Thermodynamic Formulas

Thermodynamics MTX 220 Formules Chapter 2 – Concepts & Definitions Formule 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

v=

V m

Density

m3 / kg



m ρ= V

kg / m3

1 ρ= v

Static Pressure Variation

Pa

∆P = ρ gh Absolute Temperature

↑= − , ↓ = +

T ( K ) = T (°C ) + 273.15

Chapter 3 – Properties of a Pure Substance Formule

Units

Quality

(vapour mass fraction)

x=

mvapor mtot (Liquid mass fraction)

1− x =

mliquid mtot

Specific Volume

m3 / kg

v = v f + xv fg Average Specific Volume

(only two phase

v = (1 − x )v f + xvg

m3 / kg

mixture)

Ideal –gas law

P << Pc •

Z =1

T << Tc

Equations

Pv = RT •

Universal Gas Constant

•

Gas Constant

PV = mRT = nRT kJ / kmol K

R = 8.3145 = molekulêre mass

R R= M Compressibility Factor

Z

M

kJ / kg K

Pv = ZRT

Reduced Properties

,

Pr =

P Pc

Tr =

T Tc

Chapter 4 – Work & Heat Formule

Units

Displacement Work

2

1

Integration

J

2

W = ∫ Fdx = ∫ PdV 1

J

2

W = ∫ PdV = P( V2 −V1 ) 1

Specific Work

(work per unit mass)

W w= m Power (rate of work) •

Velocity

•

Torque

J / kg

W&= FV = PV&= T ω

W

V = rω

rad / s

T = Fr

Nm

Polytropic Process

( n ≠ 1)

n n PV n = Const = PV 1 1 = PV 2 2

Pv n = C •

Polytropic Exponent

•

n=1

Polytropic Process Work

P ln  2  P1  n=  V ln  1   V2 

PV = Const = PV 1 1 =P 2 V2 W2 =

1

•

n=1

 W2 = PV 2 2 ln  

1

Adiabatic Process

1 (PV 2 2 − PV 1 1) 1− n

Q=0

V2

 V1 

n≠ 1

J

J

Conduction Heat Transfer

,

,

Q&= hA∆T Radiation Heat Transfer

W

=convection coefficient

W

k

dT Q&= −kA dx Convection Heat Transfer

=conductivity

h

4 Q&= εσ A(Ts4 − Tamb )

W

Terminology = heat

Q = heat transferred during the process between state 1 and state 2 1

Q2 = rate of heat transfer

Q& = work

W = work done during the change from state 1 to state 2

W2

1

= rate of work = Power. 1 W=1 J/s

W&

Chapter 5 – The First Law of Thermodynamics Formule

Units

Total Energy

E = U + KE + PE → dE = dU+ d (KE +) d (PE )

J

Energy

dE = δ Q − δ W → E2 − E1 =1 Q2 −1 W2

J

KE = 0.5mV

J

Kinetic Energy Potential Energy

Internal Energy

Specific Internal Energy of Saturated Steam (two-phase mass average) Total Energy

PE = mgZ → PE2 − PE1 = mg( Z2 −Z1)

u = (1 − x)u f + xu g

kJ / kg

u = u f + xu fg

m(V22 − V12 ) + mg( Z2 − Z1) = 1 Q2 −1W2 2

J

e = u + 0.5V 2 + gZ

Enthalpy

H = U + PV

Specific Enthalpy

h = u + Pv

For Ideal Gasses

Pv = RT and u = f( T)

•

Enthalpy

h = u + Pv = u + RT

•

R Constant

u = f ( t) → h = f ( T )

Specific Enthalpy for Saturation State (two-phase mass average)

J

U = U liq + U vap → mu = mliq u f + mvap ug

U 2 − U1 + Specific Energy

2

h = (1 − x )h f + xhg h = h f + xh fg

kJ / kg

kJ / kg

Specific Heat at Constant Volume

Cv =

1 δQ  1  δU   δ u    =   =  m  δ T v m  δ T v  δ T  v

→ (ue − ui ) = Cv (Te −Ti ) Specific Heat at Constant Pressure

Cp =

1  δQ  1 δH   δh    =   =  m  δ T  p m  δ T  p  δ T p

→ (he − hi ) =C p (Te −Ti ) Solids & Liquids

Incompressible, so v=constant (Tables A.3 & A.4)

C = Cc = Cp

u2 − u1 = C (T2 − T1) h2 − h1 = u2 − u1 + v( P2 − P1) Ideal Gas

h = u + Pv = u + RT u2 − u1 ≅ Cv ( T2 − T1) h2 − h1 ≅ C p (T2 − T1 )

Energy Rate

E&= Q&− W& ( rate = +in − out) → E2 − E1 = 1 Q2 − 1W2 ( change = +in −out)

Chapter 6 – First-Law Analysis for A Control Volume Formule Volume Flow Rate

Units

(using average velocity)

