Applied thermodynamics

APPLIED THERMODYNAMICS ME 320
INDEX
th
0 law of thermodynamics . 2 COP................................7 jet-propulsion cycle .....12, 13 refrigeration efficiency........ 7
1st law EER................................7 joule.....................................2 resistance
power version ................ 5 thermal ...........................7 k specific heat ratio.............7 contact.......................... 15
unit-mass version........... 6 emissivity ..........................18 k thermal conductivity......14 thermal ................... 15, 17
1st law of thermodynamics. 2, energy..................................2 Kelvin-Planck statement .....2 resistive wire ..................... 17
5 gain/loss .....................5, 6 kinetic energy ......................2 reversible process.............. 20
2nd law of thermodynamics. 2 internal .....................2, 20 latent energy......................20 Rt thermal resistance......... 15
adiabatic ......................10, 20 kinetic.............................2 Lc characteristic length .....17 R-value .............................. 15
air standard assumptions....14 latent...............................2 liter ......................................2 s entropy....................... 3, 20
area potential .........................2 lumped system analysis.....18 s entropy per unit mass....... 9
sphere ...........................19 sensible...........................2 m mass flow rate.................6 S total entropy .................... 8
atmospheric pressure .......... 4 enthalpy .....................3, 6, 20 magnitude..........................19 saturated liquid.................... 3
b time constant..................18 enthalpy of vaporization ......3 mass flow rate .....................6 saturated liquid/vapor
Bernoili equation ................ 6 entropy.................3, 9, 10, 20 mass to volume relationship 4 mixture .............................. 3
Bi Biot number .................17 total ................................8 mean effective pressure .....14 saturated state...................... 4
Biot number.......................17 entropy balance ...................8 MEP mean effective saturated vapor .................... 3
blackbody ..........................18 entropy generation...........8, 9 pressure ...........................14 sensible energy.................. 20
Boltzmann relation ............. 9 entropy in solids ..................9 methalpy..............................6 series ................................. 19
boundary work...............5, 14 entropy per unit mass ..........9 mixture chamber..................5 Sgen entropy generation....... 8
Btu ...................................... 2 Euler's equation .................19 moving boundary ..............14 sgen entropy generation per
Carnot cycle.......................10 extensive..............................8 net work ............................10 unit mass ........................... 9
characteristic length...........17 extensive properties.............3 newton.................................2 SI units ................................ 2
chemical energy.................20 force ....................................2 nozzle ..............................6, 9 single-stream ....................... 5
Clausius inequality ............. 8 Fourier’s law of heat nuclear energy ...................20 specific enthalpy ....... 3, 6, 20
Clausius statement .............. 2 conduction .......................16 open system.........................5 specific heat ........................ 7
closed system...................4, 5 g heat generation ..............17 Otto cycle ..........................11 specific heat ratio ................ 7
complex conjugate.............19 gas constant .....................3, 4 p thermodynamic probability specific internal energy ....... 3
complex numbers...............19 gas-turbine...................12, 13 ...........................................9 specific properties ............... 3
composite wall...................17 general math ......................19 Pa ........................................2 specific volume ................... 3
compressed liquid............... 3 glossary .............................20 pascal.............................2, 20 sphere ................................ 19
compression ratio...............13 H enthalpy ....................6, 20 path function ...............14, 20 state postulate...................... 2
compressor.......................6, 9 h specific enthalpy 3, 5, 6, 20 phase .................................19 steady flow system .......... 5, 6
conduction ...................14, 16 hconv convection heat transfer phases of water....................3 steady flow work............... 10
conductivity coefficient........................15 phasor notation ..................19 steam-turbine..................... 12
thermal .........................14 heat capacity......................15 pipe......................................9 Stephan-Boltzmann law .... 18
conjugate heat dissipation..................17 heat transfer..................16 superheated vapor ............... 3
complex........................19 heat engine ........................20 thermal resistance.........15 surface temperature ........... 17
conservation of energy.....2, 5 heat engines .......................10 piston-cylinder.....................5 system ................................. 4
constants ............................. 4 heat exchanger.................5, 9 point function ....................20 temperature ......................... 2
contact resistance...............15 heat flux.............................17 polytropic process .............13 temperature gradient ......... 16
control volume.................... 4 heat generation ..................17 potential energy...................2 thermal conductivity ......... 14
convection .............14, 15, 16 heat pump efficiency ...........7 pressure ...............................2 thermal diffusivity............. 15
convection heat transfer heat transfer ...............2, 4, 14 properties of saturated phases thermal efficiency ............... 7
coefficient ........................15 pipe...............................16 ...........................................3 carnot ........................... 10
COP coefficient of heat transfer limit.................8 propulsive efficiency ...12, 13 diesel............................ 11
performance...................... 7 heat transfer rate ................16 pure substance ...................20 otto............................... 11
Cp specific heat .................. 7 hfg enthalpy of vaporization 3 q heat flux.........................17 Rankin.......................... 12
critical point...................4, 20 horsepower ..........................2 Q heat transfer rate ...........16 thermal equilibrium............. 2
critical radius .....................17 hrad radiation heat transfer QL heat transfer...................7 thermal radiation ......... 14, 16
cutoff ratio .........................13 coefficient........................16 quality .............................3, 4 thermal resistance........ 15, 17
Cv specific heat .................. 7 hyperbolic functions ..........19 r compression ratio...........13 thermal time constant ........ 18
cylinder conduction ...........15 ideal gas equation ................4 R gas constant.................3, 4 thermodynamic probability . 9
density ................................ 2 increase of entropy principle9 R thermal resistance .........15 thermodynamic properties... 3
Diesel cycle .......................11 independent property ...........2 radiation ......................14, 16 time constant ..................... 18
diffuser ............................6, 9 intensive properties..............3 radiation heat transfer total entropy ........................ 8
dissipation..........................17 internal energy...................20 coefficient........................16 transfer phenomena ........... 14
duct ..................................5, 9 international system of units2 Rankin cycle......................12 turbine ............................. 6, 9
e specific energy ................ 3 irreversible process............20 rate of entropy generation....9 turbine engine.............. 12, 13
EER energy efficiency rating isentropic .....................10, 20 Rc contact resistance.........15 u internal energy........... 3, 20
.......................................... 7 isentropic relations ............13 rc cutoff ratio ....................13 unit-mass relation................ 6
efficiency isothermal..........................20 rcr critical radius ...............17 units..................................... 2

Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 1 of 1

SI 2 watt......................................2 wrev steady flow work .......10 ε emissivity ...................... 18
universal gas constant......... 4 Wb boundary work ........5, 14 x quality..........................3, 4 ηth thermal efficiency ......... 7
V average flow velocity..5, 6 wire ∆E energy ν specific volume ........... 3, 4
volume resistive ........................17 gain/loss .....................5, 6 θ methalpy.......................... 6
sphere ...........................19 wnet net work.....................10 ∆s change in entropy per unit ρCp heat capacity ............. 15
volume to mass relationship 4 Wnet,in net work input ..........7 mass...................................9
W minimum power Wnet,out net work ................10 ∆S entropy change..............9
requirement....................... 7 work ..............................2, 14 α thermal diffusivity ........15

BASIC THERMODYNAMICS
0TH LAW OF THERMODYNAMICS UNITS
Two bodies which are each in thermal equilibrium with Energy, work, heat transfer: [J]
a third body are in thermal equilibrium with each other. 2
J (joule) = N ·m = V ·C = W ·s = AV ·s = F ·V 2 = C
·
F
1ST LAW OF THERMODYNAMICS 1 kJ = 0.94782 Btu 1 Btu = 1.055056 kJ
The Conservation of Energy Principle Rate of energy, work or heat transfer: [J/s or W]
The amount of energy gained by a system is equal to 2
W (watt) = J = N ·m = C ·V = V · A = F ·V = 1 HP
the amount of energy lost by the surroundings. ss s s 746
2 2
Pressure: [Pa or N/m or kg/ms ]
2ND LAW OF THERMODYNAMICS Pa (pascal) = N = kg = J = W ·s = 1.45038×10-4 psi
Processes occur in a certain direction and energy has m 2 m·s 2 m3 m3
quality as well as quantity. For example, heat flows
Density: [kg/m3]
from a high temperature place to a low temperature
place, not the reverse. Another example, electricity Force: [N or kg·m/s2]
flowing through a resistive wire generates heat, but
N (newton) = kg ·m = J = C ·V = W ·s
heating a resistive wire does not generate electricity. s2 m m m
Kelvin-Planck statement: It is impossible for any device
Temperature: [°C or K] 0°C = 273.15K
that operates on a cycle to receive heat from a single
reservoir and produce a net amount of work. Volume: [m3] = 1000 liters
Clausius statement: It is impossible to construct a device
Note: In this class, we typically use units of KJ, KPa, and
that operates in a cycle and produces no effect other than
KW.
the transfer of heat from a lower temperature body to a
higher temperature body.
SI UNITS, International System of Units
STATE POSTULATE Length meter Temperature: kelvin
The state of a simple compressible system is Mass: kilogram Amount: mole
completely specified by two independent, intensive Time: second Light intensity: candela.
properties. Two properties are independent if one Electric current: ampere
property can be varied while the other one is held
constant. Properties are intensive if they do not ENERGY [J]
depend on size, e.g. the properties of temperature,
pressure, entropy, density, specific volume. 1 2
Kinetic energy KE = mv
2
Potential energy PE = mgz
Total energy of the system E = U + KE + PE
U = internal energy, i.e. sensible energy (translational,
rotational, vibrational), latent energy (atomic structure,
melting ice), chemical energy (bonding, separating
water into hydrogen & oxygen), nuclear.


THERMODYNAMIC PROPERTIES PROPERTIES OF WATER
Thermodynamic properties are related to the energy of the Compressed liquid: Properties for compressed
system, i.e. temperature, pressure, mass, volume. liquid are insensitive to pressure. For a given
Extensive properties depend on the size or extent of the temperature use the f subscripted values from tables
system, e.g. volume, mass, total energy. A-4 and A-5, e.g. v ≈ v f , u ≈ u f , etc. However, in the
Intensive properties are independent of size, e.g.
temperature, pressure, entropy, density, specific volume. case of enthalpy, h ≈ h f + ( P − Psat ) v .

Saturated phases: Properties for the saturated
SPECIFIC PROPERTIES phases of water are determined using tables A-4 and
A-5 in the back of the book and the formulas below.
Extensive properties per unit mass are called specific Note that the fg subscript stands for the difference
properties. between the g subscripted quantity and the f
V subscripted quantity, e.g. ug-uf = ufg, and is provided
Specific volume v = [m3/kg]* for convenience.
m
Specific volume v = (1 − x ) v f + xvg [m3/kg]
E
Specific energy e = [kJ/kg]
m Internal energy u = (1 − x ) u f + xu g = u f + xu fg [kJ/kg]

Specific internal energy u =
U
[kJ/kg] Enthalpy h = (1 − x ) h f + xhg = h f + xh fg [kJ/kg]
m
*We have to be careful with the units for specific volume. By Entropy s = (1 − x ) s f + xsg = s f + xs fg [kJ/(kg·K)]
convention, we deal in units of kJ, kW, and kPa for many
values. When specific volume or volume is included in an v − vf
Quality x= [no units]
equation, there is often a factor of 1000 involved. vg − v f

