Topic 2Topic 2

Thermodynamics

1

Why Thermodynamics?Why Thermodynamics?• To tell us the relationship between the pressure, temperature and p p , p

density of a gas as it moves. • Altogether there are 6 variables we need to describe the state of a 

gas…( )• p pressure (Pa)

• density (kg/m3), or specific volume v=1/• T temperature (K)• e internal energy per unit mass ( )• e internal energy per unit mass (          )• h enthalpy (        ), he+p/• s entropy (           )

• To relate these variables we useTo relate these variables we use 1. The Kinetic Theory of Gases, 2. The 1st Law of Thermodynamics, 3. Specific Heats3. Specific Heats 4. The 2nd Law of Thermodynamics

2

1 Kinetic Theory of Gases1. Kinetic Theory of Gases

• AssumesAssumes• Gas is a collection of molecules in random motion• Molecules bounce of each other like hard spheres

• Gives two sets of relations…

Equation of State

Neglects forces between molecules

Equation of State

Gas constant R = 287 J/kg/K for air

• Gases that obey these relations are called “Thermally 

3Perfect”

Thermally PerfectThermally Perfect

• We will always assume gases are thermally perfect in thisWe will always assume gases are thermally perfect in this class, but this isn’t always true…

Ceases to be valid at high densitiesat high densities. Why?

4

2 1st Law of Thermodynamics2. 1 Law of Thermodynamics“Energy is Conserved”

W

E QChange in 

Amount ofAmount of work d b

WQE

Collection of identified gas molecules (System)

energy Amount of heat added

done by pressure and viscous forces

Work =

p V (<0)

Surface area S

5

In terms of enthalpyIn terms of enthalpy…pdvdedq

6

3 Specific Heats3. Specific Heats• Specific heat – amount of heat needed to raise the 

temperature of 1kg by 1 Kelvin• Two kinds: 

f h l

Tq

TqC

TC )/( • Specific heat at constant volume • Specific heat at constant pressure 

vv TqC )/(

pp TqC )/(

7Pratt and Whitney JT58 Test (Powered SR71)

With the energy equationvdpdhdq pdvdedq

With the energy equation…

8

Calorically Perfect Gas dTdhCp dT

deCv

Calorically Perfect Gas

• specific heats are constant with temperature so

A gas that is thermally perfect and for which the…

…specific heats are constant with temperature, so…

• We will assume calorically perfect gases in this class, but…

9

4 2nd Law of Thermodynamics4. 2 Law of ThermodynamicsShock in a Converging Diverging Nozzle

10Bourgoing & Benay (2005), ONERA, France Schlieren visualizationSensitive to in‐plane index of ref. gradient

Processes and EntropyProcesses and Entropy

• Adiabatic Process ‐ no heat added or removedAdiabatic Process ‐ no heat added or removed.• Reversible Process – no dissipative phenomena occur.• Isentropic Process – adiabatic and reversibleIsentropic Process  adiabatic and reversible

• 2nd Law of Thermo – dissipative processes may only

Constant entropy. Entropy – measure of disorder in the gas molecules.

2 Law of Thermo  dissipative processes may only increase entropy, but adding or removing heat may increase or decrease entropy by an amount Q/T

• So…

11

The Entropy EquationThe Entropy Equation…dqTds For a reversible process…

pdvdedq

dTCd

1st Law

Th ll dTCde v

vRTp /

Thermally perfect gas

12

The Entropy Equationvp CCR vp CC /

…The Entropy Equation

13

Isentropic RelationsIsentropic Relations

1

1212 log

Tss

• If s s = 0 then

21

logTC e

v

• If s2 – s1 =  0 then…

14

SummaryCalorically

• KTG

All gases Thermally perfect Calorically perfect

RTp )(Tee • KTG

• 1st Law

RTp

pdvdedq vdpdhdq

)(Thh

• Specific heats

vdpdhdq

dTdhCdTdeCv

//

vv

ThCTeC

)/()/(

TChTCe v

heats

• 2nd Law

dTdhCp /pp ThC )/( TCh p

1

1212 log

TT

Css

e

• 2nd Law Isentropic

21 TCv

11

22

T 1

22

TpIsentropic 

1511

T 11

Tp