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