R GAS CONSTANT [kJ/(kg·K)] Superheated vapor: Properties for superheated
vapor are read directly from table A-6 in the back of
the book.
R = C p − Cv
Cp = specific heat at constant pressure [kJ/(kg·°C)]
Cv = specific heat at constant volume [kJ/(kg·°C)] PHASES OF WATER
The different states in which water exists are its
R Gas Constant of Selected Materials @300K [kJ/(kg·°C)]
phases. We are only concerned with the liquid and
Air 0.2870 Carbon monoxide 0.2968 Methane 0.5182
vapor states.
Argon 0.2081 Chlorine 0.1173 Neon 0.4119
Butane 0.1433 Helium 2.0769 Nitrogen 0.2968 compressed liquid – purely liquid, at less than saturation
Carbon dioxide 0.1889 Hydrogen 4.1240 Oxygen 0.2598 temperature (boiling point at pressure), v < vf
saturated liquid – purely liquid, but at the saturation
temperature (any additional heat will cause some
vaporization), v = vf
saturated liquid/vapor mixture – a mixture of liquid and
vapor at the temperature (and pressure) of saturation,
v f < v < vg
saturated vapor – purely vapor, but at the saturation
temperature (any loss of heat will cause some
condensation to occur), v = vg
superheated vapor – purely vapor, above the saturation
temperature, v > vg

ENTHALPY OF VAPORIZATION [hfg]
The amount of energy needed to vaporize a unit mass
of saturated liquid at a given temperature or pressure.
hfg = enthalpy of vaporization [kJ/kg]


x QUALITY CONSTANTS
The quality is a value from 0 to 1 describing the ratio Atmospheric pressure: 101.33 kPa
of vapor mass to total mass of a pure substance. It is
only applicable at saturation temperature. A quality of Boltzmann constant: 1.380658×10-23 kJ/(kmol·K)
0 denotes a saturated liquid and a quality of 1 denotes Critical point, water 22 Mpa, 374°C
saturated vapor. Gas constant R = Ru/M where M is molecular weight
(R = 287 J/(kg·K) for air)
mg
x= Temperature in Kelvin: °C + 273.15
mg + m f Universal gas constant: Ru = 8.314 kJ/(kmol·K)
mg = mass of the gas [kg]
mf = mass of the fluid (liquid) [kg]
ENERGY TRANSFER [kJ]
Whether energy transfer is described as heat or work
VOLUME TO MASS RELATIONSHIP can be a function of the location of the system
boundary. The system boundary may be drawn to
For the saturated state. suite the problem. The area enclosed is also referred
to as the control volume.
V = V f + Vg = m f v f + mg vg
System System
V = total volume [m3]
Vf = volume of the fluid (liquid) [m3]
Vg = volume of the gas [m3]
mf = mass of the fluid (liquid) [kg]
heat transfer work
vf = volume density of the fluid (liquid) [m3/kg]
mg = mass of the gas [kg]
vg = volume density of the gas [m3/kg] I I

IDEAL GAS EQUATION
The ideal gas formula assumes no intermolecular
forces are involved. The ideal gas formula may be
used as an approximation for the properties of gases
which are a high temperatures/low pressures well out
of range of their saturation (liquification) values, e.g.
air at room temperature and atmospheric pressure
can be considered an ideal gas.. Don't use this
formula for steam, especially near saturation; use the
water property tables.
Pν = RT or PV = mRT
PV1 PV2
In a closed system, m and R are constant, so
1
= 2
T1 T2
P = pressure [kPa]
ν = V/m specific volume [m3/kg]
V = volume [m3]
m = mass [kg]
R = gas constant (0.287 for air) [kJ/(kg·K)]
T = absolute temperature [K] (°C + 273.15)


1st LAW OF THERMODYNAMICS [kJ] 1st LAW, POWER VERSION [kW]
Used in piston-cylinder problems. for open systems
System +
W Differentiation of the 1st Law of Thermodynamics with
respect to time yields the power version. Used for
∆E mixture chamber, heat exchanger, heater in a duct
problems.
Q Q − W = m ( ∆h + ∆ke + ∆pe )
& & &
+

Open Systems: Q − W = ∆E = ∆H + ∆KE + ∆PE
& & &⎡ V 2 − V1 2 g ( z2 − z1 ) ⎤
Q − W = m ⎢ h2 − h1 + 2 + ⎥
where ∆H = m ( h2 − h1 ) = mC p , avg (T2 − T1 ) ⎣ 2 (1000 ) 1000 ⎦

