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Thermal Physics Topic 10.2 Processes

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Thermal Physics. Topic 10.2 Processes. The First Law of Thermodynamics. Is a statement of the Law of Conservation of Energy in which the equivalence of work and thermal energy flow is taken into account. It can be stated as - PowerPoint PPT Presentation

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Page 1: Thermal Physics

Thermal Physics

Topic 10.2 Processes

Page 2: Thermal Physics

The First Law of ThermodynamicsIs a statement of the Law of Conservation of

Energy in which the equivalence of work and thermal energy flow is taken into account.

It can be stated as

the heat added to a closed system equals the change in the internalenergy of the system plus the work done by the system.

Page 3: Thermal Physics

That is,

Q = U + W = U + pV

or

U = Q - W

Page 4: Thermal Physics

Explanation

Where `+ Q' is the thermal energy added to the system and `+ W' is the work done by the system.

If thermal energy leaves the system, then Q is negative. If work is done on the system, then W is negative.

For an isolated system, then W = Q = 0 and U = 0.

Page 5: Thermal Physics

Example

If 22 J of work is done on a system and 3.4 x 102 J of heat is added, what is the change in internal energy of the system?

Page 6: Thermal Physics

Solution

Using the formula, Q = U + W

We have that 340 J = U + (-22) J

So that U = 340 J + 22 J

= 362 J

That is, the change in internal energy of the system is 3.6 x 102 J.

Page 7: Thermal Physics

Isobaric

A graph of pressure as a function of volume change when the pressure is kept constant is shown below

Page 8: Thermal Physics

p -V diagram

p

V

p

VV1 V2

Area = work done = p (V1 - V2)

Isobaric process

Page 9: Thermal Physics

Work DoneSuch a process is said to be isobaric.

Note that the work done by the gas is equal to the area under the curve

An isobaric transformation requires a volume change at constant pressure

For this to occur, the temperature needs to change to keep the pressure constant

Page 10: Thermal Physics

Example

6.0 dm3 of an ideal gas is at a pressure of 202.6 kPa. It is heated so that it expands at constant pressure until its volume is 12 dm3. Find the work done by the gas.

Page 11: Thermal Physics

Solution Using the formula W = pV we have that W = 202.6 kPa x(12 - 6.0) dm3

= 202.6 x 103 Pa x (12 - 6.0) x 10-3m3

= 1.216 x 103 J That is, the work done by the gas in the

expansion is 1.2 x 103 J.

Page 12: Thermal Physics

Isochoric / Isovolumetric

A graph of pressure as a function of volume change when the volume is kept constant is shown below

p

V

Isochoric / Isovolumetric

Page 13: Thermal Physics

Work DoneSuch a process is said to be isochoric.

When the volume is kept fixed, the curve of the transformation is said to be an isochore.Note that the work done by the gas is equal to zero as V = 0.

There is zero area under the curve on a p-V diagram.

Page 14: Thermal Physics

p - V diagram

p

V

Isochoric / Isovolumetric

Page 15: Thermal Physics

However

The temperature and pressure can both change and so such a transformation will be accompanied by a thermal energy change.

Page 16: Thermal Physics

Example

A thermal system containing a gas is taken around the cycle as in the next slide

(Cyclic processes such as this one have important applications in heat engines that convert internal energy into useful mechanical energy).

Page 17: Thermal Physics

p

V

6

2

4 10

A

B C

D

Starting at point A on the diagram, describe the cycle, and, calculate the work done by the system in the completion of one cycle.

Page 18: Thermal Physics

SolutionFrom A to B, the volume is kept constant

(isovolumetric) as the pressure increases. This can be achieved by heating the gas. Since V = 0, then no work is done by the gas, W = 0.

Page 19: Thermal Physics

From B to C, the gas expands (volume increases) (isobaric expansion) while the pressure is kept constant. The amount of work done by the gas is given by the area under the 6 kPa isobar.

Now, we have that W = p V

So that W = 6 kPa. (10 - 4) m3

= 36,000J

Page 20: Thermal Physics

From C to D, the gas is cooled to keep the volume constant as the pressure is decreased (isovolumetric)

Again V = 0 and no work is done by the gas, W = 0

Page 21: Thermal Physics

From D to A, the gas is compressed (volume decreases) (isobaric compression) while the pressure is kept constant. The amount of work done on the gas is given by the area under the 2 kPa isobar.

