CHAPTER 1 Fundamentals of Power plant Engineering

Part 1A: THERMODYNAMICS

Thermodynamics is a branch of physical sciences that treats various phenomena of energy and the
related properties of matter, especially of the laws of transformation of heat into other forms of energy and
vice versa.

Temperature and Pressure

Temperature is an indication of degree of hotness and coldness and thereof a measure of intensity of
heat.

THE SIX TEMPERATURE SCALES

1. Celsius or Centigrade 3. Kelvin 5. Reamur

2. Fahrenheit 4. Rankine 6. Ligem

ABSOLUTE TEMPERATURE – is the temperature measured from absolute zero.

ABSOLUTE ZERO – is the temperature at which the molecules stop moving.

RELATIONS BETWEEN TEMPERATURE SCALES

9 5
F= C+32 C = (F − 32)
5 9

R = F + 460 K = C + 273

Where: R = Rankine (absolute temp scale; English)

K = Kelvin (absolute temp scale; SI)

TEMPERATURE INTERVAL – is the difference between two temperature readings from the same scale,
and the change in temperature through which the body is heated.

THE TEMPERATURE INTERVAL/CHANGE:

∆TK = ∆TC 5 4
∆𝑇𝐶 = ∆𝑇𝐹 1°Re = ℃
9 5
∆TR = ∆TF
9 1°RE = 20°L
∆TF = ∆T
5 C

Note: 1C° = 9/5 F° and degree is to be written after the scale to indicate that it is temperature change.

❖ The Absolute Pressure and Gage Pressure:

Absolute pressure – is the true pressure measured above a perfect vacuum.

Gage Pressure – is the pressure measured from the lvel of atmospheric pressure by most pressure
recording instrument like pressure gage and open-ended manometer.

Atmospheric pressure – is the pressure obtained from barometric reading.

Pabs = Pgage + Patm

Where: Pabs = absolute pressure

 Pgage = gage pressure
Patm = atmospheric pressure
Patm = 14.7 psi
= 1.032 kg/cm2
= 29.92 in. Hg
= 760 mmHg = 760 torr
= 1.013 bar
= 760 torr
= 1 atm
= 0 Kpag
= 0 psig

Note: Pgage = negative (-) when the pressure is vacuum. The pressure in perfect vacuum is -101.325 Kpa

❖ Thermodynamic System and Surroundings:

System – is the term given to the collection of matter under consideration enclosed within a boundary.
Surrounding – is the region outside the boundary or the space and matter external to a system.
Closed System – is a system in which there is no transfer of matter across the boundary.
Open System – is a system in which there is a flow of matter across the boundary.
Isolated System – is a system in which neither mass nor energy cross the boundaries and it is not
influenced by the surroundings.
Nonflow process – is a process that takes place in a closed system. Example: Compressor
Steady flow process – is a process that takes place in an open system in which the quantity of matter
within the system is constant. Example: Turbine

Non-flow work – is the work in a non-flow process

2
𝑊𝑛 = ∫1 𝑃𝑑𝑉 = 𝑃(𝑉2 − 𝑉1 )

Steady flow work – is the work in a steady flow process.

2
𝑊𝑛 = ∫1 𝑉𝑑𝑃 = 𝑉(𝑃1 − 𝑃2 )

Where: P = pressure
V = volume

❖ LAW OF CONSERVATION OF MASS

Mass (m1) entering the system is equal to the sum of the stored mass (Δm) and the mass (m 2) that
leaves the systems.

m1 = m2 + ∆m
∆m = m1 − m2

❖ STEADY FLOW SYSTEM (ΔM = 0)

m1 = m2
A1 V1 ρ1 = A2 V2 ρ2

where:
A = cross-sectional area
V = velocity
ρ = density
❖ HEAT and ENTROPY:
Heat – is a form of energy associated with the kinetic random motion of large number of molecules.
Sensible Heat – is the heat needed to change the temperature of the body without changing its
phase.

Q s = mC∆T
Where: Qs = sensible heat
m = mass
ΔT = change in temperature
C = specific heat

Example: Cwater = 1.0 Cal/gm.C° = 1.0BTU/lb.F° = 4.186 kJ/kg.K

Latent Heat – is the heat needed by the body to change its phase without changing its temperature.

