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Thermodynamics
PREPARED BY: Er.Rashwinder singh
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THERMODYNAMICS It a science that deals with energy transformation, the transformation of heat intowork or vice versa. It was derived from a Greek word therme that means Heat and dynamis that means Strength. SYSTEM : Is that portion in the universe, an atom, a galaxy, a certain quantity ofmatter or a certain volume in space in which one wishes to study. It is a region en-closed by an specified boundary, that may be imaginary, fixed or moving.OPEN SYSTEM : A system open to matter flow or a system in which there is anexchange of mass between the system and the surroundings.CLOSED SYSTEM: A system closed to matter flow or a system in which theres no
exchange of mass between the system and the surroundings.Closed System: Piston in Cylinder
boundary
Open System: Steam turbine
steam in
steam out
Work
boundary
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SURROUNDINGS OR ENVIRONMENT : It is the region all about the system.WORKING SUBSTANCE : A substance responsible for the transformation of energy.
example: steam in a steam turbine, water in a water pumpPURE SUBSTANCE : A substance that is homogeneous in nature and is homoge-neous, or a substance that is not a mixture of different specie, or a substance thatdoes not undergo chemical reaction.PROPERTY : It is a characteristic quality of a certain substance.INTENSIVE PROPERTY : Property that is independent of the mass of a system.EXTENSIVE PROPERTY : Property that is dependent upon the mass of the systemand are total values such as volume and total internal energy.
PROCESS : It is simply a change of state of a substance. If certain property of asubstance is changed, it is said to have undergone a process.CYCLE : It is a series of two or more processes in which the final and the initialstate are the same.ADIABATIC SYSTEM : A system that is impervious to heat. A system (open or closed)in which heat cannot cross its boundary.
PHASES OF A SUBSTANCEA. Solid phaseB. Liquid phaseC. Gaseous or Vapor phase
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SPECIFIC TERMS TO CHARACTERIZED PHASE TRANSITION:1. Vaporization: Change from liquid to vapor2. Condensation: Change from vapor to liquid3. Freezing: Change from liquid to solid
4. Melting: Change from solid to liquid5. Sublimation: Change from solid directly to vapor without passing the liquidstate.
MASS : It is the absolute quantity of matter in it.m mass, kg
VELOCITY: It is the distance per unit time.
secm
td
v
where:v velocity in m/secd distance in meterst time in sec
ACCELERATION: It is the rate of change of velocity with respect to time.
2secm
dtdv
a
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FORCE : Force is the mass multiplied by the acceleration.
KN 1000maF
Newtonor sec
m-kg maF 2
1 Newton = 1 kg-m/sec 2Newton : Is the force required to accelerate 1 kg mass at the rate of 1 m/secper second
WEIGHT : It is the force due to gravity.
KN1000mg
W
N mgW
Where:g gravitational acceleration, m/sec 2
At standard condition (sea level condition)g = 9.81 m/sec 2
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FORCE OF ATTRACTION : From Newtons Law of Gravitation, the force of attraction between two masses is given by the equation
Newtonr
mGmFg 2
21
Where:m1 and m 2 masses in kgr distance apart in metersG gravitational constant in N-m 2 /kg 2 G = 6.670 x 10 -11 N-m2/kg 2
PROPERTIES OF FLUIDS
DENSITY ( ): It is the mass per unit volume
3mkg
Vm
Where; - density in kg/m 3
m mass in kgV volume in m 3
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SPECIFIC VOLUME ( ): It is the volume per unit mass or the reciprocal of itsdensity.
3mKN
1000g
V1000mg
VW
Where:- specific volume in m 3 /kg
SPECIFIC GRAVITY OR RELATIVE DENSITY1. For liquids it is the ratio of its density to that of water at standard
temperature and pressure.2. For gases it is the ratio of its density to that of either air or hydrogen
at some specified temperature and pressure.
SPECIFIC WEIGHT ( ): It is the weight per unit volume.
kg3m1
mV
Where:- specific weight in KN/m 3
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CONVERSION FORMULAS
32C8.1F
8.132F
C460FR
273CK
KPaormKN
AF
P 2
ABSOLUTE SCALE
PRESSURE : Pressure is defined as the normal component of a force perunit area .
