Thermodynamics (2013 new edition) copy

THERMODYNAMICS Part 1By. Engr. Yuri G. Melliza

Terms & Definition

Properties of Fluids

Forms of Energy

Law of Conservation of Mass

Law of Conservation of Energy

(First Law of Thermodynamics)

Ideal Gas

Pure Substance

Processes of Fluids

Zeroth Law of Thermodynamics

Thermodynamics is a science that deals with energy transformation or conversion of one form of energy to another form

Therme – “Heat”

Dynamis – “Strength”

System: A portion in the universe, an Atom, a Galaxy, a certain quantity of matter, or a certain volume in space in which one wishes to study. It is a region enclosed by a specified boundary

that may Imaginary, Fixed or Moving.

System

Surrounding or Environment

Open System: A system open to matter flow.

Example: Internal Combustion Engine (ICE)

Closed System: A system close to matter flow.

Example: Piston - in - cylinder

Working Substance (Working Fluid): A fluid (Liquid or Gas) responsible for the transformation of energy.

Example: air in an air compressor Air and fuel mixture in an internal combustion

engine

Pure Substance: A substance that is homogeneous in nature and is homogeneous.

Example : Water

Phases of a Substance A phase refers to a quantity of matter that is homogeneous throughout in both chemical composition and physical structure. Solid Liquid Gas or Vapor

Specific Terms To Characterized Phase Transition

SOLIDIFYING OR FREEZING - Liquid to Solid

MELTING - Solid to Liquid VAPORIZATION - Liquid to Vapor CONDENSATION - Vapor to Liquid SUBLIMATION - a change from solid

directly to vapor phase without passing the liquid phase.

Mass : It is the absolute quantity of matter in it.

m - mass in kg

Acceleration : it is the rate of change of velocity with respect to time t.

a = dv/dt m/sec2

Velocity: It is the distance per unit time.

v = d/t m/sec

Force - it is the mass multiplied by the acceleration.F = ma/1000 KN

1 kg-m/sec2 = Newton (N) 1000 N = 1 Kilo Newton (KN)

Newton - is the force required to accelerate 1 kg mass at the rate of 1 m/sec per second.

1 N = 1 kg-m/sec2

From Newton`s Law Of Gravitation: The force of attraction between two masses m1 and m2 is given by the equation:

Fg = Gm1m2/r2 NewtonWhere: m1 and m2 - masses in kg

r - distance apart in meters G - Gravitational constant in N-m2/kg2

G = 6.670 x 10 -11 N-m2/kg2

WEIGHT - is the force due to gravity.W = mg/1000 KN

Where: g - gravitational acceleration at sea level, m/sec2 g = 9.81 m/sec2

3mkg

Vm ρ

PROPERTIES OF FLUIDS

Where: - density in kg/m3

m - mass in kg V – volume in m3

Specific Volume () - it is the volume per unit mass or the reciprocal of its density.

kgm

mV

3

υ

kgm

3

ρ1 υ

Density () - it is the mass per unit volume.

Specific Weight () - it is the weight per unit volume.

3

3

mKN

mKN

1000V

mgγ

V

Wγ

Where: - specific weight in KN/m3

m – mass in kg V – volume in m3

g – gravitational

At standard condition:g = 9.81 m/sec2

Specific Gravity Or Relative Density (S):FOR LIQUIDS: Its specific gravity or relative density is equal tothe ratio of its density to that of water at standard temperature and pressure.

w

L

w

L

L γ

γ

ρ

ρS

FOR GASES: Its specific gravity or relative density is equal to theratio of its density to that of either air or hydrogen at some specified temperature and pressure

ah

GG ρ

ρS

Where at standard condition: w = 1000 kg/m3

w = 9.81 KN/m3

Temperature: It is the measure of the intensity of heat in a body.Fahrenheit Scale:

Boiling Point = 212 FFreezing Point = 32 F

Centigrade or Celsius Scale:Boiling Point = 100 CFreezing Point = 0 C

Absolute Scale: R = F + 460 (Rankine) K = C + 273 (Kelvin)

32F8.1F8.132F

C

Conversion

Pressure: It is the normal component of a force per unit area.

KPa or 2m

KN

AF

P

Where: P – pressure in KN/m2 or KPa F – normal force in KN A – area in m2

1 KN/m2 = 1 KPa (KiloPascal) 1000 N = 1 KN

If a force dF acts on an infinitesimal area dA, the intensity of Pressure is;

KPa or 2m

KN

dAdF

P

Pascal’s Law: At any point in a homogeneous fluid at rest the pressures are the same in all directions:

y

x

z

A

BC

P1A1

P2A2

P3A3

Fx = 0 From Figure:P1A1 - P3 A3sin = 0

P1A1 = P3A3sin Eq.1 P2A2 - P3A3cos = 0

P2A2 = P3A3 cos Eq.2 sin = A1/A3

A1 = A3sin Eq.3cos = A2/A3

A2 = A3cos Eq.4 substituting eq. 3 to eq. 1 and eq.4 to eq.2

P1 = P2 = P3

Atmospheric Pressure (Pa):It is the average pressure exerted by the atmosphere. At sea level

Pa = 101.325 KPa = 0.101325 MPa= 1.01325 Bar = 760 mm Hg = 10.33 m of water= 1.033 kg/cm2

= 14.7 lb/in2

Pa = 29.921 in Hg = 33.88 ft. of water

100 KPa = 1 Bar 1000 KPa = 1 MPa

Absolute and Gauge PressureAbsolute Pressure: It is the pressure measured referred to absolute zero using absolute zero as the base.Gauge Pressure: it is the pressure measured referred to the existing atmospheric pressure and using atmospheric pressure as the base.

Pgauge – if it is above atmosphericPvacuum – negative gauge or vacuum if it is below

atmospheric

Barometer: An instrument used to determine the absolute pressure exerted by the atmosphere

Atmospheic pressure (Pa)

Absolute Zero

Pvacuum

Pgauge

Pabsolute

Pabsolute

Pabs = Pgauge + PaPabs = Pvacuum - Pa

VARIATION OF PRESSURE

PA

(P + dP)AW

dh

F = 0(P + dP)A - PA - W = 0 PA + dPA - PA - W = 0dPA - W = 0 or dPA = W Eq. 1but : W = dV dPA = - dV

where negative sign is used because distance h is measured upward and W is acting downward.

dV = Adh then dPA = -Adh, therefore

dP = - dh (Note: h is positive when measured upward and negative if measured downward)

MANOMETERSManometer is an instrument used in measuring gage pressure in length of some liquid column.1. Open Type Manometer : It has an atmospheric surface and is capable in measuring gage pressure.2. Differential Type Manometer : It has no atmospheric surface and is capable in measuring differences of pressure.

Open TypeOpen end

Manometer Fluid

Differential Type

Fluid A

Fluid B

Fluid C

ENERGY FORMSWork: It is the force multiplied by the displacement in the direction of the force.

W =∫Fdx KJ-W - indicates that work is done on the system+W - indicates that work is done by the system.

Heat: It is a form of energy that crosses a system's boundary, because of a temperature difference between the system and the surrounding.

Q - Heat KJ +Q - indicates that heat is added to the system-Q - indicates that heat is rejected from the system.

