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Fluids and Thermodynamics. Fluids Fluids are substances that can flow , such as liquids and gases, and even a few solids. In Physics B, we will limit our discussion of fluids to substances that can easily flow, such as liquids and gases. D = m/V D: density (kg/m 3 ) m: mass (kg)

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  • Fluids

    • Fluids are substances that can flow, such as liquids and gases, and even a few solids.

    • In Physics B, we will limit our discussion of fluids to substances that can easily flow, such as liquids and gases.

    • D = m/V

      • D: density (kg/m3)

      • m: mass (kg)

      • V: volume (m3)

    • You should remember how to do density calculations from chemistry!

Density can also be represented by - ρ

ρ= m/V

ρ: density (kg/m3)


  • Pressure

    • P = F/A

      • P : pressure (Pa)

      • F: force (N)

      • A: area (m2)

    • Pressure unit:

      • Pascal ( 1 Pa = 1 N/m2)

    • The force on a surface caused by pressure is always normal (or perpendicular) to the surface. This means that the pressure of a fluid is exerted in all directions, and is perpendicular to the surface at every location.


Atmospheric Pressure

  • Atmospheric pressure is normally about 100,000 Pascals.

  • Differences in atmospheric pressure cause winds to blow.

Low atmospheric pressure inside a hurricane’s eye contributes to the severe winds and the evelopment of the storm surge..


Atmospheric pressure = 100000 Pa

Pcabin = 90000 Pa

Poutside = 50000 Pa

Cabin

Outside

P = F/A

P = 90000Pa – 50000Pa = 40000Pa

A = lw

A = 0.3m · 0.43m = 0.129m2

F = AP

F = 0.129m2· 40000Pa = 5160N


  • The Pressure of a Liquid pressure is 90% of the pressure at sea level, and the external pressure is only 50% of that at sea level. Assume the window is 0.43 m tall and 0.30 m wide.

    • P = ρgh

      • P: pressure (Pa)

      • ρ : density (kg/m3)

      • g: acceleration constant (9.8 m/s2)

      • h: height of liquid column (m)

  • This type of pressure is often called gauge pressure. Why?

    • The gauges we use to measure pressure do not take into account the atmospheric pressure which is very large!

  • If the liquid is water, this is referred to as hydrostatic pressure. Why?

    • Hydro – water + static - stationary


  • Absolute Pressure pressure is 90% of the pressure at sea level, and the external pressure is only 50% of that at sea level. Assume the window is 0.43 m tall and 0.30 m wide.

    • Absolute pressure is obtained by adding the atmospheric pressure to the hydrostatic pressure.

    • Pabs = Patm + ρgh

Hydrostatic Pressure in Dam Design

The depth of Lake Mead at the Hoover Dam is 180 m. What is the hydrostatic pressure and what is the absolute pressure at the base of the dam?


Hydrostatic Pressure in Dam Design pressure is 90% of the pressure at sea level, and the external pressure is only 50% of that at sea level. Assume the window is 0.43 m tall and 0.30 m wide.

  • The depth of Lake Mead at the Hoover Dam is 180 m. What is the hydrostatic pressure and what is the absolute pressure at the base of the dam?

ρfresh water = 1000 kg / m3

P = ρgh

P = 1000 kg / m3· 9.8m/s2 · 180 m = 1764000 Pa

Pabs = Patm + ρgh

Pabs = 100000Pa + 1764000Pa = 1864000 Pa


  • Hydrostatic Pressure in Levee Design pressure is 90% of the pressure at sea level, and the external pressure is only 50% of that at sea level. Assume the window is 0.43 m tall and 0.30 m wide.

Hurricane Katrina August 2005

A hurricane’s storm

surge can overtop levees,

but a bigger problem can

be increasing the

hydrostatic pressure at

the base of the levee.


  • New Orleans Elevation Map pressure is 90% of the pressure at sea level, and the external pressure is only 50% of that at sea level. Assume the window is 0.43 m tall and 0.30 m wide.