Latent Heat of Fusion – solid to liquid

Latent Heat of Vaporization – liquid gas.

Example: melting of ice at 0°C

Q L = ±mL
QL = heat needed
m = mass
L = latent heat (fusion/vaporization)
(+) = heat entering; substance melts
(-) = heat leaving; substance freezes

Note: Latent heat of fusion of ice (Lf)

Lf = 144 BTU⁄lb = 334 kJ⁄kg = 80 Cal⁄gm

Latent heat of Vaporization of boiling water (Lv)

Lv = 970 BTU⁄lb = 2257 kJ⁄kg = 540 Cal⁄gm

ENTROPY – is the measure of the randomness of the molecules of a substance.

❖ ENTHALPY AND INTERNAL ENEGY:

Enthalpy – is the heat energy transferred to a substance at a constant pressure process.

Internal Energy – is the energy stored within the body. It is the sum of the kinetic energies of all its
constituent particles plus the sum of all the potential energies of interaction among
these particles.

H = U + PV
Where: H = enthalpy
U = internal energy
P = absolute pressure
V = volume

❖ FIRST LAW OF THERMODYNAMICS: (LAW OF CONSERVATION OF ENERGY)

Energy cannot be created nor be destroyed, it can only be transformed from form to another.

𝑆𝑢𝑚 𝐸𝑛𝑒𝑟𝑔𝑦 𝐸𝑛𝑡𝑒𝑟𝑖𝑛𝑔 = 𝑆𝑢𝑚 𝑜𝑓 𝐸𝑛𝑒𝑟𝑔𝑦 𝐿𝑒𝑎𝑣𝑖𝑛𝑔

 𝑃𝐸1 + 𝐾𝐸1 + 𝐻1 + 𝑄 = 𝑃𝐸2 + 𝐾𝐸2 + 𝐻2 + 𝑊

Where:
PE = potential energy
KE = kinetic energy
H = U + PV
H = enthalpy

❖ Zeroth Law of Thermodynamics

When any two bodies are in thermal equilibrium with the third body, they are in thermal equilibrium
with each other.

❖ Ideal Gas of Perfect Gas Law

Perfect Gas – is a theoretically ideal gas which strictly follows Boyle’s law and Charles’ law of gases.

The equation of State of an Ideal Gas:

PV = mRT ̅T
PV = nR

P = Pressure (in absolute) ̅ = universal gas constant

R
V = Volume m = mass (in kg, gm & lb)
R = gas constant n = no. of moles
MW = molecular weight (in kg/mol, gm/mol & lb/mol)

✓ The value of universal gas constant:

J ft − lb L − atm
̅ = 8.314
R = 1545 = 0.821
mol. K mol. °R mol. K

✓ Finding the value of the gas constant given the universal gas constant:

̅
R 8.314J⁄mol−K
Examples R(O2 ) = =
MW(O2 ) 32kg⁄mol
R(O2 ) = 0.268 J⁄kg − K
̅
R 1545 ft − lbf ⁄mol − °R
R(O2 ) = =
MW(O2 ) 32 lbm ⁄mol
ft − lbf
R(O2 ) = 48.28
lbm − °R

✓ The gas constant of Air:

ft − lbf kJ
R air = 53.34 = 0.287
lbm °R kg − K

❖ RELATION BETWEEN Cp, Cv, k & R:

Cp − Cv = R Cp
=k
Cv
R kR
Cv = Cp =
k−1 k−1
 Where: Cp = specific heat at constant pressure
Cv = specific heat at constant volume
K = specific ratio
R = gas constant

❖ Second Law of Thermodynamics

Heat cannot be transferred from cold body to a hot body without an input of work. It similarly states
that heat cannot be converted 100% into work. The bottom line is that an engine must operate
between a hot and cold reservoir. Also indicated, is that energy cannot naturally move from realm of
lower potential to a realm of higher potential.

CLAUSIUS STATEMENT APPLIED TO HEAT PUMP

It is impossible to construct a heat pump that operates without an input work.

KELVIN-PLANK STATEMENT APPLIED TO HEAT ENGINE

It is impossible to construct a heat engine which operates in a cycle and receives a given amount of
heat from high temperature body and does an equal amount of work.

❖ Third Law of Thermodynamics

The total entropy of pure substances approaches zero as the absolute thermodynamic temperature
approaches zero.
Note: The third body is usually a thermometer.

