Thermodynamic-Formulas.pdf
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Mechanical Engineering – GATE Exam
Mechanical Engineering – GATE Exam
Thermodynamics Symbol/Formula
Parameter
M
Molar mass (M/)
m
Mass (M)
n
Number of moles ( )
m M
E
Energy or general extensive property
e
E
e
E
Specific molar energy (energy per unit mass) or general extensive property per unit mass
m n
eM
Specific energy (energy per unit mole) or general extensive property per unit mole
P
Pressure (ML-1T-2)
V
Volume (L 3); Specific volume or volume per unit mass, v (L 3M-1) and the volume per unit mole v (L3-1)
T
Temperature ( Θ)
Density (ML-3); = 1/v.
x
Quality
U
Thermodynamic internal energy (ML 2T-2); Internal energy per unit mass, u (L 2T-2), and the internal energy per unit mole, u (ML2T-2-1)
H = U + PV
Thermodynamic enthalpy (ML 2T-2); Enthalpy per unit mass, h = u + Pv (dimensions: L 2T-2) and the internal energy per unit mole h (ML2T-2-1)
S
Entropy (ML2T-2Θ-1); Entropy per unit mass, s(L 2T-2Θ-1) and the internal energy per unit mole s (ML2T-2Θ-1-1)
W
Work (ML2T-2)
Q
Heat transfer (ML 2T-2)
: W u
The useful work rate or mechanical power (ML 2T-3)
: m
The mass flow rate (MT -1)
Mechanical Engineering – GATE Exam
V
The kinetic energy per unit mass (L 2T-2)
2
2
: The potential energy per unit mass (L 2T-2)
gz:
Etot:
The total energy = m(u +
V 2
2
+ gz) (ML2T-2)
Q :
The heat transfer rate (ML 2T-3)
dEcv : dt
The rate of change of energy for the control volume.(ml 2t-3)
M
Molar mass (M/)
m
Mass (M)
n
Number of moles ( )
m M
E
Energy or general extensive property
e
E
e
E
Specific molar energy (energy per unit mass) or general extensive property per unit mass
m n
eM
Specific energy (energy per unit mole) or general extensive property per unit mole
P
Pressure (ML-1T-2)
V
Volume (L 3); Specific volume or volume per unit mass, v (L 3M-1) and the volume per unit mole v (L3-1)
T
Temperature ( Θ)
Density (ML-3); = 1/v.
x
Quality
U
Thermodynamic internal energy (ML 2T-2); Internal energy per unit mass, u (L 2T-2), and the internal energy per unit mole, u (ML2T-2-1)
H = U + PV
Thermodynamic enthalpy (ML 2T-2); we also have the enthalpy per unit mass, h = u + Pv (dimensions: L 2T-2) and the internal energy per unit mole h (ML2T-2-1)
S
Entropy (ML2T-2Θ-1); Entropy per unit mass, s(L 2T-2Θ-1) and the internal energy per unit mole s (ML2T-2Θ-1-1)
Mechanical Engineering – GATE Exam W
Work (ML2T-2)
Q
Heat transfer (ML 2T-2)
: W u
The useful work rate or mechanical power (ML 2T-3)
: m
The mass flow rate (MT -1)
V 2
2
The kinetic energy per unit mass (L 2T-2) : The potential energy per unit mass (L 2T-2)
gz:
Etot:
The total energy = m(u +
V
2
2
+ gz) (ML2T-2)
Q :
The heat transfer rate (ML 2T-3)
dEcv : dt
The rate of change of energy for the control volume.(ml 2t-3)
M
Molar mass (M/)
m
Mass (M)
n
Number of moles ( )
m M
E
Energy or general extensive property
e
E
e
E
Specific molar energy (energy per unit mass) or general extensive property per unit mass
m n
eM
Specific energy (energy per unit mole) or general extensive property per unit mole
P
Pressure (ML-1T-2)
V
Volume (L 3); we also have the specific volume or volume per unit mass, v (L3M-1) and the volume per unit mole v (L3-1)
T
Temperature ( Θ)
Density (ML-3); = 1/v.