    • New Orleans is largely below sea level, and relies upon a system of levees to keep the lake and the river at bay

Max ht of the levee 17.5 ft

Surge elevation set at 11.5 ft

Normal lake level at 1.0ft


  • Calculate the increase in hydrostatic pressure experienced by the levee base for an expected (SPH Design) storm surge. How does this compare to the increase that occurred during Hurricane Katrina, where the water rose to the top of the levee?

  • Not to mention there is atmospheric changes during hurricanes which effect the absolute pressure

16.5ft = 12in / 1ft · 2.54 cm / 1 in · 1 m / 100 cm = 5.03m

hlevee = 16.5ft

ρsalt water = 1025 kg /m3

P= ρgh

P= 1025 kg /m3·9.8m/s2 · 5.03m = 50526.35 Pa


  • Pascal’s Principle by the levee base for an expected (SPH Design) storm surge. How does this compare to the increase that occurred during Hurricane Katrina, where the water rose to the top of the levee?

    • Applies when a pressure is applied to a container holding a fluid

    • Applies to closed systems

    • Pascal’s principle applies in hydraulic systems

    • F1 /A1 = F2 / A2


mc = 1500kg

rc = 3.1m

rp = 0.40m

Fp = ?

Ac =πr2 = π(3.1m)2 = 30.19m2

Ac =πr2 = π(0.40m)2 = 0.50m2

Fc / Ac = Fp / Ap

Fp = FcAp / Ac

Fp = (1500· 9.8m/s2) · 0.50m2 / 30.19m2 = 243N

This is equal to 54 lbs, so a child could stand on the piston and make the car move!!!


  • Calculate the pressure exerted on the bottom of a glass that moves upward in elevator with constant acceleration, increasing from rest to 1.2 m/s in 2.7 s. The height of the water is 15 cm. The radius of the cup is 2 cm. The glass is filled with water which has a density of 1000 kg /m3.Do not include atmospheric contribution in your answer.

V = Acircle · h = 0.0013 m2 · 0.15m = 0.000195 m3

ΣF = ma

h

FN - mg= ma

m = Vρ = 0.000195 m3 · 1000 kg / m3 = 0.195 Kg

a = (vf – vi) / t

FN = ma + mg

a = (1.2 m/s – 0m/si) / 2.7s = 0.44m/s2

A

FN = 0.195 Kg(0.44m/s2 + 9.8m/s2) = 1.9968 N

A = πr2

A = π(0.02m)2 = 0.0013 m2

FN = ma + mg


  • Calculate the pressure exerted on the bottom of a glass that moves upward in elevator with constant acceleration, increasing from rest to 1.2 m/s in 2.7 s. The height of the water is 15 cm. The radius of the cup is 2 cm. The glass is filled with water which has a density of 1000 kg /m3.Do not include atmospheric contribution in your answer.

P1 = FN / A

h

P2 = ρgh

P = ρgh + FN / A

A

P = 1000kg/m3 · 9.8m/s2 · 0.15m + 1.9968 N / 0.0013 m2 = 3006Pa


Buoyancy moves upward in elevator with constant acceleration, increasing from rest to 1.2 m/s in 2.7 s. The height of the water is 15 cm. The radius of the cup is 2 cm. The glass is filled with water which has a density of 1000 kg /m


An upward force counteracts the force of gravity for these objects. This upward force is called the buoyant force

  • Floating is a type of equilibrium

Fbuoyancy

mg


  • Archimedes Principle objects. This upward force is called the buoyant force

    • Archimedes’ Principle: a body immersed in a fluid is buoyed up by a force that is equal to the weight of the fluid it displaces.

  • The Buoyant Force

    • Fbuoy = ρVg

      • Fbuoy: the buoyant force exerted on a submerged or partially submerged object.

      • V: the volume of displaced liquid.

      • ρ : the density of the displaced liquid.

  • When an object floats, the upward buoyant force equals the downward pull of gravity.

  • The buoyant force can float very heavy objects, and acts upon objects in the water whether they are floating, submerged, or even sitting on the bottom.