❖ BOYLE’S LAW
In a confined gas, if the absolute temperature is held constant, the volume is inversely proportional to
the absolute pressure.

P1 V1 = P2 V2

❖ CHARLE’S LAW
In a confined gas, if the absolute pressure is held constant, the volume is directly proportional to ht
absolute temperature.

V1 V
= T2
T1 2

in a confined gas, if the volume is held constant, the absolute pressure is directly proportional to the
absolute temperature.

P1 P
= T2
T1 2

❖ GENERAL GAS LAW

(Combined Charles’ & Boyle’s Laws)
Each one of these how one quantity varies with another if the third quantity remains unchanged, but if
the three quantities change simultaneously, it is necessary to combine these laws in order to
determine the final conditions of the gas.
 P1 V1 P2 V2
=
T1 T2

❖ DALTON’S LAW OF PARTIAL PRESSURE

The pressure exerted in a vessel by a mixture of gases is equal to the sum of the pressures that each
separate gas would exert if it alone occupied the whole volume of the vessel.

P = P1 + P2 + P3 + ⋯ + Pn

Where:
P = total pressure of the mixture
P1, P2, P3, …, Pn = partial pressure of gases
Partial pressure – is the pressure exerted by each gas.

❖ AVOGADRO’S LAW
At equal volume, at the same temperature and pressure conditions, the gases contain the same
number of molecules.

m1 M R1 M
= M1 or = M1
m2 2 R2 2

Where:
m1 & m2 are masses
M1 & M2 are Molecular weights
R1 & R2 are gas constants

❖ THE INTERNAL ENERGY, ENTHALPY and ENTROPY OF AN IDEAL GAS

T
∆U = mCv ∆T ∆S = mC ln ( 2)
T1

∆H = mCp ∆T

Where:

ΔU is the change in Internal Energy

ΔH is the change in Enthalpy
ΔS is the change in Entropy
m is the mass
C is the specific heat
ΔT is the change in temperature

❖ THERMODYNAMIC PROCESSES

PROCESSES
Whenever a system undergoes any change, it is said to have described a process.

REVERSIBLE AND IRREVERSIBLE PROCESSES

Reversible process – is one in which both the system and the surroundings are returned to their
original states, otherwise it is an irreversible process.

CYCLIC PROCESS
Cyclic Process – a process which gives the same states/conditions after the system undergoes a
series of processes.

QUASI-STATIC PROCESS
 Quasi-static Process – is one in which system departs from equilibrium state only infinitesimally at
every instant.

❖ PROCESS OF IDEAL GAS

1. Isometric or Isochoric or Isovolumic Process

isometric process is an internally reversible constant volume process of a working substance.
a. PV and T relation (V=C)

P1 P2
=
T1 T2
b. PV and TS Diagram

c. Nonflow Work
2
Wn = ∫1 PdV where: dV = 0 @ V = C
Wn = 0

d. Steady Flow Work

2
Ws = ∫ VdP = V(P1 − P2 )
1

e. Change in Internal Energy

∆𝑈 = 𝑚𝐶𝑣 ∆𝑇

f. Change in Enthalpy

∆𝐻 = 𝑚𝐶𝑝 ∆𝑇

g. Heat Transferred

𝑄 = ∆𝑈 + 𝑊𝑛
𝑊𝑛 = 0 @ 𝑉 = 𝐶
𝑄 = ∆𝑈

h. Change in Entropy

𝑑𝑄
𝑓𝑟𝑜𝑚: 𝑑𝑆 =
𝑇
 𝑇2
∆𝑆 = 𝑚𝐶𝑣 ln ( )
𝑇1

2. Isobaric Process

Isobaric Process is an internally reversible constant pressure process of a working substance.

a. PV an T relation (P = C)

𝑉1 𝑉
= 𝑇2 (Charles’ Law)
𝑇1 2

b. PV and TS Diagram

c. Nonflow Work
2

Wn = ∫ Pdv
1
Wn = P(V2 − V1 )

d. Steady Flow Work

2

Wn = ∫ VdP where ∶ dP = 0 @ P = C
1
Ws = 0

e. Change in Internal Energy

∆U = mCv ∆T

f. Change in Enthalpy

∆H = mCp ∆T

g. Heat Transferred

Q = ∆U + Wn
Q = ∆U + ∆PV
Q = ∆H

h. Change in Entropy

dQ
from ∶ dS =
dT
 T2
∆S = mCp ln ( )
T1

3. Isothermal Process
Isothermal Process is an internally reversible constant temperature process of a working
substance.

