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Thermodynamics is the study of energy relationships that involve heat, mechanical work, and other aspects of energy and heat transfer. Central Heating. THERMODYNAMICS. A THERMODYNAMIC SYSTEM.

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slide1
Thermodynamics is the study of energy relationships that involve heat, mechanical work, and other aspects of energy and heat transfer.

Central Heating

THERMODYNAMICS
a thermodynamic system
A THERMODYNAMIC SYSTEM
  • A system is a closed environment in which heat transfer can take place. (For example, the gas, walls, and cylinder of an automobile engine.)

Work done on gas or work done by gas

internal energy of system
Usually the internal energy consists of the sum of the potential and kinetic energies of the working gas molecules.INTERNAL ENERGY OF SYSTEM
  • The internal energy U of a system is the total of all kinds of energy possessed by the particles that make up the system.
two ways to increase the internal energy u
+U

WORK DONE ON A GAS (Positive)

TWO WAYS TO INCREASE THE INTERNAL ENERGY, U.

HEAT PUT INTO A SYSTEM (Positive)

two ways to decrease the internal energy u
Wout

Qout

hot

hot

HEAT LEAVES A SYSTEM

Q is negative

TWO WAYS TO DECREASE THE INTERNAL ENERGY, U.

-U

Decrease

WORK DONE BY EXPANDING GAS: W is positive

thermodynamic state
THERMODYNAMIC STATE

The STATE of a thermodynamic system is determined by four factors:

  • Absolute Pressure P in Pascals
  • Temperature T in Kelvins
  • Volume V in cubic meters
  • Number of moles, n, of working gas
the first law of thermodyamics
THE FIRST LAW OF THERMODYAMICS:
  • The net heat put into a system is equal to the change in internal energy of the system plus the work done BY the system.

Q = U + W final - initial)

  • Conversely, the work done ON a system is equal to the change in internal energy plus the heat lost in the process.
application of first law of thermodynamics
Wout =120 J

Qin

400 J

Apply First Law:

Q = U + W

APPLICATION OF FIRST LAW OF THERMODYNAMICS

Example 1:In the figure, the gas absorbs400 Jof heat and at the same time does120 Jof work on the piston. What is the change in internal energy of the system?

example 1 cont apply first law
Wout =120 J

Qin

400 J



U = +280 J

Example 1 (Cont.): Apply First Law

DQ is positive: +400 J (Heat IN)

DW is positive: +120 J (Work OUT)

Q = U + W

U = Q - W

U = Q - W

= (+400 J) - (+120 J)

= +280 J

example 1 cont apply first law1
Wout =120 J

Qin

400 J



U = +280 J

Example 1 (Cont.): Apply First Law

Energy is conserved:

The 400 J of input thermal energy is used to perform 120 J of external work, increasing the internal energy of the system by 280 J

The increase in internal energy is:

four thermodynamic processes
Q = U + WFOUR THERMODYNAMIC PROCESSES:
  • Isochoric Process: V = 0, W = 0
  • Isobaric Process: P = 0
  • Isothermal Process: T = 0, U = 0
  • Adiabatic Process: Q = 0
isochoric process constant volume v 0 w 0
0

QIN

QOUT

-U

+U

HEAT IN = INCREASE IN INTERNAL ENERGY

HEAT OUT = DECREASE IN INTERNAL ENERGY

ISOCHORIC PROCESS: CONSTANT VOLUME, V = 0, W = 0

Q = U + W so that Q = U

No Work Done

isochoric example
No Change in volume:

P2

B

PA P B

=

TA T B

P1

A

V1= V2

400 J

ISOCHORIC EXAMPLE:

400 J heat input increases internal energy by 400 J and zero work is done.

Heat input increases P with const. V

isobaric process constant pressure p 0
QIN

QOUT

Work Out

Work In

HEAT IN = Wout + INCREASE IN INTERNAL ENERGY

ISOBARIC PROCESS: CONSTANT PRESSURE, P = 0

Q = U + W But W = P V

-U

+U

HEAT OUT = Win + DECREASE IN INTERNAL ENERGY

isobaric example constant pressure
B

A

VA VB

P

=

TA T B

V1 V2

400 J

ISOBARIC EXAMPLE (Constant Pressure):

400 Jheat does 120 J of work, increasing the internal energy by 280 J.

Heat input increases Vwith const. P

isobaric work
A

B

P

VA VB

=

TA T B

V1 V2

400 J

ISOBARIC WORK

PA = PB

Work = Area under PV curve

isothermal process const temperature t 0 u 0
Q = U + W AND Q = W

QIN

QOUT

Work Out

Work In

NET HEAT INPUT = WORK OUTPUT

ISOTHERMAL PROCESS: CONST. TEMPERATURE, T = 0, U = 0

U = 0

U = 0

WORK INPUT = NET HEAT OUT

isothermal example constant t
A

PA

B

PB

V2 V1

PAVA =PBVB

ISOTHERMAL EXAMPLE (Constant T):

U = T = 0

Slow compression at constant temperature: ----- No change in U.

adiabatic process no heat exchange q 0
U = -W

W = -U

Work Out

Work In

U

+U

Q = 0

Work done at EXPENSE of internal energyINPUT Work INCREASES internal energy

ADIABATIC PROCESS: NO HEAT EXCHANGE, Q = 0

Q = U + W ; W = -U or U = -W

adiabatic example
A

PA

B

PB

V1 V2

ADIABATIC EXAMPLE:

Expanding gas does work with zero heat loss. Work = -DU

Insulated Walls: Q = 0

example problem
AB: Heated at constant V to 400 K.Example Problem:

A 2-L sample of Oxygen gas has an initial temp-erature and pressure of 200 K and 1 atm. The gas undergoes four processes:

  • BC: Heated at constant P to 800 K.
  • CD: Cooled at constant V back to 1 atm.
  • DA: Cooled at constant P back to 200 K.
pv diagram for problem
B

400 K

800 K

PB

A

200 K

1 atm

2 L

PV-DIAGRAM FOR PROBLEM

How many moles of O2 are present?