A sharks body is not neutrally buoyant, so a shark must swim continuously or he will sink deeper.

  • Buoyant force on submerged object

Fbuoy = ρVg

mg

Fbouy≠ mg


SCUBA divers use a buoyancy control system to maintain neutral buoyancy (equilibrium!).

  • Buoyant force on submerged object

Fbuoy = ρVg

mg

Fbouy= mg


If the diver wants to rise, he inflates neutral buoyancy (equilibrium!).

his vest, which increases the amount

of water he displaces, and he

accelerates upward

  • Buoyant force on submerged object

Fbuoy = ρVg

mg

Fbouy= mg


If the object floats on the surface, we neutral buoyancy (equilibrium!).

know for a fact Fbuoy = mg! The volume of displaced water equals the volume of the submerged portion of the ship.

  • Buoyant force on floating object

Fbuoy = ρVg

mg

Fbouy= mg


Fbuoy = ρVg

h

y

w

l

mg

ΣF: Fbouy – mg = 0

We also know that the volume of displacement is equal to the volume of water displaced

ρglwy= (0.8ρ)glwh

ρgVsub– (0.8ρ)gVtotal = 0

y= (0.8) h

y= 0.8 · 0.1m = 0.08m

The height above the water is 0.02m


  • Buoyant Force dimensions of the raft are 6.0 meters long by 3.0 meters wide by 0.10 meter tall. How much of the raft rises above the level of the water when it floats?

    • The buoyant force can be extremely strong.

    • Incredibly massive objects can float, even when they are not intended to…


Moving Fluids dimensions of the raft are 6.0 meters long by 3.0 meters wide by 0.10 meter tall. How much of the raft rises above the level of the water when it floats?


  • When a Fluid Flows… dimensions of the raft are 6.0 meters long by 3.0 meters wide by 0.10 meter tall. How much of the raft rises above the level of the water when it floats?

    • …mass is conserved.

    • Provided there are no inlets our outlets in a stream of flowing fluid, the same mass per unit time must flow everywhere in the stream.

    • http://library.thinkquest.org/27948/bernoulli.html

  • Fluid Flow Continuity

    • The volume per unit time of a liquid flowing in a pipe is constant throughout the pipe.

    • We can say this because liquids are not compressible, so mass conservation is also volume conservation for a liquid.


  • Fluid Flow Continuity (cont.) dimensions of the raft are 6.0 meters long by 3.0 meters wide by 0.10 meter tall. How much of the raft rises above the level of the water when it floats?

    • V = Avt

      • V: volume of fluid (m3)

      • A: cross sectional areas at a point in the pipe (m2)

      • v: speed of fluid flow at a point in the pipe (m/s)

      • t: time (s)

    • A1v1 = A2v2

      • A1, A2: cross sectional areas at points 1 and 2

      • v1, v2: speed of fluid flow at points 1 and 2


A1v1 = A2v2

πr21 v1 = πr22 v2

Apipe = πr2

v2 = πr21 v1 / πr22

v2 = π(0.06m)2· (1.6 m/s) / π(0.03m)2= 6.4 m/s


  • The water in a canal flows 0.10 m/s where the canal is 12 meters deep and 10 meters across. If the depth of the canal is reduced to 6.5 meters at an area where the canal narrows to 5.0 meters, how fast will the water be moving through this narrower region?

  • What will happen to the water in an open waterway if it cannot flow as fast as it wants to through a narrow region in a channel?

    • It will rise and flow out of the channel

V2 = l1w1v1 / l2w2

V2 = A1v1 / A2

A1v1 = A2v2

V2 =(12m · 10m · 0.1 m/s) / (6.5m · 5m) = 0.37 m/s


  • Natural Waterways meters deep and 10 meters across. If the depth of the canal is reduced to 6.5 meters at an area where the canal narrows to 5.0 meters, how fast will the water be moving through this narrower region?