a. PV and T relation ( T = C )

P1 V1 = P2 V2 (Boyle′ s Law)

b. PV and TS Diagram

c. Nonflow Work
2

Wn = ∫ PdV
1

C
Where: P =
v

V2 V2
Wn = P1 V1 ln ( ) = mRT ln ( )
V1 V1

d. Steady Flow Work

2

Ws = ∫ VdP where: V = C/P

1
V2 V2
Ws = −P1 V1 ln ( ) = −mRT1 ln ( )
V1 V1
Ws = Wn

e. Change in Internal Energy

∆𝑈 = 0

f. Change in Enthalpy

∆𝐻 = 0

g. Heat Transferred

𝑄 = ∆𝑈 + 𝑊𝑛

∆𝑆 = 𝑊𝑛 (𝑓𝑜𝑟 ∆𝑈 = 0)
 h. Change in Entropy

𝑑𝑄
𝑓𝑟𝑜𝑚 ∶ 𝑑𝑆 =
𝑇
𝑉2
∆𝑆 = 𝑚𝑅 ln ( )
𝑉1

4. Isentropic Process
Isentropic Process is an internally reversible constant entropy process of a working substance. It
is also known as a reversible adiabatic process.
Adiabatic Process is a reversible process in which there is no flow of heat between a system and
its surroundings ( Q = 0 ).

a. PV and T relations ( S = C or PVk = C )

P1 V1k = P2 V2k ⇒ pressure − volume

T1 V2 k−1
=( ) ⇒ temperature − volume
T2 V1
k−1
T1 P1 k
=( ) ⇒ temperature − pressure
T2 P2

b. PV and TS Diagram

c. Nonflow Work
2
C
Wn = ∫ Pdv where: P =
Vk
1

P2 V2 − P1 V1 mR(T2 − T1 )
Wn = =
1−k 1−k

d. Steady Flow Work

2
C
Ws = ∫ VdP where: V =
Pk
1

k(P2 V2 − P1 V1 ) kmR(T2 − T1 )
Ws = =
1−k 1−k
 Ws = kWn

e. Change in Internal Energy

∆𝐻 = 𝑚𝐶𝑣 ∆𝑇 = −𝑊𝑛

f. Change in Enthalpy

∆𝐻 = 𝑚𝐶𝑝 ∆𝑇

g. Heat Transferred ( Adiabatic; Q = 0 )

𝑄=0

𝑄 = ∆𝑈 + 𝑊𝑛

∆𝑈 = −𝑊𝑛

h. Change in Entropy

∆𝑆 = 0

5. Polytropic Process
Polytropic Process is an internally reversible process during which PVn = C, where; n is constant.

a. PV and T relation (PVn = C)

𝑃1 𝑉1𝑛 = 𝑃2 𝑉2𝑛 ⇒ 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 − 𝑣𝑜𝑙𝑢𝑚𝑒

𝑇1 𝑉2 𝑛−1
=( ) ⇒ 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 − 𝑣𝑜𝑙𝑢𝑚𝑒
𝑇2 𝑉1
𝑛−1
𝑇1 𝑃1 𝑛
=( ) ⇒ 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 − 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
𝑇2 𝑃2

b. PV and TS Diagram

c. Nonflow Work
 2
C
Wn = ∫ Pdv where: P =
Vn
1

P2 V2 − P1 V1 mR(T2 − T1 )
Wn = =
1−n 1−n

d. Steady Flow Work

2
C
Ws = ∫ VdP where: V =
P𝑛
1

k(P2 V2 − P1 V1 ) kmR(T2 − T1 )
Ws = =
1−n 1−n

Ws = nWn

e. Change in Internal Energy

∆U = mCv ∆T

f. Change in Enthalpy

∆H = mCp ∆T

g. Heat Transferred

Q = mCn ∆T
k−n
Where; Cn = Cv ( )
1−n

h. Change in Entropy

dQ
from: dS =
T
T2
∆S = mCn ln ( )
T1

❖ GENERAL EQUATION FOR THERMODYNAMIC CURVES

The general equation of any process is:

