conditions to qualify a variable as a thermal property:

1. continuous, meaning no holes between molecules (note: discontinuous at high vacuum)

2. path independent (process line cannot affect value of variables)

E.G. pressure (P), temperature (T), volume (V) & specific volume (v), density (ρ),
internal energy (U) & specific internal energy (u), entropy (S) & specific entropy (s)

≥ 2 path independent properties → known state of system

gauge pressure: reading relative to Patm , always use absolute pressure for calculations (esp. important for lab work)
note: 1 bar = 105 Pa

intensive variable: mass independent, i.e. anything where mass is being divided by

1. specific volume (v): volume occupied by unit mass (mass independent quantity)
2. specific enthalpy (h): enthalpy per unit mass
3. specific entropy (s): entropy per unit mass
1
* density: v

———————————————————————————————————————————————————————

I.C.E. Principle

1. intake stroke, flow process → fluid flows into cylinder, sealed off by closing inlet valve
2. compression stroke, non–flow process → fluid compressed by piston
3. working stroke, non–flow process → heat supplied, fluid expands & forces piston downwards
4. exhaust stroke, flow process → exhaust valve is opened for fluid to flow out

steady flow processes

1. mass going past any section must be constant, i.e. mass flow rate is constant
2. properties of mass at any particular section must be constant
3. transfers of work & energy happen at uniform rates

enthalpy: measure of total energy in thermodynamic system (includes internal energy which is = energy required to create
system = energy required to make room for it)

triple point of water: 0.01℃

quasi–static process: process carried out very slowly so that system passes through continuous sequence of thermal
equilibrium states

[don’t understand slide 19]

Property of matter

 measures state of matter

 describes 2 states (e.g. P and v) required to form process
 process = path between 2 given states where there may be exchange of heat (Q) and work (W)

The sum of net change of a measurable variable must be = 0

 ∮𝜋 = ∮P = ∮T = ∮V = ∮v = ∮ρ = 0

Area under P − v diagram = W

Wnet: area enclosed Wexpansion : area of top curve Wcompression : area of bottom curve

Conditions of thermodynamic property:

Mathematical test of continuum Mathematical test of independency

∂z ∂z
total differential eqn.: if 𝑧 = f(x, y), then dz = ∂x
dx + ∂y
dy
𝜕x 𝜕y 𝜕z
( ∂z
∂x
∂z
dx: path along constant y , ∂y dy: path along constant x ) condition 2: for z = f(x, y), (𝜕y) (𝜕z) (𝜕x) = −1
shows any 2 of 3 variables are independent of each other
𝜕 𝜕z 𝜕 𝜕z 𝜕2 z 𝜕2 z
condition 1: exactness: ( )
𝜕x 𝜕y
= ( )
𝜕y 𝜕x
→ 𝜕x𝜕y
= 𝜕y𝜕x
Example:
∂T ∂T
If T = f(P, v), then dT = dP + dv
∂P ∂v
(path along constant v, path along constant P)

𝜕 𝜕T 𝜕 𝜕T 𝜕2T 𝜕2T
( )= ( )→ =
𝜕v 𝜕P 𝜕P 𝜕v 𝜕v𝜕P 𝜕P𝜕v

path dependent: Q, W (not thermal properties because value depends on path taken)

0 law: 2 systems that are in mutual equilibrium with a 3rd system will also be in equilibrium with each other
θ X
Thermometric property of sensor: θ2 = X2 (X can be ∆EMF or ∆P or ∆h or…)
1 1

———————————————————————————————————————————————————————

Process Line & Saturation Curve

Pure substance: chemical composition remains constant (i.e. no mixture, except for air with Mr = 29 & γ = 1.4)

At a specific pressure, thermal process lines are:

T−v P−v

Heating Process:
⃗⃗⃗⃗⃗ ): T < Tsat.
1. Subcooled / compressed liquid (process AB
2. Saturated liquid (instantaneous at B): moment where the 1st bubble is formed
3. Wet steam (process ⃗⃗⃗⃗⃗
BC): 2 phase mixture of saturated liquid + saturated steam
4. Dry saturated steam (instantaneous at C): moment where all liquid has converted into gas
5. Superheated steam (process at ⃗⃗⃗⃗⃗
CD): heat input to dry saturated steam
By repeating the heating process at different pressures and joining equivalent points (B1 , B2 , B3 ..critical point.. C1 , C2 , C3 ..),
a saturation curve (boundary of phase change) for each T − v and P − v is obtained:

Comparisons of: Comparisons of:

v: examine values along horizontal axis markings v: examine values along horizontal axis markings
T: examine values along vertical axis markings P: examine values along vertical axis markings
P: pressure is constant for all points on a constant pressure T: temperature is constant for all points on a constant
line temperature line

saturated (f–line): dry saturated (g–line): critical point:

subcooled: purely liquid
purely liquid (no bubbles) all liquid vaporized max. P in 2 phase mixture

