thermodynamics chapter 2

Thermodynamics

Chapter 2

Diploma in EngineeringMechanical Engineering Division, Ngee Ann Polytechnic

Properties of Liquids & Vapours

o Constant pressure process for a liquid and its vapour

o Saturation temperatureo Quality or dryness fraction of wet steamo Tabulated propertieso Properties of saturated water and steamo Properties of compressed watero Properties of wet steamo Properties of superheated steamo Interpolation of values in steam tables

Introduction• 3 existing forms of matters:

• Thermodynamics only concerns liquid and gaseous phases.

solid liquid gas

thermodynamics

Constant Pressure Process

• Boiling: a liquid heated at any one constant pressure, there is one fixed temperature at which bubbles of vapour form in the liquid.

• The higher the pressure of the liquid, the higher will be the temperature at which boiling occurs.

• The volume occupied by 1kg of a boiling liquid at a higher pressure is slightly larger than the volume occupied by 1kg of the same liquid when its boiling at a lower pressure.


• p-v diagram of a series of boiling points

PR

PQ

PPSa

tura

ted

liqui

d lin

ePressure, p

Specific volume, v

R

Q

P


• When a liquid at boiling point is heated further at constant pressure. The additional heat supplied changes the phase of the substance from liquid to vapour.

• During the change of phase, the pressure and temperatures remain constant

PR

PQ

PP

Pressure, p

Specific volume, v

R

Q

P

R’

Q’

P’


• The heat required converting 1kg of liquid at boiling point to 1kg of saturated vapour at the same temperature and pressuer is called the latent heat of vapourisation.

PR

PQ

PP

Pressure, p

Specific volume, v

R

Q

P

R’

Q’

P’

Satu

rate

d va

pour

line

The higher the pressure the smaller will be the latent heat of vapourisation

The specific volume of a saturated vapour decreases as the pressure increases


• The point where the saturated liquid line and the saturated vapour line meet is called the critical point PR

PQ

PP

Pressure, p

Specific volume, v

R

Q

P

R’

Q’

P’

Satu

rate

d va

pour

line

Satu

rate

d liq

uid

line

Critical ptCritical pressure,

At the critical pressure, the latent heat of vapourisation is zero

The substance exisiting at any point inside the loop consists of a mixture of liquid and dry vapour and is known as wet vapour

A saturated vapour is usually called dry saturated vapour.


• When a dry saturated vapour is heated at constant pressure, its temperature rises above saturation temperature and it becomes superheated vapour

• The difference between the actual temperature of the superheated vapour and the saturation temperature is called the degree of superheat

Isothermals on the Pressure-volume

Diagram

PR

PQ

PP

Pressure, p

Specific volume, v

R

Q

P

R’

Q’

P’

Critical pressure

Tc

Critical point • Isothermal: constant temp.

• The temperature lines become horizontal between the saturated liquid line and the saturated vapour line. Thus there is a corresponding saturation temperature for each pressure

• The critical temp. line just touches the top of the loop at the critical point

T3

T2

T1

Quality or Dryness Fraction of Wet

Vapour• The saturated liquid exists in equilibrium with dry

vapour• The pressure and temperature alone are not

sufficient to define completely the state of mixture

• At pressure PQ and temperature T2, could be a saturated liquid, a wet vapour or a dry saturated vapour.

• The state can’t be defined until one other property, e.g. specific volume is known.

Quality or Dryness Fraction of Wet

Vapour• The condition or quality of a wet vapour is most

frequently defined by its dryness fraction

,mass of vapourdryness fraction x

mass of vapour mass of liquid

Tabulated Properties• The properties are tabulated as functions of

pressure and temperature for a wide variety of substances to facilitate computation.

• The water and steam properties are tabulated in the steam tables.

Properties of Saturated Water and

Steam• The specific properties of saturated water is

tabulated in Saturated Water and Steam Table from page 8-3 to 8-6.

• The suffix “f” is used to denote the saturated water state.

Example• Determine the temperature, specific enthalpy,

specific internal energy and specific volume of saturated water at p=0.34 bar.


Steam

1 • For most practical purpose, the value of vf is generally taken as 0.001 m3/kg.

Subscript “f” for saturated water Subscript “s” for saturated water and

steam

Page 8-4


Steam• The suffix “g” is used to denote the dry

saturated state (i.e. saturated vapour).