  • Flash flooding can be explained by fluid flow continuity


  • Artificial Waterways meters deep and 10 meters across. If the depth of the canal is reduced to 6.5 meters at an area where the canal narrows to 5.0 meters, how fast will the water be moving through this narrower region?

Flooding from the Mississippi River Gulf Outlet was responsible for

catastrophic flooding in eastern New Orleans and St. Bernard during Hurricane Katrina.


Mississippi River Gulf Outlet levees are overtopped by Katrina’s storm surge.

  • Fluid Flow Continuity in Waterways

A hurricane’s storm surge can be “amplified” by waterways that become narrower or shallower as they move inland.


Moving Fluids and Bernoulli Effect Katrina’s storm surge.


  • Bernoulli’s Theorem Katrina’s storm surge.

    • The sum of the pressure, the potential energy per unit volume, and the kinetic energy per unit volume at any one location in the fluid is equal to the sum of the pressure, the potential energy per unit volume, and the kinetic energy per unit volume at any other location in the fluid for a non-viscous incompressible fluid in streamline flow.

    • All other considerations being equal, when fluid moves faster, the pressure drops.


  • Bernoulli’s Theorem Katrina’s storm surge.

    • p + ρgh + ½ ρv2 = Constant

    • p : pressure (Pa)

    • ρ : density of fluid (kg/m3)

    • g: gravitational acceleration constant (9.8 m/s2)

    • h: height above lowest point (m)

    • v: speed of fluid flow at a point in the pipe (m/s)


Longer d, higher v, lower p Katrina’s storm surge.

  • Knowing what you know about Bernoulli's principle, design an airplane wing that you think will keep an airplane aloft. Draw a cross section of the wing.

Flift

Fthurst

Fdrag

Flift = ΔPA

mg


po + ρgh + ½ ρv2 = Constant

h1 = 2m

po + ρgh1 + ½ ρv12 = po + ρgh2 + ½ ρv22

ρgh1 + ½ ρv12 = ρgh2 + ½ ρv22

h2 = 0 m

v1 = 0 m/s

ρgh1 + = ½ ρv22

gh1 = ½v22

What about energy

mgh = ½ mv2

v2 = √(2gh1)

gh = ½ v2

v2 = √(2·9.8 m/s2 · 2.0m) = 6.26 m/s

v2 = √(2gh1)

Same equation!!!

Bernoulli’s principle


v2 = 6.26 m/s

h1 = 2m

V = Avt

V/t = Av

V/t = πr2v

V/t = π(0.001 m)2 · (6.26m/s) = 1.97 x 10-5 m3/s


  • Water travels through a 9.6 cm diameter fire hose with a speed of 1.3 m/s. At the end of the hose, the water flows out of a nozzle whose diameter is 2.5 cm. What is the speed of the water coming out of the nozzle? If the pressure in the hose is 350 kPa, what is the pressure in the nozzle?

A1v1 = A2v2

v2 = A1v1 / A2

v2 = πr21v1 / πr22

p1 = 350000 Pa

v1 = 1.3 m/s

p2 = ? Pa

v2 = ? m/s

v2 = π(0.096m)21· 1.3 m/s / π(0.025m)22 =19.17 m/s


  • Water travels through a 9.6 cm diameter fire hose with a speed of 1.3 m/s. At the end of the hose, the water flows out of a nozzle whose diameter is 2.5 cm. What is the speed of the water coming out of the nozzle? If the pressure in the hose is 350 kPa, what is the pressure in the nozzle?

p1 + ρgh1 + ½ ρv12 = p2 + ρgh2 + ½ ρv22

h1 = 0m h2 = 0m

p1 = 350000 Pa

v1 = 1.3 m/s

p2 = ? Pa

v2 = 19.17 m/s

p1 + ½ ρv12 = p2 + ½ ρv22

p2 = p1 + ½ ρv12 - ½ ρv22

p2 = p1 + ½ ρ(v12 - v22)

p2 =350000Pa + ½ · 1000 kg / m3((1.3 m/s)2 – (19.17m/s)2) = 167100.55 Pa


  • Bernoulli’s Principle and Hurricanes speed of 1.3 m/s. At the end of the hose, the water flows out of a nozzle whose diameter is 2.5 cm. What is the speed of the water coming out of the nozzle? If the pressure in the hose is 350 kPa, what is the pressure in the nozzle?

