Download ENGINEERING_THERMODYNAMICS

1
UNIT - I
DEFINITIONS AND BASIC CONCEPTS
1.1 What is Thermodynamics?
The science, which deals the analysis of various machines by quantity, which involves the
transfer of energy into useful work, is called thermodynamics. Many energy conversion devices
require the transfer of energy into work. Thermodynamics is applied in various thermal equipments
like steam turbine, boiler, condenser, cooling tower, heat exchanger, reciprocating engines, heat
pump etc. It is also used in internal combustion engines, turbo jets and rockets. In all these cases,
design of thermal equipments essentially requires the in depth knowledge of thermodynamics.
Work transfer maybe in any form – Mechanical work, electrical work or from surface tension or
magnetic field. Hence the study of work and energy transfers is very important in the field of
engineering.
1.2 UNITS
SI unit system has three divisions: 1.Base units 2. Derived units 3. Supplementary unit
1.2.1 Base units:
The base unit is the first division of SI units in which seven kinds of fundamental quantities
are measured. They are given in the table 1.1
Metre (m):
Kilogram (Kg
Ampere (A
Kelvin (K):It
Candela (L) :
Mole (mol):
1.2.2. Derived units:
1.2.3. Supplementary units:
Radian (rad):
Steradian (Sr) :
1.3 Important units
1.3.1.Mass and weight:
1.3.2. Pressure:
There are various units followed for pressure. They are given below.
1 bar
Pa
=
=
10 5 N / m 2
1N/m2
2
1 Atmospheric pressure
Atmospheric pressure
1mm of Hg
1N/m2
=
=
=
=
1.013 x 10 5 N / m 2
760 mm of Hg
1013 x 10 2 / 760
=
760 / 1013 x10 2
=
1ata (atmospheric technical absolute) =
133.3 N / m 2
7.5 x 10 -3 mm of Hg
736 mm of Hg
1.3.2.1. Elements of Measurements of Pressure :
There are three elements in measurement of pressure.
1. Atmospheric pressure 2. Gauge pressure 3. Absolute pressure
Atmospheric pressure (Patm
Gauge pressure (Pg):
Absolute pressure:
The relation between above three pressures is given below :
pabs = patm


+
pg
* Normal temperature and pressure ( NTP ): Normal temperature and pressure refers to the
conditions of atmospheric pressure at 760 mm of Hg and 20 ºC
Standard temperature and pressure (STP): STP refers that the temperature and pressure of any
gas in standard atmospheric pressure are taken at 760 mm of Hg and 0 ºC.
1.3.2.2 Pressure measurement:
(i) The Piezometer Tube:
(ii) U-Tube manometer:
Manometer is commonly used for measuring pressure. It consists of a glass U-tube. Mercury
is filled in the U-tube. One end of the manometer is connected to the tank pipe in which the
pressure of fluid is to be measured and the other end is open to the atmosphere. The gauge pressure
and vacuum pressure can be measured using U tube manometers (Fig 1.4).
(iii) Mechanical gauge:
1.3.3.ZEROTH LAW OF THERMODYNAMICS :
Temperature:
Two scales are used to measure the temperature
1. Centigrade scale
2. Fahrenheit scale
Centigrade or Celsius scale:
Fahrenheit scale:
Absolute temperature:
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Measurement of Temperature:
(i) Gas Filled Thermometer:
(ii)
Electrical Resistance Thermometer:
(iii) Thermocouple:
A few thermocouple metal pairs that are commonly used for temperature measurements are Iron –
Constantan, Copper – Constantan, Chromel – Alumel, Platinum/Platinum rhodium etc.
(iv) Pyrometers:
1.3.4 FORCE:
1.3.5 ENERGY:
A capacity to do work is called energy. Energy may be given in two forms: a) Stored
energy, b) Transit energy
Stored energy:
Transit energy:
Kinetic energy:
Potential energy:
Flow energy:
Energy is a thermodynamic property of system. The energy is a point function and the value of it
depends on the end states only and not on the path by which the change of state occurs in a process.
LAW OF CONSERVATION OF ENERGY:
Concepts Of Thermodynamics :
Thermodynamic system:
Boundary
Surroundings
Boundary
System
System
Gas filled in a vessel
Fig 1.9 (a)
Fig 1.9 (b)
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Types of thermodynamic systems:
Thermodynamic systems are generally classified in to three as given below:
Thermodynamic systems
Closed system
Open system
Isolated system
Closed system: A system, which has a fixed mass in it and allows only energy transfer in to the
system and out of the system, is referred as closed system.