Again, using the fact that W = p V we have that W = 2 kPa. (4 -10) m3

= - 12,000 J

Page 22: Thermal Physics

That is, the net work done by the gas is therefore 36,000 J – 12,000 J =24,000 J.

Because the cycle is traced in a clockwise direction, the net work done on the surroundings is positive. If the cycle was traced in a counter-clockwise direction, the work done would be negative

Page 23: Thermal Physics

Isothermal Processes

A thermodynamic process in which the pressure and the volume are varied while the temperature is kept constant is called an isothermal process

Page 24: Thermal Physics

In other words, when an ideal gas expands or is compressed at constant temperature, then the gas is said to undergo an isothermal expansion or compression.

Page 25: Thermal Physics

The figure below shows three isotherms for an ideal gas atdifferent temperatures where

Tl < T2 < T3p

V

T1 T2 T3

Isothermal process

Page 26: Thermal Physics

Isothermal Processes

The curve of an isothermal process represents a Boyle's Law relation

pV =constant = nRT • the moles of gas n, • the molar gas constant R, • the absolute temperature T are constant

Page 27: Thermal Physics

How?

In order to keep the temperature constant during an isothermal process • the gas is assumed to be held in a thin container

with a high thermal conductivitythat is in contact with a heat reservoir - an ideal body of large mass whosetemperature remains constant when heat is exchanged with it.e.g.. a constant-temperature water bath.

Page 28: Thermal Physics

And

• the expansion or compression should be done slowly so that no eddies are produced to create hot spots that would disrupt the energy equilibrium of the gas.

Page 29: Thermal Physics

An Ideal GasConsider an ideal gas enclosed in a thin

conducting vessel that is in contact with a heat reservoir, and is fitted with a light, frictionless, movable piston

If an amount of heat Q is added to the system which is at point A of an isotherm, then the system will move to another point on the graph, B.

Page 30: Thermal Physics

The heat taken in will cause the gas to expand isothermally and willbe equivalent to the mechanical work done by the gas

Because the temperature is constant, there is no change in internal energy of the gas

Page 31: Thermal Physics

That is,

T = 0

and U = 0

Q = W

The work done by gas is equal to the Area under the curve between A and B

Page 32: Thermal Physics

Isothermally

If the gas expands isothermally from A to B and then returns from B to A following exactly the same path during compression

Then the isothermal change is said to be reversible.

Page 33: Thermal Physics

Adiabatic Processes

An adiabatic expansion or contraction is one in which no heat Q is allowed to flow into or out of the system

For the entire adiabatic process, Q = 0

Page 34: Thermal Physics

How

To ensure that no heat enters or leaves the system during an adiabatic process it is important to:• make sure that the system is extremely

well insulated • carry out the process rapidly so that the

heat does not have the time to leave the system

Page 35: Thermal Physics

p

V

T1 T2

Adiabatic compression

Adiabatic Expansion

Page 36: Thermal Physics

Example

The compression stroke of an automobile engine is essentially an adiabatic compression of the air-fuel mixture

The compression occurs too rapidly for appreciable heat transfer to take place

Page 37: Thermal Physics

What happensIn an adiabatic compression the work

done on the gas will lead to an increase in the internal energy resulting in an increase in temperature.

U=Q - W

but Q=O

therefore U= - W

Page 38: Thermal Physics

In an adiabatic expansion the work done by the gas will lead to a decrease in the internal energy

Resulting in a decrease in temperature.

Page 39: Thermal Physics

Work done

p

V

Area = work done by the gas expanding

p

VArea = work done on the gas as it is compressed

p

VArea between curves = Net work done by the gas

Page 40: Thermal Physics

ExampleFor the compression stroke of an experimental diesel

engine, the air is rapidly decreased in volume by a factor of 15, the compression ratio.

The work done on the air-fuel mixture for this compression is measured to be 550 J

(a) What type of thermodynamic process is likely to have occurred?

(b) What is the change in internal energy of the air-fuel mixture?

(c) Is the temperature likely to increase or decrease?