Internal Energy: It is the energy acquired due to the overall molecular interaction, or the total energy that a molecule has.

U = mu KJU - total internal energy KJu - specific internal energy KJ/kgU- change of internal energy

Flow Energy Or Flow Work: It is the energy required in pushing a fluid usually into the system or out from the system.

System orControl Volume

P1

P2

A1

A2

L1

L2

Ef1 = F1L1

F1 = P1A1

Ef1 = P1A1L1

A1L1 = V1

Ef1 = P1V1

Ef2 = F2L2

F2 = P2A2

Ef2 = P2A2L2

A2L2 = V2

Ef2 = P2V2

Ef = Ef2 – Ef1

Ef = P2V2 – P1V1

Ef = PV

PV = P2V2 - P1V1 KJ

P = P22 - P11 m3/kg

Where: P – pressure in KPaV – volume in m3

- specific volume in m3/kgEf = PV – Flow energy or flow work

Kinetic Energy: It is the energy acquired due to the motion of a body or a system.

1 2

m mF

d x

dxFdKE

dxFKE

KJ/kg

2(1000)ΔKE

KJ 2(1000)

mΔKE

dtdv

dx1000

mKE

dtdv

1000m

1000ma

F

FdxKE

vv

vv2

1

2

2

2

1

2

2

2

1

21000m

ΔKE

dvv1000m

ΔKE

dtdx

dv1000m

ΔKE

vv2

1

2

2

2

1

2

1

Where: m – mass , kg v – velocity , m/sec

kg

KJ

10002vv

KE

KJ 10002

vvmKE

21

22

21

22

Potential Energy: It is the energy required by virtue of its configuration or elevation.

m

m

dZ

Reference Datum

kgKJ

1000

ZZgPE

KJ 1000

ZZmgPE

dZ1000mg

PE

dZWPE

12

12

Where:W – WorkQ – HeatU – Internal EnergyPV – Flow Energy or flow workKE – Kinetic EnergyPE – Potential Energy

Note: +Z – if measured upward - Z –if measured downward

Law of Conservation of Mass

Mass is indestructible: In applying this law we must except nuclear processes during which mass is converted into energy.

The verbal form of the law is:Mass Entering - Mass Leaving = Change of Mass stored in the systemIn equation Form:

m1 - m2 = m

1 2

m1 m2m = 0

ab c

d

For a steady-state, steady-flow system m = 0, thereforem1 - m2 = 0 or m1 = m2

For one dimensional flow, where1 = 2 = Let m1 = m2 = m

Continuity Equation:

υ

AvAvρm

Where:m - mass flw rate in kg/sec - density in kg/m3

- specific volume inm3/kgA - cross sectional area in m2

v - velocity in m/sec

υυυ

ρ

ΑvvΑvΑΑvvΑρvΑρ

mmm

2

22

1

11

222111

21

Zeroth Law of ThermodynamicsIf two bodies are in thermal equilibrium with a third body, they are in thermal equilibrium with each other, and hence their temperatures are equal.

Specific Heat or Heat Capacity: It the amount of heat required to raise the temperature of a 1 kg mass of a substance 1C or 1K.

tmCTmCQ

m; gConsiderin

Cdt CdT dQ

C;constant ForK-kg

KJ or

C-kgKJ

dtdQ

dTdQ

C

SENSIBLE HEAT: The amount of heat per unit mass that must be transferred (added or remove) when a substance undergoes a change in temperature without a change in phase.

Q = mC(t) = mC(T)where: m - mass , kgC - heat capacity or specific heat, KJ/kg-C or KJ/kg- Kt - temperature in CT - temperature in K

HEAT OF TRANSFORMATION: The amount of heat per unit mass that must be transferred when a substance completely undergoes a phase change without a change in temperature.

Q = mL

A. Heat of Vaporization: Amount of heat that must be added to vaporize a liquid or that

must be removed to condense a gas.Q = mL

where L - latent heat of vaporization, KJ/kgB. Heat of Fusion : Amount of heat that must be added to melt a solid or that must be removed to freeze a liquid.

Q = mLwhere L - latent heat of fusion, KJ/kg

THE FIRST LAW OF THERMODYNAMICS (The Law of Conservation of (Energy)

“Energy can neither be created nor destroyed but can only be converted from one form to another.”Verbal Form:

Energy Entering – Energy Leaving = Change of Energy stored in the systemEquation Form:

E1 – E2 = Es1. First Corollary of the First Law: Application of first Law to a Closed System

U

Q

WFor a Closed System (Non FlowSystem), PV, KE and PE are negligible, therefore the changeof stored energy Es = U

Q – W = U 1Q = U + W 2

By differentiation:dQ = dU + dW 3

where:

dQ Q2 – Q1

dW W2 – W1

Work of a Closed System (NonFlow)

P

V

W = PdV

P

dV

5 Eq. dVPUQ

4 Eq. dVPdUdQ

3 Eq. From

dVPdW

dVPW

dVAdx dxPAW

PAF dxFW

2. Second Corollary of the First Law: Application of First Law to an Open System

System orControl volume

Datum Line

Q

W

1

2

U1 + P1V1 + KE1 + PE1

U2 + P2V2 + KE2 + PE2

For an Open system (Steady state, Steady Flow system) Es = 0, thereforeE1 – E2 = 0 or E1 = E2 orEnergy Entering = Energy Leaving

Z1

Z2

U1 + P1V1 + KE1 + PE1 + Q = U2 + P2V2 + KE2 + PE2 + W 1Q = (U2 – U1) + (P2V2 – P1V1) + (KE2 – KE1) + (PE2 – PE1) + W 2Q = U + (PV) + KE + PE + W 3By differentiationdQ = dU + d(PV) + dKE + dPE + dW 4But dQ Q2 – Q1 and dW W2 – W1

Enthalpy (h)h = U + PVdh = dU + d(PV) 5dh = dU + PdV + VdP 6But: dQ = dU + PdV dh = dQ + VdP 7From Eq. 3Q = h + KE + PE + W 8dQ = dh + dKE + dPE + dW 9dQ = dU + PdV + VdP + dKE + dPE + dW 10dQ = dQ + VdP + dKE + dPE + dW 0 = VdP + dKE + dPE + dW dW = -VdP - dKE - dPE 11By IntegrationW = - VdP - KE - PE 12

If KE = 0 and PE = 0Q = h + W 13W = Q - h 14W = - VdP 15

PEKEhVdP-W

PEKEhQW

WPEKEhQ

SYSTEM OPEN an For .B

PdVW

dWdUdQ

WUQ

SYSTEM CLOSED a For .A

SUMMARY

IDEAL OR PERFECT GAS

Prepared By: Engr Yuri G. Melliza

1. Ideal Gas Equation of StatePV = mRTP = RT

2T2V

2P

1T1V

1P

CTPV

RTP

ρ

Where: P – absolute pressure in KPa V – volume in m3

m – mass in kg R – Gas Constant in KJ/kg-°K T – absolute temperature in°K

IDEAL OR PERFECT GAS

2. Gas Constant

K-m

kgKJ

8.3143R

K-kgKJ

MR

R

Where:R- Gas Constant in KJ/kg-K

Km

kgKJ

constant gas universal R

M – Molecular weight kg/kgm

3. Boyle’s Law If the temperature of a certain quantity of gas is held constant the volume V is inver- sely proportional to the absolute pressure P.

