A. Thermodynamic Properties : 1. Every system has certain A. Dalton’s Law of Partial Pressure : 1. According to this law, “The PMM-1.

his law, “The PMM-1. 1. The first law states the general principle of the A. Heat Reservoir : 1. A heat reservoir is a body with a very large heat
characteristics by which its physical condition may be described e.g., total pressure exerted by a mixture of gases is the algebraic sum of conservation of energy. Energy is neither created nor destroyed, but capacity to which, and from which, heat can be transferred without
volume, temperature, pressure etc. Such characteristics are called partial pressures exerted by the individual constituents when they only gets transformed from one form to another. 2. There can be no any change in its temperature. 2. Such a body at high temperature is
properties of the system. They are macroscopic in nature. 2. In other occupy the same volume and temperature of the mixture”. 2. machine which would continuously supply mechanical work without referred to as a high temperature reservoir. 3. If heat is transferred
words, “properties are the coordinates to describe the state of a Consider two individual gases A and B which occupy volume V at some other form of energy disappearing simultaneously (Fig. 1.28.1). from it, then it is considered as a heat source. 4. A body at low
system”. 3. There are two types of thermodynamic properties : a. temperature T are mixed together and kept in a third container of 3. Such a fictitious machine is called a perpetual motion machine of temperature is referred to as a low temperature reservoir. 5. If heat is
Intensive Property : These are independent of mass and size of volume V and temperature T. Volume of gas A = Volume of gas B = the first kind, or in brief, PMM1. A PMM1 is thus impossible. 4. The transferred to it, then it is considered as a heat sink. 6. By definition,
system. e.g., pressure, density, temperature. b. Extensive Property : Volume of mixture (A + B) From mass balance equation, m = mA + mB converse of the above statement is also true, i.e., there can be no a heat reservoir is a closed system with no work interaction. 7. The
These properties depend on mass and size of system. e.g., volume, From Dalton’s law of partial pressure, (p) V, T = (pA ) V, T + (pB ) V, T machine which would continuously consume work without some environment constitutes the largest heat reservoir operating without
energy, entropy. B. Path : Succession of states passed through during Above equations can be extended for any number of non-reactive other form of energy appearing simultaneously any change in temperature usually employed as a heat sink, and
a change of state is called ‘path’ of the change of state. C. Process : 1. gases. p = pA + pB + ..... = piB. Gibbs Dalton’s Law : 1. According to sometimes as a heat source. 8. The mediums in the environment
When the path is completely specified, the change of state is called this law, “The internal energy, enthalpy and entropy of mixture of A. Assumptions Made in the Analysis of SFEE : 1. There is no which are used as such are generally the following : a. Atmosphere
‘process’. 2. In other words, “A process is the cause of change of state gases are equal to the algebraic sum of internal energies, enthalpies accumulation or decrease of mass in the control volume at any time air, b. Ocean, river or well water, and c. Ground. 9. The characteristic
of a system”. e.g. Isothermal expansion, Isochoric, Isobaric etc. D. and entropies of individual gases where they occupy the same i.e., there is no other source or sink of mass in the control volume. 2. which remains constant for a heat reservoir is its temperature.
Open SystemE. Concept of Continuum : 1. According to this concept, volume and temperature of mixture of gases.” 2. From this law, Rate of mass flow in and out of the control volume is equal and Hence, a heat reservoir is characterised by its temperature. B. Heat
there is a minimum limit of volume upto which properties of the following relations can be written mu = mA uA + mB uB + mC uC + ... constant with respect to time. 3. State, velocity and elevation of fluid Engine : 1. A heat engine is a thermodynamic system operating in a
systems will remain in continuum. But below this volume, there is an = mi ui mh = mA hA + mB hB + mC hC + ... = mi hi ms = mA s A + mass entering and leaving the surface do not change with time. 4. cycle to which net positive amount of heat is added, and from which
abrupt change in the value of property. 2. Such a region where mB s B + mC s C + ... = mi s i 3. Although Gibbs Dalton’s law is Rate of heat and work transfers across the control volume is net positive amount of work is obtained.
