Thermo & Fluids3 - Flipped PhysicsFluids.pdf · Molecules obey classical mechanics laws regarding collisions, energy ... Boyle’s Law (V, P), Charles’ Law ... with a specific amount

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Assumptions: There are many molecules moving in random

directions at a variety of speeds. The molecules are, on average, far apart from

each other. Their separation >> their diameter. Molecules obey classical mechanics laws

regarding collisions, energy etc. And they interact only through collision, not through attractive forces (PE).

Collisions with other molecules or the container wall are perfectly elastic. Time of collision << time between collisions.




Equations: pV = nRT pV = nkT PV/T = const. Boyle’s Law (V, P), Charles’ Law (V, T), Gay-

Lussac’s Law (P, T). KE = 3/2 kT = ½ mv2

Vrms = 3kT/m




Linear Expansion: Volume Expansion:

You’re having trouble opening a glass jar, how can you make it easier?




Linear Expansion: Volume Expansion:

You’re having trouble opening a glass jar, how can you make it easier?




AP Physics B




Density, pressure, volume…




Two systems are said to be in thermal equilibrium if there is no net flow of heatbetween them when they are brought into thermal contact.




Two systems individually in thermal equilibrium with a third system are in thermal equilibrium with each other.




Change in internal energy of a system is equal to the heat transferred to/from the system plus the work done on/by the system: ΔU = Q + W

Comes from the law of conservation of energy.

Sign convention: heat energy Q/W is positivewhen the system gains heat and negativewhen the system loses heat.




Suppose you had a piston filled with a specific amount of gas. As you add heat, the temperature rises and thus the volume of the gas expands. The gas then applies a force on the piston wall pushing it a specific displacement. Thus it can be said that a gas can do WORK.







a) Jogging along the beach one day you do 4.3 x 105 J of work and give off 3.8 x 105 J of heat. What is the change in your internal energy?

b) Switching over to walking, you give off 1.2 x 105 J of heat and your internal energy decreases by 2.6 x 105 J. How much work have you done while walking?




Sketch a PV diagram and find the work done by the gas during the following stages.

(a) A gas is expanded from a volume of 1.0 L to 3.0 L at a constant pressure of 3.0 atm.

(b) The gas is then cooled at a constant volume until the pressure falls to 2.0 atm




c) The gas is then compressed at a constant pressure of 2.0 atm from a volume of 3.0 L to 1.0 L.

d) The gas is then heated until its pressure increases from 2.0 atm to 3.0 atm at a constant volume.




What is the NET WORK?




Internal Energy is a function of state – it depends only on the state of a system, not on the method by which the system arrives at a given state

Quasi-static – a process that occurs slowly enough that a uniform pressure and temperature exist throughout all regions of the system at all times. There is no friction nor dissipative




To keep the temperature constant both the pressure and volume change to compensate. (Volume goes up, pressure goes down)

“BOYLES’ LAW”




Heat is added to the gas which increases the Internal Energy (U) Work is done by the gas as it changes in volume.

The path of an isobaric process is a horizontal line called an isobar.

∆U = Q - W can be used since the WORK is NEGATIVE in this case







ADIABATIC- (GREEK-adiabatos-"impassable")

In other words, NO HEAT can leave or enter the system.







Example A gas expands from an initial volume of 0.40 m3 to a final

volume of 0.62 m3 as the pressure increases linearly from 110 kPa to 230 kPa. Find the work done by the gas.




Heat will only flow spontaneously from a body of higher temperature to a body of lower temperature.

For the reverse to happen, work must be done.

Example?




Disorder in the universe can only increase.




Heat flows from a HOT reservoir to a COLD reservoir

QH = remove from, absorbs = hotQC= exhausts to, expels = cold




In order to determine the thermal efficiency of an engine you have to look at how much ENERGY you get OUT vshow much you energy you take IN.

Eff. = Wout = 1 - QoutQin Qin




A heat engine with an efficiency of 24.0% performs 1250 J of work. Find (a) the heat absorbed from the hot reservoir, and (b) the heat given off to the cold reservoir.