If a force dF acts normally on an infinitesimal area dA, the intensity of pres-sure is equal to
dAdFP
where;P pressure, KPaF- force KNA area, m 21 KPa = 1 KN/m 2 1MPa = 1000 KPa
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PASCALS LAW : At any point in a homogeneous fluid at rest the pressures arethe same in all directions.
x
y
z
A
BC
P 1 A1
P 2 A2
P 3 A3
Fx = 0 and Fy = 0P 1A1 P 3A3 sin = 0 1P 2A2 P 3A3cos = 0 2
From Figure:A1 = A3sin 3A
2= A3cos 4
Eq. 3 to Eq. 1P 1 = P 3Eq. 4 to Eq. 2P 2 = P 3Therefore:
P 1 = P 2 = P 3
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Atmospheric Pressure : It is the absolute pressure exerted by the atmosphere.
At Standard ConditionPa = 101.325 KPa
= 1.033 kg/cm2 = 0.101325MPa
= 1.01325 Bar= 760 mm Hg= 76 cm Hg= 14.7 lb/in 2
= 10.33 m of H 2O= 29.921 in of Hg= 33.88 ft of H 2O
Barometer: an instrument used determine the absolute pressure exerted bythe atmosphere.
ABSOLUTE AND GAGE PRESSUREAbsolute Pressure is the pressure measured referred to absolute zero
and using absolute zero as the base.Gage Pressure is the pressure measured referred to the existing atmospheric
pressure and using atmospheric pressure as the base.
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Atmospheric Pressure
Absolute Zero
P gage
P vacuum
P abs
P abs
P abs = P a P vacuum
P abs = P a + P gage
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VARIATION OF PRESSURE WITH ELEVATION
(P + dP) A
(P) A
dh
W
Fx = 0(P + dP)A PA W = 0PA + dPA PA = WdPA = WW = dVdV = AdhdPA = - Adh
dP = - dh
Note: Negative sign is used becausePressure decreases as elevation in-creases and increases as elevationdecreases
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Manometer: it is a device used in measuring gage pressure in length of someliquid column.
Open Type It has an atmospheric surface and is capablein measuring gage pressure.
Differential Type it has no atmospheric surface and is capablein measuring differences of pressure.
open
Open Type
close
close
Differential Type
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FORMS OF ENERGY
Work: It is the force multiplied by the displacement in the direction of the force.W = Fdx+W indicates that work is done by the system-W indicates that work is done on the system
Heat: It is an energy that crosses a systems boundary because of a temperature difference between the system and the surrounding.+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 it is the total energy that a molecule has.
U total internal energy, KJu specific internal energy, KJ/kg
Flow Energy or Flow Work: It is the work required in pushing a fluid usually into
the system or out from the system.
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System orControl volume
P 1A1
L1
P 2A2
L2
E f1 = F 1L1 = P 1A1L1A1L1 = V1E f1 = P 1V1
E f2 = F 2L2 = P 2A2L2A2L2 = V2E f2 = P 2V2
E f = E f2 E f1
Ef = P 2V2 P 1V1 KJ
E f = (PV)
E f = E f2 E f1
E f = P 2 2 P 1 1 KJ/kg
E f = (P )
Where:P pressure in KPaV volume in m 3
- specific volume in m 3 /kg
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Kinetic Energy: It is the energy or the work required due to the motion of abody or a system.
kgKJ
)1000(2
vvKE
KJ )1000(2
vvmKE
Joules 2
vvmvdvmKE
dt
dxdvmdx
dt
dvmdxmaKE
FdxKE
21
22
21
22
2
1
21
22
2
1
2
1
2
1
2
1
Where:m mass in kgv velocity in m/sec
m mF
x1 2
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Potential Energy: It is the energy or work required by a system by virtue ofits configuration or elevation.
m
m
1
2
Reference Datum Or Datum LIne
Z
kg
KJ ZZ
1000
gPE
KJ ZZ1000mg
PE
Joules ZZmgPE
ZZmgdZmgdZWPE
12
12
12
2
1 12
2
1
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Enthalpy: It a thermodynamic property that is equal to the sumof the internal energy and the flow energy of a substance.