Page 41: Thermal Physics

Solution(a) Because the compression occurs

rapidly appreciable heat transfer does not take place, and the process can be considered to be adiabatic, Q = 0

Page 42: Thermal Physics

(b) U = Q - W

= 0 - (-550) J

Therefore, the change in internal energy is 550J

Page 43: Thermal Physics

(c) The temperature rise will be very large resulting in the spontaneous ignition of the air-fuel mix

Page 44: Thermal Physics

Thermodynamic Cycle and Engines

A thermodynamic engine is a device that transforms thermal energy to mechanical energy

Cars, Steam trains, jets and rockets have engines that transform fuel energy (chemical energy) into kinetic energy of their motion

Page 45: Thermal Physics

Efficiency

In all engines the conversion is accompanied by the emission of exhaust gases that waste some of the thermal energy

These engines are not very efficient as only part of the thermal energy is converted to mechanical energy

Page 46: Thermal Physics

The EngineBy Vija and Bombo

Internal CombustionBy Adil and Suhayb

Schematic of a Heat Engine By Ham and Heisei

Heat PumpsBy Alex and Said

Carnot´s TheoremBy Martin and Mitul

Presenters

Schematic for a Refrigerator by Karim and Chilla

RefrigeratorsBy Aleem and Bhavik

Page 47: Thermal Physics

The EngineBy Vija and Bombo

Has two crucial features:

It must work in cycles to be useful

the cyclic engine must have more than one heat reservoir

Page 48: Thermal Physics

The Thermodynamic Cycle

Is the process in which the system is returned to the same state from which it started

That is, the initial and final states are the same in the cyclic process

Page 49: Thermal Physics

Combustion Engines

The steam engine and the steam turbine are examples of external combustion engines

The fuel is burnt outside the engine and the thermal energy is transferred to the piston or a turbine chamber by means of steam

Page 50: Thermal Physics

The Steam Engine

Water is heated in a boiler heat reservoir (high temperature) to produce steam

As the piston returns to its original position, the intake valve is closed and the exhaust valve opens

Page 51: Thermal Physics

The Steam Engine cont...

The steam passes into an open intake valve where it expands causing a piston to move outwards

This allows the exhaust gas to be forced out into a condenser heat reservoir (low temperature)

Page 52: Thermal Physics

The Steam Engine

Page 53: Thermal Physics

Steam Turbines

Most electricity is produced using steam engines

The piston is replaced with a rotating turbine that contains blades

The rotating turbine converts mechanical energy to electrical energy via a generator

The steam is cooled in cooling towers

Page 54: Thermal Physics

Internal CombustionBy Adil and Suhayb

Page 55: Thermal Physics

The Intake Stroke

With the exhaust valve closed, a mixture of petrol vapour and air from the carburetor is sucked into the combustion chamber through the inlet valve as the piston moves down

Page 56: Thermal Physics

The Compression Stroke

Both valves are closed and the piston moves up to squeeze the mixture of petrol vapour and air to about 1/8th its original volume

Page 57: Thermal Physics

The Power Stroke

Both valves are closed the mixture is ignited by a spark from the spark plug

The mixture burns rapidly and the hot gases then expand against the piston

Page 58: Thermal Physics

The Exhaust Stroke

1. The exhaust valve opens as the piston moves upwards

2. Exhaust gases are expelled

3. Then the cycle begins again

Page 59: Thermal Physics

Hot reservoir at TH

Cold reservoir at TL

Engine

Schematic of a Heat Engine BBy Ham and Heisei

QL

W

QH

Fluids leave carrying energy

QL = QH - W

Fluids enter carrying energy

QH Engine does useful work, W

Page 60: Thermal Physics

Efficiency

The thermal efficiency of a heat engine, , is defined as the ratio of work it does, W to the heat input QH

= W / QH

= (QH - QL) / QH

It is clear that the efficiency will be greater if QL can be made small

Page 61: Thermal Physics

QuestionAn engine absorbs 230J of thermal

energy from a high temperature reservoir, does work and exhausts 140J to a cold temperature reservoir.

What is the efficiency?

How much work is done? Answer 39% and 90J

Page 62: Thermal Physics

RefrigeratorsBy Aleem and Bhavik

A refrigerator is a device operating in a cycle that is designed to extract heat from its interior so as to achieve or maintain a lower temperature inside

The heat is exhausted to the surroundings normally at a higher temperature

Page 63: Thermal Physics

How the Refrigerator Works

A motor driving a compressor pump provides the means by which a net amount of work can be done by the system for a cycle

Even though the cabinets are well insulated, heat from the surroundings leaks back inside

Page 64: Thermal Physics

Heat PumpsBy Alex and Said1. Any device that can pump heat from a low

temperature reservoir to a high temperature reservoir is called a heat pump

2. Examples of heat pumps include refrigerator and reverse cycle air-conditioning devices used for space heating and cooling

Page 65: Thermal Physics

How..

