1206 - Concepts in physics Monday, November 2

Notes • Midterm tonight at 4:30 pm in this room • 7:30 pm group meets at my office (F511) • Next assignment due November 11th

This will be displayed If you have studied, you can do this - be confident

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Read carefully, don’t rush Start with the parts you know well If you get stuck - do something else first Remember that units are important Remember to only use numbers at the end Work alone - be honest If you know your result can’t be correct, comment on it (if you don’t have time to redo it)

From last time: BERNOULLI’S EQUATION In the steady flow of a nonviscous, incompressible fluid of density ρ, the pressure P, the fluid speed v, and the elevation h at any two points (1 and 2) are related by: P1 + 1/2ρv12 + ρgh1 = P2 + 1/2ρv22 + ρgh2

Example: An aneurysm is an abnormal enlargement of a blood vessel such as the aorta. Suppose that, because of an aneurysm, the cross-sectional area A1 of the aorta increases to a value A2 = 1.7*A1. The speed of the blood (ρ = 1060 kg/m3) through a normal portion of the aorta is v1 = 0.40 m/s. Assuming that the aorta is horizontal (the person is lying down), determine the amount by which the pressure P2 in the enlarged region exceeds the pressure P1 in the normal region. Bernoulli’s equation may be used to find the pressure difference between two points in a fluid moving horizontally. However, in order to use this relation we need to know the speed of the blood in the enlarged region of the artery, as well as the speed in the normal section. The speed in the enlarged region can be obtained by using the equation of continuity, which relates the blood speeds in the enlarged and normal regions to the crosssectional areas of these regions. We define the region with the aneurysm as region 2. In form for horizontal flow the equation is: P1 + 1/2ρv12 = P2 + 1/2ρv22 and therefore: P2 - P1 = 1/2ρ (v12 - v22). From the equation of continuity, the speed v2 of the blood in the aneurysm is given by v2 = (A1/A2)v1. We can substitute this into Bernoulli’s equation and obtain: P2 - P1 = 1/2ρ (v12 - (A1/A2)2v12) = 1/2ρv12 (1- (A1/A2)2) = = 1/2(1060 kg/m3)(0.40 m/s)2 (1- (A1/1.7*A1)2) = 55 Pa This result is positive, indicating that P2 is greater than P1. The excess pressure puts added stress on the already weakened tissue of the arterial wall at the aneurysm.

Example: The tank in the figure is open to the atmosphere at the top. Find an expression for the speed of the liquid leaving the pipe at the bottom.

We assume that the liquid behaves as an ideal fluid. Therefore, we can apply Bernoulli’s equation, and in preparation for doing so, we locate two points in the liquid. Point 1 is the center of the opening in the bottom just outside the efflux pipe, and point 2 is at the top surface of the liquid. The pressure at each of these points is equal to the atmospheric pressure, a fact that will be used to simplify Bernoulli’s equation. Since the pressures at points 1 and 2 are the same, we have P1 = P2, and Bernoulli’s equation becomes 1/2ρv12 + ρgy1 = 1/2ρv22 + ρgy2 The density ρ can be eliminated algebraically and we can solve for v1: v12 = v22 + 2g(y2-y1) = v22 + 2gh

where h is y2-y1

If the tank is very large, the liquid level changes only slowly, and the speed at point 2 can be set equal to zero, so that v1 = sqrt(2gh)

Temperature scales Everybody is familiar with the instrument we use to measure temperature - a thermometer. Many thermometers make use of the fact that materials usually expand with increasing temperature. Commonly used is mercury-inglass. When the mercury is heated, it expands into the tube (see picture). The expansion of the mercury is proportional to the change in temperature. We use a number of different scales: Celsius, Fahrenheit or Kelvin. See the comparison between Celsius and Fahrenheit, anchored to the freezing and steam point (boiling) of water. For the Fahrenheit scale we have 180 degrees between the two points, for the celsius scale we have 100 degrees.