Example• Determine the temperature, specific enthalpy,

specific internal energy and specific volume of saturated steam at p=0.34 bar.


Steam

1

Subscript “g” for saturated steam

Page 8-4


SteamProperties of compressed water

• Water at a temperature below the boiling point (ts)corresponding to the pressure.

• Assumption: For normal pressure, it is sufficiently accurate to take the properties of compressed water, as being equal to those of saturated water at the same temperature.


SteamExample• Determine the specific internal energy, specific

enthalpy and specific volume of compressed water at p=20 bar and t=179.9ºC Page 8-5

1

Same as “f” saturated water at the same temp.


SteamProperties of wet steam• The specific properties of wet steam can be

obtained using the dryness fraction, x:

The dryness fraction of the wet steam is x

Then 1 kg of wet steam contains:X kg of dry saturated steam, and

(1-x) kg of saturated water


SteamTherefore, the specific enthalpy of the wet stream

hx=xhg+(1-x)hf

Similarly, the specific internal energy of the wet steam

ux=xug+(1-x)uf

And the specific volume of the wet steamvx=xvg+(1-x)vf

Since (1-x)vf≈0Hence vx=xvg

5


SteamExample• Determine the temperature, specific enthalpy,

specific internal energy and specific volume of wet steam at p=10 bar with a dryness fraction of 0.5.


SteamApplying hx=xhg+(1-x)hf

Hence hx=0.5=0.5×2778+(1-0.5)×763 =1770.5 kJ/kg

Applying ux=xug+(1-x)uf

Hence u0.5=0.5×2584+(1-0.5)×762 =1673 kJ/kg

Applying vx=xvg

Hence v0.5=0.5×0.1944 =0.0972 m3/kg


SteamExample• A cylinder contains 0.113 m3 of wet steam at 11

bar. If the mass of the steam is 0.9 kg, determine its dryness fraction, enthalpy and internal energy.

Solution:

30.113 0.1256 /0.9x

Vvm

v m kg


SteamPage 8-5


SteamApplying

Applying hx=xhg+(1-x)hf

Hence hx=0.708=0.708×2781+(1-0.708)×781 =2197 kJ/kgHence, the enthalpy of the wet steam

0.1256 0.7080.1774

x g

x

g

v xv

vxv

0.708 0.708 0.9 2197 1977.3xH m h kJ


SteamApplying ux=xug+(1-x)uf

Hence u0.708=0.708×2586+(1-0.708)×781 =2058.94 kJ/kg

Hence, the internal energy of the wet steam0.708 0.708 0.9 2058.94 1853.05xU m u kJ

Properties of Superheated Steam

• Tabulated separately

• Temperature and pressure are independent properties. i.e. only when the temperature and pressure are given for superheated steam, then the state is defined and all the other properties can be found.

Properties of Superheated Steam

Example• Determine the specific enthalpy, specific internal

energy and specific volume of superheated steam at p=2 bar and t=200°C. Page 8-7

Interpolation of Values in Steam Tables

• Steam tables for the superheated steam as well as the saturated water & steam give the values of properties at selected pressures and temperatures.

• To obtain values of properties at conditions other than those tabulated, interpolation is used.


Example• Determine the saturation temperature of the

steam at pressure p=9.6 bar

Solution: refer to saturated water and steam table

Pressure, P Saturation temperature, ts

9 bar 175.4 °C10 bar 179.9 °C


Using straight line interpolationThe slope of AC=the slope of AB

A

B

C

( 175.4)t

(179.9 175.4)

(9.6

9)

(10

9)

10

9.6

9

175.4 t 179.9 ts(°C)

p (bar)

1 0.64.5 175.4175.4 0.6 4.52.7 175.4

178.1

ttt

C


Example• Superheated steam at p=15 bar and t=380ºC.

Determine its specific volume, enthalpy and internal energy.