Open system:
Isolated system:
Thermodynamic Properties:
Thermodynamic property is classified in to two types:
1. Intensive property 2. Extensive property
Intensive property:
Extensive property:
State:
Change of state:
Thermodynamic Process:
Path:
Thermodynamic cycle:
Equilibrium:
Mechanical Equilibrium:.
Chemical equilibrium:
Thermal equilibrium:
Thermodynamic equilibrium:
Reversible process
It is defined as a process, which proceeds in such a manner that the system remains ,
almost close to an equilibrium state at all time . It is also called quasi static process: It is shown in
Fig 1.13.
2
1
Equilibrium states
Fig 1.13
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1.3.6 WORK :
Work developed or work output
Work done or work input
T
C
work done from turbine
(Positive work)
work to the compressor
(Negative work)
In thermodynamics, the work done by the system is considered as positive work transfer. The work
done on the system is considered as negative work transfer.
There are a few forms of work transfer considered in thermodynamics. They are given below:
1. Displacement work 2. Shaft work 3. Electrical work 4. Paddle wheel work 5. Flow work 6.
Work done of a stretching wire
7. Work done in changing the area of surface film 8.
Magnetization of paramagnetic solid 9. Free expansion
Form of work
Compression or
expansion work
Shaft work
Electrical work
Flow work
Magnetic work
Surface tension
Paddle wheel work
Work in a
stretching wire
Various forms of work transfer
Force
Type of
displacement
Pressure
Change in volume
Work
Torque
Electric field
pressure
Magnetic field
Surface tension
Torque
Rotation
Polarization
Volume
Magnetic moment
Peripheral area
Change in volume
Pdv (flow process)
-vdp (non flow process)
2NT
EI
pv
-H.dI
-dA
-Tdv
Stress
Strain
- AL2 / 2
1.3.7 HEAT :
It is a form of energy in transit, which occurs due to the temperature difference between
two points in a substance or a body. Heat can be transferred across the boundary only when a
system at a higher-level temperature to a system at a lower level temperature. It can never happen
in the reverse direction by nature. This contrary will be discussed in the Second law of
thermodynamics. Heat is measured in Joules or Kilo Joules. Heat can only be sensed due to the
temperature difference. Heat can be transferred by any of the following modes:
(i) Conduction (ii) Convection (iii) Radiation
1) Conduction:
It is the transfer of heat within a body or more than a body by the physical contact but in the
absence of fluid motion. Example for conduction is a metal rod heated up at one end by an external
source of heat. Heat is transferred from the source end to the other end of the rod by movement of
molecules from one end to the other end.
2) Convection:
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It is the transfer of heat within a fluid or gas or between two mediums by the motion of
molecules from one region to another region. Water heated up in a vessel is an example for
convection heat transfer.
3) Radiation :
It is the transfer of heat from hot body to cold body by movement particles in the space or
between gas mediums.
1.3.7.1 Heat may be classified into two forms :
1. Sensible heat
2. Latent heat
1. Sensible heat:
2. Latent heat :
Three types of latent heat are: a) Latent heat of melting or fusion b) Latent heat of freezing or
sublimation c) Latent heat of vaporization or evaporation
1.3.7.2 Specific heat capacity:
The values of specific heat for heating and cooling the gas may be taken in two ways :
1) Specific heat at constant pressure (Cp)
2) Specific heat at constant volume (Cv) .
It is to be noted that the Cp value is always greater than Cv.