    • In a hurricane or tornado, the high winds traveling across the roof of a building can actually lift the roof off the building.

    • http://video.google.com/videoplay?docid=6649024923387081294&q=Hurricane +Roof&hl=en



Thermodynamics them the information and they must predict where the water will hit the floor.


  • Thermodynamics them the information and they must predict where the water will hit the floor.

    • Thermodynamics is the study of heat and thermal energy.

    • Thermal properties (heat and temperature) are based on the motion of individual molecules, so thermodynamics is a lot like chemistry.

  • Total energy

    • E = U + K + Eint

      • U: potential energy

      • K: kinetic energy

      • Eint: internal or thermal energy

    • Potential and kinetic energies are specifically for “big” objects, and represent mechanical energy.

    • Thermal energy is related to the kinetic energy of the molecules of a substance.


  • Temperature and Heat them the information and they must predict where the water will hit the floor.

    • Temperature is a measure of the average kinetic energy of the molecules of a substance. Think of it as a measure of how fast the molecules are moving. The unit is ºC or K.

    • Temperature is NOT heat!

    • Heat is the internal energy that is transferred between bodies in contact. The unit is Joules (J), or sometimes calories (cal).

    • A difference in temperature will cause heat energy to be exchanged between bodies in contact. When two bodies are at the same temperature, no heat is transferred. This is called Thermal Equilibrium.


  • Thermal Expansion them the information and they must predict where the water will hit the floor.

    • Most substances expand when their temperature goes up.

    • ΔL = αLoΔT

    • Δ L is change in length

    • α is coefficient of linear expansion

    • Lo is original length of substance

    • ΔT is change in temperature


L0 = 301m

T1 =22º C

T2 =0º C

α =11x10-6 /ºC

ΔL = αL0ΔT

ΔL = 11x10-6 /ºC · 301m · (0 ºC - 22 ºC ) = 0.073 m


  • Ideal Gas Law when the temperature is 22

    • P1 V1 / T1 = P2 V2 / T2

    • P1, P2: initial and final pressure (any unit)

    • V1, V2: initial and final volume (any unit)

    • T1, T2: initial and final temperature (in Kelvin!)

    • Temperature in K is obtained from temperature in oC by adding 273.


P1 = 2.3 atm

V1 = 4.0 L

T1 = 23ºC + 273 = 296 K

P2 = 3.1 atm

V2 = ? L

T2 = 0ºC + 273 = 273 K

P1V1/T1 = P2V2/T2

V2 = (P1V1 T2)/(T1P2)

V2 = (2.3 atm · 4.0 L · 273 K)/ (296 K · 3.1 atm) = 2.737 L


  • Ideal Gas Equation when the temperature is 22

    • P V = n R T

      • P: pressure (in Pa)

      • V: volume (in m3)

      • n: number of moles

      • R: gas law constant

      • 8.31 J/(mol K)

      • T: temperature (in K)

PV = N/m2· m3 = Nm = J

  • nRT = mol · J/(mol · K) · K = J

Both sides = J


25 when the temperature is 22ºC + 273 = 298K

  • Determine the number of moles of an ideal gas that occupy 10.0 m3 at atmospheric pressure and 25oC.

n = PV / RT

PV = nRT

n = 100000 Pa · 10 m3 / (8.31J/ (mol · K) · 298K) = 403.82 mols


Kinetic Theory of Gases when the temperature is 22


  • Ideal Gas Equation when the temperature is 22

    • PV = nRT (using moles)

    • PV = NkBT (using molecules)

      • P: pressure (Pa)

      • V: volume (m3)

      • N: number of molecules

      • kB: Boltzman’s constant

        • 1.38x10-23 J/K

      • n: Number of moles

      • R: Universal Gas constant

        • 8.31 J / (mol · K)

      • T: Temperature in Kelvin


PV = NkBT

P = NkBT / V

P = 25000 · 1.38 x 10-23 J/k 273k / 1m3 = 9.42 x 10 -17 Pa


NA = 6.02 x 1023 mol-1

R = 8.31 J/(mol K)

KB = R/ NA

KB = 8.31 J / (mol K) / 6.02 x 1023 mol-1 = 1.38 x 10-23 J/K


  • The kinetic theory of gases uses mechanics to describe the motion off each single molecule in a sample off an ideal gas..