Consider point A: PV = nRT

process ab isochoric
B

400 K

800 K

PB

A

200 K

1 atm

PA P B

=

2 L

TA T B

1 atmP B

P B = 2 atm

=

200 K400 K

or 203 kPa

PROCESS AB: ISOCHORIC

What is the pressure at point B?

process bc isobaric
B

400 K

800 K

PB

C

200 K

VB V C

1 atm

D

=

TB T C

2 L

4 L

2 LV C

V C = V D = 4 L

=

400 K800 K

PROCESS BC: ISOBARIC

What is the volume at point C (& D)?

finding w for process bc
B

400 K

800 K

2 atm

C

200 K

1 atm

2 L

4 L

W = +400 J

FINDING W FOR PROCESS BC.

Work depends on change in V.

P = 0

Work = PV

W = (2 atm)(4 L - 2 L) = 4 atm L = 400 J

process cd isochoric
B

400 K

800 K

PB

C

A

200 K

D

1 atm

PC P D

=

2 L

TC T D

2 atm1 atm

T D = 400 K

=

800 KTD

PROCESS CD: ISOCHORIC

What is temperature at point D?

finding w for process da
400 K

800 K

2 atm

A

200 K

400 K

1 atm

D

2 L

4 L

W = -200 J

FINDING W FOR PROCESS DA.

Work depends on change inV.

P = 0

Work = PV

W= (1 atm)(2 L - 4 L) = -2 atm L = -200 J

net work for complete cycle is enclosed area
+404 J

B

C

B

C

-202 J

2 atm

2 atm

Neg

1 atm

1 atm

2 L

4 L

2 L

4 L

B

C

2 atm

Area = (1 atm)(2 L)

Net Work = 2 atm L = 200 J

1 atm

2 L

4 L

NET WORK FOR COMPLETE CYCLE IS ENCLOSED AREA
heat engines
Hot Res. TH

Qhot

Wout

Engine

Qcold

Cold Res. TC

HEAT ENGINES

A heat engine is any device which through a cyclic process:

  • Absorbs heat Qhot
  • Performs work Wout
  • Rejects heat Qcold
the second law of thermodynamics
Hot Res. TH

It is impossible to construct an engine that, operating in a cycle, produces no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.

Qhot

Wout

Engine

Qcold

Cold Res. TC

THE SECOND LAW OF THERMODYNAMICS
the second law of thermodynamics1
Hot Res. TH

Hot Res. TH

400 J

400 J

100 J

400 J

Engine

Engine

300 J

Cold Res. TC

Cold Res. TC

  • An IMPOSSIBLE engine.
  • A possible engine.
THE SECOND LAW OF THERMODYNAMICS
efficiency of an engine
Hot Res. TH

QH

W

W

QH

QH- QC

QH

Engine

e = =

QC

Cold Res. TC

QC

QH

e = 1 -

EFFICIENCY OF AN ENGINE

The efficiency of a heat engine is the ratio of the net work done W to the heat input QH.

efficiency example
Hot Res. TH

800 J

W

QC

QH

Engine

e = 1 -

600 J

600 J

800 J

Cold Res. TC

e = 1 -

e = 25%

EFFICIENCY EXAMPLE

An engine absorbs 800 J and wastes 600 J every cycle. What is the efficiency?

Question: How many joules of work is done?

efficiency of an ideal engine carnot engine
Hot Res. TH

QH

W

Engine

TH- TC

TH

QC

e =

Cold Res. TC

TC

TH

e = 1 -

EFFICIENCY OF AN IDEAL ENGINE (Carnot Engine)

For a perfect engine, the quantities Q of heat gained and lost are proportional to the absolute temperatures T.

slide35
TC

TH

e = 1 -

W

QH

e =

300 K

500 K

e = 1 -

Work = 120 J

Example 3:A steam engine absorbs 600 J of heat at 500 K and the exhaust temperature is 300 K. If the actual efficiency is only half of the ideal efficiency, how much work is done during each cycle?

Actual e = 0.5ei = 20%

W = eQH = 0.20 (600 J)

e = 40%

refrigerators
Hot Res. TH

Qhot

Win

Engine

Qcold

Cold Res. TC

WIN = Qhot - Qcold

REFRIGERATORS

A refrigerator is an engine operating in reverse: Work is done on gas extracting heat from cold reservoir and depositing heat intohot reservoir.

Win + Qcold = Qhot

the second law for refrigerators
Hot Res. TH

Qhot

Engine

Qcold

Cold Res. TC

THE SECOND LAW FOR REFRIGERATORS

It is impossible to construct a refrigerator that absorbs heat from a cold reservoir and deposits equal heat to a hot reservoir with W = 0.

If this were possible, we could establish perpetual motion!

summary
TheFirst Law of Thermodynamics:The net heat taken in by a system is equal to the sum of the change in internal energy and the work done by the system.

Q = U + W final - initial)

Summary
  • Isochoric Process: V = 0, W = 0
  • Isobaric Process: P = 0
  • Isothermal Process: T = 0, U = 0
  • Adiabatic Process: Q = 0
summary cont
TheSecond Law of Thermo:It is impossible to construct an engine that, operating in a cycle, produces no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.

Hot Res. TH

Qhot

Wout

Engine

Qcold

Cold Res. TC

Summary (Cont.)
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