Specific heat at constant pressure ( Cp ):
Specific heat at constant pressure ( Cv ):
1.3.7.2 Molar specific heats of gases :
1.4 FIRST LAW OF THERMODYNAMICS:
[Qalgebraic] = [Walgebraic ]
cycleQ =  cycleW
Qb
Qa
System
Wa
Boundary
Wb
Fig 1.22
1.4.1 Perpetual motion machine of first kind (PMM – I ):
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1.4.2 VARIOUS THERMODYNAMIC PROCESSES :
Non-Flow Process
Fig 1.24
Reversible
Non-flow
process
Irreversible
Non-Flow
process
Flow Process
Fig 1.25
Steady flow
Process
Unsteady
flow process
REVERSIBLE NON-FLOW PROCESS
Constant
volume
Process
Isochori
c
Constant
Pressure
Constant
Temperature
Process
Process
Isobari
c
Isothermal
Hyperbolic
ReversibleAdi
abatic
Process
Process
Isentropi
c
Fig 1.26
1.4.2.1 Constant volume process:
1. P-V-T relationship:
We know that the general gas equation is
PV = m RT
2
P/T = Constant
P/T
=
P1/T1
= P2/T2
P
1
V
1.4.2.2 Constant pressure process:
1. P-V-T relationship:
Polytropic
Process
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We know that the general gas equation is
PV
= m RT
V/T = Constant
V/T
=
V1/T1
= V2/T2
p
1
2
WORK
TRANSFER
W1-2
V
1.4.2.3. Constant Temperature process:
1. P-V-T relationship:
We know that the general gas equation is
PV = m RT
As the temperature is constant during the process the gas equation becomes pV = constant
P
V
Fig 1.29
1.4.2.4 Reversible Adiabatic process:
1. P-V-T relationship:
We know that the general gas equation is
PV = m RT
pV  = Constant
pV  =
P1V1 =
P2V2
pV  = Constant
P
Work
transfer W1-2
V
1.4.2.5. Polytropic process
:
1. P-V-T relationship:
We know that the general gas equation is
PV
=
m RT
pV n = Constant
pV n =
P1V1n =
p2V2n
pV n = Constant
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
It is important to note that the sign conversions for heat and work should be carefully followed
in case of heating or expansion and cooling or compression of gases. The expansion of gases is
considered to be positive work and cooling of gas is considered as negative work. Similarly the
expansion or heating of gas in which positive heat transfer is considered whereas cooling or
compression of gas is negative heat transfer.
Flow processes or open systems :
Steady flow energy equation ( SFEE )
As per energy balance at both the sides of inlet and outlet
Sum of all energies at inlet
=
Sum of all energies at outlet
Heat transfer + [ Kinetic energy + Potential energy
+ flow energy + Internal energy ] inlet
=
Work transfer + [ Kinetic energy + Potential energy
+ flow energy + Internal energy ] outlet
Applications of steady flow processes :
1. Boiler 2. Evaporator 3. Condenser 4. Heat exchanger 5. Steam nozzle 6.Turbine
7. Reciprocating compressor 8. Rotary compressor
1) Boiler : Boiler is a closed vessel in which steam is generated by adding heat to water. Boiler is
assumed as a steady flow heat absorbing system.
Assumptions made :
1. No work transfer (W = 0)
2. Change in kinetic energy is negligible
3. Change in potential energy is negligible
2)Evaporator:
3) Condenser :
4) Turbine :
5) Rotary compressor:
6) Reciprocating compressor :
7) Steam nozzle
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SOLVED PROBLEMS IN FIRST LAW OF
THERMODYNAMICS , NON FLOW
PROCESSES AND FLOW PROCESSES
Summary :
First law of thermodynamics applied to
A ) Thermodynamic cycle :  Q =  W
B ) During a change of state in a process : Q = W + U
Table: 1.5
Type of
P-V-T
Work
Change
Heat
Change P-V
Process
relationship transfer
in
transfer
in
diagram
internal
enthalp
energy
y
1
Constant
P
volume
P/T = C
2
process
0
m Cv T m Cv T m Cp
V
P/T = P1/T1
T
( Isochoric
= P2/T2
)
Constant
1
2
pressure
V/T = C
P
process
V/T = V1/T1 P (V2- V1)
m Cp
m Cv T m Cp T
=
T
V
(Isobaric) V2/T2
Constant
Temperatu
re process
pV = C
pV = p1V1
= pV2
2.3 P1V1
log10
(V2/V1 )
0
2.3 P1V1
log10
(V2/V1 )
P
0
Reversible
adiabatic
process
1
pV = C
pV = p1V1
= pV2
2.3 P1V1
log10
(V2/V1 )
0
2.3 P1V1
log10
(V2/V1 )
P
=C
pV = p1V
T
2
P
2
T
=
pV
m Cv
T
0
2
V
P
[ P2V2 P1V1 ] / (1)
1
1
2
m Cp
T
V
2
PVn = C
PVn = p1V
1
n
= pV
2
n
[ P2V2 P1V1 ] / (n1)
m Cv
T
[ P2V2 P1V1 ] / (n1)
+
mCv ( T2 –
T1)
1
m Cp
T
2
1
m Cv
T
pV
T
1
P
V
1
Polytropic
process
2
P
1
1
(Isotherma
l)
Hyperboli
c process
P-T
Diagram
P
2
V
P
2
T
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Problem 1:
A system undergoes a thermodynamic cycle . Four processes are involved in the cycle. The values
of work transfer , heat transfer and internal energy in each process are tabulated in the table . Find
rate of work. Check the work and heat transfer follows first law.