Solution: refer to superheated steam tablePressure, p Temperature, t

350°C 400°C

15 bar

v=0.1865 m3/kg

v=0.2029 m3/kg

u=2868 kJ/kg u=2952 kJ/kgh=3148 kJ/kg h=3256 kJ/kg


400

380

350

t (°C)

0.1865 v 0.2029 v(m3/kg)

Using straight line interpolationThe slope of AB=the slope of AC

(400

350)

(380

350)

( 0.1865)v

(0.2029 0.1865)

3

400 350 380 3500.2029 0.1865 0.186550 30

0.0164 0.186530 0.1640.1865

500.1963 /

v

v

v

m kg


400

380

350

t (°C)

2868 u 2952 u(kJ/kg)


(400

350)

(380

350)

( 2868)u

(2952 2868)

400 350 380 3502952 2868 286850 3084 2868

30 84286850

2918.4 /

u

u

u

kJ kg


400

380

350

t (°C)

3148 h 3256 h(kJ/kg)


(400

350)

(380

350)

( 3148)h

(3256 3148)

400 350 380 3503256 3148 314850 30108 3148

30 108314850

3212.8 /

h

h

h

kJ kg

Non-flow Processes with Steam

o Introduction

o Constant volume process

o Constant pressure process

o Isothermal process

o Adiabatic non-flow process

o Polytropic process

Introduction• What steam can do?

expand

compress

receive heat

reject heat

do external work

have work done

Introduction

p1 T1

v1 h1

u1

p2 T2

v2 h2

u2

m kg of steam

Constant Volume Process

• The system is contained in a rigid vessel. Hence the boundary of the system is immoveable and no work can be done on or by the system.

• v1=v2

• W12=0

Specific volume, v

Pressure, p

system

boundary


• Applying the non-flow energy equation

• The net heat supplied in a constant volume process goes to increasing the internal energy of the steam.

2 1 12 12

12

12 2 1

12 2 1

U -U =Q -WSince W =0

Q =U -Uor Q =m(u -u )


Example• During a constant volume process, the specific

internal energy of the steam increased from 120 kJ/kg to 180 kJ/kg. Determine the amount of heat energy required to bring about this increase for 2 kg of steam.

Solution:2 1 12 12

12

12 2 1

12 2 1


Q =U -Uor Q =m(u -u )=2 (180-120)=120kJ


Example• Dry saturated steam at pressure 3 bar is

contained in a rigid vessel of volume 0.9 m3. if the vessel is cooled until the pressure is reduced to 1.2 bar. Determine:

A) the mass of steam in the vessel B) the final dryness fraction of the steam C) the amount of heat transferred during the cooling process


Solution:• A)

31 g at p=3bar

1

1

v =v =0.6057m /kg

0.9 1.4860.6057

VvmVm kgv


B) for steam undergoing a constant volume process

32 1

32 1.2

22

2

0.6057 /

1.428 /

0.6057 0.42421.428

g g at p bar

x g

g

v v m kg

v v m kg

v xv

vxv


• C)1 g at p=3bar

2 2 2

2 2 2 2 2

2 1 12 12

12

12 2 1

12 2 1

=u =2544kJ/kg

at p 1.2 439 / 2512 /

(1 )

(1 ) 0.4242 251 (1 0.4242) 439 1318.37 /


Q =U -Uor Q =m(u -u )=1.486 (1318.37-2544

f g

x g f

g f

u

bar u kJ kg u kJ kg

u xu x u

u x u x u kJ kg

)=-1821.29kJ


• The boundary of the system expands as heat is added to the system or contracts as heat is removed from the system . Hence work is done either by the system on its surrounding or by the surrounding on the system.

• p1=p2=p Specific volume, v

Pressure, p

v1 v2

1 2p1= p2


• The shaded area on the p-v diagram represents the work done by the steam.

Hence

Applying the non-flow energy equation

12 2 1

12 2 1

( )( )

W p V VorW mp v v

2 1 12 12

12 2 1 2 1 2 2 1 1

12 2 1

12 2 1

U -U =Q -W( ) ( ) ( )

since

( )

Q U U p V V U pV U pVH U pV

Q H HorQ m h h

The net amount of heat energy supplied to or taken from the steam during a constant pressure non-flow process is equal to the change of enthalpy of the steam during the process


Example• 2.5 kg of wet steam at pressure 15 bar and

dryness fraction 0.8 receives heat at constant pressure until its temperature become 250ºC. Determine the heat received by the steam

Solution:• At p1=15 bar

1 15

1 15

1 15

198.3

845 /

2792 /

s at p bar

f f at p bar

g g at p bar

t t C

h h kJ kg

h h kJ kg


Since t2=250ºC and t2 > ts2 (ts2=ts at p=15bar=198.3°C)