1. A volatile liquid called Freon is circulated in a closed system of pipes by the compressor pump

2. By the process of evaporative cooling, the vaporised Freon is used to remove the heat

3. The compressor maintains a high pressure difference across a throttling valve

Page 66: Thermal Physics

How..Evaporation of the Freon occurs in

several loops called the evaporator pipes that are usually the coldest part of the fridge

As the liquid evaporates on the low

pressure, low temperature side, heat is added to the system

Page 67: Thermal Physics

How..• In order to turn from a liquid to a gas the

Freon requires thermal energy equal to the latent heat of vaporization

• This energy is obtained from the fridge

Page 68: Thermal Physics

How..• On the high pressure, high temperature

side of the throttling valve, thermal energy is removed from the system

• The vaporized Freon in the compressor pipes is compressed by the compressor pump

• It gives up its latent heat of vaporization to the air surrounding the compressor pipes

Page 69: Thermal Physics

How..

• The heat fins act as a heat sink to radiate the thermal energy to the surroundings at a faster rate

• • The fins are painted black and have a

relatively large surface area for their size

Page 70: Thermal Physics

A Typical Small Refrigerator

Page 71: Thermal Physics

Schematic for a Refrigerator Karim and Chilla

Hot reservoir at TH

Cold reservoir at TL

QL

W

QH

Page 72: Thermal Physics

Refrigerator

• The previous figure shows the energy transfers that occurs in a refrigerator cycle

• By doing work on the system, heat QL is added from the low temperature TL reservoir

• A greater amount of energy QH is exhausted to the high temperature TH

Page 73: Thermal Physics

Reverse Cycle Heat Pump

room

evaporatorcondenser

TL TH

WINTER

room

condenserTH TL

evaporator

SUMMER

Page 74: Thermal Physics

Winter• The evaporator heat exchanger is

outside the room • It exhausts its heat to the inside air

Page 75: Thermal Physics

Summer

• The evaporator heat exchanger is inside the room

• It exhausts its heat to the outside air

Page 76: Thermal Physics

In Both Cases

• Thermal energy is pumped from a low-temperature reservoir to a high-temperature reservoir

• The efficiency of refrigerators and heat pumps is defined in a different manner to heat engines

Page 77: Thermal Physics

Coefficient of Performance, K

• For a refrigerator and a heat pump used for cooling

K = QL / W

• Ratio of heat absorbed from inside the refrigerator and the work supplied to the refrigerator in one cycle

Page 78: Thermal Physics

Coefficient of Performance, K

• For a heat pump used for heating

• K = QH / W

• Ratio of the total positive heat exhausted to the inside and the work supplied to the heat pump in one cycle

Page 79: Thermal Physics

Carnot´s Theorem By Martin and Mitul

• No engine working between two heat reservoirs can be more efficient than a reversible engine between those reservoirs

• He argued that if thermal energy does flow from a cold body to a hot body then work must be done

• Therefore no engine can be more efficient than an ideal reversible one

Page 80: Thermal Physics

And...• All such engines have the same

efficiency • Therefore only a simple equation is

needed to calculate the efficiency • Consider an ideal perfectly insulated,

frictionless engine that can work backwards as well as forwards

• The p-V diagram would have the form...

Page 81: Thermal Physics

The Carnot Engine

Page 82: Thermal Physics

Explain

• The net work is the area enclosed by ABCDA

• Thermal energy is absorbed by the system at the high temperature TH and is expelled at the low temperature reservoir TL

Page 83: Thermal Physics

and• Work is done by the system as it expands

along the top isotherm from A to B • And also along the adiabat from B to C • Work is then done on the system to compress

it along the bottom isotherm from C to D • And also along the left adiabat from D to A

Page 84: Thermal Physics

Carnot’s Efficiency• Carnot was able to obtain an expression

for the efficiency in terms of temperature

• = 1 - TL / TH

• Therefore the efficiency of the Carnot cycle depends only on the absolute temperatures of the high and low temperature reservoirs

• The greater the temperature difference the greater the efficiency will be

Page 85: Thermal Physics

• But engine efficiency is

• = 1 - QL / QH

• Therefore TC/TH = QC/QH

• Or QC/TC = QH/TH

• What would happen if the

temperature difference was not maintained?

Page 86: Thermal Physics

Real Life

• In practice, friction and other mechanical losses and thermal losses reduce the overall efficiency

• Nearly 2/3 of the heat generated by modern power stations is released into the environment

Page 87: Thermal Physics

Thermal PhysicsTopic 10.3 Second Law of Thermodynamics and Entropy