fluids, gases as well as liquids. It occurs faster in gases because the speed of the molecules is faster. It plays an important part in many biological processes. Oxygen released in capillaries diffuses from blood to cells near the capillaries, and muscles receive oxygen that diffuses from the surrounding material. Since muscles need oxygen to move, the rate at which muscles can do work is limited by the rate of diffusion of oxygen into muscle tissue.

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Liquid in Contact With Air |
Temperature (Degrees Celsius) |
Surface Tension () |

Benzene |
20 |
28.9 |

Carbon Tetrachloride |
20 |
26.8 |

Ethanol |
20 |
22.3 |

Glycerin |
20 |
63.1 |

Mercury |
20 |
465 |

Olive Oil |
20 |
32 |

Soap Solution |
20 |
25 |

Water |
0 |
75.6 |

Water |
20 |
72.8 |

Water |
60 |
66.2 |

Water |
100 |
58.9 |

Oxygen |
-193 |
15.7 |

Neon |
-247 |
5.15 |

Helium |
-269 |
0.12 |

Surface tension usually decreases as the temperature increases, because as the temperature increases, molecules become more widely separated and the intermolecular forces decrease.

Surface tension can be changed by mixing solutions. For instance the surface tension of water can be lowered by mixing it with soap. This also makes your clothes cleaner, because lowering the surface tension makes it easier for water to penetrate into the fibres and dissolve the dirt there.

]]>In many circumstances, drag is a nuisance. To maintain a constant speedsome other external force must be applied. In a powered vehicle this force is supplied by the engine: If the drag force has magnitudeand no other forces act, the engine must work at a rateMany efforts are made to reduce drag, by streamlining for example.

]]>If the flow is smooth, the drag force is proportional to the relative speedof object and fluid.

(1)

whereis the drag coefficient. Stoke's Lawis an example of this.

The flow is smooth when the speedis small. Ifincreases, the flow becomes turbulent. At higher speed, the drag force depends on

(2)

This means that at higher speeds the drag force increases very rapidly.

Both of these expressions are approximations. The drag force may be more accurately described as a sum of expressions (1) and (2).

At higher speeds still, even this expression becomes inadequate, and higher powers ofmust be taken into account.

In general the coefficientsare decreasing, so that each term becomes important only as increases sufficiently.

]]>Water entering the pipe at 1 m/s.Water must entr and leave the pipe at the same rate of mass or volume per second, since water is virtually in compressible. The volume of water that enters the pipe each second is the cross sectional area of the entry point times the speed at which water enters the pipe and the rate at which water leaves at the exit point is equal to the cross sectional area at the exit point times the speed with which water exits the pipe. These entry and exit rates are equal.

{jatex options:inline}A_{Entry}v_{Entry}= A_{Exit}v_{Exit} \rightarrow v_{Exit}= \frac{A_{Entry}}{A_{Exit}} v_{Entry}=\frac{10^{-3}}{10^{-4}} \times 1=10 m/s{/jatex}

The force exerted on the pipe at the entry point is equal to the rate of change of momentum at that point:

{jatex options:inline}\begin{equation} \begin{aligned}F_{Entry} &= m_{Entry}v_{Entry} \\ &= ( \rho A_{Entry} v_{Entry})v_{Entry} \\ &= (1000 \times 10^{-3} \times 1) \times 1 \\ &= 10^{-2} N \end{aligned} \end{equation}{/jatex}.

The force exerted on the pipe at the exit point is equal to the rate of change of momentum at that point:

{jatex options:inline}\begin{equation} \begin{aligned}F_{Exit} &= m_{Exit}v_{Exit} \\ &= ( \rho A_{Exit} v_{Exit})v_{Exit} \\ &= (1000 \times 10^{-4} \times 10) \times 10 \\ &= 10^{-1} N \end{aligned} \end{equation}{/jatex}

The net force on the pipe is the difference between these two and is equal to {jatex options:inline}9 \times 10^{-2} N{/jatex}.]]>

The flow is greatest at the centre and diminishes towards the periphery. This makes the laminar flow describe a bullet shaped velocity profile shown in red below:

Turbulent flow occurs in rough tubes and at higher flow rates.

The flow is not streamlined. There is a lot of swirling of the fluid and a lot of energy is wasted overcoming the internal forces of the fluid.

The flow is not greatest at the centre. Thus, as shown in red below, the velocity profile of turbulent flow is more regular than that caused by laminar flow.

]]>Using the first of these, {jatex options:inline}1.29=A \times e^{-k(0)}=A \rightarrow A=1.29{/jatex}

Using the second, {jatex options:inline}0.43=1.29 \times e^{-k(10000)}=A \rightarrow k=- \frac{1}{10000} ln(\frac{0.43}{1.29}=1.0986 \times 10^{-4}{/jatex}

Hence {jatex options:inline}\rho (h)=1.29 e^{-1.0986 \times 10^{-4} h}{/jatex}.

Above every square metre of the Earth's surface there is

{jatex options:inline}\begin{equation} \begin{aligned} M &= \int^{\infty}_0 1.29 e^{-1.0986 \times 10^{-4} h} dh \\ &= 1.29[- \frac{1}{1.0986 \times 10^{-4}} e^{- 1.0986 \times 10^{-4} h} ]^{\infty}_0 \\ &= 1.29(0-- \frac{1}{1.0986 \times 10^{-4}}) \\ &= 1,174 \times 10^4 kg \end{aligned} \end{equation}{/jatex}

The pressure at the Earth's surface according to this model is then {jatex options:inline}p=\frac{Mg}{1}=1.15 \times 10^{5} N/m^2{/jatex}.]]>

In the diagram above, on the left of the membrane is only water and on the right is a mixture of water and glucose molecules. The two types of molecules act much like molecules of gas. The sugar molecules create a pressure on the membrane we would expect from a group of atoms of gas hitting a wall:

Next consider the pressure exerted by the water molecules. More water molecules per unit volume are on the left than the right hand side, so water molecules will hit the left hand side of the membrane at a greater rate than the right hand side, so there will be a net passage of water from the left to the right hand side. The fluid level on the left will fall and the level on the right will rise.

The concentration of sugar molecules isso {jatex options:inline}p=cRT{/jatex}.

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