Therefore, the steam at state 2 is in the superheated region

h2=hat p=15bar, t=250ºC=2925 kJ/kg

1 1 1 1 1

(1 )

(1 ) 0.8 2792 (1 0.8) 845 2402.6 /x g f

g f

h xh x h

h x h x h kJ kg


• For steam undergoing constant pressure process

Heat received

12 2 1

2 1( )2.5 (2925 2402.6) 1307.5

Q H Hm h h

kJ

Isothermal Process• Temperature

remains constant

• An isothermal process for steam is shown on the right. The initial and final sate has been chosen in the wet region and superheated region respectively. Specific volume,

v

Pressure, p

v1 v2

1 Ap1

p2

2

Constant temp. line

Isothermal Process• From state 1 to state

A, the pressure remains at p1, since the wet region the pressure and temperature are the corresponding saturation values. It can be seen therefore that an isothermal process for wet steam is also a constant pressure process.

Specific volume, v

Pressure, p

v1 v2

1 Ap1

p2

2

Constant temp. line

Isothermal Process• Hence Q1A=HA-H1

Or Q1A=m(hA-h1)

• From state A to state 2, the steam is in the superheated region and the pressure fall from p1 to p2.

Specific volume, v

Pressure, p

v1 v2

1 Ap1

p2

2

Constant temp. line

Isothermal Process• When state 1 and

state 2 are fixed, then the specific internal energy u1 and u2 may be obtained from steam tables. The work done is represented by the shaded area.

Specific volume, v

Pressure, p

v1 v2

1 Ap1

p2

2

Constant temp. line

Isothermal ProcessExample• In a closed system, 3 kg of wet steam at 10 bar

has an initial dryness fraction 0.9 and is expanded isothermally to a pressure of 2 bar. Determine:

A) the change of internal energy and the change of

enthalpy of the steamB) if the heat supplied during the process is found

to be 1000 kJ, determine the work done by the steam.

Isothermal Process

Page 8-5

Solution:

Isothermal Process

1 1 1 1 1

1 1 1 1 1

(1 )

(1 ) 0.9 2778 (1 0.9) 763 2576.5 /

(1 )

(1 ) 0.9 2584 (1 0.9) 762 2401.8 /

x g f

g f

x g f

g f

h xh x h

h x h x h kJ kg

alsou xu x u

u x u x u kJ kg

Isothermal ProcessAt p2=2 bart2=t1=179.9°CFrom superheated steam table

Using straight line interpolation

T [°C] u[kJ/kg] h [kJ/kg]150 2578 2770200 2655 2871

2

2

2655 25782578 (179.9 150) 2624 /200 150

2871 27702770 (179.9 150) 2830 /200 150

u kJ kg

h kJ kg

Isothermal ProcessHence

And

B) Applying the non-flow energy equation

2 1 2 1

2 1 2 1

( ) 3 (2624 2401.8) 666.6( ) 3 (2830 2576.5) 760.5

U U m u u kJH H m h h kJ

2 1 12 12

12 12 2 1

U -U =Q -W( ) 1000 760.5 239.5W Q U U kJ

Adiabatic Non-flow Process

• An adiabatic process is one in which no heat is transferred to or from the system during the process.

Hence Q12=0

Applying the non-flow energy equationU2-U1=Q12-W12

Since Q12=0W12= -(U2-U1)

• In an adiabatic expansion process: the work done by the system is at the expense of a reduction in the internal energy of the system

• In an adiabatic compression process: the work done on the system goes into increasing the internal energy of the system.


Example• In a closed system, 2 kg of steam at initial

pressure 50 bar and temperature 400°C undergoes adiabatic process until the pressure is 15 bar and the steam is then dry saturated. Determine the work done by the steam .

Solution:u1=uat p=50bar, t=400˚C=2907 kJ/kg

u2=uat p=15bar=2595 kJ/kg


For steam undergoing an adiabatic processQ12=0

Applying the non-flow energy equationU2-U1=Q12-W12

Hence W12=-(U2-U1) =-m(u2-u1) =-2×(2595-2907) =624 kJ

Polytropic Process• The most general type of process, in which both

heat energy and work energy cross the boundary of the system. It is expressed by means of the equation

pVn=constantOr pvn=constant Where n known as the index of expansion or compression and is constant.