x
Quality
U
Thermodynamic internal energy (ML 2T-2); we also have the internal energy per unit mass, u (L 2T-2), and the internal energy per unit mole, u (ML2T-2-1)
H = U + PV
Thermodynamic enthalpy (ML 2T-2); we also have the enthalpy per unit mass, h = u + Pv (dimensions: L 2T-2) and the internal energy per unit mole h (ML2T-2-1)
Mechanical Engineering – GATE Exam S
Entropy (ML2T-2Θ-1); we also have the entropy per unit mass, s(L 2T2 -1 2 -2 -1 -1 Θ ) and the internal energy per unit mole s (ML T Θ )
W
Work (ML2T-2)
Q
Heat transfer (ML 2T-2)
: W u
The useful work rate or mechanical power (ML 2T-3)
: m
The mass flow rate (MT -1)
The kinetic energy per unit mass (L 2T-2)
V 2
:
2
The potential energy per unit mass (L 2T-2)
gz:
Etot:
The total energy = m(u +
V 2
2
+ gz) (ML2T-2)
Q :
The heat transfer rate (ML 2T-3)
dEcv : dt
The rate of change of energy for the control volume. (ml 2t-3)
Unit conversion factors For metric units
Basic:
2
1 N = 1 kg·m/s ; 1 J = 1 N·m; o 1 W = 1 J/s; o 2 o 1 Pa = 1 N/m . Others: 3 1 kPa·m = 1 kJ; o o T(K) = T( C) + 273.15; o 3 o 1 L (liter) = 0.001 m ; 2 2 1 m /s = 1 J/kg. o Prefixes (and abbreviations) : -9 nano(n) – 10 ; o -6 micro() – 10 ; o -3 o milli(m) – 10 ; 3 kilo(k) – 10 ; o 6 mega(M) – 10 ; o 9 giga(G) – 10 . o A metric ton (European word: tonne) is 1000 kg. o o
For engineering units
Mechanical Engineering – GATE Exam
Energy: 3 1 Btu = 5.40395 psia·ft = 778.169 ft·lbf = (1 kWh)/3412.14 = (1 hp·h )/2544.5 = o 2 2 25,037 lbm·ft /s . Pressure : 2 2 o 1 psia = 1 lbf /in = 144 psfa = 144 lbf /ft . Others: o T(R) = T( F) + 459.67; o 2 1 lbf = 32.174 lbm·ft/s ; o o 1 ton of refrigeration = 200 Btu/min.
Concepts & Definitions Formula Pressure
Units
P
Units
Pa
F A
1 Pa 1 N / m2 1 bar 105 Pa 0.1 Mpa 1 atm 101325 Pa
Specific Volume Density
v
Static Pressure Variation Absolute Temperature
m3 / kg
V m m
V
kg / m3
1 v
, P gh T ( K ) T (C ) 273.15
Pa
Properties of a Pure Substance Formula Quality
x
mvapor mtot
1 x
(vapour mass fraction)
mliquid mtot
(Liquid mass fraction)
Specific Volume
v v f xv fg
Average Specific Volume
v (1 x)v f xvg
Ideal –gas law
P P c
Equations
Pv RT
Units
m3 / kg
T T c
(only two phase mixture)
Z 1
PV mRT nRT
m3 / kg
Mechanical Engineering – GATE Exam
Universal Gas Constant
Gas Constant
Compressibility Factor Z Reduced Properties
R 8.3145
R
R
kJ / kmol K kJ / kg K
M = molecular mass
M Pv ZRT P P r , P c
T r
T T c
Work & Heat Formula Displacement Work Integration
W W
2
1 2
1
Specific Work Power (rate of work)
Velocity
Torque
Polytropic Process ( n 1)
Polytropic Exponent
n=1
Polytropic Process Work
w
n=1
Fdx
Conduction Heat Transfer
1
m
J J / kg
W FV PV T
W rad / s Nm
V r T Fr n n PV n Const PV 1 1 PV 2 2
Pv n C
P ln 2
P 1 n V ln 1 V 2 PV Const PV 1 1 PV 2 2 W2
1 1 n
( PV 2 2 PV 1 1)
W2 PV 2 2 ln
V 2
n 1
V 1
J J
Q 0
Q kA
dT dx
, k =conductivity