C2V

2P

1V

1P

CPVP1

CV

PV

α

4.Charle’s LawA. At Constant Pressure (P = C) If the pressure of a certain quantity ofgas is held constant, the volume V is directly proportional to the temperature T during a qua-sistatic change of state

2

2

1

1

T

V

T

V

CTV

T;CV; T α V

B. At Constant Volume (V = C)If the volume of a certain quantity of gas isheld constant, the pressure P varies directlyas the absolute temperature T.

2

2

1

1

T

P

T

P

CTP

; TCPT α P

;

5. Avogadro’s LawAll gases at the same temperature and

pressure have the same number of molecules per unit of volume, and it follows that thespecific weight is directly proportional toits molecular weight M.

M

6.Specific HeatSpecific Heat or Heat Capacity is the amountof heat required to raise the temperature of a 1 kg mass 1C or 1KA. SPECIFIC HEAT AT CONSTANT PRESSURE (Cp)

From: dh = dU + PdV + VdPbut dU + VdP = dQ ; therefore

dh = dQ + VdP 1

but at P = C ; dP = O; thereforedh = dQ 2

and by integrationQ = h 3

considering m, h = m(h2 - h1) 4Q = h = m (h2 - h1) 5

From the definition of specific heat, C = dQ/TCp = dQ /dt 6

Cp = dh/dT, then dQ = CpdT 7

and by considering m,dQ = mCpdT 8

then by integration Q = m Cp T 9

but T = (T2 - T1)Q = m Cp (T2 - T1) 10

B SPECIFIC HEAT AT CONSTANT VOLUME (Cv)At V = C, dV = O, and from dQ = dU + PdV dV = 0, therefore

dQ = dU 11 then by integration

Q = U 12then the specific heat at constant volume Cv is;

Cv = dQ/dT = dU/dT 13 dQ = CvdT 14

and by considering m, dQ = mCvdT 15

and by integration Q = mU 16Q = mCvT 17 Q = m(U2 - U1) 18 Q = m Cv(T2 - T1) 19

From: h = U + P and P = RT h = U + RT 20

and by differentiation, dh = dU + Rdt 21 but dh =CpdT and dU = CvdT,

therefore CpdT = CvdT + RdT 22and by dividing both sides of the

equation by dT, Cp = Cv + R 23

7. Ratio Of Specific Heatsk = Cp/Cv 24k = dh/du 25k = h/U 26

From eq. 32,Cp = kCv 27

substituting eq. 27 to eq. 24Cv = R/k-1 28

From eq. 24,

Cv = Cp/k 29substituting eq. 29 to eq. 24

Cp = Rk/k-1 30

8. Entropy Change (S)Entropy is that property of a substance that determines the amount of randomness and disorder of a substance. If during a process, an amount of heat is taken and is by divided by the absolute temperature at which it is taken, the result iscalled the ENTROPY CHANGE.

dS = dQ/T 31and by integration

S = ∫dQ/T 32and from eq. 39

dQ = TdS 33

2

2

1

1

2

2

1

1

2

2

1

1

2211

2

22

1

11

MM

LAW SAVOGADRO' .4

CTV

TV

C P At b.

CTP

TP

C V At a.

LAW CHARLES .3

CVPVP

C) T ( LAW BOYLES 2.

mRTPV

CTVP

TVP

State of Equation .1

SUMMARY

T

dQS

CHANGE ENTROPY .8CvCp

k

HEAT SPECIFIC OF RATIO .7

RCvCp 1-k

RCv ;

1-kRk

Cp

HEAT PECIFICS 6.

kgkg

R

8.3143M

K-kgKJ

M

8.3143R

CONSTANT GAS .5

mol

GAS MIXTURE

Total Mass of a mixture

inn

mm

x ii

imm Mass Fraction

Total Moles of a mixture

nn

y ii

Mole Fraction

Where:m – total mass of a mixturemi – mass of a componentn – total moles of a mixtureni – moles of a componentxi – mass fraction of a componentyi - mole fraction of a component

Equation of StateMass Basis

A. For the mixture

iiiii TRmVP

mRTPV

TRnPV

B. For the components

iiiii TRnVP

Mole Basis

A. For the mixture

B. For the components

Where:R – Gas constant of a mixture

in KJ/kg-K - universal gas constant in

KJ/kgm- KR

AMAGAT’S LAW The total volume of a mixture V is equal to the volume occupied by each component at the mixture pressure P and temperature T.

1n1

V1

2n2

V2

3n3

V3

P,T

P = P1 = P2 = P3

T = T1 = T2 = T3

For the components:

TR

PVn ;

TR

PVn ;

TR

PVn 3

32

21

1

The mole fraction:

V

Vyi

TR

PVTR

PV

y

n

ny

i

i

i

ii

321

321

321

321

VVVV

P

TR

TR

PV

TR

PV

TR

PV

TR

PV

TR

PV

TR

PV

TR

PV

TR

PV

nnnn

The total moles n:

DALTON’S LAW The total pressure of a mixture P is equal to the sum of the partial pressure that each gas would exert at mixture volume V and temperature T.

1n1

P1

2n2

P2

3n3

P3

MIXTURE

nP

T1 = T2 = T3 = TV1 = V2 = V3 = V

For the mixture

For the components

TR

VPn

TR

VPn

TR

VPn

33

22

11

TR

PVn

321

321

321

321

PPPP

V

TR

TR

VP

TR

VP

TR

VP

TR

PV

TR

VP

TR

VP

TR

VP

TR

PV

nnnn

The total moles n: The mole fraction:

P

Pyi

TR

PVTR

VP

y

n

ny

i

i

i

ii

Molecular Weight of a mixture

R

RM

MyM ii

M

RR

RxR ii

Gas Constant of a mixture

Specific Heat of a mixture

RCC

CxC

CxC

vp

viiv

piip

Ratio of Specific Heat

u

h

C

Ck

v

p

Gravimetric and Volumetric AnalysisGravimetric analysis gives the mass fractions of the components

in the mixture. Volumetric analysis gives the volumetric or molal fractionsof the components in the mixture.

i

i

i

i

i

ii

iii

Mx

Mx

y

My

Myx

PROPERTIES OF PURE SUBSTANCE

a - sub-cooled liquidb - saturated liquidc - saturated mixtured - saturated vapore - superheated vapor

Considering that the system is heated at constant pressure where P = 101.325 KPa, the 100C is the saturation temperature corresponding to 101.325 KPa, and 101.325 KPa is the saturation pressure correspon-ding 100C.