Polytropic Processhence

p1V1n=p2V2

n

Or p1v1n=p2v2

n

And the work energy crossing the boundary (or work transferred) is given by

Or

1 1 2 212

p V -p VW =

n-1

1 1 2 212

m(p v -p v )W =

n-1

Polytropic Process• Applying the non-flow energy equation

Or

2 1 12 12

1 1 2 22 1 12

p V -p Vn-1

U U Q W

U U Q

1 1 2 22 1 12

m(p v -p v )n-1

U U Q

Polytropic ProcessExample• 2.5 kg of wet steam has an initial pressure 6 bar

and a dryness fraction of 0.85 and is expanded according to the law pV1.2=constant to a final pressure of 1.2 bar. Determine:

A) the work done by the steam. B) the heat energy transferred between the steam and surroundings stating the direction of transfer.

Polytropic ProcessSolution:At p1=6 bar:

vg1=vg at p=6bar=0.3156 m3/kguf1=uf at p=6bar=699 kJ/kg

ug1=ug at p=6bar=2568 kJ/kg


Hence u1=x1ug1+(1-x)uf1

=0.85×2568+(1-0.85)×669 =2182.8 kJ/kg

Applying vx=xvg

Hence v1=x1vg1=0.85×0.3156=0.2683 m3/kg

Polytropic ProcessFor steam undergoing a polytropic process

1.2 1.21 1 2 2

11.2

12 1

2

11.2 3

p v =p v

pv = ×v

p

6= ×0.2683=1.0259m /kg1.2

Polytropic ProcessAt p2=1.2 bar:

vg2=vg at p=1.2bar=1.428 m3/kguf2=uf at p=1.2bar=439 kJ/kg

ug2=ug at p=1.2bar=2512 kJ/kg

Applying vx=xvg

Hence


Hence u2=x2ug2+(1-x)uf2

=0.72×2512+(1-0.72)×439 =931.56 kJ/kg

Polytropic ProcessFor steam undergoing a polytropic process

Applying the non-flow energy equation U2-U1=Q12-W12

Hence Q12=(U2-U1)+W12

=m(u2-u1)+W12

=2.5×(1931.56-2182.8)+473.4 =-154.7 kJOr Q12=154.7 kJ (heat lost)

1 1 2 212

5 5

m(p v -p v )W =

n-12.5 (6 10 0.2683 1.2 10 1.0259)

1.2 1473400 437.4 ( )J kJ work output

Properties of Perfect Gas

• The characteristic gas equation• Specific heats• Joule’s law• Relationships between the specific heats• Enthalpy of a perfect gas• Ratio of specific heats

The Characteristic Gas Equation

• One important type of fluid which has many applications in thermodynamics is the type in which the working temperature of the fluid remains well above the critical temperature of the fluid.

• The vapor of the fluid tends to obey the equation:

p: pressure [N/m2] v: specific volume [m3/kg] T: temperature [K]


• In practice, no gasses obey this law rigidly, but many tend towards it. An imaginary ideal gas, which obeys the law is called a perfect gas.

• R is the characteristic gas constant in J/kg∙K, each perfect gas has a different value of gas constant.

• Or pv RT


• For m kg of the perfect gas, occupying V m3

• Hence

R: the characteristic gas constant [J/kg∙K] p: pressure [N/m2] m: mass [kg] v: specific volume [m3/kg] V: volume [m3] T: temperature [K]

Vvm

pV mRT Characteristic gas equation


• Anther form of the characteristic gas equation can be derived using the kilogram-mole.

• The Kilogram-mole is defined as a quantity of a gas equal to M kg of the gas where M is the molecular weight of the gas.


E.g. 1 kilogram-mole of oxygen is equal to 32 kg of oxygen. Since the molecular weight of oxygen is 32.

For m kg of a gasWhere: n is the number of moles

Hence,

m nM

pV mRTpV nMRT

pV MRnT


• Avogadro’s Hypothesis states that the volume of 1 mole (a simplified term for kg-mole) of any gas is the same as the volume of 1 mole of any other gas. When the gases have at the same temperature and pressure.

• i.e.