  • When a very large number off molecules is considered,, the mechanical properties off the individual molecules are summed in a statistical way to predict the behavior off the gas sample


  • Gases consist of a large number of molecules that make elastic collisions with each other and the walls of the container.

  • Molecules are separated, on average, by large distances and exert no forces on each other except when they collide.

  • There is no preferred position for a molecule in the container, and no preferred direction for the velocity.


  • Simulations elastic collisions with each other and the walls of the container.

  • http://comp.uark.edu/~jgeabana/mol_dyn/KinThI.html

  • http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm


  • K elastic collisions with each other and the walls of the container.ave = 3/2 kBT

    • Kave: average kinetic energy (J)

    • kB: Boltzmann’s Constant (1.38 x 10-23 J/K)

    • T: Temperature (K)

  • The molecules have a range of kinetic energies; kave is just the average of that range

  • Need to change this make connection

  • PV = 3/2 N KB T

  • PV = to work which is equal to KE we are looking at 1 molecule for N


Use atomic mass number to get the mass per mole elastic collisions with each other and the walls of the container.

  • What is the average kinetic energy and the average speed of oxygen molecules in a gas sample at 0oC?

atomic mass number = number of Neutrons + number of Protons

O2 32g/mol · 1 mol / 6.02 x 1023 · 1 kg / 1000g = 5.315 x 10-26 kg

Kave = 3/2 kBT

½ mvrms2 = 3/2 kBT

vrms =√(3 kBT / m)

vrms =√((3 · 1.38 x 10-23 J/K · 273K) / 5.32 x 10-26kg) = 460.1 m/s

vrms =√( J / kg)

vrms =√( J/K · K / kg)

vrms =√( kg · m2/s2 / kg)

vrms =√( kg · m/s2 · m / kg)

vrms =√(m2/s2)


Boundary elastic collisions with each other and the walls of the container.

Environment

For our purposes,

the system will

almost always be

an ideal gas.

System

(gas)


  • The system boundary controls how the environment affects the system.

  • If the boundary is “closed to mass”, that means that mass can’t get in or out.

  • If the boundary is “closed to energy”, that means energy can’t get in or out.

  • Consider the earth as a system. What type of boundary does it have?

    • What is the boundary?

    • Is the boundary closed to mass?

    • Is the boundary closed to energy?

Atmosphere, crust, where gravity is negligible

Yes, the increase in mass each day is negligible

No, the sun adds energy to the system


System system.

  • First Law of Thermodynamics

ΔU

Q

W

Work

Heat

U = ΔTotal energy

ΔU = W + Q


  • Awkward notation WARNING! system.

    • We all know that U is potential energy in mechanics. However…

    • U is Eint (thermal energy) in thermodynamics!

    • This means when we are in thermo, U is thermal energy, which is related to temperature. When we are in mechanics, it is potential energy, which is related to configuration or position.



  • 1 system (or gas).st Law of Thermodynamics

    • ΔU = Q + W

      • Δ U: change in internal energy of system (J)

      • Q: heat added to the system (J). This heat exchange is driven by temperature difference.

      • W: work done on the system (J). Work will be related to the change in the system’s volume.

    • This law is sometimes paraphrased as “you can’t win”.


Q = 200 J

U

W = -100 J

ΔU = Q + W

ΔU = 200J – 100J = 100 J


Q = 0 J and does 100 J of work on the environment. What is its change in internal energy?