P P P P P

Q

30°C100°C

100°C 100°CT100°C

(a) (b) (c) (d) (e)

Q Q Q Q

Saturation Temperature (tsat) - is the highest temperature at a given pressure in which vaporization takes place.Saturation Pressure (Psat) - is the pressure corresponding to the temperature.Sub-cooled Liquid - is one whose temperature is less than the saturation temperature corresponding to the pressure.Compressed Liquid - is one whose pressure is greater than the saturation pressure corresponding to the temperature. Saturated Liquid - a liquid at the saturation temperatureSaturated Vapor - a vapor at the saturation temperatureSaturated Mixture - a mixture of liquid and vapor at the saturation temperature.Superheated Vapor - a vapor whose temperature is greater than the saturation temperature.

a

b c de

T

F

Saturated Vapor

Saturated Vapor

30°C

100°C

t 100°C

Saturated Mixture

P = C

Critical Point

T- Diagram

a

b c de

T

S

F

Saturated Vapor

Saturated Vapor

30°C

100°C

t 100°C

Saturated Mixture

P = C

Critical Point

T-S Diagram

F(critical point)- at the critical point the temperature and pressure is unique.For Steam: At Critical Point, P = 22.09 MPa; t = 374.136C

a

b c de

T

S

F

Saturated Vapor

Saturated Vapor

ta

tsat

te

Saturated Mixture

P = C

Critical Point

T-S Diagram

tsat - saturation temperature corresponding the pressure Pta - sub-cooled temperature which is less than tsatte - superheated vapor temperature that is greater than tsat

h-S (Enthalpy-Entropy Diagram)

h

S

t = C (constant temperature curve)

P = C (constant pressure curve)F

I

II

III

I - subcooled or compressed liquid regionII - saturated mixture regionIII - superheated vapor region

Quality (x):

Lv

v

Lv

v

mmm

m

m

mm

mx

Where:mv – mass of vapormL – mass of liquidm – total massx- quality

The properties at saturated liquid, saturated vapor, superheatedvapor and sub-cooled or compressed liquid can be determined from tables. But for the properties at saturated mixture (liquid and vapor) they can be determined by the equation

rc = rf + x(rfg) rfg = rg – rf

Where: r stands for any property (, U, h and S)rg – property at saturated vapor (from table)rf – property at saturated liquid

Note: The properties at siub-cooled or compressed liquid is approximately equal to the properties at saturated liquidcorresponding the sub-cooled temperature.

Throttling Calorimeter

Main Steam Line

P1 – steam line pressure

To main steam line

P2 -Calorimeter pressure

h1 = h2

h1 = hf1 + x1(hfg1)Where:

1 – main steam line2 - calorimeter

thermometer

P1

P2

1

2

T

S

h = C

T-S Diagram Throttling Process

P1 – steam line pressureP2 – calorimeter pressure

1. Isobaric Process ( P = C): An Isobaric Process is an internally reversible constant pressure process. A. Closed System:(Nonflow)

P

V

21P

dV

Q = U + W 1 any substanceW = PdV 2 any substanceU = m(U2 - U1) 3 any substanceW = P(V2 - V1) 4 any substanceQ = h = m(h2-h1) 5 any substance

T

S

1

2

dS

TP = C

PROCESSES OF FLUIDS

For Ideal Gas:PV = mRTW =mR(T2-T1) 5U = mCv(T2-T1) 6 Q = h = mCP (T2-T1) 7Entropy ChangeS = dQ/T 8 any substancedQ = dhFor Ideal Gasdh = mCPdTS = dQ/TS = mCP dT/TS = mCP ln(T2/T1) 9B. Open System:Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substance- VdP = 0

Q = h 12W = - KE - PE 13If KE = 0 and PE = 0W = 0 14Q = mCP(T2-T1) 15 Ideal Gas

2. Isometric Process (V = C): An Isometric process is internally reversible constant volume process.

A. Closed System: (Nonflow)

P

V1

2T

S

T

dS

1

2V = C

Q = U + W 1 any substanceW = PdV at V = C; dV = 0W = 0 Q = U = m(U2 - U1) 2 any substanceh = m(h2-h1) 3 any substance

For Ideal Gas:Q = U = mCv(T2-T1) 4 h = mCP(T2-T1) 5Entropy Change:S = dQ/T 6 any substancedQ = dUdU = mCvdT for ideal gasS = dU/T = mCvdT/TS = mCv ln(T2/T1) 6

B. Open System:Q = h + KE + PE + W 7 any substanceW = - VdP - KE - PE 8 any substance-VdP = -V(P2-P1) 9 any substanceQ = U = m(U2 - U1) 10 any substanceh = m(h2-h1) 11 any substanceFor Ideal Gas:-VdP = -V(P2-P1) = mR(T1-T2)Q = U = mCv(T2-T1) 12 h = mCP(T2-T1) 13If KE = 0 and PE = 0Q = h + W 14 any substanceW = - VdP 15W = -VdP = -V(P2-P1) 16 any substance W = mR(T1-T2) 16 ideal gash = mCP(T2-T1) 17 ideal gas

3. Isothermal Process(T = C): An Isothermal process is a reversible constant temperature process. A. Closed System (Nonflow)

dS

T

S

T1 2

P

V

1

2P

dV

PV = C orT = C

Q = U + W 1 any substanceW = PdV 2 any substanceU = m(U2 - U1) 3 any substanceFor Ideal Gas:dU = mCv dT; at T = C ; dT = 0Q = W 4

W = PdV ; at PV = C ; P1V1 = P2V2 = C; P = C/VSubstituting P = C/V to W = PdV W = P1V1 ln(V2/V1) 5Where (V2/V1) = P1/P2

W = P1V1 ln(P1/P2) 6P1V1 = mRT1 Entropy Change:dS = dQ/T 7S = dQ/TdQ = TdS ;at T = CQ = T(S2-S1)(S2-S1) = S = Q/T 8S = Q/T = W/T 9 For Ideal Gas

B. Open System (Steady Flow)Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substance-VdP = -V(P2-P1) 12 any substanceh = m(h2-h1) 13 any substanceFor Ideal Gas:-VdP = -P1V1ln(P2/P1) 14 -VdP = P1V1ln(P1/P2) 15 P1/P2 = V2/V1 16dh = CPdT; at T = C; dT = 0h = 0 16 If KE = 0 and PE = 0Q = h + W 17 any substanceW = - VdP = P1V1ln(P1/P2) 18For Ideal Gash = 0 19Q = W = - VdP = P1V1ln(P1/P2) 20

4. Isentropic Process (S = C): An Isentropic Process is an internally“Reversible Adiabatic” process in which the entropy remains constantwhere S = C and PVk = C for an ideal or perfect gas.