h = U + PV
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2
22
1
11
222111
21
vAvA
vAvA
mm
LAW OF CONSERVATION OF MASS : Mass is indestructible. In applyingthis law we must except nuclear processes during which mass is convertedinto energy.Verbal Form: Mass Entering Mass Leaving = change of mass stored within
the systemEquation Form: m 1 m2 = mFor an Open System (steady state, steady flow system) the m = 0.
m = 0m1 m2
m 1 m 2 = 0m 1 = m 2
For one dimensional flow, the mass rate of flow entering or leaving a system is
AvAvm where:m mass flow rate, kg/secA cross sectional area, m 2 v- velocity, m/sec
- density, kg/m 3
- specific volume, m3
/kg
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PROPERTIES OF PURE SUBSTANCE
a - sub-cooled liquid b - saturated liquid c - saturated mixture d - saturated vapor e - superheated vapor
Considering that the system is heated at constantpressure where P = 101.325 KPa, the 100C is thesaturation temperature corresponding to 101.325 KPa,and 101.325 KPa is the saturation pressure correspon-ding 100 C.
P P PP
P
Q
30C100C
100C 100CT 100C
(a) (b) (c) (d) (e)
Q Q Q Q
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Saturation Temperature (tsat) - is the highest temperature at a given pressure in whichvaporization takes place.Saturation Pressure (Psat) - is the pressure corresponding to the temperature.Sub-cooled Liquid - is one whose temperature is less than the saturation temperaturecorresponding to the pressure.Compressed Liquid - is one whose pressure is greater than the saturation pressurecorresponding 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
30C
100C
t 100C
Saturated Mixture
P = C
Critical Point
T- Diagram
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a
b c de
T
S
F
Saturated Vapor
Saturated Vapor
30C
100C
t 100C
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
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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
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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
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Quality (x):
Lv
v
Lv
v
mmm
mm
mmm
x
Where:mv mass of vapormL mass of liquid
m total massx- qualityThe 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 (liquidand vapor) they can be determined by the equationrc = r f + x(r fg)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 isapproximately equal to the properties at saturated liquid
corresponding the sub-cooled temperature.
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P 1
P 2
1
2
T
S
h = C
T-S Diagram Throttling Process
P1 steam line pressureP2 calorimeter pressure
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KJ )T-(TmC)t-mC(tQ
m,massthegConsiderin
kgKJ
)T-(TC)t-C(tQ
tionintegraby
CdTCdTdQdTdQ
dtdQ
C
1212
1212
Zeroth Law of Thermodynamics:If two bodies are in thermal equilibrium with a third body, they are in thermalequilibrium with each other and hence their temperatures are equal.Specific Heat:
It is the amount of heat required to raise the temperature of a 1 kg mass 1 C or1 K.
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Heat of Transformation: It is the amount of heat that must be transferred whena substance completely undergoes a phase change without a change in
temperature.a. Heat of Vaporization: It is the amount of heat added to vaporize a
liquid or amount of heat removed to condense a vapor or gas.Q = mLwhere: L latent heat of vaporization, KJ/kg
m mass, kg, kg/secb. Heat of Fusion: It is the amount of heat added to melt a solid or the
amount of heat removed to freeze a liquid.Q = mLwhere: L latent heat of fusion, KJ/kg
Sensible Heat: It is the amount of heat that must be transferred (added orremoved) when a substance undergoes a change in temperature withouta change in phase.