• The constant is called the universal gas constant, and is given the symbol, R0

constant (for all gases)pVnT

0

0

0

pV MR RnTMR R

RRM


• Experiment has shown that the volume of 1 mole of any perfect gas at 1 bar and 0°C is approximately 22.71 m3

• Therefore0

51 10 22.711 273.15

8314.3 /

pVRnT

J kg mole K


• Hence, the gas constant, R for any gas can be found when the molecular weight of the gas is known.

• Since the molecular weight of oxygen is 32,• Hence, the gas constant for oxygen

0

831432

259.8 /

RRM

J kg K

Specific Heats• The specific heat of a solid or liquid is usually

defined as the amount of heat energy required to raise the temperature of unit mass of the substance through one degree

• c: the specific heat [J/kg∙K]• m: the mass [kg]• (T2-T1): the increase in temperature [K]• Q: the heat required [J]

2 1( )Q mc T T

Specific HeatsIn case of gas, there are infinite numbers of ways in which heat may be added between any two temperatures, and hence a gas could have infinite number of specific heats. However, only two specific heats for gasses are defined.

The specific heat at constant volume, cv. It is an amount of heat energy required to raise the temperature of one kg mass of the gas by one degree whilst the volume of the gas remain constant.

The specific heat at constant pressure, cp. It is an amount of heat energy sufficient to raise the temperature of one kg mass of the gas by one degree whilst the pressure of the gas remains constant.

Specific Heats• For a perfect gas, the values of cp and cv are

constant for any given gas at all pressure and temperature.

Noted: cp air=1.005 kJ/kg∙K cv air=0.718 kJ/kg∙K

Hence Q12=m∙cv(T2-T1) for constant volume non-flow process Q12=m∙cp(T2-T1) for constant pressure non-flow process

Joule’s Law• Joule’s law states that the internal energy of a

perfect gas is a function of the temperature only. i.e. u=f(T)

• To evaluate the function, u=f(T), consider m kg of a perfect gas be heated at constant volume from state 1 to state 2

U2-U1=Q12-W12

Since W12=0 Hence Q12=U2-U1

Joule’s LawFor a perfect gas undergoing constant volume non-flow process Q12=mcv(T2-T1) Therefore, U2-U1=mcv(T2-T1)

Hence, it is true to say that for a perfect gas undergoing any non-flow process between state 1 and state 2 U2-U1=mcv(T2-T1)

Relationships between the Specific Heats

• Consider m kg of a perfect gas at pressure, p1 and temperature T1, volume V1 undergoes a constant pressure process to a final pressure p2, temperature T2 and volume V2

Applying W12=p(V2-V1)

And p1V1=mRT1

p2V2=mRT2

Hence W12=mR(T2-T1)

Relationships between the Specific Heats

For a perfect gas undergoing constant pressure non-flow process Q12=mcp(T2-T1)And U2-U1=mcv(T2-T1)

Applying the non-flow energy equation U2-U1=Q12-W12

Hence, mcv(T2-T1)=mcp(T2-T1)+mR(T2-T1) cv=cp-ROr cp-cv=R

Enthalpy of a Perfect Gas

• The enthalpy of a perfect gas is given by h=u+pv Since u=cvT And pv=RTHence, h=cvT+RT =(cv+R)T =cpT

For m kg of perfect gas H=mcpTHence, H2-H1=mcp(T2-T1)

Ratio of Specific Heats• The ratio of the specific heat at constant

pressure, cp to the specific heat at constant volume cv is given the symbol

p

v

cc

Ratio of Specific Heats

Ratio of Specific HeatsExample

• A quantity of certain perfect gas has an initial pressure of 150 kN/m2, volume 0.2 m3 and temperature 30°C. It is then compressed to a pressure of 600 kN/m2 and temperature 65°C. Determine the final volume of the gas.

Ratio of Specific HeatsSolution:Given: p1=150 kN/m2, V1=0.2 m3, t1=30°C, p2=600 kN/m2, t2=65°C

Applying

1 1 2 2

1 2

pV p VT T

1 2 12

2 1

33

3

150 10 (65 273) 0.2 0.0552600 10 (30 273)

p T VVp T

m

Ratio of Specific HeatsExample

• A certain perfect gas has an initial pressure of 360 kN/m2, volume of 0.004 m3, and temperature of 32°C. if the value of R=0.29 kJ/kg∙K. Determine:

(a) the mass of the gas (b) if the pressure of the gas is now increased to 10 bar while the volume remains constant. Calculate the final temperature of the gas.