𝐏𝐕 𝐧 = 𝐂

If:

n = 0; Isobaric Process
n = 1; Isothermal Process
n = k; Isentropic Process
n = −∞ to +∞; Polytropic Process
n = ∞; Isometric Process

Note: PVk is steeper than PV curve;

❖ MIXTURE INVOLVING IDEAL GASES:

 1. Mass or Gravimetric Analysis:

𝑚 = 𝑚1 + 𝑚2 + 𝑚3 + ⋯ + 𝑚𝑛

2. Volumetric or Molal Analysis

𝑣 = 𝑣1 + 𝑣2 + 𝑣3 + ⋯ + 𝑣4

3. Dalton’s Law of Partial Pressure

𝑃 = 𝑃1 + 𝑃2 + 𝑃3 + ⋯ + 𝑃𝑛

4. Specific Heat of the Mixture

𝑚𝐶𝑝 = 𝑚𝐶𝑝1 + 𝑚𝐶𝑝2 + 𝑚𝐶𝑝3 + ⋯ + 𝑚𝐶𝑝𝑛

𝑚𝐶𝑣 = 𝑚𝐶𝑣1 + 𝑚𝐶𝑣2 + 𝑚𝐶𝑣3 + ⋯ + 𝑚𝐶𝑣𝑛

Where: Cp = Specific heat at constant pressure

Cv = Specific heat at constant volume

❖ TWO PHASE SYSTEM (PURE – SUBSTANCE)

1. Pure Substance – is a working substance that has homogeneous and invariable chemical
composition even though there is a change of phase. Example, H2O and NH3
2. Working Substance – is a working substance which energy can be stored or from which energy
can be removed.

❖ THE PV and TS DIAGRAM OF PURE SUBSTANCE

❖ PROPERTIES OF PURE SUBSTANCE WITH ILLUSTRATIONS:

Saturation Temperature is the temperature at which liquid start to boil and vapour start to condense.
Example, water boils @ 200°C at a pressure of 1.55 MPa

Subcooled Liquid is a liquid whose temperature is lower than the saturation temperature
corresponding to the given pressure.
Example, Liquid water @ 50°C and 97.78 kPa is Subcooled liquid since 50°C is less than 99°C (t sat
@ 97.78 kPa).
Compressed Liquid is a liquid whose pressure is higher than the saturation pressure corresponding
to the given temperature.
Example, Liquid water @ 105 kPa and 90°C is a compressed since 105 kPa is greater than
70.14kPa (Psat @ 90°C).

Saturated Liquid is a liquid at the saturation temperature or pressure whose temperature is equal to
the boiling point corresponding to the given pressure. It has no vapour content.
Example, Liquid water @ 50°C and 12.35 kPa is a saturated liquid since the given temperature,
50°C is equal to 50°C (tsat @ 12.35 kPa).

Vapor is the term given to a gaseous phase that is in contact with the liquid phase.
Example, steam

Saturated Vapor is a vapor whose temperature and pressure. It has no liquid or moisture content.
Example, Steam at 300°C and 8.58 MPa

Superheated Vapor is a vapor whose temperature is higher than the saturation temperature
corresponding to the given pressure.
Example, Steam @ 350°C and 11.8 MPa is a superheated vapor since 350°C is greater than
324°C (tsat @ 11.88 Mpa)

Degrees Superheat is the difference between the actual superheated temperature and the saturation
temperature.

Example, find the degrees superheated of steam at 350°C and 8.58 MPa.

Solution:

°SH = actual temp − sat. temp

Where: sat temp = 300°C (tsat @ 8.58 MPa)
Actual temp. = 50 °C

Thus; °SH = 20°C

Degrees Subcooled is the difference between the saturation temperature and the actual Subcooled
temperature.

Example, Find the degrees Subcooled of liquid water @ 50°C and 19.94 MPa.

Solution:

°SC = sat. temp − actual temp

Where: sat temp = 60°C ( tsat @ 19.94 kPa )

Actual temp. = 50 °C

Thus; °SC = 10°C

Wet Vapor is a mixture of saturated vapor and saturated liquid.