Steam Tables

a) terminology

regular symbols:
u: specific internal energy h: specific enthalpy s: specific entropy

subscripts:
s: saturation temperature f: saturated liquid g: dry saturated vapor fg: difference between saturated liquid & vapor

b) which table to refer to?

temperature table (left most column is temperature): for compressed liquid (only T value required → look for uf / hf / sf as
thermal properties of both compressed & saturated liquid are assumed to be identical)
* if T < 100℃, look under the temperature column of the temperature table
* if T > 100℃, look under the temperature column of the pressure tables

pressure table (left most column is pressure): for saturated liquid (f), dry saturated vapor (g) & 2 phase mixture (ALL @ Tsat. )
vf = 0.001 m3 ∙ kg −1  specific volume of (incompressible) water

superheated steam table: need values of P and T to determine other properties

c) 2 phase mixture calculations

mvapor
within 2 phase mixture, P and T are dependent → phase is defined by dryness factor (x), where x=
mliquid + mvapor
 u = x ∙ ug + (1 − x)uf = uf + x ∙ ufg h = x ∙ hg + (1 − x)hf = hf + x ∙ hfg s = x ∙ sg + (1 − x)sf = sf + x ∙ sfg
s − s f h − hf u − uf v − vf v
x= = = = ≈
sfg hfg ufg vg − vf vg

1st STEP  refer PRESSURE TABLE

2nd STEP:
* u < uf / h < hf / s < s f
* saturated liquid: u = uf / h = hf / s = sf
* 2 phase mixture: uf < u < ug / hf < h < hg / sf < s < sg
* dry superheated vapor: u = ug / h = hg / s = sg
v
* superheated vapor: v > 1
g

y −y
Linear interpolation: y − y1 = x2 −x1 (x − x1 )
2 1

Gases

· assumed to be ideal (incompressible) · STP: 1 mole of gas occupies 22.4 L

R
· PV = mRT = m M0 T (gravimetric basis) · PV = nR 0 T (volumetric basis)
r

———————————————————————————————————————————————————————

1st Law Of Thermodynamics

A) Closed system

“total energy of isolated system is constant; energy can only be transformed from 1 form to another”
𝑑U = 𝑑Q − 𝑑W
𝑑U: change in system’s internal energy 𝑑Q: heat input to system 𝑑W: energy output from system

D B

Q = ∫ T dS W = ∫ P dV
C A
(+) : heat supplied (–) : heat lost (+) : volume expansion (–) : compression

Special cases: Special cases:

Q = 0 (vertical line) → thermally insulated W = 0 (vertical line) → constant volume process

A closed system also undergoes cyclic processes.

Cycle: 2 or more processes that return to initial point (i.e. 𝑑U = 0)

∮ Q = ∮ W → ∮(Q − W) = 0

(Q − W) is path independent and is a thermodynamic property

For any arbitrary process i, (Q − W)i = ∆E where ∆E = energy accumulated / dissipated in the system

Equivalently, Q ⃗⃗⃗⃗⃗⃗
AB − W⃗⃗⃗⃗⃗⃗
AB = UB − UA when process proceeds from state A to state B
For constant pressure process, Q ⃗⃗⃗⃗⃗⃗
AB = H⃗⃗⃗⃗⃗⃗
AB and that Q ⃗⃗⃗⃗⃗⃗
(T )
AB = m ∙ Cp B − TA as well as W = P(V2 − V1
)
For constant vol. process, Q ⃗⃗⃗⃗⃗⃗ (T
CD = m ∙ Cv D − TC
)
For other process, calculate U → calculate W → therefore Q = U + W

1 ∂Q
Cv = ( ) where the volume is kept constant and is a thermal property
m ∂T v
R0
Cp − Cv = R =
MR

Joule’s Law: applicable to any process for closed system of ideal gas to calculate internal energy  U2 − U1 = mCv (T2 − T1 )
* to calculate enthalpy: dH = mCp dT = dU + P ∙ dV

General Ideal Gas Relations

n = 0: isobaric PA VA − PB VB mR
WAB
⃗⃗⃗⃗⃗⃗ = = (T − T2 )
polytropic process: Pv = k n n = 1: isothermal (∆U = 0)* n − 1n−1 n−1 1
reversible only for these powers → n = γ: isentropic (reversible adiabatic) T2 V n−1 P n
n = ∞: isochoric / vertical line (W = 0) T1
= (V1 ) = (P2 ) (given limited information)
2 1