Q = mC T = mC t
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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 thesystem
Equation Form:E1 E2 = Es
1. First Corollary of the First Law: Application of first Law to a Closed System
U
Q
W For a Closed System (Non FlowSystem),PV, KE and PE are negligible, thereforethe changeof stored energy Es = U
Q W = U 1Q = U + W 2
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By differentiation:dQ = dU + dW 3
where:
dQ Q2 Q1dW W2 W1
Work of a Closed System (NonFlow)
P
V
W = PdV
P
dV
W = FdxF = PAW = PAdxAdx = dVW = PdVdW = PdVFrom Eq. 3dQ = dU + dWdQ = dU + PdV 4
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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 + P 1V1 + KE 1 + PE 1
U2 + P 2V2 + KE 2 + PE 2
For an Open system (Steady state, Steady Flow system)
Es = 0, thereforeE1 E2 = 0 orE1 = E 2 or Energy Entering = Energy Leaving
Z1
Z2
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U1 + P 1V1 + KE 1 + PE 1 + Q = U 2 + P 2V2 + KE 2 + PE 2 + W 1Q = (U 2 U1) + (P 2V2 P 1V1) + (KE 2 KE1) + (PE 2 PE 1) + W 2Q = U + (PV) + KE + PE + W 3By differentiationdQ = dU + d(PV) + dKE + dPE + dW 4 But dQ Q2 Q1 and dW W2 W1
Enthalpy (h)h = U + PVdh = dU + d(PV) 5 dh = 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 + dW0 = VdP + dKE + dPE + dWdW = -VdP - dKE - dPE 11By IntegrationW = - VdP - KE - PE 12
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If KE = 0 and PE = 0Q = h + W 13W = Q - h 14 W = - VdP 15
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IDEAL OR PERFECT GAS
Prepared By:Rashwinder singh
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1. Ideal Gas Equation of StatePV = mRTP = RT
2T
2 V
2P
1T
1 V
1P
CT
PV
RT
P
Where: P absolute pressure in KPaV volume in m 3 m mass in kgR Gas Constant in KJ/kg- KT absolute temperature in K
IDEAL OR PERFECT GAS
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2. Gas Constant
K-m
kg
KJ 8.3143R
K-kg
KJ
M
RR
Where:
R- Gas Constant in KJ/kg-K
Km
kg
KJ constant gas universalR
M Molecular weight kg/kg m
3. Boyles Law If the temperature of a certain quantity ofgas is held constant the volume V is inver-sely proportional to the absolute pressure P.
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C2
V 2
P1
V 1
P
CPV
P
1
C V
P V
4.Charles Law A. At Constant Pressure (P = C)If the pressure of a certain quantity ofgas is held constant, the volume V is directlyproportional to the temperature T during a qua-sistatic change of state
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2
2
1
1T
V
T
V
CT
V T;C V ;T 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
CT
P
;TCPTP ;
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5. Avogadros Law All 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 ofa 1 kg mass 1 C or 1 K
A. SPECIFIC HEAT AT CONSTANT PRESSURE (Cp) From: dh = dU + PdV + VdPbut dU + VdP = dQ ; therefore
dh = dQ + VdP 1
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but at P = C ; dP = O; thereforedh = dQ 2
and by integrationQ = h 3
considering m,h = m(h 2 - h 1 ) 4
Q = h = m (h 2 - h 1 ) 5From the definition of specific heat, C = dQ/T
Cp = dQ /dt 6
Cp = dh/dT, thendQ = CpdT 7
and by considering m,dQ = mCpdT 8
then by integration
Q = m Cp T 9
but T = (T 2 - T 1 )Q = m Cp (T 2 - T 1 ) 10
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B SPECIFIC HEAT AT CONSTANT VOLUME (Cv) At V = C, dV = O, and from dQ = dU + PdVdV = 0, therefore
dQ = dU 11then by integration
Q = U 12then the specific heat at constant volumeCv is;
Cv = dQ/dT = dU/dT 13
dQ = CvdT 14and by considering m,
dQ = mCvdT 15and by integration
Q = m U 16
Q = mCv T 17Q = m(U 2 - U 1) 18Q = m Cv(T 2 - T 1 ) 19
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From:h = U + P and P = RTh = U + RT 20
and by differentiation, dh = dU + Rdt 21 but dh =CpdT and dU = CvdT,
therefore CpdT = CvdT + RdT 22
and by dividing both sides of theequation by dT,
Cp = Cv + R 23
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7. Ratio Of Specific Heats
k = Cp/Cv 24 k = dh/du 25 k = h/ U 26
From eq. 32, Cp = kCv 27
substituting eq. 27 to eq. 24 Cv = R/k-1 28
From eq. 24, Cv = Cp/k 29
substituting eq. 29 to eq. 24 Cp = Rk/k-1 30
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8. Entropy Change ( S) Entropy is that property of a substance thatdetermines the amount of randomness and disorder
of a substance. If during a process, an amount ofheat is taken and is by divided by the absolutetemperature at which it is taken, the result iscalled the ENTROPY CHANGE .