For Ideal Gas

1

2

1

1

1

2

1

2

2

22

1

kk

k

k22

k11

11

k

V

V

P

P

T

T

VPVP and T

VP

T

VP

C PV and CTPV Using

A. Closed System (Nonflow)

T

S

1

2

P

V

1

2

dV

P

S = C orPVk = C

Q = U + W 1 any substanceW = PdV 2 any substanceU = m(U2 - U1) 3 any substanceQ = 0 4W = - U = U = -m(U2 - U1) 5

For Ideal Gas U = mCV(T2-T1) 6 From PVk = C, P =C/Vk, and substituting P =C/Vk to W = ∫PdV, then by integration,

11

11

1

1

1

211

1

1

21

kk

VP

kk

12

1122

P

P

kPdV

P

P

k

mRT

k-1

T-TmRPdV

k

VP-VPPdV W 7

8

9

Q = 0

Entropy ChangeS = 0S1 = S2

B. Open System (Steady Flow)Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substanceh = m(h2-h1) 12 any substanceQ = 0W = -h - KE - PE 13From PVk = C ,V =[C/P]1/k, substituting V to-∫VdP, then by integration,

11

11

1

1

1

211

1

1

21

kk

kk

12

1122

P

P

k

VkPVdP

P

P

k

kmRT

k-1

T-TkmRVdP

k

VP-VPkVdP

PdV kVdP

14

15

16

If KE = 0 and PE = 00 = h + W 17 any substanceW = - VdP = - h 18 any substanceh = m(h2-h1) 19 any substanceQ = 0

12P

kk

kk

12

1122

T-TmChW

P

P

k

VkPW

P

P

k

kmRT

k-1

T-TkmRW

k

VP-VPkPdV kVdPW

11

11

1

1

1

211

1

1

21

20

22

21

23

P

V

dP

V

Area = -VdP

S = C

1

2

1

1

1

2

1

2

2

22

1

nn

n

n22

n11

11

n

V

V

P

P

T

T

VPVP and T

VP

T

VP

C PV and CTPV Using

5. Polytropic Process ( PVn = C): A Polytropic Process is an internally reversible process of an Ideal or Perfect Gas in which PVn = C, where n stands for any constant.

A. Closed System: (Nonflow)

Q = U + W 1 W = PdV 2 U = m(U2 - U1) 3 Q = mCn(T2-T1) 4U = m(U2 - U1) 5

P

V

1

2

dV

P

PVn = C

T

S

2

1

dS

T

PVn = C

K-kgKJ

or C-kg

KJ heat specific polytropic C

n1nk

CC

n

vn

From PVn = C, P =C/Vn, and substituting P =C/Vn to W = ∫PdV, then by integration,

11

11

1

1

1

211

1

1

21

nn

VP

nn

12

1122

P

P

nPdVW

P

P

n

mRT

n-1

T-TmRPdVW

n

VP-VPPdV W

Entropy ChangedS = dQ/TdQ = mCndTS = mCnln(T2/T1)

6

8

9

10

B. Open System (Steady Flow)Q = h + KE + PE + W 11 W = - VdP - KE - PE 12 h = m(h2-h1) 13 Q = mCn(T2-T1) 14dQ = mCn dTW = Q - h - KE - PE 15From PVn = C ,V =[C/P]1/n, substituting V to-∫VdP, then by integration,

n

VP-VPnVdP

PdV nVdP

1122

1 16

11

11

1

1

211

1

1

21

nn

nn

12

P

P

n

VnPVdP

P

P

n

nmRT

n-1

T-TnmRVdP

If KE = 0 and PE = 0Q = h + W 19 any substanceW = - VdP = Q - h 20 any substanceh = m(h2-h1) 21 any substanceh = mCp(T2-T1)Q = mCn(T2-T1) 22

17

18

11

11

1

1

211

1

1

21

nn

nn

12

P

P

n

VnPW

P

P

n

nmRT

n-1

T-TnmRW

24

23

6. Isoenthalpic or Throttling Process: It is a steady - state, steady flow process in which Q = 0; PE = 0; KE = 0; W = 0 and the enthalpy remains constant.

h1 = h2 or h = C

Throttling valve

Main steam line

thermometer

Pressure Gauge

Pressure Gauge

To main steam line

Throttling Calorimeter

Irreversible or Paddle Work

m

W

Q

UWp

Q = U + W - Wp

where: Wp - irreversible or paddle work

THERMODYNAMICS Part 2By. Engr. Yuri G. Melliza

2nd Law of Thermodynamics

Carnot Cycles

Steam Cycles

Fuels and Combustion

ICE Cycles

2nd Law of Thermodynamics

• Second Law of Thermodynamics• Kelvin – Planck Statement• Carnot engine• Carnot Refrigerator• Sample Problems

Second Law of Thermodynamics:Whenever energy is transferred, the level of energy cannot be conserved and some energy must be permanently reduced to a lower level. When this is combined with the first law of thermodynamics, the law of energy conservation, the statement becomes:

Second Law of Thermodynamics:Whenever energy is transferred, energy must be conserved, but the level of energy cannot be conserved and some energy must be permanently reduced to a lower level.

Kelvin-Planck statement of the Second Law:No cyclic process is possible whose sole result is the flow of heat from a single heat reservoir and the performance of an equivalent amount of work.For a system undergoing a cycle: The net heat is equal to the net work. QW dWdQ Where:

W - net workQ - net heat

CARNOT CYCLENicolas Leonard Sadi Carnot 1796-1832 1.Carnot Engine Processes:

1 to 2 - Heat Addition (T = C)2 to 3 - Expansion (S = C)3 to 4 - Heat Rejection (T = C)4 to 1 - Compression (S = C)

P

V

2

1

3

4

T = C

S = CS = C

T = C

T

S

21

34

T H

T L

Q A

Q R

Heat Added (T = C)

QA = TH(S) 1

Heat Rejected (T = C)QR = TL(S) 2S = S2 - S1 = S3 – S4 3

Net WorkW = Q = QA - QR 4W = (TH - TL)(S) 5

% xQ

Qe

% xQ

QQe

% x QW

e

A

R

A

RA

A

1001

100

100

6

7

8

Substituting eq.1 and eq. 5 to eq 6

% xT

Te

% x T

TTe

H

L

H

LH

1001

100

9

10

TH

TL

W

QA

QR

E

Carnot Engine

2. Carnot Refrigerator: Reversed Carnot CycleProcesses: 1 to 2 - Compression (S =C) 2 to 3 - Heat Rejection (T = C) 3 to 4 - Expansion (S = C) 4 to 1 - Heat Addition (T = C)

Q

R

Q

A

1

2

4

3

S

T

TH

TL

Heat Added (T = C)QA = TL(S) 1

Heat Rejected (T = C)QR = TH(S) 2S = S1 - S4 = S2 - S3 3

Net WorkW = Q 4

W = QR - QA 5W = (TH - TL)(S) 6

Coefficient of Performance(COP)

LH

L

A

TT

TCOP

W

QCOP

7

8

1H

L

T

TCOP 9

Tons of Refrigeration211 KJ/min = 1 TR3. Carnot Heat Pump:A heat pump uses the same components as therefrigerator but its purpose isto reject heat at high energy level.