Example, Steam @ 15 MPa and 310°C is a wet vapor since 310°C is less than 342.24°C (tsat @ 15
MPa).

Quality of Wet Vapor is the fraction or percentage by weight that is saturated vapor. It is the ratio of
the mass of saturated vapor to the total mass of the mixture.
 mg
x= mt
Where: x = quality of wet vapor
mg = mass of vapor
mt = mass of the mixture (wet vapor)

Percentage moisture is the fraction or percentage by weight that is saturated liquid. It is the ratio of
the mass of saturated liquid to the total mass of the mixture.

m
y = mf
t

Where: y = percentage moisture

mf = mass of liquid
mt = mass of the mixture ( wet vapor)

Critical Point is a point that represents the pressure and temperature at which liquid and vapor can
coexist in equilibrium.

❖ PROPERTIES OF MIXTURE (WET VAPOR / STEAM)

1. 𝑣 = 𝑣𝑓 + 𝑥𝑣𝑓𝑔
2. 𝑢 = 𝑢𝑓 + 𝑥𝑢𝑓𝑔
3. ℎ = ℎ𝑓 + 𝑥ℎ𝑓𝑔
4. 𝑠 = 𝑠𝑓 + 𝑥𝑠𝑓𝑔
Note: 𝑥 = 1 − 𝑦
5. 𝑣𝑓𝑔 = 𝑣𝑔 − 𝑣𝑓
6. 𝑢𝑓𝑔 = 𝑢𝑔 − 𝑢𝑓
7. ℎ𝑓𝑔 = ℎ𝑔 − ℎ𝑓
8. 𝑠𝑓𝑔 = 𝑠𝑔 − 𝑠𝑓

Guide to determine the phase of the working substance

𝐼𝑓 𝑃𝑔𝑖𝑣𝑒𝑛 > 𝑃𝑠𝑎𝑡 → 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑙𝑖𝑞𝑢𝑖𝑑

𝐼𝑓 𝑇𝑔𝑖𝑣𝑒𝑛 < 𝑇𝑠𝑎𝑡 → 𝑠𝑢𝑏𝑐𝑜𝑜𝑙𝑒𝑑 𝑙𝑖𝑞𝑢𝑖𝑑
𝐼𝑓 𝑃𝑔𝑖𝑣𝑒𝑛 < 𝑃𝑠𝑎𝑡 → 𝑠𝑢𝑝𝑒𝑟ℎ𝑒𝑎𝑡𝑒𝑑 𝑣𝑎𝑝𝑜𝑟
𝐼𝑓 𝑇𝑔𝑖𝑣𝑒𝑛 > 𝑇𝑠𝑎𝑡 → 𝑠𝑢𝑝𝑒𝑟ℎ𝑒𝑎𝑡𝑒𝑑 𝑣𝑎𝑝𝑜𝑟

Where:

v = specific volume
u = specific internal energy
h = specific enthalpy
s = specific entropy
x = quality
f = fluid
g = vapor
fg = mixture change by evaporation
y = % moisture

Note: The properties of liquid and vapor at a given temperature and pressure can be found in Steam table
and Mollier chart.
❖ PROCESS INVOLVING PURE SUBSTANCE
Vapors (steam) and Ideal gases have similarity in their processes but ideal gas equations are not
generally applicable to vapors.

1. Isometric or Isochoric or Isovolumic Process

Isometric process is an internally reversible constant volume process of a pure substance.
a. PV and TS Diagram

b. Nonflow Work
2

Wn = ∫ PdV
1
Where: dV = 0 @ V
Wn = 0

c. Steady Flow Work

2

Ws = ∫ VdP = V(P1 − P2 )
1

Ws = Q − ∆KE − ∆PE − ∆H

d. Change in Internal Energy

∆U = m(u2 − u1 )

e. Change in Enthalpy

∆H = m(h2 − h1 )

f. Heat Transferred

Q = ∆U
Q = m(u2 − u1 )

g. Change in Entropy
∆S = m(s2 − s1 )

2. Isobaric Process
Isobaric process is a constant pressure process of a pure substance.

a. PV and TS Diagram
 b. Nonflow Work
2

Wn = ∫ PdV = P(V2 − V2 ) = mP(v2 − v1 )