dS = dQ/T 31 and by integration
S = dQ/T 32 and from eq. 39
dQ = TdS 33
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GAS MIXTURE
Total Mass of a mixture
inn
mm
x ii
immMass Fraction
Total Moles of a mixture
nn
y ii
Mole Fraction
Where:m total mass of a mixturem i mass of a componentn total moles of a mixturen
i moles of a component
xi mass fraction of a componentyi - mole fraction of a component
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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 mixturein KJ/kg- K- universal gas constant in
KJ/kg m- KR
AMAGATS LAW
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AMAGATS LAW The total volume of a mixture V is equal to the volume occupied by eachcomponent at the mixture pressure P and temperature T.
1n1V1
2n2V2
3n3V3
P,T
P = P1
= P2
= P3
T = T 1 = T 2 = T 3
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For the components:
TR
PVn ;
TR
PVn ;
TR
PVn 33
22
11
321
321
321
321
VVVV
P
TR
TR
PV
TR
PV
TR
PV
TR
PV
TRPV
TRPV
TRPV
TRPV
nnnn
The total moles n:
V
Vyi
TRPVTR
PV
y
nn
y
i
i
i
ii
The mole fraction
DALTONS LAW
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The total pressure of a mixture P is equal to the sum of the partial pressurethat each gas would exert at mixture volume V and temperature T.
1n1P1
2n2P2
3n3P3
MIXTURE
n P
T1 = T 2 = T 3 = TV1 = V 2 = V 3 = V
For the mixture
For the components
TR
VPn
TRVPn
TR
VPn
33
22
11
TR
PVn
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321
321
321
321
PPPP
V
TR
TR
VP
TR
VP
TR
VP
TR
PV
TR
VP
TRVP
TRVP
TRPV
nnnn
The total moles n: The mole fraction:
PP
yi
TRPV
TRVP
y
nny
i
i
i
ii
M l l W i ht f i t
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Molecular Weight of a mixture
RRM
MyM ii
MRR
RxR iiGas Constant of a mixture
Specific Heat of a mixture
RCC
CxC
CxC
vp
viiv
piip
Ratio of Specific Heat
uh
C
C
k v
p
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Gravimetric and Volumetric AnalysisGravimetric analysis gives the mass fractions of the components
in the mixture. Volumetric analysis gives the volumetric or molal fractions
of the components in the mixture.
i
i
i
i
i
ii
iii
Mx
Mx
y
MyMy
x
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1. Isobaric Process ( P = C): An Isobaric Process is an internallyreversible constant pressure process.
A. Closed System :(Nonflow)
P
V
21P
dV
Q = U + W 1 any substanceW = PdV 2 any substance
U = m(U 2 - U 1 ) 3 any substanceW = P(V 2 - V 1 ) 4 any substance
Q = h = m(h 2 -h 1 ) 5 any substance
T
S
1
2
dS
TP = C
PROCESSES OF FLUIDS
For Ideal Gas:
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For Ideal Gas:PV = mRTW =mR(T 2 -T 1 ) 5
U = mCv(T
2-T
1) 6
Q = h = mC P (T 2 -T 1 ) 7
Entropy ChangeS = dQ/T 8 any substance
dQ = dhFor Ideal Gasdh = mC PdT
S = dQ/TS = mC P dT/T
S = mC P ln(T 2 /T 1 ) 9B. Open System:Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substance- VdP = 0
Q = h 12
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QW = - KE - PE 13If KE = 0 and PE = 0W = 0 14Q = mC P(T 2 -T 1 ) 15 Ideal Gas