Closed Systems: Q − W = ∆E = ∆U + ∆KE + ∆PE where h2 − h1 = C p , avg (T2 − T1 )
where ∆U = m ( u2 − u1 ) = mCv , avg (T2 − T1 ) &
Q = net heat transfer per unit time across system
(Closed System means that mass does not enter or leave boundaries, positive when flowing inward [kW or kJ/s]
the system.) W& = net work done per unit time in all forms, positive when
m ( V2 − V1
2 2
), flowing outward [kW or kJ/s]
∆KE = ∆PE = mg ( z2 − z1 ) &
m = mass flow rate through the control volume [kg/s]
2000 Note that to obtain this value, typically the ideal gas
NOTE: Since the piston-cylinder is a closed system, we equation (p4) and the mass flow rate (p6) formulas will
normally use the Closed System version of the law. An be used.
exception occurs when the piston is allowed to move as ∆h = net change in enthalpy [kJ/kg]
the gas expands under constant pressure. In this case, ∆ke = net change in the kinetic energy per unit mass [kJ/kg]
there is boundary work Wb, which can be included on the ∆pe = net change in the potential energy per unit mass
right-hand side of the equation by using the Open
[kJ/kg]
Systems version since ∆U + Wb = ∆H.
V = average flow velocity (note 1 kJ/kg = 1000 m2/s2) [m/s]
Q = net heat transfer across system boundaries, positive g = acceleration of gravity 9.807 m/s2
when flowing inward [kJ] z = elevation to some reference point [m]
W = net work done in all forms, positive when flowing Cp,avg = specific heat at constant pressure, averaged for the
outward [kJ] two temperatures [kJ/(kg·°C)]
∆E = net change in the total energy of the system [kJ] T1, T2 = temperature of the fluid at the inlet and outlet
∆U = net change in the internal energy of the system [kJ] respectively [°C or K]
∆KE = net change in the kinetic energy of the system [kJ]
∆PE = net change in the potential energy of the system [kJ]
m = mass [kg]
u = internal energy [kJ/kg]
h = enthalpy [kJ/kg]
Cp,avg = specific heat at constant pressure, averaged for the
two temperatures [kJ/(kg·°C)]
Cv,avg = specific heat at constant volume, averaged for the
two temperatures [kJ/(kg·°C)]


1st LAW, UNIT-MASS VERSION [kJ/kg] h SPECIFIC ENTHALPY [kJ/kg]
st
The division of the power version of the 1 Law of The per unit mass version of enthalpy (see previous)
Thermodynamics equation by the flow rate yields the and often referred to as simply enthalpy, the exact
unit-mass version. Used in nozzle, diffuser, turbine, meaning to be determined from context.
and compressor problems.
∆h = ∆u + ν∆P incompressible substance
Open Systems: q − w = ∆h + ∆ke + ∆pe
∆h = ∫ C p ( T ) dT
2

g ( z2 − z1 ) C p , avg ∆T ideal gas
V − V1 2 2
q − w = h2 − h1 + 2 +
1

2 (1000 ) 1000 h = u + RT ideal gas

where h2 − h1 = C p , avg (T2 − T1 ) u = internal energy [kJ/kg]
Closed Systems: q − w = ∆u + ∆ke + ∆pe P = pressure [kPa]
Cp,avg = specific heat at constant pressure, averaged for the
V2 2 − V1 2 g ( z2 − z1 ) two temperatures [kJ/(kg·°C)]
q − w = u2 − u1 + +
2 (1000 ) 1000 T = absolute temperature [K] (°C + 273.15)
R = gas constant (287 for air) [J/(kg·K)]
where u2 − u1 = Cv , avg (T2 − T1 )
q = heat transfer per unit mass [kJ/kg] θ METHALPY [kJ/kg]
w = work done per unit mass [kJ/kg]
Methalphy means "beyond enthalpy". The factor of
see also BERNOULI EQUATION next. 1000 is used to convert m2/s2 to kJ/kg.
V2
θ = h + ke + pe = h + + gz
BERNOULI EQUATION 2 ×1000
For the steady flow of liquid through a device that
involves no work interactions (such as a nozzle or
pipe section), the work term is zero and we have the &
m MASS FLOW RATE [kg/s]
expression known as the Bernouli equation. The rate of flow in terms of mass.
V2 − V1
2
g ( z2 − z1 )
2
1
0 = υ ( P2 − P ) + + m=
& V1 A1
2 (1000 ) ν
1
1000
ν = V/m specific volume [m3/kg] ν = V/m specific volume [m3/kg]
P = pressure [kPa] V = average flow velocity (note 1 kJ/kg = 1000 m2/s2) [m/s]
V = average flow velocity (note 1 kJ/kg = 1000 m2/s2) [m/s] A = cross-sectional area [m2]
g = acceleration of gravity 9.807 m/s2
z = elevation to some reference point [m]

H ENTHALPY [kJ]
The sum of the internal energy and the volume-
pressure product. If a body is heated without
changing its volume or pressure, then the change in
enthalpy will equal the heat transfer. We see this
more often in its per unit mass form (see next) called
specific enthalpy but still referred to as enthalpy.
H = U + PV
U = internal energy [kJ]
P = pressure [kPa]
V = volume [m3]