Sometimes it is useful to express the energy usage of an engine as a RATE.

For example:

The RATE at which heat is absorbed!

The RATE at which heat is expelled.

The RATE at which WORK is DONEPOWER

tW

tQ

tQ

C

H







Our goal is to figure out just how efficient such a heat engine canbe: what’s the most work we can possibly get for a given amount of fuel?

The efficiency question was first posed—and solved—by SadiCarnot in 1820, not long after steam engines had become efficient enough to begin replacing water wheels, at that time the main power sources for industry. Not surprisingly, perhaps, Carnot visualized the heat engine as a kind of water wheel in which heat (the “fluid”) dropped from a high temperature to a low temperature, losing “potential energy” which the engine turned into work done, just like a water wheel.




Carnot a believed that there was an absolute zero of temperature, from which he figured out that on being cooled to absolute zero, the fluid would give up all its heat energy. Therefore, if it falls only half way to absolute zero from its beginning temperature, it will give up half its heat, and an engine taking in heat at T and shedding it at ½T will be utilizing half the possible heat, and be 50% efficient. Picture a water wheel that takes in water at the top of a waterfall, but lets it out halfway down. So, the efficiency of an ideal engine operating between two temperatures will be equal to the fraction of the temperature drop towards absolute zero that the heat undergoes.

Carnot Eff. = TH – TCTH




Carnot temperatures must be expressed in KELVIN!!!!!!

The Carnot model has 4 parts•An Isothermal Expansion•An Adiabatic Expansion•An Isothermal Compression•An Adiabatic Compression

The PV diagram in a way shows us that the ratio of the heats are symbolic to the ratio of the 2 temperatures




If the heat engine from the example before is operating at a maximum efficiency, and its cold reservoir is at a temperature of 295 K, what is the temperature of the hot reservoir?




A particular engine has a power output of 5000 W and an efficiency of 25%. If the engine expels 8000 J of heat in each cycle, find (a) the heat absorbed in each cycle and (b) the time for each cycle




The efficiency of a Carnot engine is 30%. The engine absorbs 800 J of heat per cycle from a hot temperature reservoir at 500 K. Determine (a) the heat expelled per cycle and (b) the temperature of the cold reservoir




The relationship for a Carnot engine, can be rearranged to

. The quantity Q/T is the change in entropy, ΔS:

The temperature, again, must be in Kelvins, the subscript R refers to reversible process.

Units for entropy = J/KEntropy is a function of state (like internal energy) – only the state

of the system determines the entropy

H

C

H

C

TT

QQ

H

H

C

C

TQ

TQ

RTQΔS




Calculate the change in entropy when a 0.125 kg chunk of ice melt at 0ºC. Assume the melting occurs reversibly.




The entropy of a Carnot Engine: as the engine operates, the entropy of the hot reservoir

decreases, since heat QH leaves. the change in the entropy of the hot reservoir is

(minus indicates decrease in S)

the change in the entropy of the cold reservoir is

Thus, the total change in entropy is

(equals zero because )

Thus, ΔS = 0 for a Carnot Engine. This is also true for any reversible process: the total entropy of the universe does not change

Reversible processes do not change the total entropy of the universe. (The entropy of one part of the universe may change, but if so, the entropy of

another part must change in the opposite way by the same amount.)Irreversible processes increase the entropy of the universe. ΔS > 0

H

HH T

QΔS

C

CC T

QΔS

0TQ

TQ

SΔSH

H

C

CHC

H

H

C

C

TQ

TQ




A hot reservoir at the temperature 576 K transfers 1050 J of heat irreversibly to a cold reservoir at the temperature 305 K. Find the change in entropy of the universe.




The Second Law of Thermodynamics in terms of Entropy:

The total entropy of the universe does not change when a reversible process occurs and increases when an irreversible process occurs.




AP Physics B




By definition, a fluid is a substance that has no fixed shape and yields easily to external pressure.