Your turn: Converting Fahrenheit to Celsius: 98.6 F Converting Celsius to Fahrenheit: -20 C

Your turn: Converting Fahrenheit to Celsius: 98.6 F 98.6 F - 32 F = 66.6 F (it is 66.6 F above ice point) Now 180/100 = 9/5 = 1.8 which leads to (66.6 F) (1C/1.8 F) = 37.0 C (typical body temperature) Converting Celsius to Fahrenheit: -20 C (-20 C) = -1.8F/1C + 32 F = -4 F

Kelvin scale Scientifically more significant: Kelvin temperature scale a temperature scale having an absolute zero below which temperatures do not exist. Absolute zero , or 0°K, is the temperature at which molecular energy is a minimum, and it corresponds to a temperature of -273.15° on the Celsius temperature scale . The Kelvin degree is the same size as the Celsius degree; hence the two reference temperatures for Celsius, the freezing point of water (0°C), and the boiling point of water (100°C), correspond to 273.15°K and 373.15°K, respectively. When writing temperatures in the Kelvin scale, it is the convention to omit the degree symbol and merely use the letter K.

Scottish physicist William Thompson (Lord Kelvin, 1824-1907)

The number 273.15 is an experimental result, obtained in studies that utilize a gas-based thermometer. When a gas confined to a fixed volume is heated, its pressure increases. Conversely, when the gas is cooled, its pressure decreases. For example the pressure in a typical car tire can rise by 20% when the tires get warm due to driving. This change in gas pressure with temperature is the basis for the constant-volume gas thermometer.

A constant-volume gas thermometer consists of a gas-filled bulb to which a pressure gauge is attached. the gas is often hydrogen or helium at low density and the pressure gauge can be a U-tube manometer filled with mercury. The bulb is placed in thermal contact with the substance whose temperature is being measured. The volume of the gas is held constant by raising or lowering the column (right) of the U-tube manometer on order to keep the mercury level (left column) at the same reference level. The absolute pressure is proportional to the height h of the mercury on the right. As the temperature changes, the pressure changes and can be used to indicate the temperature, once the constant-volume gas thermometer has been calibrated.

We have learned two things in the previous example which we will use later on (not today). a) temperature and pressure influence each other b) heat gets transferred from one material to another

Also, note that it is a very common concept in physics and other sciences to find a relationship between parameters and use this to measure “something”. Over the next few weeks we will see many examples. Note, that a lot of the parameters we will talk about are measured quantities

Thermometers There are many different versions. All you need is some physical property that changes with temperature and you can make a thermometer. A property that changes with temperature is called a thermometric property.

Description: Sitting cat. Real infrared (thermic) photo in blue. A thermographic camera is a device that forms an image using infrared radiation and making visible the heat of the model. This image is not a digital effect.

Thermographs are used to diagnose cancer (here breast cancer) and are often an earlier indication than other methods. They can also be used to monitor the progress of treatment.

Linear thermal expansion This effect is likely very familiar to you. Solids expand when they are heated. In the summer some doors expand so much for example, that is gets harder to open and close them. There are many other examples. But we want to quantify this effect. temperature T0

For modest temperature changes ΔL∝ΔT By using a proportionality constant α, which is called the coefficient of linear expansion, we can write down:

temperature T0 + ΔT

length: L0

ΔL

LINEAR THERMAL EXPANSION OF A SOLID The length L0 of an object changes by an amount ΔL when its temperature changes by an amount ΔT: ΔL = α L0 ΔT where α is the coefficient of linear expansion. Common unit for the coefficient of linear expansion is 1/C°

Example: buckling of a sidewalk ... A concrete sidewalk is constructed between two buildings on a day when the temperature is 25 C. The sidewalk consists of two slabs, each three meters in length and of negligible thickness. As the temperature rises to 38 C, the slabs expand, but no space is provided for thermal expansion. The buildings do not move, so the slabs buckle upward. Determine the vertical distance y given in the drawing. The coefficient for linear expansion of concrete is 12 x 10-6 y The expanded length of each slab is equal to its original length plus the change in length ΔL due to the rise in temperature. We know the original length, an we can find the change in length with using the definition of linear thermal expansion. Once this has been determined we can use the Pythagorean theorem to find the vertical distance y. Let’s calculate the change in temperature first ΔT = 38 C - 25 C = 13 C. The change in length of each slab for this temperature change is: ΔL = ΔTαL0, where L0 is the original length of each slab and α the coefficient for linear expansion of concrete. ΔL = ΔTαL0 = (13C)(12 x 10-6)(3.0m) = 0.00047 m Pythagorean theorem: y = sqrt[(3.00047 m)2 +(3.00000 m)2] = 0.053 m = 5.3 cm