Cp, Cv SPECIFIC HEAT [kJ/(kg·°C)] COP COEFFICIENT OF PERFORMANCE
Describes the energy storage capability of a material. A unitless value describing the efficiency of a
The energy required to raise the temperature of a unit refrigerator, of a heat pump.
of mass of a substance by one degree under constant & &
QL QH
pressure (Cp), or under constant volume (Cv). This COPRefrig. = COPH.P. =
can be confusing since Cp can be used in problems &
Wnet,in &
Wnet,in
involving changing pressure and Cv can be used in
problems involving changing volume. Note that Cp is QL QH
used in calculations involving open systems, and Cv COPRefrig. = COPH.P. =
is used for closed systems. Cp > Cv because at
Q H − QL Q H − QL
constant pressure, the system is allowed to expand
Maximum possible COP for a refrigerator, for a heat pump:
when heated, requiring additional energy. The values
for specific heat increase slightly with increased 1 1
temperature. COPRefrig. = COPH.P. =
TH / TL − 1 1 − TL / TH
C p = Cv + R
&
QL = heat transfer [kW]
u2 − u1 = Cv (T2 − T1 ) h2 − h1 = C p (T2 − T1 ) & & &
Wnet,in = QH − QL = net work input [kW]
R = gas constant (287 for air) [J/(kg·K)] QH = heat transfer from a high temperature source [kJ]
u = internal energy [kJ/kg] QL = heat transfer from a low temperature source [kJ]
h = enthalpy [kJ/kg] TH = temperature of high-temperature source [K]
Example: The Cp of water at room temperature is 4.18 TL = temperature of low-temperature source [K]
kJ/(kg·°C), for iron it's 0.45 kJ/(kg·°C). Therefore it takes
about nine times as much energy to heat water as it does to
heat iron. EER ENERGY EFFICIENCY RATING
specific heat × mass × ∆temp = energy An efficiency rating system used in the United States.
The amount of heat removed in Btu’s for 1 Wh of
Cp Specific Heat of Selected Materials @300K [kJ/(kg·°C)] electricity consumed. Since 1 Wh = 3.412 Btu, this
Air 1.005 Concrete 0.653 Iron 0.45 works out to:
Aluminum
Brass
0.902
0.400
Copper
Glass
0.386
0.800
Steel
Wood, hard
0.500
1.26
EER = 3.412 COPRefrig.
Wh = watt-hour, a unit of electrical energy
Btu = British thermal unit, a unit of thermal energy
k SPECIFIC HEAT RATIO [no units] COPRefrig. = coefficient of performance for the refrigeration
An ideal gas property that varies slightly with cycle, an efficiency rating [no units]
temperature. For monatomic gases, the value is
essentially constant at 1.667; most diatomic gases,
including air have a specific heat ratio of about 1.4 at &
W MINIMUM POWER REQUIREMENT
room temperature. [kW]
Cp = specific heat at constant pressure C The amount of power required to operate a heat
[kJ/(kg·°C)] k= p pump/refrigerator.
Cv = specific heat at constant volume Cv
[kJ/(kg·°C)] &
Q
&
W= L
COP
ηth THERMAL EFFICIENCY &
QL = heat transfer [kW]
The efficiency of a heat engine. The fraction of the & & &
W = QH − QL = net work input [kW]
heat input that is converted to net work output.
Wnet,out Qout wnet
ηth = = 1− = = 1 − r1− k
Qin Qin qH
Wnet,out = QH - QL = net work output [kW]
Qin = heat input [kJ]
Qout = heat output [kJ]


HEAT TRANSFER LIMIT [kW] Sgen ENTROPY GENERATION [kJ/K]
This expression is an equality for a reversible cycle The entropy change of a closed system during an
engine (a theoretical device not realizable in practice). irreversible process is always greater than the entropy
QH TH transfer. The entropy generation is the entropy
≤ created within the system boundaries due to
QL TL irreversibilities. Note that it may be necessary to
extend the boundaries of a system in order to
QH = magnitude of the heat transferred from a high-
consider it a closed system.
temperature source [kJ]
QL = magnitude of the heat transferred to a low- 2 δQ
temperature source [kJ] S2 − S1 = ∫ + Sgen
1 3 { {
2 1 T
TH = temperature of high-temperature source [K] Entropy
Entropy
Entropy
change of generation
TL = temperature of low-temperature source [K] the system transfer within the
with heat system

S1 = initial entropy [kJ/K]
CLAUSIUS INEQUALITY S2 = final entropy [kJ/K]
The cyclic integral of the change in heat transfer δQ = the change in heat transfer [kJ]
divided by the absolute temperature is always less T = absolute temperature at the inside surface of the
than or equal to zero. system boundary [K]
δQ
∫ T
≤0
ENTROPY BALANCE FOR
CONTROL VOLUMES
∫ = the integration is to be performed over a full cycle
For a control volume, we must consider mass flow
δQ = the change in heat transfer [kJ] across the control volume boundary
T = absolute temperature at the inside surface of the
dSCV &
Qk
system boundary [K]
dt
= ∑ T + ∑ mi si − ∑ me se + {
&
1442443
& &
Sgen,CV
{ { k
Rate of entropy Rate of entropy
Rate of entropy Rate of entropy transport with generation
S TOTAL ENTROPY [kJ/K] change of CV transfer with heat mass within CV

The term entropy is used both for the total entropy
and the entropy per unit mass s [kJ/(kg·K)]. Entropy is SCV = entropy within the control volume [kJ/K]
an intensive property of a system (does not depend &
mi = inlet mass flow rate [kg/s]
on size).
&
me = exit mass flow rate [kg/s]
⎛ δQ ⎞
dS = ⎜ ⎟
si = inlet entropy [kJ/(kg·K)]
⎝ T ⎠ se = exit entropy [kJ/(kg·K)]
δQ = the change in heat transfer [kJ] &
Qk = rate of heat transfer through the boundary at internal
T = absolute temperature at the inside surface of the boundary temperature Tk [kJ/s]
system boundary [K]
Tk = absolute temperature at the inside surface of the
system boundary [K]


&
Sgen RATE OF ENTROPY GENERATION s ENTROPY PER UNIT MASS [kJ/(kg·K)]
[kW/K] Entropy change is caused by heat flow, mass flow,
and irreversibilities. Irreversibilities always increase
Most steady-state processes such as turbines, entropy.
compressors, nozzles, diffusers, heat exchangers,
pipes, and ducts, experience no change in their s2 = s1 Isentropic process
entropy. This gives the relation:
T2
& s2 − s1 = Cavg ln incompressible substances
Q
&
Sgen = ∑ me se − ∑ mi si − ∑ k
& & T1
Tk T2 ν
∆s = Cv ,avg ln + R ln 2 ideal gas
For a single-stream (one inlet, one exit) steady-flow device: T1 ν1
Q& T2 P
Sgen = m ( se − si ) − ∑ k
& & ∆s = C p ,avg ln − R ln 2 ideal gas
Tk T1 P1