Performance Factor (PF)

AR

R

R

QQ

QPF

W

QPF

10

11

1

1

1

COPPF

T

TPF

Q

QPF

TT

TPF

L

H

A

R

LH

H 12

13

14

15

TH

TL

W

QA

QR

R

Carnot Refrigerator

A Carnot engine operating between 775 K and 305K produces 54 KJ of work. Determine the change of entropy during heat addition.TH = 775 K ; TL = 305 KW = 54 KJ

TH

TL

W

QA

QR

E

K

KJ 0.015

775

89.04

T

QS-S

)S-(STQ

KJ 89.040.606

54

e

WQ

Q

We

0.606775

305775

T

TTe

H

A12

12HA

A

A

H

LH

A Carnot heat engine rejects 230 KJ of heat at 25C. The net cycle work is 375 KJ. Determine the cycle thermal efficiency and the cycle high temperature .Given:QR = 230 KJTL = 25 + 273 = 298KW = 375 KJ

TL = 298K

TH

WE

QR = 230 KJ

QA

K87.783772.0

605

)S-(S

QT

KKJ/ -0.772)S-(S

KKJ/ 772.0)SS(

)SS(298230

)SS(SS

)SS(TQ

)SS(TQ

62.0605

375

QA

We

KJ 605QA

)230375(QWQ

QQW

12

AH

12

34

34

1234

34LR

12HA

RA

RA

A Carnot engine operates between temperature reservoirs of 817C and 25C and rejects 25 KW to the low temperature reservoir. The Carnot engine drives the compressor of an ideal vapor compres-sion refrigerator, which operates within pressure limits of 190 KPa and 1200 Kpa. The refrigerant is ammonia. Determine the COP and the refrigerant flow rate.(4; 14.64 kg/min)TH = 817 + 273 = 1090 KTL = 25 + 273 = 298 KQR = 25 KW

Internal Combustion Engine Cycles

1. Air Standard Otto Cycle (Spark Ignition Engine Cycle)Processes1 to 2 - Isentropic Compression (S = C)2 to 3 - Constant Volume Heat Addition ( V = C)3 to 4 - Isentropic Expansion (S =C)4 to 1 - Constant Volume Heat Rejection (V = C)

P

PmV

VD

CVD

W1

42

3

S = C

S = C

3QAT

S

1

24

V = C

V = C

QR

Compression Ratio

3

4

2

1

V

V

V

Vr

where: r - compression ratioV1 = V4 and V2 = V3

1

Heat Added (V = C) QA = mCV(T3 - T2) 2

Heat RejectedQR = mCV(T4 - T1) 3

Net Cycle WorkW = QA - QR

W = mCV[(T3 - T2) - (T4 - T1)] 4

Thermal Efficiency

100%r

11e

100%TT

TT1e

100%x Q

Q1e

100%Q

QQe

100%xQ

We

1-k

23

14

A

R

A

RA

A

x

x

x

5

6

7

8

9

Mean Effective Pressure

KPa V

WP

Dm

where:W - net work, KJ, KJ/kg, KWVD - Displacement Volume, m3, m3/kg, m3/secVD = V1 - V2 m3

VD = 1 - 2 m3/kg

10

Percent Clearance

100% x V

VC

D

2

V2 = CVD

V1 = Vd + CVD

C

C1

V

Vr

2

1 11 12

2. Diesel Cycle: (Compression Ignition Engine Cycle)

Processes1 to 2 - Isentropic Compression (S = C)2 to 3 - Constant Pressure Heat Addition (P = C)3 to 4 - Isentropic Expansion (S = C)4 to 1 - Constant Volume Heat Rejection (V = C)

P

V

T

S

2 3

4

1

S = C

S = C

VDCVD

1

2

3

4V = C

P = C

QR

QA

Heat Added (P = C) QA = mCP(T3 - T2) 3 QA = mkCV(T3 - T2) 4

Heat Rejected (V = C)QR = mCV(T4 - T1) 5

Net Cycle WorkW = QA - QR

W = mCV[k(T3 - T2) - (T4 - T1)] 6

Compression Ratio

3

4

2

1

V

V

V

Vr 1

Cut - Off Ratio

2

3c V

Vr 2

Thermal Efficiency

100%1)k(r

1)(r

r

11e

100%TT

TT1e

100%x Q

Q1e

100%Q

QQe

100%xQ

We

c

kc

1-k

23

14

A

R

A

RA

A

x

x k

x

7

8

9

10

11

where:W - net work, KJ, KJ/kg, KWVD - Displacement Volume, m3, m3/kg, m3/secVD = V1 - V2 m3

VD = 1 - 2 m3/kg

KPa V

WP

Dm

Mean Effective Pressure

12

3. Air Standard Dual CycleProcesses:1 to 2 - Compression (S = C)2 to 3 -Heat Addition (V = C)3 to 4 - Heat Addition (P = C)4 to 5 -Expansion (S = C)5 to 1 _ Heat Rejection (V = C)

11

2

2

33

44

5

5S = C

S = C

P = C

V = C

V = C

QA1

QA2

QR

P

V

T

S

VDCVD

Comprssion Ratio

4

5

3

5

2

1

V

V

V

V

V

Vr

V5 = V1 ; V2 = V3

1

Pressure Ratio

2P

4P

2

3p P

Pr 3

2

4

3

4c V

Vr

V

V 2

Cut-Off Ratio

Heat Added (V = C) QA1 = mCV(T3 - T2) 4Heat Added (P =C)QA2 = mCP(T4 - T3) 5QA2 = mkCV(T4 - T3) 6

Heat Rejected (V = C)QR = mCV(T5 - T1) 7

Net Cycle WorkW = QA - QR 8QA = QA1 + QA2 9QA = mCV[(T3 - T2) +k (T4 - T3)] 10W = mCV[(T3 - T2) + k(T4 - T3) - (T5 - T1) ]

100%x Q

Q1e

100%Q

QQe; 100%x

Q

We

A

R

A

RA

A

x

12

13

Thermal Efficiency

11

100%1-(rkr1)-(r

1)-r(r

r

11e

100% )]T - (Tk )T - [(T

)T - (T1e

cpp

kcp

1-k

3423

15

x

x

Mean Effective PressurePm = W/VD KPa

where: VD = V1 - V2 m3 ; W in KJ

V1 = V5 ; V2 = V3

VD = 1 - 2 m3/kg ; W in KJ/kg

1 = 5 ; 2 = 3

 For Cold Air Standard: K = 1.4For Hot Air Standard: K = 1.3

Vapor Power Cycle

RANKINE CYCLEProcesses:

1 to 2 - Expansion (S = C)2 to 3 - Heat Rejection (P = C)3 to 4 - Compression or Pumping (S = C)4 to 1 - Heat Addition (P = C)

Boiler or SteamGenerator

Turbine

Condenser

Pump

WP

QA

QR

Wt

1

2

3

4

Major Components of a Rankine Cycle1. Steam Generator or Boiler: The working substance absorbs heat from products of combustion or other sources of heat at constant pressure which in turn changes the state of the working substance (water or steam) from sub-cooled liquid and finally to superheated vapor whence at this point it enters the turbine. 2. Steam Turbine: A steady state, steady flow device where steam expands isentropically to a lower pressure converting some forms of energy (h, KE, PE) to mechanical work that finally be converted into electrical energy if the turbine is used to drive an electric gene- rator.3. Condenser: Steam exiting from the turbine enters this device to re- ject heat to the cooling medium and changes its state to that of the saturated liquid at the condenser pressure which occurred at a cons- tant pressure process.