2. Isometric Process (V = C): An Isometric process is internallyreversible constant volume process.
A. Closed System: (Nonflow)
P
V 1
2T
S
T
dS
1
2V = C
b
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Q = U + W 1 any substanceW = PdV at V = C; dV = 0W = 0
Q = U = m(U 2 - U 1 ) 2 any substanceh = m(h 2 -h 1 ) 3 any substance
For Ideal Gas:Q = U = mC v (T 2 -T 1 ) 4
h = mC P(T 2 -T 1 ) 5Entropy Change:
S = dQ/T 6 any substancedQ = dU
dU = mC v dT for ideal gasS = dU/T = mC v dT/TS = mC v ln(T 2 /T 1 ) 6
B Open System:
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B. Open System:Q = h + KE + PE + W 7 any substanceW = - VdP - KE - PE 8 any substance- VdP = -V(P
2-P
1) 9 any substance
Q = U = m(U 2 - U 1 ) 10 any substanceh = m(h 2 -h 1 ) 11 any substance
For Ideal Gas:- VdP = -V(P 2 -P 1 ) = mR(T 1 -T 2 )
Q = U = mC v (T 2 -T 1 ) 12h = mC P(T 2 -T 1 ) 13
If KE = 0 and PE = 0Q = h + W 14 any substance
W = - VdP 15W = - VdP = -V(P 2 -P 1 ) 16 any substanceW = mR(T 1 -T 2 ) 16 ideal gas
h = mC P(T 2 -T 1 ) 17 ideal gas
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3. Isothermal Process(T = C): An Isothermal process is reversibleconstant 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 substance
U = m(U 2 - U 1 ) 3 any substanceFor Ideal Gas:dU = mC v dT; at T = C ; dT = 0
Q = W 4
W = PdV ; at PV = C ;
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W PdV ; at PV C ;P 1 V1 = P 2 V2 = C; P = C/VSubstituting P = C/V to W = PdVW = P
1V
1ln(V
2/V
1) 5
Where (V 2 /V 1 ) = P 1 /P 2 W = P 1 V1 ln(P 1 /P 2 ) 6P 1 V1 = mRT 1 Entropy Change:
dS = dQ/T 7S = dQ/T
dQ = TdS ;at T = CQ = T(S 2 -S 1 )
(S 2 -S 1 ) = S = Q/T 8S = Q/T = W/T 9 For Ideal Gas
B Open System (Steady Flow)
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B. Open System (Steady Flow)Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substance- VdP = -V(P
2-P
1) 12 any substance
h = m(h 2 -h 1 ) 13 any substanceFor Ideal Gas:- VdP = -P 1 V1 ln(P 2 /P 1 ) 14- VdP = P 1 V1 ln(P 1 /P 2 ) 15
P 1 /P 2 = V 2 /V 1 16dh = C PdT; at T = C; dT = 0
h = 0 16If KE = 0 and PE = 0
Q = h + W 17 any substanceW = - VdP = P 1 V1 ln(P 1 /P 2 ) 18For Ideal Gas
h = 0 19Q = W = - VdP = P 1 V1 ln(P 1 /P 2 ) 20
4 Isentropic Process (S = C): An Isentropic Process is an internally
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4. Isentropic Process (S = C): An Isentropic Process is an internallyReversible Adiabatic process in which the entropy remains constant where S = C and PV k = C for an ideal or perfect gas.