Convection Heat Transfer
Q hAT
Radiation Heat Transfer
4 Q A(T s4 Tamb )
, h =convection coefficient
Terminology: = heat Q2 = heat transferred during the process between state 1 and state 2 Q1
J
PdV
(work per unit mass)
1
Adiabatic Process
2
PdV P (V2 V1 )
W
1
Units
W
W W
Mechanical Engineering – GATE Exam = rate of heat transfer W = work 1W 2 = work done during the change from state 1 to state 2 W = rate of work = Power. 1 W=1 J/s Q
The First Law of Thermodynamics Formula
Units
Total Energy
E U KE PE dE dU d (KE ) d (PE )
Energy
dE Q W E2 E1 1 Q2 1W2
J J
Kinetic Energy Potential Energy
KE 0.5mV 2 PE mgZ PE2 PE1 mg ( Z 2 Z 1 )
J J
Internal Energy
U U liq U vap mu mliqu f mvapu g
Specific Internal Energy of Saturated Steam (two-phase mass average) Total Energy
u (1 x)u f xug
kJ / kg
u u f xu fg
U 2 U1
m(V22 V 12 ) 2
mg ( Z 2 Z1 ) 1 Q2 1W2
Specific Energy
e u 0.5V 2 gZ
Enthalpy
H U PV h u Pv Pv RT and u f (T )
Specific Enthalpy For Ideal Gasses
Enthalpy
R Constant
J
kJ / kg
h u Pv u RT u f (t ) h f (T ) h (1 x) h f xhg
Specific Enthalpy for Saturation State (two-phase mass average) Specific Heat at Constant Volume
C v
Specific Heat at Constant Pressure
C p
h h f xh fg
1 Q 1 U u m T v m T v T v (ue ui ) Cv (Te Ti ) 1 Q 1 H h m T p m T p T p
( he hi ) C p (Te Ti ) Solids & Liquids
Incompressible, so v=constant
C Cc C p
(Tables A.3 & A.4)
u2 u1 C (T2 T1 )
h2 h1 u 2 u1 v (P2 P1 )
kJ / kg
Mechanical Engineering – GATE Exam h u Pv u RT u2 u1 Cv (T2 T1 )
Ideal Gas
h2 h1 C p (T2 T1 )
E Q W
Energy Rate
( rate in out )
E2 E1 1 Q2 1W2 (change in out )
First-Law Analysis for Control Volume Formula Volume Flow Rate Mass Flow Rate Power
Units
V V dA AV (using average velocity)
m VdA AV A
V v
W mCv T
W mC p T
m V
W flow PV mPv
Flow Direction
From higher P to lower P unless significant KE or PE
Total Enthalpy
W
v
Flow Work Rate
kg / s
(using average values)
htot h 1 V 2 gZ 2
Instantaneous Process
m m
Continuity Equation
mC.V .
Energy Equation
EC .V . QC .V . WC .V .
Steady State Process
i
e
m h
me htot e
i tot i
Q mi (hi 1 2 V 2 gZ i )
dE dt
First
Law
me he 1 2V 2 gZ e W
A steady-state has no storage effects, with all properties constant with time
No Storage
mC .V . 0, E C .V . 0
Continuity Equation
Energy Equation
m m (in = out) Q m h W m h (in = out) First Law Q m (h 1 2 V gZ ) W m h 1 2 V gZ i
e
C .V .
i tot i
Specific Heat Transfer Specific Work
SS Single Flow Eq. Transient Process
e tot e
2
i
C .V .
q
QC .V .
w
W C .V .
i
2
i
e
e
e
kJ / kg
m kJ / kg
m
q htot i w htot e
(in = out)
Change in mass (storage) such as filling or emptying of a container.
Mechanical Engineering – GATE Exam
m2 m1
Continuity Equation Energy Equation
QC .V
m m i
e
E2 E1 QC.V WC.V .
m h
mehtot e
i tot i
E2 E1 m2 u2 1 V22 gZ 2 m1 u1 1 V12 gZ 1 2 2
mh
i tot i
me htot e m2 u2 1 2 V22 gZ 2 m1 u1 1 2 V22 gZ1 WC .V . C .V .