V&= ∫ V dA = AV Mass Flow Rate

(using average values)

kg / s

V m&= ∫ ρVdA = ρ AV = A v Power

& p VT W&= mC

& v VT W&= mC

Flow Work Rate

& W&flow = PV&= mPv

Flow Direction

From higher P to lower P unless significant KE or PE

•

Total Enthalpy

Instantaneous Process • Continuity Equation •

Energy Equation

W

& m&= V v

htot = h + 1 V 2 + gZ 2

m&C .V . = ∑ m&i − ∑ m&e  First Law

E&C .V . = Q&C .V . − W&C .V . +∑ & mi htot i −∑ & me htot e

(

)

dE → Q&+ ∑ m&i (hi + 1 V 2+ gZ i )= + ∑ m&e he+ 1 V 2+ gZe − W & 2 2 dt Steady State Process •

No Storage

•

Continuity Equation

A steady-state has no storage effects, with all properties constant with time

m&C .V . = 0, E&C .V . = 0 (in = out)

∑ m& = ∑ m& i

e

•

Energy Equation

(in = out)  First Law

& +∑m &i htot i = W &e htot e Q&C .V . + ∑ m C .V .

(

→ Q&+ ∑ m&i (hi + 1 V 2 + gZ i ) = W&+ ∑ m&e h e+ 1 V 2+ gZ e 2 2 •

•

•

Specific Heat Transfer Specific Work

SS Single Flow Eq.

Transient Process

kJ / kg

Q& q = C .V . m&

w=

)

W&C .V . m&

kJ / kg

(in = out)

q + htot i = w + htot e Change in mass (storage) such as filling or emptying of a container.

•

Continuity Equation

m2 − m1 = ∑ mi − ∑ me

•

Energy Equation

E2 − E1 = QC V. − WC V. . +∑ mi htot i −∑ me htot e

(

)

(

)

E2 − E1 = m2 u2 + 1 V22 + gZ2 − m1 u1 +1 V12 +gZ1 2 2 

(

)

(

)

QC .V + ∑ mi htot i = ∑ me htot e +  m2 u2 + 1 V2 2 + gZ2 − m1 u1 + 1 V2 2 + gZ1  − W 2 2   C .V . C .V .

Chapter 7 – The Second Law of Thermodynamics Formule All

Units

can also be rates W, Q

W&, Q&

Heat Engine

WHE = QH − QL •

Thermal efficiency

•

Carnot Cycle

η HE =

WHE Q = 1− L QH QH

ηThermal = 1 − •

Real Heat Engine

η HE =

QL T = 1− L QH TH

WHE T ≤ ηCarnot HE = 1 − L QH TH

Heat Pump

WHP = QH − QL •

Coefficient of Performance

•

Carnot Cycle

•

Real Heat Pump

Refrigerator •

Coefficient of Performance

•

Carnot Cycle

′ = β HP

QH QH = WHP QH − QL

′ = β HP

QH TH = QH − QL TH − TL

β HP =

QH TH ≤ βCarnot HP = WHP TH − TL

WREF = QH − QL

β REF =

β=

QL QL = WREF QH − QL

QL TL = QH − QL TH − TL

•

Real Refrigerator

Absolute Temp.

β REF =

QL TL ≤ βCarnot REF = WREF TH − TL

TL QL = TH QH

Chapter 8 – Entropy Formule Inequality of Clausis

Entropy

Change of Entropy

Specific Entropy

δQ

Ñ ∫ T

≤0

 δQ  dS ≡    T rev

kJ / kgK

 δQ  S 2 − S1 = ∫   T  rev 1

kJ / kgK

s = (1 − x ) s f + xsg

kJ / kgK

2

s = s f + xs fg Entropy Change

Units

•

Carnot Cycle

Isothermal Heat Transfer: 2

S 2 − S1 =

1 Q δQ = 1 2 ∫ TH 1 TH

Reversible Adiabatic (Isentropic Process):

 δQ  dS =    T rev Reversible Isothermal Process:

 δQ  3 Q4 S 4 − S3 = ∫   = T  rev TL 3 4

Reversible Adiabatic (Isentropic Process): Entropy decrease in process 3-4 = the entropy increase in process 1-2. •

Reversible Heat-Transfer Process

Gibbs Equations

s2 − s1 = s fg =

2 2 h 1  δQ  1 q = δ Q = 1 2 = fg   ∫ ∫ m 1  T  rev mT 1 T T

Tds = du + Pdv Tds = dh − vdP

Entropy Generation

dS =

δQ + δ Sgen T

δ Wirr = PdV − T δ Sgen 2

δQ + 1 S2 gen T 1 2

S 2 − S1 = ∫ dS = ∫ 1

Entropy Balance Eq.