Thermodynamic
process
A-B
B-C
C-D
D-A
Work transfer
KJ / min
250
-800
-600
---
Change in internal
energy kJ / min
-400
100
Heat transfer
KJ / min
300
-200
50
Process A- B :
Q A- B =
300
=
U A- B
W A- B + U A- B
250 + U
=
50 kJ/min
Process B - C :
Q B- C =
Q B- C
Q B- C
W B - C+ U B-C
=
-800 +400
=
-400 kJ/min
Process C - D :
Q C-D =
200
=
U
=
W C - D + U C - D
-600 +U C - D
800 kJ/min
Process D - A :
Q D-A =
50
=
W
=
W D - A + U D - A
W +100
-50 kJ/min
Thermodynamic cycle :  Q
=
 W
Qalgebric = Walgebric
Q A-B + Q B-C + Q C-D + QD-A
=
W A-B +W B-C + W C-D +W D-A
300 –400+200 + 50
=
250 +400 - 600 +100
CONSTANT VOLUME PROCESS:
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Problem 2:
A cylindrical vessel has a diameter of 1.1 m contains 1kg mole of nitrogen at 100 º C. If he
gas is cooled to 60 o Cat constant volume , calculate the Work transfer , change in internal energy ,
change in enthalpy and heat transfer. The ratio of the specific heats is 1.4 and 1 kg of mole
nitrogen is 28. Assume that the stroke of the cylinder is equal to diameter of the cylinder.
4. Change in enthalpy
H =
m cp (T2 -T1 ) kJ
H
=
1 x 1.036 x (333-373 )
H
=
1 x 1.036 x (-40)
H
=
-41.44 kJ/mole
CONSTANT PRESSURE PROCESS:
Problem 3:
0.5 kg of a perfect gas is heated from 100 º to 300 º C at a constant pressure of 2.8 bar..
Determine the work transfer , change in internal energy , change in enthalpy , heat transfer. Take
gas constant R = 0.28 kJ/kg K. Cp = 1 kJ/kg K & Cv = 0.72 kJ/kg K
Given:
m
=
T1
=
0.5 kg;
100 + 273 = 373 K;
p
T2
=
=
2.8 bar = 2.8 x 10 5 N/m2;
300 + 273 = 573 K;
CONSTANT TEMPERATURE PROCESS:
Problem 4:
A certain quantity of a perfect gas is heated in a reversible isothermal process from 1 bar
and 40 o C to 10 bar. Find the work done per kg of gas . Take R = 287 J/Kg K.
T2
P
T1
T
Given:
T
=
p1
=
40 + 273 = 313 K;
1 bar = 1 x 10 5 N/m2;
m
=
1 kg;
p2 = 10 bar = 10 x 10 5 N/ m2;
POLYTROPIC PROCESS:
Problem 5:
1 kg of air at a pressure of 7 bar and a temperature of 363 K undergoes a reversible
polytropic process which may be represented by p v 1. 1 = constant. The final pressure is 1.4 bar.
Evaluate a) the final specific volume, temperature . b) the work done and the heat transfer during
the process. Assume R = 287 J/Kg K and  = 1.4.
Given:
T1
=
p1
=
R
=
363 K;
7 bar = 7 x 10 5 N/m2;
287 J/Kg K;
m

p2
=
=
=
1Kg;
1.4;
1.4 bar = 1.4 x 10 5 N/m2;
13
n
=
1.1.