Typically, liquids are considered to be incompressible. That is once you place a liquid in a sealed container you can DO WORK on the FLUID as if it were an object. The PRESSURE you apply is transmitted throughout the liquid and over the entire length of the fluid itself.




Can exert pressure in any direction. Pressure always acts perpendicular to the surface.




Pat is a direct result of the weight of the air above us.




Suppose a Fluid (such as a liquid) is at REST, we call this HYDROSTATIC PRESSURE.

Notice that the arrows on TOP of the objects are smaller than at the BOTTOM. This is because pressure is greatly affected by the DEPTH of the object. Since the bottom of each object is deeper than the top the pressure is greater at the bottom.




Suppose we had an object submerged in water with the top part touching the atmosphere. If we were to draw an FBD for this object we would have three forces




But recall, pressure is force per unit area.

Note: The initial pressure in this case is atmospheric pressure, which is a CONSTANT.

Po=1x105 N/m2

FINAL EQUATION:

DP = rgh




a) Calculate the absolute pressure at an ocean depth of 1000 m. Assume that the density of water is 1000 kg/m3 and that Po= 1.01 x 105 Pa (N/m2).

b) Calculate the total force exerted on the outside of a 30.0 cm diameter circular submarine window at this depth.




Therefore: PA = PB = PC = PD (because they all have the same depth)




Mercury Barometer: measures atmospheric pressure

Open Tube Manometer: measures pressure in a container

Po = 0P = PatmPatm = 0 + ρghPatm = ρgh

P = Patm + ρghExample: blood pressure cuff




If you take a liquid and place it in a system that is CLOSED like plumbing for example or a car’s brake line, the PRESSURE is the same everywhere.

Since this is true, if you apply a force at one part of the system the pressure is the same at the other end of the system. The force, on the other hand MAY or MAY NOT equal the initial force applied. It depends on the AREA.

You can take advantage of the fact that the pressure is the same in a closed system as it has MANY applications.

The idea behind this is called PASCAL’S PRINCIPLE










To inspect a 14,000 N car, it is raised with a hydraulic lift. If the radius of the small piston is 4.0 cm, and the radius of the large piston is 17cm, find the force that must be exerted on the small piston to lift the car.




When an object is immersed in a fluid, such as a liquid, it is buoyed upwards by a force called the buoyant force.




" An object is buoyed up by a force equal to the weight of the fluid displaced."

In the figure, we see that the difference between the weight in AIR and the weight in WATER is 3 lbs. This is the buoyant force that acts upward to cancel out part of the force. If you were to weight the water displaced it also would weigh 3 lbs.




*V = A(h2 – h1)




A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weight in air to be 7.84 N. She then weighs the crown while it is immersed in water (density of water is 1000 kg/m3) and now the scale reads 6.86 N. Is the crown made of pure gold if the density of gold is 19.3 x 103 kg/m3?




A piece of wood with a density o 706 kg/m3 is tied with a string to the bottom of a water-filled flask. The wood is completely immersed, and has a volume of 8.00 x 10-6 m3. What is the tension in the string?




Steady – velocity of the fluid particles at any point is constant as time passes

Compressible – density of the fluid changes as pressure changes (usually gases)

Viscous – “a large viscosity” –doesn’t readily flow: the viscosity hinders the neighboring layers of fluid from sliding past each other.

Rotational – a part of the fluid has rotational as well as translational motion. Place a paddle wheel in the fluid, if it rotates, flow is rotational.

Unsteady – velocity at a point in the fluid changesas time passes (ex: Turbulent flow: extremely unsteady flow)

Incompressible – densityof the fluid remains constant as pressure changes (usually liquids)

Nonviscous – “a low viscosity” – flows readily –layers are not hindered from sliding past each other.

Irrotational – fluid has only translational motion. The paddle wheel will not turn.







A

A

v

v

L

L

Consider a pipe with a fluid moving within it.