The Second Law of Thermodynamics Formula All W , Q can also be rates W , Q Heat Engine
Thermal efficiency
Real Heat Engine
W HE
HE
HP HP
Q H W HP
Q H
QH
1
T L T H T L T H
QH QH QL
Q H
Q H QL Q H W HP
T H TH TL
Carnot HP
T H TH T L
W REF QH QL
Coefficient of Performance
REF
Carnot Cycle
Absolute Temperature
Q L
QL
W HP QH QL
Carnot Cycle
Real Refrigerator
Q H
1
Carnot HE 1
Q H
HP
Real Heat Pump
W HE
Thermal 1
Coefficient of Performance
Refrigerator
HE
Carnot Cycle
Heat Pump
W HE QH QL
T H
W REF Q L
Q H QL
REF T L
Q L
Q L W REF
QL QH
QL QH QL T L TH TL
Carnot REF
T L TH T L
Units
Mechanical Engineering – GATE Exam Entropy Formula Inequality of Clausis Entropy
Change of Entropy
Specific Entropy
Q T
Units
0
Q T rev
kJ / kgK
dS
kJ / kgK
2
Q 1 T rev s (1 x)s f xsg S 2 S 1
kJ / kgK
s s f xs fg Entropy Change
Carnot Cycle
Isothermal Heat Transfer: S 2
S1
1 T H
2
Q
1
1
Q2
T H
Q T rev
Reversible Adiabatic (Isentropic Process): dS 4
Reversible Isothermal Process: S4
Q Q S 3 3 4 T rev T L 3
Reversible Adiabatic (Isentropic Process): Entropy decrease in process 3-4 = the entropy increase in process 1-2.
Reversible HeatTransfer Process
Gibbs Equations Entropy Generation
2 2 h fg 1 Q 1 1 q2 s2 s1 s fg Q m 1 T rev mT 1 T T
Tds du Pdv Tds dh vdP Q dS S gen T Wirr PdV T Sgen 2
1
1
S 2 S1 dS Entropy Balance Equation Principle of the Increase of Entropy Entropy Change
Solids & Liquids
2
Q T
1 S 2 gen
Entropy in out gen
dSnet dSc.m. dS surr
s2 s1 c ln Reversible Process: ds gen
0
Adiabatic Process: dq 0
S
T 2 T 1
0
gen
Mechanical Engineering – GATE Exam
Ideal Gas
2
Constant Volume: s2
dT
s1 Cv0
T
1 2
Constant Pressure: s2
s1 Cp0
R ln v2
dT
1
T
R ln P 2
1
s2 s1 Cp0 ln T
s 0 T
T 0
Change in Standard Entropy
C p 0 T
T2
T1
1
R ln P 2
P 1 kJ / kgK
dT
s2 s1 sT0 2 sT 0 1 R ln
Ideal Gas Undergoing an Isentropic Process
P 1
s1 Cv0 ln T2 T R ln v2 v
Constant Specific Heat: s2
Standard Entropy
v1
P 2
kJ / kgK
P 1
s2 s1 0 Cp0 ln
T2
P R ln 2 T1 P 1 R
P 2 Cp0 T1 P 1 C C v 0 k 1 R , p 0 T2
but
k
C p 0
C p 0 C v 0
Cp0
k
= ratio of specific heats
v 1 T1 v2 T2
k 1
,
v 1 P1 v2
P2
k
Special case of polytropic process where k = n: Pv n n PV n const PV 1 1 PV 2 2
Reversible Polytropic Process for Ideal Gas
n
V 1 , P1 V2 P2
Work
2
1
Values for n
2
Isobaric process: n 0,
dV
V
W2 PdV const
1
const
k
1
n
P 2 T1 P1
T2
PV 2 2 PV 1 1 1 n
P const T const Isothermal Process: n 1, s const Isentropic Process: n k , Isochronic Process: n , v const
n 1
n
V 1 V 2
mR(T2 T 1) 1 n
n 1
Mechanical Engineering – GATE Exam Second-Law Analysis for Control Volume Formula nd
2 Law Expressed as a Change of Entropy Entropy Balance Equation
dS c.m .
Q
S gen dt T rateof change in out generation
dSC .V .
mi si me se
dt
where SC.V . and S gen Steady State Process
dS C .V . dt
Continuity equation
QC .V . T
sgen dV S gen. A S gen .B ...
0
m s m s e e
QC .V .
i i
T
Sgen
m( se si )
mi me m
QC .V . T
C .V .