Problem 6:
A steam turbine in a steam power plant receives steam at 40 bar. The specific enthalpy and
specific volume of steam at inlet of turbine are 3214 kJ/kg and 0.07m3/kg, respectively. Steam after
expanding on turbine has the pressure at exit of 35 bar. The enthalpy and volume of steam are
3202.6 kJ/Kg and 0.084 m3/Kg, respectively. If the steam is carried in a 0.2m diameter pipe the heat
loss in the pipeline is 8.5 kJ/Kg. Calculate the mass flow rate of steam.
Given:
p1
=
v1
=
h2
=
Q
=
40 bar;
0.073 m3/kg;
3202.6KJ/kg;
8.5 kJ/kg.
h1
p2
v2
=
=
=
3214 kJ/kg;
35 bar;
0.084 m3/Kg;
Problem 7:
Air enters a compressor with a velocity of 60m/s, pressure of 100 kpa, temperature of 40 C and
leaves the compressor with a velocity of 90m/s, pressure of 500 kpa, temperature of 120 C .
Consider the system is adiabatic. Find the power of the motor for a mass rate of flow of 40 kg/min.
Write the assumptions made.
Given:
V1
=
T1
=
p2
=
m
=
60m/s;
p1
=
313 K;
V2
500 kpa = 500 x 1000 N/ m2; T2
40 kg/min=0.66 kg/s.
100 kpa=100 x 1000 N/ m2;
=
90m/s;
=
393 K;
Problem 8:
Air is compressed from 100 kpa and at 22 C to a pressure of 1Mpa while being cooled at
the rate of 16kJ/kg by circulating the water through the compressor casing. The volume flow rate
of air inlet condition is 150m3/min and the power input to the compressor is 500Kw. Determine
Mass flow rate. b) Temperature of air exit. Neglect the datum head.
Given:
p1
=
q
=
w
=
Problem 9:
100 kpa;
16 kJ/s;
-500 kw;
p2
v1
T1
=
=
=
1Mpa;
150 m3/min =2.5m3/s;
22 + 273 = 295K.
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Steam enters an adiabatic turbine at 10Mpa and 500 oC at the rate of 3 kg/s and
leaves at 50Kpa. If the power output of the turbine is 2Mw, determine the temperature of steam at
exit of turbine. Neglect the kinetic energy changes.
Given:
P1
=
m
=
w
=
10Mpa = 10 x 10 6 N/m2;
3 kg/s;
p2
2Mw = 2 x 10 3Kw.
T1
=
=
500 + 273 = 773 K;
50 kpa = 50 x 10 3 N/m2;
Problem 10:
A steady flow system in which 1 kg/s of air at 20ºC and consumes a power of 15 kw. The
inlet and outlet velocities of air are 100 m/s and 150 m/s respectively. Find the exit of air
temperature, assuming adiabatic conditions. Take the cp of air is 1.005 kJ/Kg K.
Given:
T1
=
V1
=
20 + 273 = 293 K;
100 m/s;
w
V2
=
=
15 kw;
150 m/s;
Problem 11:
In an heat exchanger, oil is cooled from 90oC to 30oC. Water is used to cool the oil. Inlet
temperature of water is 25oC and the outlet temperature is 70oC. The enthalpies of oil and water at
inlet and outlet conditions are given below.
Water : hinlet = 593.64 kJ/kg; houtlet = 699.77 kJ/kg;
Oil
: hoilin = 748.19 kJ/kg;
hoilout = 605.43 kJ/kg;
Calculate the mass flow rate of water required to cool the oil for cooling 2.78 kg/s of oil ?
Given:
Tw1
=
25 + 273=298 K;
Tw2
=
90 + 273=363 K;
Toil1 =
70 + 273=343 K;
Toil2 =
30 + 273=303 K;
moil
=
2.78 kg/s.
OBJECTIVE TYPE QUESTIONS :
1.Which one is not the extensive property of a thermodynamic system ?
(i)
Mass (ii)Total volume (iii) Total internal energy (iv) Density
2.One bar in SI unit is equal to
(a) 1x10 5 N/m2 (b) 1x10 4 N/m2 (c) 1x10 3 N/m2 (d) 1 N/m2
3. Unit of power is
(a) Watt (b)Joule (c) Joule-metre (d) None of the above
4. The law that states that heat and work are mutually convertible is known as
(a) Zeroth law of thermodynamics (b) first law of thermodynamics
(c) Second law of thermodynamics (d) None of the above