  • How much work does the environment do on a system if its internal energy changes from 40,000 J to 45,000 J without the addition of heat?

ΔU = 45000J – 40000J = 5000J

W = ? J

ΔU = Q + W

5000 J = W therefore W = 5000 J


W and does 100 J of work on the environment. What is its change in internal energy?

ΔU

Q

-53 J

A  B

a.

0 J

b.

-53 J

B  C

130 J

c.

-150 J

-280 J

d.

+203 J

C  A

e.

+353 J

-150 J

ΔU = Q + W

U = N Kave


  • The and does 100 J of work on the environment. What is its change in internal energy?thermodynamic state of a gas is defined by pressure, volume, and temperature.

  • A gas process describes how gas gets from one state to another state.

  • Processes depend on the behavior of the boundary and the environment more than they depend on the behavior of the gas.


Isothermal Process and does 100 J of work on the environment. What is its change in internal energy?

P

Constant Temperature

T1

T2

T3

PV = nRT

Initial state of the gas

Isothermal Process

Final state of the gas

V

ΔT = 0 ( constant T)


Isobaric Process and does 100 J of work on the environment. What is its change in internal energy?

P

Constant Pressure

T1

T2

T3

PV = nRT

Isobaric Expansion

Isobaric Contraction

V

ΔP = 0 ( constant P)


Isometric Process and does 100 J of work on the environment. What is its change in internal energy?

P

Constant Volume

T1

T2

T3

PV = nRT

V

ΔV = 0 ( constant V)


Adiabatic Process and does 100 J of work on the environment. What is its change in internal energy?

P

Insulated

T1

T2

PV = nRT

Isotherm

Temperature, pressure and volume all change in an adiabatic process

adiabat

V

Q = 0 ( No heat enters or leaves the system)


  • Work and does 100 J of work on the environment. What is its change in internal energy?

    • Calculation of work done on a system (or by a system) is an important part of thermodynamic calculations.

    • Work depends upon volume change.

    • Work also depends upon the pressure at which the volume change occurs.


W and does 100 J of work on the environment. What is its change in internal energy?gas = PΔV

  • Work Done BY gas

Positive Work

Wenv = -PΔV

Negative Work

ΔV


W and does 100 J of work on the environment. What is its change in internal energy?gas = PΔV

  • Work Done ON gas

Negative since ΔV is negative

Wenv = -PΔV

Positive since ΔV is negative

ΔV


V and does 100 J of work on the environment. What is its change in internal energy?1 = 0.020 m3

V2 = 0.80 m3

Wgas = PΔV

  • Calculate the work done by a gas that expands from 0.020 m3 to 0.80 m3 at constant atmospheric pressure.

  • How much work is done by the environment when the gas expands this much?

Wgas = 100000 Pa·0.078m3 = 78000J

Wgas = - 78000J


Δ and does 100 J of work on the environment. What is its change in internal energy?U = -230 J

  • What is the change in volume of a cylinder operating at atmospheric pressure if its internal energy decreases by 230 J when 120 J of heat are removed from it?

Q = -120 J

ΔU = Q+W

W = ΔU - Q

W = -230 J – (-120 J) = 110 J

Wgas = PΔV

ΔV = 110 J / 100000 Pa

ΔV = 0.0011 m3


W and does 100 J of work on the environment. What is its change in internal energy?AB > WCD

P

Where we are considering work done BY the gas

  • Work (isobaric)

A

B

P2

WAB = P2ΔV

C

D

P1

WCD = P1ΔV

V

V1

V2


P and does 100 J of work on the environment. What is its change in internal energy?

Where we are considering work done BY the gas

  • Work is path dependent

WABD > WACD

A

B

P2

WABD

C

P1

D

WACD

V1

V2

V


  • Work done by a cycle and does 100 J of work on the environment. What is its change in internal energy?

    • When a gas undergoes a complete cycle, it starts and ends in the same state. The gas is identical before and after the cycle, so there is no identifiable change in the gas.

    • DU = 0 for a complete cycle.

    • The environment, however, has been changed.


