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/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/Vρ: 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..

• Calculate the net force on an airplane window if cabin 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.

Pcabin = 90000 Pa

Atmospheric pressure = 100000 Pa

Poutside = 50000 Pa

Cabin Outside P = F/A

A = lw

A = 0.3m · 0.43m = 0.129m2

F = AP

F = 0.129m2 · 40000Pa = 5160N

P = 90000Pa – 50000Pa = 40000Pa

• The Pressure of a Liquid– 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– 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• 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

Hurricane Katrina August 2005

A hurricane’s stormsurge can overtop levees,but a bigger problem canbe increasing thehydrostatic pressure atthe base of the levee.

• New Orleans Elevation Map– New Orleans is largely below sea level, and relies upon a

system of levees to keep the lake and the river at bay

Normal lake level at 1.0ft

Surge elevation set at 11.5 ft

Max ht of the levee 17.5 ft

• 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

hlevee = 16.5ft

P = ρgh

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

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

ρsalt water = 1025 kg /m3

• Pascal’s Principle– 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

• A person stands on a piston. The system is full of hydraulic fluid. The car with a mass of 1500 kg is sitting on a piston that has a radius of 3.1 m. A person is standing on the other piston that has a radius of 0.40 m. What must the weight of the person be to cause the car to rise?

mc = 1500kg

Fp = ?

rc = 3.1m

rp = 0.40m

Fc / Ac = Fp / Ap

Ac =πr2 = π(3.1m)2 = 30.19m2

Ac =πr2 = π(0.40m)2 = 0.50m2

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.

h

A

ΣF = ma

FN - mg= ma

a = (vf – vi) / t

FN = ma + mg

A = πr2

A = π(0.02m)2 = 0.0013 m2

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

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

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

FN = ma + mg

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

• 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.

h

A

P1 = FN / A

P2 = ρgh

P = ρgh + FN / A

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

Buoyancy

• Floating is a type of equilibrium

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

Fbuoyancy

mg

• Archimedes Principle– 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.

• Buoyant force on submerged object

Fbuoy = ρVg

mg

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

Fbouy ≠ mg

• Buoyant force on submerged object

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

Fbuoy = ρVg

mg

Fbouy = mg

• Buoyant force on submerged object

If the diver wants to rise, he inflateshis vest, which increases the amountof water he displaces, and heaccelerates upward

Fbuoy = ρVg

mg

Fbouy = mg

• Buoyant force on floating object

If the object floats on the surface, weknow for a fact Fbuoy = mg! The volume of displaced water equals the volume of the submerged portion of the ship.

Fbuoy = ρVg

mg

Fbouy = mg

• Assume a wooden raft has 80.0% of the density of water. The 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?

l

w

h

y

Fbuoy = ρVg

mgΣF: Fbouy – mg = 0

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

ρgVsub – (0.8ρ)gVtotal = 0 ρglwy = (0.8ρ)glwh

y = (0.8) h y = 0.8 · 0.1m = 0.08mThe height above the water is 0.02m

• Buoyant Force– The buoyant force can be extremely strong.– Incredibly massive objects can float, even when they are

not intended to…

Moving Fluids

• When a Fluid Flows…– …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.)– 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

• A pipe of diameter 6.0 cm has fluid flowing through it at 1.6 m/s. How fast is the fluid flowing in an area of the pipe in which the diameter is 3.0 cm? How much water per second flows through the pipe?

A1v1 = A2v2

Apipe = πr2

πr21 v1 = πr2

2 v2

v2 = πr21 v1 / πr2

2

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

A1v1 = A2v2V2 = A1v1 / A2

V2 = l1w1v1 / l2w2

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

• Natural Waterways

• Flash flooding can be explained by fluid flow continuity

• Artificial Waterways

Flooding from the Mississippi River Gulf Outlet was responsible forcatastrophic flooding in eastern New Orleans and St. Bernard during Hurricane Katrina.

• Fluid Flow Continuity in Waterways

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

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

Moving Fluids and Bernoulli Effect

• Bernoulli’s Theorem– 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– 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)

• 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.

mg

Fthurst

Flift

Fdrag

Longer d, higher v, lower p

Flift = ΔPA

• Problem: An above-ground swimming pool has a hole of radius 0.10 cm in the side 2.0 meters below the surface of the water. How fast is the water flowing out of the hole? How much water flows out each second?

h1 = 2m po + ρgh + ½ ρv2 = Constant

po + ρgh1 + ½ ρv12 = po + ρgh2 + ½ ρv2

2

ρgh1 + ½ ρv12 = ρgh2 + ½ ρv2

2

v1 = 0 m/s h2 = 0 m

ρgh1 + = ½ ρv22

v2 = √(2gh1)

gh1 = ½v22

v2 = √(2·9.8 m/s2 · 2.0m) = 6.26 m/s

Bernoulli’s principle

What about energy

mgh = ½ mv2

gh = ½ v2

v2 = √(2gh1)

Same equation!!!