VEntropy = + in − out + gen

Principle of the Increase of Entropy

dSnet = dSc. m. + dSsurr =∑ δ S gen ≥0

Entropy Change •

Solids & Liquids

s2 − s1 = c ln

T2 T1

Reversible Process:

dsgen = 0 Adiabatic Process:

dq = 0 •

Ideal Gas

Constant Volume: 2

s2 − s1 = ∫ Cv0 1

dT v + R ln 2 v1 T

Constant Pressure: 2

s2 − s1 = ∫ Cp0 1

dT P − R ln 2 P1 T

Constant Specific Heat:

s2 − s1 = Cv0 ln

s2 − s1 = Cp0 ln Standard Entropy

T

s =∫ 0 T

T0

Change in Standard Entropy

C p0 T

T2

T1

− R ln

P2

T1

+ R ln

v2

v1

P1 kJ / kgK

dT

s2 − s1 = ( sT02 − sT01 ) − R ln P2

T2

kJ / kgK

P1

Ideal Gas Undergoing an Isentropic Process

s2 − s1 = 0 = Cp0 ln T2

T1

− R ln P2

but

T P  → 2 = 2  T1  P1 

P1

,

C − Cv 0 k − 1 R = p0 = C p0 C p0 k

= ratio of

k=

C p0 Cv 0

specific heats

T v  ⇒ 2 = 1  T1  v 2 

k −1

,

P2  v 1  =  P1  v 2 

k

Special case of polytropic process where k = n:

Pv k = const Reversible Polytropic Process for Ideal Gas

n n PV n = const = PV 1 1 = PV 2 2

n

P V  → 2 = 1  , P1  V2 

•

•

Work

T2  P2  =  T1  P1 

2

2

1

1

1W2 = ∫ PdV = const ∫

Values for n

n

V  = 1  V2 

n−1

dV PV − PV mR (T2 − T1 ) = 2 2 1 1= n V 1− n 1− n

Isobaric process:

Isothermal Process:

Isentropic Process:

n −1

R

Cp 0

Isochronic Process:

n = ∞,

v = const

Chapter 9 – Second-Law Analysis for a Control Volume Formule 2nd Law Expressed as a Change of Entropy

dSc.m. Q& = ∑ + S&gen dt T

Entropy Balance Eq.

rate of change = + in − out + generation

→

Unit s

& dSC .V . Q = ∑ m&i si − ∑ m&e se + ∑ C .V . +S&gen dt T

where

SC .V . = ∫ ρ sdV = mc.v.s = m As A + m Bs B + ... and

S&gen = ∫ ρ s&gen dV = S&gen .A + S& gen .B + ... Steady State Process

dSC .V . =0 dt •

→

∑ m&e se −

Q&C .V . & + Sgen C .V . T

∑ m&i si = ∑

Continuity eq.

m&i = m&e = m&

Q&C .V . & + Sgen C .V . T

⇒ m&( se − si ) = ∑

•

Adiabatic process

Transient Process

se = si + sgen ≥ si

Q& d ( ms ) C .V . = ∑ m&i si − ∑ m&e se + ∑ C.V . + S&gen dt T

→ ( m2 s2 − m1 s1 ) C .V . = Reversible Steady State Process • If Process Reversible & Adiabatic

∑ mi si − ∑ me se +

Q&C .V . dt + 1 S& 2 gen T 0 t

∫

se = si e

he − hi = ∫ vdP i

Vi 2 − Ve 2 + g ( Zi − Ze ) 2 e Vi 2 − Ve 2 = − ∫ vdP + + g ( Zi − Ze ) 2 i

w = ( hi − he ) +

•

If Process is Reversible and Isothermal

m&( se − si ) =

Q 1 Q&C .V . = C .V . ∑ T C .V . T

or

T ( se − si ) = •

Incompressible Fluid

e Q&C .V . = q → T ( se − si ) = ( he − hi ) − ∫ vdP m& i

 Bernoulli Eq.

v ( Pe − Pi ) +

V − Vi + g ( Ze − Z i) = 0 2 2 e

2

•

Reversible Polytropic Process for Ideal Gas

e

w = − ∫ vdP and i

e

e

i

i

w = − ∫ vdP = −C ∫ =− •

Isothermal Process (n=1)

Principle of the Increase of Entropy

Pv n = const = C n

dP P

1

n

n nR ( Pe ve − Pv ( Te − Ti ) i i) = − n −1 n −1 e

e

i

i

w = − ∫ vdP = −C ∫

Pe dP = − Pv i i ln P Pi

dSnet dS C.V . dS surr = + = ∑ S&gen ≥ 0 dt dt dt

Efficiency •

Turbine

Turbine work is out

η= •

•

•

Compressor (Pump)

Cooled Compressor

wa hi − he = ws hi − hes Compressor work is in

η=

ws hi − hes = wa hi − he

η=

wT w

Nozzle

Kinetic energy is out

1 V2 e η= 2 2 1 V 2 es