P and does 100 J of work on the environment. What is its change in internal energy?

Work done by the gas is equal to the area circumscribed by the cycle

  • Work done by a cycle

Work done by gas is positive for clockwise cycles, negative for counterclockwise cycles.

B

A

P2

WABCD

P1

C

D

V1

V2

V

Work done by environment is negative of work done by gas.


  • Second Law of Thermodynamics and does 100 J of work on the environment. What is its change in internal energy?

    • No process is possible whose sole result is the complete conversion of heat from a hot reservoir into mechanical work. (Kelvin-Planck statement.)

    • No process is possible whose sole result is the transfer of heat from a cooler to a hotter body. (Clausius statement.)


Heat Engines and does 100 J of work on the environment. What is its change in internal energy?

Heat Engines and Carnot Cycle

Efficiency


  • Heat Engines and does 100 J of work on the environment. What is its change in internal energy?

    • Heat engines can convert heat into useful work.

    • According to the 2nd Law of Thermodynamics. Heat engines always produce some waste heat.

    • Efficiency can be used to tell how much heat is needed to produce a given amount of work.

    • NOTE: A heat engine is not something that produces heat. A heat engine transfers heat from hot to cold, and does mechanical work in the process.


Heat Source (High Temperature) and does 100 J of work on the environment. What is its change in internal energy?

QH

  • Heat Transfer

QH = QC

QC

Heat Sink (Low Temperature)

QH = input heat

QC = magnitude of rejected heat


Heat Source (High Temperature) and does 100 J of work on the environment. What is its change in internal energy?

QH

  • Heat Engines

QH = QC + W

QC

Heat Sink (Low Temperature)


Heat Source (High Temperature) and does 100 J of work on the environment. What is its change in internal energy?

  • QH = W + QC

  • QH: Heat that is put into the system and comes from the hot reservoir in the environment.

  • W: Work that is done by the system on the environment.

  • QC: Waste heat that is dumped into the cold reservoir in the environment.

QH

  • Work and Heat Engine

W

QC

Heat Sink (Low Temperature)


  • Efficiency of Heat Engine and does 100 J of work on the environment. What is its change in internal energy?

    • Efficiency = W/QH = (QH - QC)/QH

      • W: Work done by engine on environment

      • QH: Heat absorbed from hot reservoir

      • QC: Waste heat dumped to cold reservoir

    • Efficiency is often given as percent efficiency.


QH = 3600J

QC = 1500J

QH = W + QC

W = QH - QC

W = 3600 J – 1500 J = 2100J


Adiabatic during a complete cycle. How much work does it do during the cycle?Vs Isothermal Expansion

In an adiabatic expansion, no heat energy can enter the gas to replace energy being lost as it does work on the environment. The temperature drops, and so does the pressure.

P

Initial State

Adiabatic Expansion

Isothermal Expansion

V


Carnot Cycle during a complete cycle. How much work does it do during the cycle?

P

Heat In

QH = QC + W

Eff = W / QH

Isothermal expansion

Work

Adiabatic compression

Adiabatic expansion

Isothermal compression

V

Heat Out


  • Efficiency of Carnot Cycle during a complete cycle. How much work does it do during the cycle?

  • For a Carnot engine, the efficiency can be calculated from the temperatures of the hot and cold reservoirs.

  • Carnot Efficiency = (TH - TC)/TH

    • TH: Temperature of hot reservoir (K)

    • TC: Temperature of cold reservoir (K)


TC = 60 ºC

TH = 1500 ºC

Efficiency = (TH - TC)/TH

Efficiency = ((1500 ºC + 273) – (60 ºC + 273))/ (1500 ºC + 273) = 0.81 or 81 %


Entropy between the temperatures of 60 and 1500


  • Entropy… between the temperatures of 60 and 1500

    • Entropy is disorder, or randomness.

    • The entropy of the universe is increasing. Ultimately, this will lead to what is affectionately known as “Heat Death of the Universe”.

    • Does the entropy in your room tend to increase or decrease?



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