For Ideal Gas
1
2
1
1
1
2
1
2
222
1
kkk
k22k1111
k
V
V
P
P
T
T
VPVPand T
VP
T
VP
CPVandCT
PV Using
A Closed System (Nonflow)
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A. Closed System (Nonflow)
T
S
1
2
P
V
1
2
dV
PS = C orPV k = C
Q = U + W 1 any substanceW = PdV 2 any substance
U = m(U 2 - U 1 ) 3 any substanceQ = 0 4W = - U = U = -m(U 2 - U 1 ) 5
For Ideal Gas
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For Ideal GasU = mC V(T 2 -T 1 ) 6From PV k = C, P =C/V k , and substituting P
=C/V k to W = PdV, then by integration,
11
11
1
1
1
211
1
1
21
kk
VP
kk
12
1122
PP
kPdV
P
P
k
mRT
k-1
T-TmRPdV
k
VP-VPPdVW 7
8
9
Q = 0
E t Ch
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Entropy ChangeS = 0
S 1 = S 2
B. Open System (Steady Flow)Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substance
h = m(h 2 -h 1 ) 12 any substance
Q = 0W = - h - KE - PE 13From PV k = C ,V =[C/P] 1/k , substituting V to- VdP, then by integration,
PdVkVdP
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11
11
1
1
1
211
1
1
21
kk
kk12
1122
P
P
k
VkPVdP
P
P
k
kmRT
k-1
T-TkmRVdP
k
VP-VPkVdP 14
15
16
If KE = 0 and PE = 00 = h + W 17 any substanceW = - VdP = - h 18 any substance
h = m(h 2 -h 1 ) 19 any substanceQ = 0
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12P
k
k
kk
12
1122
T-TmChW
P
P
k
VkPW
P
P
k
kmRT
k-1
T-TkmRW
k
VP-VPkPdVkVdPW
11
11
1
1
1
211
1
1
21
20
22
21
23
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P
V
dP
V
Area = - VdP
S = C
5. Polytropic Process ( PVn = C): A Polytropic Process is an
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1
2
1
1
1
2
1
2
2
22
1
nnn
n22
n11
11
n
V
V
P
P
T
T
VPVPand T
VP
T
VP
CPVandCT
PV Using
5. Polytropic Process ( PVn C): A Polytropic Process is aninternally reversible process of an Ideal or Perfect Gas in whichPV n = C, where n stands for any constant.
A. Closed System: (Nonflow)
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y ( )
Q = U + W 1
W = PdV 2U = m(U 2 - U 1 ) 3
Q = mC n (T 2 -T 1 ) 4U = m(U 2 - U 1 ) 5
P
V
1
2
dV
P
PV n = C
T
S
2
1
dS
T
PV n = C
From PV n = C, P =C/V n , and substituting
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gP =C/V n 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-VPPdVW
Entropy ChangedS = dQ/TdQ = mC n dT
S = mC n ln(T 2 /T 1 )
6
8
9
10
B. Open System (Steady Flow)
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Q = h + KE + PE + W 11W = - VdP - KE - PE 12
h = m(h 2 -h 1 ) 13Q = mC n (T 2 -T 1 ) 14dQ = mC n dTW = Q - h - KE - PE 15From PV n = C ,V =[C/P] 1/n , substituting V to
- VdP, then by integration,
n
VP-VPnVdP
PdVnVdP
1122
1
16
1n
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11
11
1
1
211
1
21
nn
n12
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 substance
h = m(h 2 -h 1 ) 21 any substanceh = mC p (T 2 -T 1 )
Q = mC n (T 2 -T 1 ) 22
17
18
1n
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11
11
1
1
211
1
21
nn
n12
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
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p g y , yflow process in which Q = 0; PE = 0; KE = 0; W = 0 and theenthalpy remains constant.
h 1 = h 2 or h = C
Throttling valve
Main steam line
thermometer
Pressure Gauge
Pressure Gauge
To main steam line
Throttling Calorimeter
Irrversible and Paddle Work
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m
W
Q
UWp
Q = U + W - Wp
where: Wp - irreversible or paddle work
Second Law of Thermodynamics:
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Kelvin-Planck statement of the second law:No cyclic process is possible whose sole resultis 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 isequal to the network.
Whenever energy is transferred, the level of energy cannot beconserved and some energy must be permanently reduced to a lowerlevel.
When this is combined with the first law of thermodyna-mics,the law of energy conservation, the statementbecomes:Whenever energy is transferred, energy must be conserved,but thelevel of energy cannot be conserved and some energy must bepermanently reduced to a lower level.
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P T Q
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Pm W
1
2
34
T = C
T = C
S = CS = C
VD
V S
3
21
4
TH
TL
QA
QR
Heat Added (T = C) QA = T H( S) 1
Heat Rejected (T = C) Q
R= T
L( S) 2
S = S 2 - S 1 = S 4 - S 3 3 Net Work
W = Q = Q A - Q R 4 W = (T H - T L)( S) 5
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2. Carnot Refrigerator: Reversed Carnot CycleP
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Processes: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)
QR
QA
1
2
4
3
S
T
TH
TL
Heat Added (T=C) Q T ( S) 1
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QA = T L( S) 1 Heat Rejected (T=C)
QR = T H( S) 2 S = S 1 - S 4 = S 2 - S 3 3
Net Work W = Q 4
W = Q R - Q A 5
W = (T H - T L)( S) 6Coefficient of Performance
LH
L
A
TTTCOP
W
QCOP 7
8
LT
9
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1H
L
TCOP 9
Tons of Refrigeration211 KJ/min = 1 TR
3. Carnot Heat Pump:A heat pump uses the samecomponents as the refrigerator but its pur-pose is to reject heat at high energy level.