• Problem: An above-ground swimming pool has a hole of radius 0.10 cm in the side 2.0 meters below the surface of the water. How fast is the water flowing out of the hole? How much water flows out each second?

h1 = 2m v2 = 6.26 m/s

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?

p1 = 350000 Pav1 = 1.3 m/s

p2 = ? Pav2 = ? m/s

A1v1 = A2v2

v2 = A1v1 / A2

v2 = πr21v1 / πr2

2

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 = 350000 Pav1 = 1.3 m/s

p2 = ? Pav2 = 19.17 m/s

p1 + ρgh1 + ½ ρv12 = p2 + ρgh2 + ½ ρv2

2

h1 = 0m h2 = 0m

p1 + ½ ρv12 = p2 + ½ ρv2

2

p2 = p1 + ½ ρv12 - ½ ρv2

2

p2 = p1 + ½ ρ(v12 - v2

2)

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

• Bernoulli’s Principle and Hurricanes– 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

• Make a device just like the above problem with a whole give them the information and they must predict where the water will hit the floor.

Thermodynamics

• Thermodynamics– 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– 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– Most substances expand when their temperature goes

up.– ΔL = αLo ΔT– Δ L is change in length– α is coefficient of linear expansion– Lo is original length of substance– ΔT is change in temperature

• Sample problem: If the Eiffel tower is 301 m tall on a day when the temperature is 22º C, how much does it shrink when the temperature drops to 0º C? The coefficient of linear expansion is 11 x 10-6 ºC-1 for the iron the tower is made from.

ΔL = αL0 ΔT

L0 = 301m T1 =22º C T2 =0º C α =11x10-6 /ºC

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

• Ideal Gas Law– 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.

• Suppose an ideal gas occupies 4.0 liters at 23oC and 2.3 atm. What will be the volume of the gas if the temperature is lowered to 0oC and the pressure is increased to 3.1 atm.

P1V1/T1 = P2V2/T2

V1 = 4.0 L T1 = 23ºC + 273 = 296 K P1 = 2.3 atm

V2 = ? L T2 = 0ºC + 273 = 273 K P2 = 3.1 atm

V2 = (P1V1 T2) /(T1P2)

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

• Ideal Gas Equation– 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

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

PV = nRT

25ºC + 273 = 298K

n = PV / RT

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

Kinetic Theory of Gases

• Ideal Gas Equation– 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

• Suppose a near vacuum contains 25000 molecules of helium in one cubic meter at 0ºC. What is the pressure?

PV = NkBT

P = NkBT / V

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

• What is the relationship between the Universal Gas Law constant (8.31 J / (mol K) and Boltzman’s constant kB (1.38 x 10-23 J/K)?

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• http://comp.uark.edu/~jgeabana/mol_dyn/KinTh

I.html• http://intro.chem.okstate.edu/1314F00/Laborato

ry/GLP.htm

• Kave = 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

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

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

Use atomic mass number to get the mass per mole

atomic mass number = number of Neutrons + number of Protons

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/K · K / kg) vrms =√( J / kg)

vrms =√( kg · m/s2 · m / kg)

vrms =√(m2/s2)

vrms =√( kg · m2/s2 / kg)

Boundary

For our purposes,the system willalmost always bean ideal 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

• First Law of Thermodynamics

ΔU

W

U = Δ Total energy

Q

ΔU = Q – W

• Awkward notation WARNING!– 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.

• U is the sum of the kinetic energies of all molecules in a system (or gas).

• U = N Kave

• U = N (3/2 kBT)• U = n (3/2 R T)

– Since kB = R /NA

• 1st 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”.

• A system absorbs 200 J of heat energy from the environment and does 100 J of work on the environment. What is its change in internal energy?

ΔU = Q – W

ΔU = 200J – 100J = 100 J

U

Q = 200 J

W = 100 J

• 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 = Q - W

5000 J = W therefore W = -5000 J

ΔU = 45000J – 40000J = 5000J

Q = 0 J

W = ? J

• The 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.