Adiabatic process
Transient Process
S gen
sdV mc.v. s mA sA mB sB ...
C .V .
Units
S gen
se si s gen s i
d dt
ms C.V . mi si me se
QC .V . T
Sgen t
m2 s2 m1s1 C .V . mi si me se 0
If Process Reversible & Adiabatic
Reversible Steady State Process se si e
he hi vdP i
w hi he e
vdP
Vi 2 V e 2
Vi 2 V e 2
i
If Process is Reversible and Isothermal
m se si or T se
2
1 T
si
2
g ( Zi Z e )
g ( Zi Ze )
QC .V .
C .V .
QC .V . m
QC .V . T
q e
T se si he hi vdP i
QC .V . T
dt 1 S 2 gen
Mechanical Engineering – GATE Exam Incompressible Fluid Reversible Polytrophic Process for Ideal Gas
v Pe Pi
Ve2 V i 2 2
g Ze Zi 0
Bernoulli Eq.
e
w vdP
Pv n const C n
and
i e
e
dP w vdP C 1 n i i P n nR Pe ve Pv Te T i i i n 1 n 1 Isothermal Process (n=1)
e
e
w vdP C i
Principle of the Increase of Entropy
Turbine
Compressor (Pump) Cooled Compressor
Nozzle
dSnet dt
wa
w s
w s wa
dP
P
Pv i i ln
i
dSC .V . dt
dSsurr dt
P e P i
S gen 0
Efficiency h h i e Turbine work is out hi hes h h i es Compressor work is in hi he
wT w
1 V 2 e 2 2 1 V 2 es
Kinetic energy is out
Note:
°F = (°C x 9/5) + 32 °C = (°F - 32) x (5/9) °K = °C + 273 Q = mC∆T thermal energy = mass x specific heat x change in T Q = mHf thermal energy = mass x heat of fusion Q = mHv thermal energy = mass x heat of vaporization ∆L = αLi∆T change in length = coefficient of expansion x initial length x change in T ∆V = βVi∆T change in volume = coefficient of expansion x initial volume x change in T ∆U = Q – W internal energy = heat energy - work
Mechanical Engineering – GATE Exam Plausible Physical Name Situations Insulated Add weight Adiabatic sleeve to or push compression or rapid down on process piston Insulated Remove Adiabatic sleeve or weight from expansion rapid or pull up process on piston Heat gas Locked Isochoric piston or rigid container Cool gas Locked Isochoric piston or rigid container Heat gas Piston free Isobaric to move, expansion load unchanged Cool gas Piston free Isobaric to move, compression load unchanged Immerse Add weight Isothermal gas in to piston compression large bath Immerse gas in large bath
Remove weight from piston
Isothermal expansion
Unknown
Unknown
No Name
State Variables PV γ = Const; TV γ-1 = Const PV γ = Const; TV γ-1 = Const
P
V
Up
Down
T
Down
Up
Up
Down
V fixed;
PαT
Up
Fixed
Up
V fixed;
PαT
Down
Fixed
Down
P fixed;
V α T
Fixed
Up
Up
P fixed;
V α T
T fixed at temperature of bath, PV = Const T fixed at temperature of bath, PV = Const PV/T = Const
Fixed
Up
Down
?
Down
Down
Up
?
Down
Fixed
Fixed
?
ΔEth
Q
W s
W
nC v ΔT > 0
0
-nC v ΔT < 0
nC v ΔT > 0
nC v ΔT < 0
0
-nC v ΔT > 0
nC v ΔT < 0
nC v ΔT > 0
nC v ΔT >0
0
0
nC v ΔT < 0
nC v ΔT 0
nCpΔT >0
PΔV > 0
-PΔV < 0
nC v ΔT < 0
nCpΔT 0
nC v ΔT = nRT*ln( nRT*ln(V f/ 0 V f /V )i V )i < 0 0
nC v ΔT = nRT*ln( nRT*ln(V f/ 0 V f /V )i V )i > 0 >0
-nRT* ln(V f/ )i < V 0
nC v ΔT
- PdV = ± area under curve in PV diagram
ΔEth +
PdV = ±
W s
area under curve in PV diagram
ALL THE BEST for GATE 2016 Exam!!!
Mechanical Engineering – GATE Exam
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