&
mi = inlet mass flow rate [kg/s] R = gas constant (287 for air) [J/(kg·K)]
&
me = exit mass flow rate [kg/s] Cavg = the specific heat at average temperature [kJ/(kg·K)]
Cp = specific heat at constant pressure [kJ/(kg·°C)]
Cv = specific heat at constant volume [kJ/(kg·°C)]
se = exit entropy [kJ/(kg·K)] T1, T2 = initial and final temperatures [K]
&
Qk = rate of heat transfer through the boundary at P1, P2 = initial and final pressure [Pa]
temperature Tk [kJ/s]
Tk = absolute temperature at the system boundary [K] sgen ENTROPY GENERATION PER UNIT
MASS [kJ/(kg·K)]
INCREASE OF ENTROPY PRINCIPLE Applies to a single-stream, steady-flow device such as
The entropy of an isolated system during a process a turbine or compressor.
always increases or, in the limiting case of a q
reversible process, remains constant. sgen = ( se − si ) − ∑
T
∆Sisolated ≥ 0
+ se = exit entropy [kJ/(kg·K)]
System W There is no entropy
transfer with work. q = heat transfer per unit mass [kJ/kg]
T = absolute temperature at the inside surface of the
∆E system boundary [K]
Heat transfer is
+ Q accompanied by
entropy transfer.
p THERMODYNAMIC PROBABILITY
Molecular randomness or uncertainty. The
thermodynamic probability is the number of possible
microscopic states for each state of macroscopic
equilibrium of a system. It is related to the entropy
(disorder) of the system by the Boltzmann relation:
S = k ln p ⇒ p = eS / k
S = entropy [kJ/K]
k = Boltzmann constant 1.3806×10-23 [kJ/(kmol·K)]


wrev STEADY-FLOW WORK [kJ/kg] CARNOT CYCLE
One needs to know ν as a function of P in order to Introduced in 1824 by French engineer Sadi Carnot,
perform the integration, but when the working fluid is the Carnot cycle is a combination of four reversible
an incompressible fluid, the specific volume ν remains processes that are the basis for the theoretical Carnot
constant during the process and can be taken out of heat engine. The cycle forms a rectangle on the T-s
the integration. For the steady flow of a liquid through plot. Use Cv for specific heat.
a device that involves no work (such as nozzle or a The numbered corners T 1-2 • Volume expansion
• Reversible isothermal process
pipe section), the work term is zero represent the four QH • Heat is added from a
states. high temperature area.
2 qin
wrev = − ∫ ν dP − ∆ke − ∆pe The area enclosed is TH
equal to the work.
1 2
1

wrev = ν ( P − P2 ) − ∆ke − ∆pe
wnet 2-3 • Volume expansion
• Reversible adiabatic
Temperature isentropic process
1
PV k = a constant
TL
ν = V/m specific volume [m3/kg] 4 qout 3
4-1 3-4 • Volume compression
• Reversible isothermal process
• Volume compression QL • Heat sinks to a low
HEAT ENGINES • Reversible adiabatic,
isentropic process
temperature area.

PV k = a constant
s1 Entropy s2 s
HEAT ENGINES
The conversion of heat to work requires the use of
special devices; these are called heat engines and P constant s

have the following characteristics: qin
constant T 1
• They receive heat from a high-temperature source.
2
• They convert part of this heat to work. Pressure

• They reject the remaining waste heat to a low- 4
temperature sink such as the atmosphere or a body of qout 3
water.
• They operate on a cycle.
Volume v
• They usually involve a fluid used in the transfer of heat;
TL
this is called the working fluid.
Thermal efficiency: ηth,Carnot = 1 −
TH
Wnet,out NET WORK [kJ] Heat transfer and work:
The work produced by a heat engine. The net work is qin = TH ( s2 − s1 ) , qout = TL ( s2 − s1 ) [kJ/kg]
equal to the area bounded by the cycle as plotted on a
T-S diagram. It is also the difference between the heat wnet = qin − qout [kJ/kg]
consumed by a heat engine and its waste heat, that
Work occurs in all 4 processes of the Carnot cycle (work is 0
is, the difference between heat taken from the high- for constant volume processes).
temperature source and the heat deposited in the low-
temperature sink. wnet = w12 + w23 + w34 + w41 [kJ/kg]
Net work: Wnet = QH − QL [kJ]
per unit mass: wnet = qin − qout [kJ/kg]
per unit time: & & &
Wnet = QH − QL [kW]
QH = magnitude of the heat transferred from a high-
QL = magnitude of the heat transferred to a low-
qin = magnitude per kilogram of the heat transferred from a
high-temperature source [kJ/kg]
qout = magnitude per kilogram of the heat transferred to a
low-temperature source [kJ/kg]


IDEAL OTTO CYCLE IDEAL DIESEL CYCLE
The Otto cycle is the model for the spark-ignition The Diesel cycle is the model for the compression-
reciprocating engine. It consists of four internally ignition reciprocating engine. It consists of four
reversible processes: 1) isentropic compression, 2) internally reversible processes: 1) isentropic
heat addition, 3) isentropic expansion, and 4) heat compression, 2) heat addition, 3) isentropic
rejection. expansion, and 4) heat rejection.
T 2-3 • Heat addition T 2-3 • Heat addition
•w =0 3 • Constant pressure
TH 3
The area
The area enclosed is qin
enclosed is equal to the
equal to the qin work.
work.
2 4
Temperature 4 Temperature
2 3-4 • Isentropic
expansion 3-4 • Isentropic
1-2 • Isentropic qout expansion
1-2 • Isentropic qout compression
compression 4-1 • Heat
1 rejection TL 4-1 • Heat
•w =0 1 rejection
•w =0