Ratio of Specific HeatsSolution:(a)Applying the characteristic gas equation

pV=mRT

(b)Applying

And V1=V2

Hence

3

3

360 10 0.04 0.16280.29 10 (32 273)

pVmRT

kg

1 1 2 2

1 2

pV p VT T

1 2

1 2

52 1

2 31

10 10 (32 273) 847.2360 10

p pT T

p TT Kp

Non-flow Processes with Perfect Gas

• Constant volume process• Constant Pressure process• Isothermal or constant temperature

process• Polytropic process• Adiabatic process


• Consider m kg of a perfect gas contained in a rigid vessel at pressure p1, temperature T1 and volume V1, undergoes a constant volume process to a final pressure p2, temperature T2, and volume V2.

1

2P

P2

P1

V1=V2 V


• For a perfect gas undergoing constant volume process

• Applying 1 1 2 2

1 2

1 2

1 2

pV p VT Tp pT T


• For a perfect gas undergoing constant volume process

• For a perfect gas undergoing any process


12 0W

2 1 2 1( )vU U mc T T

2 1 12 12

12 2 1

12 2 1( )v

U U Q WQ U U

or Q mc T T

The net amount of heat energy supplied to or taken from a perfect gas during a constant volume process is equal to the change in internal energy of the perfect gas during that process


Example• 0.5 kg of air with an initial temperature of 29°C

and a pressure of 1.5 bar is heated to a final temperature of 350°C at constant volume. Determine the heat energy supplied to the air and its final pressure. Take cp=0.287 kJ/kg∙K, cv=0.718 kJ/kg∙K.

Solution• For a perfect gas undergoing constant volume

process12 0W


• Applying

• Since


• Consider m kg of a perfect gas at pressure p1, temperature T1 and volume V1, undergoes a constant pressure process to a final pressure p2, temperature T2, and volume V2.

1 2P

V2V1

P1=P2

V


• For a perfect gas undergoing constant pressure process

• Applying

1 1p p p

1 1 2 2

1 2

1 2

1 2

pV p VT TV VT T




• Applying the non-flow energy equation2 1 2 1( )vU U mc T T

12 2 1( )W p V V


• Since

• Hence

• Or 12 2 1( )pQ mc T T

12 2 1Q H H

The net amount of heat energy supplied to or taken from a perfect gas during a constant pressure process is equal to the change of enthalpy of the perfect gas during that process


Example• A mass of air has an initial pressure, 2.5 bar,

volume 0.08 m3 and temperature 175°C. It is then undergoes a constant pressure process until its final temperature become 30°C. Take R=0.287 kJ/kg∙K, cv=0.718 kJ/kg∙K as the mass of the air. Determine:

• A) the work done on the air, and • B) the heat transferred from the air during the

process.


Solution• Applying

• Applying





2 1 2 1

32 1

( )

0.1556 0.718 10 (30 175) 16199.5vU U mc T T

U U J

2 1 12 12

12 2 1 12( )16199.5 ( 6475) 22674.5

U U Q WQ U U W

J


• Alternatively,

• And 2 1 2 1

12 2 1

3 3

( )

( ) ( )

0.1556 (0.718 10 0.287 10 ) (30 175)22674.5

p

p v

v

H H mc T T

c c R

Q m c R T T

J

Isothermal or Constant Temp.

Process• An isothermal expansion or compression

process is defined as a process carried out while the temperature remains constant.

• Consider m kg of a perfect gas at pressure p1, temperature T1

and volume V1 undergoes a constant temperature process to a final pressure p2, temperature T2 and volume V2.

1

2

P

p2

p1

V1 VV2

pV=c


Process• For a perfect gas undergoing isothermal process

• Applying

• Hence constantpV


Process• For a perfect gas undergoing isothermal process

• and

• Hence


Process• For a perfect gas undergoing any process

• since


• Hence

• or

2 1 2 1( )vU U mc T T

2 1T T

2 1 12 12

12 12

U U Q WQ W

212

1

lnVQ pVV

During an isothermal process expansion all the heat energy received is converted into external work. Conversely, during an isothermal compression, all the work done on the gas is rejected by the gas as heat energy.