Mass flow rate is:m = Vr = ALr = AvrDt Dt Dt

(m = rV)




A1

A2

v1

v2

L1=v1t

L2=v2t

The first thing you MUST understand is that MASS is NOT CREATED OR DESTROYED!IT IS CONSERVED.

The MASS that flows into a region = The MASS that flows out of a region.

Using the Mass Flow rate equation and the idea that a certain mass of water is constant as it moves to a new pipe section:

We have the Fluid Flow Continuity equation




The speed of blood in the aorta is 50 cm/s and this vessel has a radius of 1.0 cm. If the capillaries have a total cross sectional area of 3000 cm2, what is the speed of the blood in them? (Equation of continuity – the last equation we derived last class.)




What happens to the roof of the “wind tunnel”?




The Swiss Physicist Daniel Bernoulli, was interested in how the velocity changes as the fluid moves through a pipe of different area. He especially wanted to incorporate pressure into his idea as well. Conceptually, his principle is stated as: "an increase in velocity of a stream of fluid results in a decrease of pressure in the fluid”




Assumptions:

Laminar flow.Steady flow.Incompressible fluid.




Work is done by a section of water applying a force on a second section in front of it over a displacement. According to Newton’s 3rd law, the second section of water applies an equal and opposite force back on the first. Thus is does negative work as the water still moves FORWARD. Pressure*Area is substituted for Force.

X = L

F1 on 2

-F2 on 1




A1

A2

v1

v2

L1=v1t

L2=v2t

y2

ground

Work is also done by GRAVITY as the water travels a vertical displacement UPWARD. As the water moves UP the force due to gravity is DOWN. So the work is NEGATIVE.

The fluid in section 1 flows towards section 2 a distance L1 and in doing so pushes the fluid in section 2 a distance L2.

y1




The fluid in the first section is pushed by the fluid to the left of it, and work is done.

W1 = F1L1 = P1A1L1

(Since P = F/A)





W1 = F1L1 = P1A1L1

(Since P = F/A)

The fluid in the second section is held back by the fluid to the right of it, and negative work is done.

W2 = -F2L2 = -P2A2L2





W1 = F1L1 = P1A1L1

(Since P = F/A)

The fluid in the second section is held back by the fluid to the right of it, and negative work is done.

W2 = -F2L2 = -P2A2L2

And negative work is also done by gravity, resisting the motion:

W3 = -mg (y2 – y1)




Total work done:W = W1 + W2 + W3





And this work is equal to the change in kinetic energy of the system, so:

½ mv22 – ½ mv1

2 = P1A1L1 - P2A2L2 - mgy2 + mgy1






½ mv22 – ½ mv1

2 = P1A1L1 - P2A2L2 - mgy2 + mgy1

Realize that m = rV = rAL






½ mv22 – ½ mv1

2 = P1A1L1 - P2A2L2 - mgy2 + mgy1

Realize that m = rV = rAL

Plug this value of m in and cancel the AL in each term, using the fact that A1L1 = A2L2 to get:

½rv22 – ½ rv1

2 = P1 – P2 – rgy2 + rgy1




½rv22 – ½ rv1

2 = P1 – P2 – rgy2 + rgy1

Rearrange to get:

P1 + ½ rv12 + rgy1 = P2 + ½rv2

2 + rgy2




½rv22 – ½ rv1

2 = P1 – P2 – rgy2 + rgy1

Rearrange to get:

P1 + ½ rv12 + rgy1 = P2 + ½rv2

2 + rgy2

Or more simply:

P + ½ rv2 + rgy = const.Notice, if v = 0, this becomes the hydrostatic equation.




Water circulates throughout the house in a hot-water heating system. If the water is pumped at a speed of 0.50 m/s through a 4.0 cm diameter pipe in the basement under a pressure of 3.0 atm, what will be the flow speed and pressure in a 2.6 cm-diameter pipe on the second floor 5.0 m above?




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Thermo & Fluids3 - Flipped PhysicsFluids.pdf · Molecules obey classical mechanics laws regarding collisions, energy ... Boyle’s Law (V, P), Charles’ Law ... with a specific amount