s1 Entropy s3 s s1 s3 s
Entropy

P
3 Isentropic P qin
P2
2 3
Isentropic
qin
Constant volume,
Pressure so no work
4 Pressure
2
qout 4
Constant volume 1 qout
so no work
1
ν2 Volume density ν1 v
Volume density ν1 v
The otto cycle is more efficient than the diesel for equal
compression ratios, but the compression ratio is limited due Although less efficient than the otto cycle at a given
to spontaneous ignition of the fuel at higher temperatures. compression ratio, higher compression ratios are possible in
the diesel engine, enabling greater thermal efficiency than in
1 gasoline engines.
Thermal efficiency: ηth,Otto = 1 −
r k −1 1 ⎡ rck − 1 ⎤
V1 ν1 Thermal efficiency: ηth,Diesel = 1 − ⎢ ⎥
Otto cycle compression ratio is: r= = r k −1 ⎣ k ( rc − 1) ⎦
V2 ν 2
r = compression ratio [no units]
r = compression ratio [no units] rc = cutoff ratio [no units]
k = Cp/Cv = specific heat ratio k = Cp/Cv = specific heat ratio


IDEAL BRAYTON CYCLE IDEAL JET-PROPULSION CYCLE
The Brayton cycle is the model used for modern gas- The jet-propulsion cycle is the model used for aircraft
turbine engines. Although the turbine is an open gas-turbine engines. It consists of six internally
system, it can be modeled by the Brayton cycle, which reversible processes:1) isentropic compression in a
is a closed cycle. It consists of four internally diffuser, 2) isentropic compression in a compressor, 3)
reversible processes:1) isentropic compression, 2) heat addition, 4) isentropic expansion in a turbine, 5)
heat addition under constant pressure, 3) isentropic isentropic expansion in a nozzle, and 6) heat
expansion, and 4) heat rejection under constant rejection.
pressure. 3-4 • Heat addition 4
• Constant pressure
T 2-3 • Heat addition T •w =0
4-5 • Isentropic
•w =0 3 expansion
in turbine
qin
The area
2-3 • Isentropic
5 V5 ≈ 0
enclosed is
equal to the qin compression
work. in compressor
5-6 • Isentropic
3 expansion
Temperature 4 Temperature 6 in nozzle

2 3-4 • Isentropic
expansion V2 ≈ 0

1-2 • Isentropic qout 2 qout
compression 4-1 • Heat 6-1 • Heat rejection
1 rejection 1-2 • Isentropic 1 • Constant pressure
•w =0 compression •w = 0
by diffuser

s1 Entropy s3 s s1 Entropy s6 s
P qin
0 − V1 2
2 3 1 − 2 0 = C p (T2 − T1 ) +
Isentropic 2000
P3
Pressure
wturb,out 2 − 3 rp =
wcomp,in
P2
qout
4 − 5 wcomp,in = wturb,out → C p (T3 − T2 ) = C p (T4 − T5 )
1 4
V6 2 − 0
ν2 Volume density ν4 v 5 − 6 0 = C p (T6 − T5 ) +
2000
qin = q23 = C p (T3 − T2 ) , qout = q41 = C p (T4 − T1 ) W & w
Propulsive efficiency: ηp = & P = P
Compressor work: wcomp,in = C p (T2 − T1 ) Qin qin
Turbine work: wturb,out = C p (T3 − T4 ) m ( V6 − V1 ) V1
& ( V6 − V1 ) V1
where &
WP = , wP =
Net work: wnet,out = wturb,out − wcomp,in = qin − qout 1000 1000
& = mC (T − T ) , q = C (T − T )
Qin & p 4 3
1 P2 in p 4 3
Thermal efficiency: ηth,Brayton = 1 − ( k −1) / k
, rp =
rp P 1
1 m=
& V1 A1
Cp = specific heat at constant pressure (1.005 @ 300k) ν
[kJ/(kg·°C)] Cp = specific heat at constant pressure (1.005 @ 300k)
q = heat transfer per unit mass [kJ/kg] [kJ/(kg·°C)]
w = work per unit mass [kJ/kg] rp = pressure ratio [no units]
rp = pressure ratio [no units] &
WP = propulsive power [kW]
k = Cp/Cv = specific heat ratio
&
Q = heat transfer rate to the working fluid [kW]
in
&
m = mass flow rate [kg/s]
T = temperature [K]
P = pressure [kPa]
V = air velocity [m/s]


IDEAL RANKIN CYCLE rc CUTOFF RATIO
The Rankin cycle is the model used for vapor power The ratio of cylinder volumes after and before the
plants such as steam-turbine engines. It consists of combustion process. Applies to the diesel cycle.
four internally reversible processes: 1) isentropic
V3 ν 3
compression in a pump (The vertical distance rc = =
between 1 and 2 is actually greatly exaggerated on V2 ν 2
the diagram below.), 2) heat addition in a boiler at
constant pressure, 3) isentropic expansion in a V = volume [m3]
turbine, and 4) heat rejection in a condenser at ν = V/m specific volume [m3/kg]
constant pressure.
T Saturated
liquid/vapor
Constant
pressure
ISENTROPIC RELATIONS
mixture lines
2-3 • Heat addition 3 3-4 • Isentropic Isentropic means that the entropy does not change.
in boiler expansion
• Constant in turbine s2 = s1. The following relations apply to ideal gases:
pressure
k
wturb,out ⎛ P2 ⎞ ⎛ ν1 ⎞
Temperature qin ⎜ ⎟=⎜ ⎟ → Pν k = a constant
wpump,in
2 ⎝ P ⎠ ⎝ ν2 ⎠
1
4
k −1
1 ⎛ T2 ⎞ ⎛ ν1 ⎞
1-2 • Isentropic qout 4-1 • Heat rejection
in condenser ⎜ ⎟=⎜ ⎟ → T ν k −1 = a constant
⎝ T1 ⎠ ⎝ ν 2 ⎠
compression
in pump • Constant pressure