ProcessExample0.2 kg of air has an initial pressure of 30 bar and a volume of 0.015 m3. it is then expanded to a pressure of 4 bar while the temperature remains constant. Determine:• A) the temperature of the air• B) the final volume of the air• C)the work done by the air, and• D) the heat transferred to the air during the process

Take R=0.287 kJ/kg∙K and cv=0.718 kJ/kg∙K


ProcessSolution:Applying

For a perfect gas undergoing isothermal processpV=constant

Hence

51 1

1 3

30 10 0.015 783.970.2 0.287 10

pV mRTpVT KmR

1 1 2 2

531

2 1 52

30 10 0.015 0.11254 10

pV p V

pV V mp


ProcessFor a perfect gas undergoing isothermal process

For a perfect gas undergoing any process

Applying the non-flow energy equation

212

1

512

ln

0.112530 10 0.015 ln 906710.015

VW pVV

W J

2 1 2 1

2 1

( )0vU U mc T T

U U

2 1 12 12

12 12 90671U U Q WQ W J

Polytropic Process• In polytropic process, the gas is compressed or expanded

according to the law pVn=constant

• Where n is the index of expansion or compression and both work and heat energy may be transferred across the boundary.



1

2

P

p2

p1

V1 VV2

pVn=c

Polytropic Process• Since

• Applying

• Hence

1 1 2 2

1 2

2 1 2

1 2 1

pV p VT Tp V Tp V T

1 1 2

2 2 1

1

2 1

1 2

n

n

V V TV V T

T VT V

Polytropic Process• Substitute into

• For a perfect gas undergoing polytropic process

• And pV=mRT• Hence

1

2 2

1 1

nnT p

T p

1 1 2 212 1

pV p VWn

1 212

( )1

mR T TWn

Polytropic Process• For a perfect gas undergoing any process


2 1 2 1( )vU U mc T T

Polytropic ProcessExample0.008 kg of air initially at a pressure of 1.2 bar and a volume of 0.005 m3 is compressed to a volume of 0.001 m3 according to the law pV1.25=constant. Determine:• A) the final pressure of the air• B) the initial and final temperature of the air• C)the work transferred to the air, and• D)the heat transferred from the air during the

processTake R=0.287 kJ/kg∙K and cv=0.718 kJ/kg∙K

Polytropic ProcessSolution:• Applying

• Applying 5

1 11 3

1.2 10 0.005 261.30.008 0.287 10

pV mRTpVT KmR

Polytropic Process• Applying



1 1 2 2

1 2

52 2

2 1 51 1

8.912 10 0.001261.3 390.71.2 10 0.005

pV p VT T

p VT T Kp V

2 1 2 1

32 1

( )

0.008 0.718 10 (390.7 261.3) 743vU U mc T T

U U J

Polytropic Process• Applying the non-flow energy equation

2 1 12 12

12 2 1 12 743 ( 1189) 446U U Q WQ U U W J

Adiabatic Process• In an adiabatic process, the gas is compressed or

expanded according to the law pVγ=constant

• Where



1

2

P

p2

p1

V1 VV2

pVγ=c

Q12=0

Adiabatic Process• Since

• Applying

• Hence

1 1 2 2

1 2

2 1 2

1 2 1

pV p VT Tp V Tp V T

Adiabatic Process• Substitute into


• And pV=mRT• Hence

1 212

( )1

mR T TW

1 1 2 212 1

pV p VW

Adiabatic Process• For a perfect gas undergoing any process


2 1 2 1( )vU U mc T T

Q12=0

Adiabatic ProcessExampleIn a thermally insulated cylinder 0.01 m3 of air has an initial pressure of 1.0 bar and temperature of 35°C. It is compressed adiabatically to a final volume of 0.002 m3. Determine:• A) the mass of the air• B) the final pressure of the air• C) the work done to the air during the process

Take R=0.287 kJ/kg∙K, cv=0.718 kJ/kg∙K, γ=1.4 and cp=1.005 kJ/kg∙K,

Adiabatic Process• Solution:• Applying pV=mRT

• Applying

Adiabatic Process• For a perfect gas undergoing adiabatic process

Thank youQ & A

thermodynamics chapter 2

Education