Performance Factor:
AR
R
R
QPF
W
QPF 10
11
TH
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1
1
1
COPPF
T
TPF
Q
QPF
TTPF
L
H
A
R
LH
H 12
13
14
15
Carnot Refrigerator
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TH
TL
W
QA
QR
R
Vapor Power Cycle
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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
QAQR
W t
1
2
34
Major Components of a Rankine Cycle
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j p y1. 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 superheatedvapor whence at this point it enters the turbine.
2. Steam Turbine: A steady state, steady flow device where steamexpands isentropically to a lower pressure converting some formsof energy (h, KE, PE) to mechanical work that finally be convertedinto 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 thesaturated 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
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condensate leaving the condenser at lower pressure be pumpedback to the boiler in an isentropic process in order to raise thepressure 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 Work ) Id l C l
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a) Ideal CycleW t = (h 1 - h 2) KJ/kgW t = m s(h1 - h 2) KW
b) Actual CycleW t = (h 1 - h 2) KJ/kgW t = m s(h1 - h 2) KW
where: m s - steam flow rate in kg/sec
Turbine Efficiency
100%xhhhh
100%xWW
t
21
2'1t
t
t'
Pump Work ) Id l C l
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a) Ideal CycleWP = (h 4 - h 3) KJ/kgWP = m s(h4 - h 3) KW
b) Actual CycleWP = (h 4 - h 3) KJ/kgWP = m s(h4 - h 3) KW
Pump Efficiency
100%xhh
hh
100%xWW
34'
34
p
p'
pp
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Heat Rejecteda) Ideal Cycle
QR = (h 2 - h 3) KJ/kgQR = m s(h2 - h 3) KWQR = m s(h2 - h 3) KW = mwCpw(two - twi) KW
b) Actual Cycle
QR = (h 2 - h 3) KJ/kgQR = m s(h2 - h 3) KW = m wCpw(two - twi) KW
Where: m w - cooling water flow rate in kg/sectwi - inlet temperature of cooling water in C
two - outlet temperature of cooling water in CCpw - specific heat of water in KJ/kg- C or KJ/kg- KCpw = 4.187 KJ/kg- C or KJ/kg- K
Heat Added:a) Ideal Cycle
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a) Ideal CycleQA = (h 1 - h 4) KJ/kgQA = m s (h1 - h 4) KW
b) Actual CycleQA = (h 1 - h 4) KJ/kgQA = m s (h1 - h 4) KW
Steam Generator or boiler Efficiency
100%x(HV)m
)h(hm
100%xQ
Q
f
41sB
S
AB
Where: Q A - heat absorbed by boiler in KWQS - heat supplied in KWm f - fuel consumption in kg/secHV - heating value of fuel in KJ/kg
Steam Rate
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KW-seckg
ProducedKW
rateFlowSteamSR
Heat Rate
KW-secKJ
ProducedKWSuppliedHeat
HR
Reheat CycleA 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 itsinitial temperature and expands finally in the turbine to the con-
denser pressure.
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Reheater
QA
WP
QR
W t
1 kg
1 23
4
56
Regenerative CycleIn 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 fromthe condenser or drains from the previous heater causing a decrease inheat absorbed by steam in the boiler which result to an increase in
thermal efficiency of the cycle.
W1 kg
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QA
W P1
QR
W t1
2
3
456
7
W P2
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 heatingto an open or closed type heaters after its partial expansion in theturbine. It will result to a further increase in thermal efficiency of the
plant.
1-m
1-m
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QA
WP1
QR
W t
1 kg
1
2
4
5678
WP2
m
23
For a 1 kg basis of circulating steam, m is the fraction of steamextracted for feed-water heating as shown on the schematic diagramabove where the reheat and bled steam pressure are the same