1

c. Steady Flow Work

2

Ws = − ∫ VdP
1
Ws = 0 where: dP = 0 ( P = C )

d. Change in Internal Energy

∆U = m(u2 − u1 )

e. Change in Enthalpy

∆H = m(h2 − h1 )

f. Heat Transferred

Q = u2 − u1 + Wn
Q = h2 − h1

g. Change in Entropy

∆S = m(s2 − s1 )

3. Isothermal Process
Isothermal process is an internally reversible constant temperature process of a pure substance.

a. PV and TS Diagram
 b. Nonflow Work

Wn = Q − ∆U

c. Steady Flow Work

Ws = Q − ∆KE − ∆PE − ∆H

d. Change in Internal Energy

∆U = m(u2 − u1 )

e. Change in Enthalpy

∆H = m(h2 − h1 )

f. Heat Transferred

Q = T∆S = T(S2 − S1 )

g. Change in Entropy

∆S = m(s2 − s1 )

4. Isentropic or Adiabatic Process

Isentropic process is a constant entropy process of a pure substance. It is a process in which no
heat is transferred.

a. PV and TS Diagram

b. Nonflow Work

Wn = Q − ∆U where ; Q = 0 (adiabatic)
Wn = −∆U
 c. Steady Flow Work

Ws = −∆KE − ∆PE − ∆H

d. Change in Internal Energy

∆U = m(u2 − u1 )

e. Change in Enthalpy

∆H = m(h2 − h1 )

f. Heat Transferred

Q=0

g. Change in Entropy

∆S = 0

5. Isenthalpic or Throttling Process

Isenthalpic process is a constant enthalpy process of a pure substance. It is a process in which
no work is being done and is utilized in getting the quality of steam.

a. PV, TS and HS Diagram

h2 = h1 = hf1 + x1 hfg1
h2 = hg + Cp ∆T (Approx. )

b. Steady Flow Work

Ws = 0
 c. Change in Internal Energy

∆U = m(u2 − u1 )

d. Change in Enthalpy

∆H = 0

e. Heat Transferred

Q=0

f. Change of Entropy

∆S = m(s2 − s1 )

6. Polytropic Process
Polytropic process is reversible process during which PVn = C, where; n is a constant.

a. PV and TS Diagram

b. Nonflow Work
2
P2 V2 − P1 V1
Wn = ∫ PdV =
1−n
1

c. Steady Flow Work

2
n(P2 V2 − P1 V1 )
Ws = ∫ PdV =
1−n
1
Ws = nWn ; Ws = Q − ∆KE − ∆PE − ∆H (with ∆KE)

d. Change in Internal Energy

∆U = m(u2 − u1 )

e. Change in Enthalpy

∆H = m(h2 − h1 )

f. Heat Transferred
 𝑄 = ∆𝑈 + 𝑊𝑛

g. Change in Entropy

∆𝑆 = 𝑚(𝑠2 − 𝑠1 )

❖ CARNOT CYCLE
Carnot Cycle – is the most efficient thermodynamic cycle. It consists of two isothermal and two
isentropic processes.

Process 1 – 2: Isentropic Expansion

Process 2 – 3: Isothermal Compression
Process 3 – 4: Isentropic Compression
Process 4 – 1: Isothermal Expansion
➢ Analysis of Carnot Cycle:

a. Heat Added (QA)

𝑄𝐴 = 𝑇1 (𝑆1 − 𝑆2 )

b. Heat Rejected

𝑄𝑅 = 𝑇2 (𝑆2 − 𝑆3 )
𝑄𝑅 = 𝑇2 (𝑆1 − 𝑆4 )

c. Net Work or Work Done:

𝑊𝑛𝑒𝑡 = 𝑄𝐴 − 𝑄𝑅
𝑊𝑛𝑒𝑡 = (𝑇1 − 𝑇2 )(𝑆1 − 𝑆4 )
 d. Cycle Efficiency

𝑊𝑛𝑒𝑡 𝑄𝑅
𝑒= = 1−
𝑄𝐴 𝑄𝐴
𝑇1 − 𝑇2 𝑇2
𝑒= =1−
𝑇1 𝑇1

e. Heat-temperature relations

𝑄𝑅 𝑇2
=
𝑄𝐴 𝑇1

 Note: Carnot Cycle Efficiency can be increased by increasing T 1 and lowering T2.