4. Pump: It is also a steady state, steady flow machine where the condensate leaving the condenser at lower pressure be pumped back to the boiler in an isentropic process in order to raise the pressure of the condensate to that of the boiler pressure.

h

S S

T

3

42

1

3

4

1

2

P1

P2

P1

P2

4’2’

2’

4’

Turbine Worka) Ideal Cycle

Wt = (h1 - h2) KJ/kgWt = ms(h1 - h2) KW

b) Actual CycleWt’ = (h1 - h2’) KJ/kgWt’ = ms(h1 - h2’) KW

where: ms - steam flow rate in kg/secTurbine Efficiency

100%x hhhh

η

100%x WW

tη

21

2'1t

t

t'

Pump Worka) Ideal Cycle

WP = (h4 - h3) KJ/kgWP = ms(h4 - h3) KW

b) Actual CycleWP’ = (h4’ - h3) KJ/kgWP’ = ms(h4’ - h3) KW

Pump Efficiency

100%xhh

hhη

100%xW

Wη

34'

34p

p'

pp

Heat Rejecteda) Ideal Cycle

QR = (h2 - h3) KJ/kgQR = ms(h2 - h3) KW QR = ms(h2 - h3) KW = mwCpw(two - twi) KW

b) Actual CycleQR = (h2’ - h3) KJ/kgQR = ms(h2’ - h3) KW = mwCpw(two - twi) KW

Where: mw - cooling water flow rate in kg/sec twi - inlet temperature of cooling water inC two - outlet temperature of cooling water inC Cpw - specific heat of water in KJ/kg- C or KJ/kg-K Cpw = 4.187 KJ/kg- C or KJ/kg- K

Heat Added:a) Ideal Cycle

QA = (h1 - h4) KJ/kgQA = ms (h1 - h4) KW

b) Actual CycleQA = (h1 - h4’) KJ/kgQA = ms (h1 - h4’) KW

Steam Generator or boiler Efficiency

100%x(HV)m

)h(hmη

100%xQ

Qη

f

41sB

S

AB

Where: QA - heat absorbed by boiler in KWQS - heat supplied in KWmf - fuel consumption in kg/secHV - heating value of fuel in KJ/kg

Steam Rate

KW-sec

kg

ProducedKW

rate Flow SteamSR

Heat Rate

KW-sec

KJ

ProducedKW

SuppliedHeat HR

Reheat Cycle A steam power plant operating on a reheat cycle improves the thermalefficiency of a simple Rankine cycle plant. After partial expansion of the steam in the turbine, the steam flows back to a section in the boiler which is the re-heater and it will be reheated almost the same to its initial temperature and expands finally in the turbine to the con-denser pressure.

Reheater

QA

WP

QR

Wt

1 kg

12 3

4

56

Regenerative Cycle In a regenerative cycle, after partial expansion of the steam in theturbine, some part of it is extracted for feed-water heating in an open orclose type feed-water heater. The bled steam heats the condensate from the condenser or drains from the previous heater causing a decrease in heat absorbed by steam in the boiler which result to an increase in thermal efficiency of the cycle.

QA

WP1

QR

Wt

1 kg

1

2

3

456

7

WP2

m

Reheat-Regenerative Cycle For a reheat - regenerative cycle power plant, part of the steam is re-heated in the re-heater and some portion is bled for feed-water heating to an open or closed type heaters after its partial expansion in the turbine. It will result to a further increase in thermal efficiency of theplant.

QA

WP1

QR

Wt

1 kg

1

2

4

5678

WP2

m

23

1-m

1-m

For a 1 kg basis of circulating steam, m is the fraction of steam extracted for feed-water heating as shown on the schematic diagram above, where the reheat and bled steam pressure are the same.

FUELS and

COMBUSTION

By. Engr. Yuri G. Melliza

FUELS AND COMBUSTION Fuels and Combustion Types of Fuels Complete/Incomplete Combustion Oxidation of Carbon Oxidation of Hydrogen Oxidation of Sulfur Air composition Combustion with Air Theoretical Air Hydrocarbon fuels Combustion of Hydrocarbon Fuel

Fuels and Combustion

Fuel: Substance composed of chemical elements which in rapid chemical union with oxygen produced combustion.

Combustion: Is that rapid chemical union with oxygen of an element, whose exothermic heat of reaction is sufficiently great and whose rate of reaction is suffi-ciently fast whereby useful quantities of heat are liberated at elevated temperature.

TYPES OF FUELS Solid Fuels

ex: Wood, coal, charcoal Liquid Fuels

ex: gasoline, diesel, kerosene Gaseous Fuels

ex: LPG, Natural Gas, Methane Nuclear Fuels

ex: Uranium

Combustible Elements1. Carbon (C) 3. Sulfur (S)2. Hydrogen (H2)

Complete Combustion: Occurs when all the combustible elements has been fully oxidized.

Ex:C + O2 CO2

Incomplete Combustion: Occurs when some of the combustible elements has not been fully oxidized.

Ex:C + O2 CO

Common Combustion GasesGAS MOLECULAR

Weight (M)

C 12

H 1

H2 2

O 16

O2 32

N 14

N2 28

S 32

THE COMBUSTION CHEMISTRY Oxidation of Carbon

11 83

44 3612

32)1(12 1(16)1(12)

Basis Mass

1 11

Basis Mole

CO OC 2

2

Oxidation of Hydrogen

9 81

18 162

2)1(16 (32)1(2)

Basis Mass

1 1

Basis Mole

OH OH

2

1

21

2

22 21

Oxidation of Sulfur

2 11

64 3232

32)1(32 (32)1(32)

Basis Mass

1 11

Basis Mole

OS OS

1

22

Composition of AIRa. Percentages by Volume (by

mole)O2 = 21%N2 = 79%

b. Percentages by MassO2 = 23%N2 = 77%

76321

79.

2

2

O of MoleN of Moles

Combustion with AirA. Combustion of Carbon with air

C + O2 + 3.76N2 CO2 + 3.76N2

Mole Basis:1 + 1 + 3.76 1+ 3.76

Mass Basis:1(12) + 1(32) + 3.76(28) 1(44) + 3.76(28)12 + 32 + 3.76(28) 44 + 3.76(28) 3 + 8 + 3.76(7) 11+ 3.76(7)

kg of air per kg of Carbon:

C of kgair of kg

11.44=3

3.76(7)+8=

C of kgair of kg

B. Combustion of Hydrogen with air H2 + ½ O2 + ½ (3.76)N2 H2O + ½(3.76)N2 Mole Basis:

1 + ½ + ½(3.76) 1 + ½(3.76)Mass Basis:

1(2) + ½ (32) + ½(3.76)(28) 1(18) + ½ (3.76)(28)

2 + 16 + 3.76(14) 18 + 3.76(14) 1 + 8 + 3.76(7) 9 + 3.76(7)

kg of air per kg of Hydrogen:

22 H of kgair of kg

34.32=1

3.76(7)+8=

H of kgair of kg

C. Combustion of Sulfur with airS + O2 + 3.76N2 SO2 + 3.76N2

Mole Basis:1 + 1 + 3.76 1 + 3.76

Mass Basis:1(32) + 1(32) + 3.76(28) 1(64) +

3.76(28) 32 + 32 + 105.28 64 + 105.28

kg of air per kg of Sulfur:

S of kgair of kg

4.29=32105.2832

=S of kgair of kg

Theoretical AirIt is the minimum amount of air required to oxidize the reactants or the combustible elements found in the fuel. With theoretical air no O2 is found in products.

Excess AirIt is an amount of air in excess of the Theoretical requirements in order to influence complete combustion. With excess air O2 is present in the products.