s1 Entropy s4 s
( k −1) / k
⎛ T2 ⎞ ⎛ P2 ⎞
1 − 2 wpump,in = ν1 ( P2 − P ) = h2 − h1 TP (
1− k ) / k
1 ⎜ ⎟=⎜ ⎟ → = a constant
where h1 = h f @ P , ν1 = ν f @ P ⎝ T1 ⎠ ⎝ P ⎠
1
1 1
2 − 3 qin = h3 − h2
T1, T2 = initial and final temperatures [K]
3 − 4 wturb,out = h3 − h4 P1, P2 = initial and final pressure [Pa]
Read h3 and s3 from Superheated Water Table k = Cp/Cv = specific heat ratio
based on T3 and P3. So s3 = s4 and x4 = (s4-sf)/sfg.
Then find h4 = hf+x4hfg.
4 − 1 qout = h4 − h1 POLYTROPIC PROCESS
A process in which the compression and expansion of
q w real gases have the following pressure/volume
Thermal efficiency: η p = 1 − out = net
qin qin relationship.

where wnet = wturb,out − wpump,in = qin − qout PV n = a constant and Pν n = another constant
where n is also a constant
w = work per unit mass [kJ/kg]
mR (T2 − T1 )
q = heat transfer per unit mass [kJ/kg] Work is W= , n ≠1
h = enthalpy [kJ/kg] 1− n
P = pressure [kPa]

r COMPRESSION RATIO
Vmax ν max
r= =
Vmin ν min
V = volume [m3]


Wb BOUNDARY WORK [kJ] HEAT TRANSFER
The work done in a moving boundary process such as HEAT TRANSFER
piston-cylinder expansion (positive W) and
Energy transport due to temperature difference.
compression (negative W). The boundary work
depends on the initial and final states as well as the o Conduction – diffusion in a material. In liquids and
gases, conduction is due to molecular collisions. In
path taken between them. In practice, it is often
solids, it is due to 1) molecular vibration in the lattice
easier to measure work than to calculate it. and 2) energy transport by free electrons.
2
Wb = ∫ P dV o Convection – by bulk motion of the fluid
1 o Thermal radiation – by electromagnetic waves; doesn't
mR (T2 − T1 )
require a medium
For a polytropic process: W= , n ≠1 The three mechanisms for heat transfer cannot all operate
1− n simultaneously in a medium.
In order to include the boundary work in a closed system Gases - are usually transparent to radiation, which can
piston-cylinder operating under constant pressure, it may be occur along with either conduction or convection but not
necessary to use the open system equation since both. In gases, radiation is usually significant compared
to conduction, but negligible compared to convection.
∆H = ∆U + Wb Solids - In opaque solids, heat transfer is only by
V = volume [m ] 3 conduction. In semitransparent solids, heat transfer is by
m = mass [kg] conduction and radiation.
n = a constant Fluids - In a still fluid, heat transfer is by conduction; in a
T1, T2 = initial and final temperatures [K] flowing fluid, heat transfer is by convection. Radiation
may also occur in fluids, usually with a strong absorption
P = pressure [kPa]
factor.
R = gas constant (0.287 kJ/(kg·K) for air) [kJ/(kg·K)]
Vacuum - In a vacuum, heat transfer is by radiation only.
∆U = net change in the internal energy of the system [kJ]

AIR STANDARD ASSUMPTIONS HEAT TRANSFER AND WORK
Heat transfer and work are interactions between a
Since air is composed mostly of nitrogen, which
system and its surroundings. Both are recognized as
undergoes few changes in the combustion chamber,
they cross the boundaries of a system. Heat and
internal combustion engines can be modeled as
work are transfer phenomena, not properties. They
containing air only.
are associated with a process, not a state. Both are
1) The working fluid is air that continuously circulates in a path functions, meaning that their magnitudes
closed loop and behaves as an ideal gas. depend on the path taken as well as the end states.
2) All the processes are reversible.
3) The combustion process is replaced by a heat addition
process from an external source k THERMAL CONDUCTIVITY [W/(m·°C)]
4) The exhaust process is replaced by a heat rejection A measure of the ability of a material to conduct heat.
process that restores the working fluid to its initial state. k varies with temperature, but we will consider it
constant in this class. The conductivity of most solids
falls with increasing temperature but for most gases it
MEP MEAN EFFECTIVE PRESSURE rises with increasing temperature. For example, the
A fictitious pressure which, if it acted on the piston conductivity of copper at 200K is 413, and at 800K is
during the entire power stroke, would produce the 366. The conductivity of air at 200K is 0.0181, and at
same amount of work that is produced during the 800K is 0.0569. The change in conductivity becomes
actual cycle. more dramatic below 200K.
Thermal Conductivity of Selected Materials @300K [W/(m·°C)]
Wnet wnet RT
MEP = = , where ν= Air 0.0261 Copper 401 Human skin 0.37
Vmax − Vmin ν max − ν min P Aluminum 237 Diamond 2300 Iron 80.2
Brick 0.72 Fiberglass insul. 0.04 Mercury 8.9
Carbon dioxide 0.0166 Glass 1.4 Plywood 0.12
Concrete 1.4 Gypsum 0.17 Water 0.608
Concrete block 1.1 Helium 0.150 Wood (oak) 0.17


Applied thermodynamics

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