P

V

T1 T2 T3

Initial state of the gas

Isothermal Process

Constant Temperature

PV = nRT

Final state of the gas

Isothermal Process

ΔT = 0 ( constant T)

P

V

T1 T2 T3

Isobaric Expansion

Isobaric Process

Constant Pressure

PV = nRT

Isobaric Contraction

ΔP = 0 ( constant P)

P

V

T1 T2 T3

Isometric Process

Constant Volume

PV = nRT

ΔV = 0 ( constant V)

P

V

T1 T2

Adiabatic Process

Insulated

PV = nRT

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

Temperature, pressure and volume all change in an adiabatic process

Isotherm

adiabat

• Work– 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.

• Work Done BY gas

ΔV

Wgas = PΔV

Positive Work

Wenv = -PΔV

Negative Work

• Work Done ON gas

ΔV

Wgas = PΔV

Negative since ΔV is negative

Wenv = -PΔV

Positive since ΔV is negative

• 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 = PΔVV1 = 0.020 m3 V2 = 0.80 m3

Wgas = 100000 Pa·0.078m3 = 78000J

Wgas = - 78000J

• 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?

Wgas = PΔV

Q = -120 JΔU = -230 J

ΔU = Q+W

ΔV = 110 J / 100000 Pa

ΔV = 0.0011 m3

W = ΔU - Q

W = -230 J – (-120 J) = 110 J

A B

B C

C A

Q W ΔU

-53 J

-280 J -130 J

150 J

a. 0 J b.

c.

d.e.

a.

WAB = PVAB = P ∙ 0 = 0 J

A B

B C

C A

Q W ΔU

-53 J

-280 J -130 J

150 J

a. 0 J b.

c.

d.e.

-53 J

b.

UAB = QAB - WAB

= -53 - 0 = -53 J

A B

B C

C A

Q W ΔU

-53 J

-280 J -130 J

150 J

a. 0 J b.

c.

d.e.

-53 J

-150 J

c.

UBC = QCB - WBC

= -280 – (-130) = -150 J

A B

B C

C A

Q W ΔU

-53 J

-280 J -130 J

150 J

a. 0 J b.

c.

d.e.

-53 J

-150 J

+203 J

d.

UAB + UBC + UCA = 0

UCA = - UAB – UBC

= -53 – (-150) = +203 J

A B

B C

C A

Q W ΔU

-53 J

-280 J -130 J

150 J

a. 0 J b.

c.

d.e.

-53 J

-150 J

+203 J+353 J

e.

QCA = UCA + WCA

UCA = - 203 + 150 = 353 J

• Work (isobaric)P

VV1 V2

P1

P2

WAB > WCD

Where we are considering work done BY the gas

WAB = P2ΔV

WCD = P1ΔV

D

B

C

A

• Work is path dependentP

VV1 V2

P1

P2

Where we are considering work done BY the gas

WABD

WABD > WACD

WACD

D

B

C

A

• Work done by a cycle– 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.

• Work done by a cycleP

VV1 V2

P1

P2

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

WABCD

BA

C D

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

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

• Second Law of Thermodynamics– 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 EnginesHeat Engines and Carnot Cycle

Efficiency

• Heat Engines– 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 TransferHeat Source (High Temperature)

Heat Sink (Low Temperature)

QH

QH = QC

QC

QC = magnitude of rejected heat

QH = input heat

• Heat EnginesHeat Source (High Temperature)

Heat Sink (Low Temperature)

QH

QH = QC + W

QC

• Work and Heat EngineHeat Source (High Temperature)

Heat Sink (Low Temperature)

QH

W

QC

• 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.

• Efficiency of Heat Engine– 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.

• A piston absorbs 3600 J of heat and dumps 1500 J of heat during a complete cycle. How much work does it do during the cycle?

QH = W + QC

QH = 3600J QC = 1500J

W = QH - QC

W = 3600 J – 1500 J = 2100J

P

V

Isothermal expansion

Adiabatic expansion

Isothermal compression

Adiabatic compression

Carnot Cycle

QH = QC + W

Eff = W / QH

• Efficiency of Carnot 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)

• Calculate the Carnot efficiency of a heat engine operating between the temperatures of 60 and 1500 oC.

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

Efficiency = (TH - TC)/TH

TC = 60 ºC TH = 1500 ºC

Entropy

• Entropy…– 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?