HYDROCARBON FUELSFuels containing the element s Carbon and Hydrogen. Chemical Formula: CnHm

Family Formula Structure Saturated

Paraffin CnH2n+2 Chain Yes

Olefin CnH2n Chain No

Diolefin CnH2n-2 Chain No

Naphthene CnH2n Ring Yes

Aromatic

Benzene CnH2n-6 Ring No

Naphthalene CnH2n-12 Ring No

Alcohols Note: Alcohols are not pure hydrocarbon, because one of its hydrogen atom is replace by an OH radical. Sometimes it is used as fuel in an ICE.

Methanol CH3OH

Ethanol C2H5OH

Saturated Hydrocarbon: All the carbon atoms are joined by a single bond.Unsaturated Hydrocarbon: It has two or more adjacent Carbon atoms joined by a double or triple bond.Isomers: Two hydrocarbons with the same number of carbon and hydrogen atoms but atdifferent structures.

H H H H H C C C CH H H H H

Chain structure Saturated

H H HC C=C C H H H H H

Chain Structure Unsaturated

Ring structure Saturated H H H C H C C H C H H H

Theoretical Air: It is the minimum or theoretical amount of air required to oxidized the reactants. With theoretical air no O2 is found in the products. Excess Air: It is an amount of air in excess of the theo-retical air required to influence complete combustion. With excess air O2 is found in the products.

Combustion of Hydrocarbon Fuel(CnHm)

A. Combustion with 100% theoretical air CnHm + aO2 + a(3.76)N2 bCO2 + cH2O + a(3.76)N2

fuel

air

t kg kg

m12n

)a(3.76)(28a(32)FA

Combustion of Hydrocarbon FuelFormula: (CnHm)

A. Combustion with 100% theoretical air CnHm + aO2 + a(3.76)N2 bCO2 + cH2O + a(3.76)N2

fuel

air

t kg kg

m12n

)a(3.76)(28a(32)FA

fuel

air

a kg kg

m12n

)a(3.76)(28a(32)e)(1

FA

B. Combustion with excess air e CnHm +(1+e) aO2 + (1+e)a(3.76)N2 bCO2 +

cH2O + dO2 + (1+e)a(3.76)N2

Actual Air – Fuel Ratio

fuel

air

ta kg kg

FA

e)(1FA

Where: e – excess air in decimalNote: Sometimes excess air is expressible in terms of theoretical air. Example: 25% excess air = 125% theoretical air

Orsat Analysis: Orsat analysis gives the volumetric or molal analysis of the PRODUCTS on a DRY BASIS, (no amount of H2O given).

Proximate Analysis: Proximate analysis gives the amount of Fixed Carbon, Volatiles, Ash and Moisture, in percent by mass. Volatiles are those compounds that evaporates at low temperature when the solid fuel is heated.

ULTIMATE ANALYSIS: Ultimate analysis gives the amount of C, H, O, N, S in percentages by mass, and sometimes the amount of moisture and ash are given.

SOLID FUELSComponents of Solid Fuels:

1. Carbon (C) 2. Hydrogen (H2)3. Oxygen (O2)4. Nitrogen (N2)5. Sulfur (S)6. Moisture (M)7. Ash (A)

A.Combustion with 100% theoretical airaC + bH2 + cO2 + dN2 + eS + fH2O + gO2 + g(3.76)N2 hCO2 + iH2O + jSO2 + kN2

B.Combustion with excess air x: aC + bH2 + cO2 + dN2 + eS + fH2O +

(1+x)gO2 +(1+x)g(3.76)N2 hCO2 + iH2O + jSO2 + lO2 + mN2

WHERE: a, b, c, d, e, f, g, h, I, j, k, x are the number of moles of the elements.x – excess air in decimal

fuel kgair kg

18f32e28d32c2b12a

3.76(28)g32gFA

t

Theoretical air-fuel ratio:

Actual air-fuel ratio:

fuel kgair kg

18f32e28d32c2b12a

3.76(28)g32gx)(1

a

F

A

MASS FLOW RATE OF FLUE GAS (Products)

Air +Fuel Products

A. Without considering Ash loss

1

F

Amm Fg

B. Considering Ash loss

lossAsh 1

F

Amm Fg

Heating Value

Heating Value - is the energy released by fuel when it is completely burned and the products of combustion are cooled to the original fuel temperature.Higher Heating Value (HHV) - is the heating value obtained when the water in the products is liquid.Lower Heating Value (LHV) - is the heating value obtained when the water in the products is vapor.

For Solid Fuels with the presence of Fuel’s ULTIMATE ANALYSIS

kg

KJ S9304

8

OH212,144C820,33HHV 2

2

where: C, H2, O2, and S are in decimals from the ultimate analysis

HHV = 31 405C + 141 647H KJ/kgHHV = 43 385 + 93(Be - 10) KJ/kg

For Liquid Fuels

where: Be - degrees Baume

For Coal and Oils with the absence of Ultimate Analysis

fuel of kg

air of Kg

3041

HHV

F

A

t

For Gasoline

kgKJ )API(93639,38LHV

kgKJ )API(93160,41HHV

kgKJ )API(93035,39LHV

kgKJ )API(93943,41HHV

For Kerosene

For Fuel Oils

Institute Petroleum AmericanAPI

kgKJ )API(6.139105,38LHV

kgKJ )API(6.139130,41HHV

For Fuel Oils (From Bureau of Standard Formula)

).t(.St@S 561500070API131.5

141.5S

HHV = 51,716 – 8,793.8 (S)2 KJ/kgLHV = HHV - QL KJ/kg

QL = 2442.7(9H2) KJ/kg

H2 = 0.26 - 0.15(S) kg of H2/ kg of

fuel

WhereS - specific gravity of fuel oil at 15.56 CH2 - hydrogen content of fuel oilQL - heat required to evaporate and superheat the water vapor formed bythe combustion of hydrogen in the fuelS @ t - specific gravity of fuel oil at any temperature tOxygen Bomb Calorimeter - instrument used in mea-suring heating value of solid and liquid fuels.Gas Calorimeter - instrument used for measuring heating value of gaseous fuels.

Properties of Fuels and Lubricantsa) Viscosity - a measure of the resistance to flow that a lubricant offers when it is subjected to shear stress.b) Absolute Viscosity - viscosity which is determined by direct measurement of shear resistance.c) Kinematics Viscosity - the ratio of the absolute viscosity to the densityd) Viscosity Index - the rate at which viscosity changes with temperature.e) Flash Point - the temperature at which the vapor above a volatile liquid forms a combustible mixture with air.f) Fire Point - The temperature at which oil gives off vapor that burns continuously when ignited.

g) Pour Point - the temperature at which oil will no longer pour freely.h) Dropping Point - the temperature at which grease melts.i) Condradson Number(carbon residue) - the percentage amount by mass of the carbonaceous residue remaining after destructive distillation.j) Octane Number - a number that provides a measure of the ability of a fuel to resist knocking when it is burnt in a gasoline engine. It is the percentage by volume of iso-octane in a blend with normal heptane that

matches the knocking behavior of the fuel.

k) Cetane Number - a number that provides a measure of the ignition characteristics of a diesel fuel when it is burnt in a standard diesel engine. It is the percentage of cetane in the standard fuel.

Prepared By: ENGR YURI G. MELLIZA, RME