2 v
Isothermal: Q = W → Q AB
⃗⃗⃗⃗⃗⃗ = mRT ln (v )
1

B) Open system
dm dx
= ρAv = ρA
dt dt
dE dQ 𝑑min v2 dW 𝑑mout v2 dU
= +[ × (h + + gz) ] − [ + × (h + + gz) ] ≈
dt dt 𝑑t 2 in
dt 𝑑t 2 out
dt

v2
E = m (u + + gz)
2

Steady Flow Energy Equation

dQ dW dm v2 dm v2
− + ∑ ( ) (h + + gz) − ∑ ( ) (h + + gz) = 0
dt dt dt 2 dt 2
in out

Alternatively, in intensive differential form: dq − dw = dh + v ∙ dv + g ∙ dz

non–moving boundary: W = 0

———————————————————————————————————————————————————————

2ND Law Of Thermodynamics

· Quality of energy is considered, and refers to the disorder of the system (disorder here refers to the ‘heat’)

· Hot reservoir @ constant temperature TH ; Cold reservoir @ constant temperature TL

· Heat engine  undergoes cyclic process and produces work

· Reverse heat engine (e.g. refrigerator & heat pump)  heat transfer from low to high temperature reservoir with the help
of external work
· Kelvin–Planck statement: impossible to construct cyclic device where the only external effects are the transfer of heat from
a single reservoir and the transfer from it, an equivalent amount of work (i.e. no 100% efficiency)

· Clausius statement: impossible to construct a cyclic device that will produce no effect other than the transfer of heat from
cooler to hotter object (i.e. W input to Tcold needed to obtain Thot )
W Q T
· heat engine: ηideal = Q = 1 − Q L = 1 − T L
in H H

desired output
· reverse heat engine, COP = Wnet input
(refrigerator → desired output = Q H ; heat pump → desired output = Q L )

For all reversible (ideal) engines, quantity of heat rejected to the reservoirs (|Q|) ∝ absolute temperature (T)
|Q| |Q H | |Q L |
= constant → = =𝑘
T TH TL
Wreal TL
ηreal < ηideal → <1−
Q1, real TH

Clausius–Clapeyron Equation
hfg dT hg − hf dT
T= ( )= ( )
vfg dP vg − vf dP

Clausius Inequality
dQ
∮ ≤0
T
Reversible process Non–reversible process
dQ Q H Q L dQ
∮ = + =0 ∮ <0
T TH TL T
note: apply sign convention (heat loss: Q L < 0)

Entropy

Q = ∫ T dS

· measure of molecular disorder / randomness of a system

· only associated with heat transfer (Q), and not W or U

· portion of heat is reduced to disorder (cannot be used) due to irreversibility e.g. friction → ‘quality of heat’ drops
2 dQ
· reversible process: ∆S = S2 − S1 = ∫1 ( T )  path independent
rev.

2 dQ 2 dQ
· irreversible process: ∫1 ( T ) = ∫1 ( T ) + Sgen.
rev. irrev.

· isentropic process: adiabatic & reversible (constant entropy)  ∆S = S2 − S1  vertical line on T − S diagram
∆Q
· special case 1 – constant temperature (hot / cold reservoir, isothermal process):∆S = T0

· special case 2 – isolated system containing all sub–systems (S1 , S2 …): n o external heat exchange (only internal) so s2 = 0
and ∆S = ∆SA + ∆SB + ∆SC + ⋯ ∆Sn = Sgen. ≥ 0 ← used to test if process undertaken by system is possible

· special case 3 – for a cycle, sgen. = sum of sgen. for each process
Entropy Balance

∆Ssystem = Sin − Sout + Sgen.

General case
* Sin − Sout : net entropy transfer by heat & mass
n
Qn
∆Ssystem = ∑ + Sgen.
Closed system Tn
1
Qn
Tn
: entropy by heat transfer across the boundary
n
Qn
∆Ssystem = ∑ + [∑ min × sin − ∑ mout × sout ] + Sgen.
Tn
1
∑ min × sin − ∑ mout × sout : entropy transfer by mass flow

n
Open system d(Ssystem ) Q ṅ
=∑ + [∑ ṁin × sin − ∑ ṁout × sout ] + Ṡgen.
dt Tn
1
as entropy is only associated with heat transfer (Q), all other energies (u, Pv, w, etc.) are not considered
dm gen. dm d(S )
Steady flow: 0 = ∑in (s ∙ dt ) − ∑out (s ∙ dt ) + dt
e.g. turbines, compressors, nozzles, diffusers, heat exchangers, pipes, ducts

· Carnot cycle (reversible): 2 isothermal + 2 adiabatic

TH
· reverse heat engine: COPheat pump = T = COPrefrigerator + 1
H −TL

Entropy changes of water, steam & gas

· ∆s of water & steam: wet steam (refer to pressure table) → s = x ∙ sg + (1 − x)sf = sf + x ∙ sfg

· ∆s of gas (I): Gibbs Equation → T ∙ ds = P ∙ dv + du = dh − v ∙ dP = Cp ∙ dT − v ∙ dP (intensive units!)

T P T V
· ∆s of gas (II): s2 − s1 = Cp ln (T2 ) − R ln (P2 ) = Cv ln (T2 ) + R ln (V2 )
1 1 1 1

Isentropic pump / compressor: dw = v ∙ dp → w = vf × ∆P = CP × ∆T = (Cv + R)∆T (do not calculate 𝐡𝟐 − 𝐡𝟏 )

Within rigid tank,

1. mass is constant
2. there is work done in expanding after partition breaks
 PART 2

Entropy is increased by:

1. transfer of thermal energy

2. disturbance induced to fluid during flow process

scope: all compression + expansion processes are assumed isentropic (∆s = 0 / straight line T − s diagram)

In a cycle, there are 4 elements:

Compressor (Win ) Boiler (Q H ) Turbine (Wout ) Condenser (Q L )

causes ∆T → results in ∆P @ constant temperature perfectly insulated perfectly insulated
 Wnet Wnet
MEP = =
Vmax. − Vmin. Vdisplacement

Vmax.
rv =
Vmin.
 T1
ηmax. = 1 − = 1 − rv1−γ
T2

rc γ − 1
ηmax. = 1 − rv1−γ
γ(rc − 1)

rc γ rp − 1
ηmax. = 1 − rv1−γ
rp − 1 + γ ∙ rp (rc − 1)

ηdiesel < ηdual < ηotto

 heat exchange occurs @ constant pressure
1−γ
Wturbine − Wcompressor PBγ
ηth = =1−( )
QH PA

Brayton Cycle: Aircraft Gas Turbine

Turbine: Wt ≈ Wc → h4 − h5 = h3 − h2
v6 2
Velocity of air: h5 = h6 + 2

Brayton Cycle w/ Intercooling & Regeneration

Refrigeration

Carnot refrigeration cycle

vapor compression cycle

For both carnot & vapor compression cycles,

 Work input (compressor): h2 − h1 = vg (P2 − P1 )

 Orifice / throttling device: from qH to qL , specific enthalpy (h) is constant i.e. h = hf  from 1st law

Cascade refrigeration cycle

Gas Mixtures

Assumption: non–reacting
Ki
Terminology: k̅ i =
ni
m
ni ( ̅̅̅i )
M
Important relation: xi = n
= m
(̅̅̅)
M

1) Mass fraction: mmixture = m1 + m2 + m3 + ⋯

2) Avogadro’s hypothesis: equal volumes of all gases at the same temperature contain the same number of molecules

ni Pi Vi
3) Mole fraction: xi = = =
n P V

4) Partial pressure: Pmixture = P1 + P2 + P3 + ⋯

5) Partial volumes: Vtotal = V1 + V2 + V3 + ⋯

U U1 +U2 +U3 +⋯ ∑ ni ∙u
̅i
6) Internal energy: = = = ∑ xi ∙ u̅i
n n n
 H
7) Enthalpy: n = ∑ xi ∙ h̅i

8) Specific heats: C̅v = ∑ xi ∙ C̅v,i and that C̅p = ∑ xi ∙ C̅p,i

̅ = ∑ xi ∙ R
9) Universal gas constant: R ̅

10) Entropy (I): Smixture = ∑(ni ∙ S̅i ) and that S̅ = ∑(xi ∙ S̅i )

dTi dPi 1 dTi dPi

11) Entropy (II): dS̅ = ∑ (xi C̅p Ti
) ̅
− ∑ (xi R Pi
) = m [∑ (xi ∙ Cp,i ∙ Ti
)− ∑ (xi ∙ R i ∙
Pi
)]

Psychometric Properties

Air–water vapor mixtures

saturated air: 𝑔 line on T − s diagram

unsaturated air: mixture of dry air vapor (vapor region)
dew point: state at which dew begins to form when unsaturated air is cooled (𝑔 line, but follow isobar)

mvapor P R P
specific humidity / humidity ratio: ω = mair
= Rv Pa ≈ 0.622 Pv
v a a
* water vapor + air assumed to be ideal gases
ωP
alternative equation: Pv =
0.622+ω

P v
relative humidity: ϕ = Pv = vg where 𝑔 is at the saturated state and v is the actual state considered
g v

H mvapor
enthalpy for air–water vapor mixture: m = hair + mair
hvapor = hair + ω ∙ hvapor
air
to find the change, = Cp,air ∆T

entropy for air–water vapor mixture: s = sair + ω ∙ svapor

dh v dT R
ds = − ∙ dP = Cp ∙ − ∙ dP
T T T P

adiabatic saturation temperature: ω1 hv1 + (ω2 − ω1 )hf2 = Cp,air (T2 − T1 ) + ω2 (hf2 + hfg2 )