properties remain in continuum is known as “region of continuum” applicable to perfect gases, it can be applied to real gases for constant. B. General Energy Equation For Steady Flow Processes : 1. A. Heat Pump : 1. A heat pump is a reversed heat engine. It receives
and region in which properties change abruptly is called “region of approximate engineering calculations involving mixture of gases at Let’s consider the flow of a fluid through a control volume as shown heat from a low temperature reservoir (source) and rejects it to a
discrete particles”. 3. Limiting volume upto which continuum lower pressure. C. Amagat’s Law : 1. According to this law, “The total in Fig. 1.29.1. In the time interval “dt”, there occurs a flow (or flux) of high temperature reservoir (sink). 2. This transfer of heat from a low
properties are maintained is called “continuum limit”. 4. According to volume occupied by a mixture of gases is equal to the sum of mass and energy into the control volume. 2. Section 1-1 is the inlet temperature body to a high temperature one is essentially a non-
“concept of continuum” density can be defined as volumes which would be occupied by each constituent, when they side and section 2-2 is the outlet side. 3. At the inlet side (section 1-1) spontaneous process. And that calls for the help of an external work
are at the same pressure and temperature as that of the mixture”. 2. to control volume following are the fluidparameters : Average which is supplied to the heat pump3. A heat pump extracts Q2
A. Reversible Process : 1. A process is called reversible if after the It follows that (V) p, T = (VA ) p, T + (VB ) p, T + (VC ) p, T + ... = (Vi ) velocity = C1 Pressure = p1 Specific volume = v1 Internal energy = amount of heat from the low temperature (T2 ) source and delivers
conclusion of reversed process the initial states of system and p, T u14. At the outlet (section 2-2), following are fluid parameters : Q1 amount of heat to the high temperature (T1 ) sink by consuming
surroundings are restored without any extraordinary changes either Velocity = C2 Pressure = p2 Specific volume = v2 Internal energy = u2 W amount of external work. 4. Now, the first law of efficiency of a
in the system or surroundings. 2. A reversible process is a quasi-static Van der Waal’s equation of state1. Real gases differ from ideal gas 5. During the flow of fluid through the control volume, heat ‘Q’ and heat pump cycle is usually called the coefficient of performance. 5. It
process, a process carried out infinitely slowly with infinitesimal because of presence of intermolecular forces and also due to finite mechanical work ‘Ws ’ are also supposed to cross the control surface. is the desired effect upon the external work supplied for obtaining
gradient with the system passing through a series of equilibrium molecular volumes. Van der Waal’s equation of state has been While writing the energy balance equation on the sides of the control that desired effect, COP = Desired effect Work input ...(2.2.1) 6. Now,
states. 3. Consider a process in which the system undergoes a change established by incorporating the following two corrections in the volume, following energies are taken into consideration : a. Internal the desired effect for a heat pump is to supply heat Q1 to the hot
of state 1 to state 2 and during the process, system does work ‘W’ equations of state, pV = RT. Correction 1 : i. Let equation of state pV = energy stored in the fluid. b. Potential energy and kinetic energy. c. body. Therefore
and it transfers heat ‘Q’ to the surroundings. 4. If at the completion RT be represented in the form V = RT p . When pressure increases, Flow energy (or flow work) required to push the fluid in or out of
of its reverse process, i.e., from state ‘2’ to state ‘1’ and the system is volumes decreases ii. But in case of real gases, molecules occupy a control volume. d. Heat and shaft (mechanical) work which may cross B. Refrigerator : 1. A refrigerator is similar to a heat pump. 2. It
restored to its initial state then the work ‘W’ must be done on the finite volume and an allowance is to be made for the volume of voids the control volume. 6. Since the energy is conserved therefore energy operates as a reversed heat engine. 3. Its duty is to extract heat as
system and the heat Q must be transferred from thesurroundings to existing between the molecules. iii. Thus the free volume available balance for the control volume mentioned above can be written in much as possible from the cold body/space and deliver the same to
the system so that there may be no outstanding changes either in the for molecular motion will be equal to [V – b] where b = Vmolecule + the following form7. The eq. (1.29.1) is a general energy equation high temperature body/surroundings. 4. The desired effect of a
system or in the surroundings. 5. After the completion of reversed Vvoid ; where ‘b’ represents the smallest volume upto which the gas and can be applied to all fluids compressible or incompressible, ideal refrigerator, under a steady state, is to pump out the heat in the
process no traces will be left in the universe and such a process is can be compressed. Correction 2 : i. Because of intermolecular forces or real fluids, liquids and gases. same rate as is infiltrating into the system (Q2 ). And in order to do
referred to as reversible processB. Irreversible Process : 1. A process in real gases, some impacts will be there on the walls of vessel. It is so, the refrigerator or an air conditioner takes up W amount of
is called irreversible if the initial state of the system cannot be due to the fact that molecules positioned at the walls are attracted A. Application of SFEE to Engineering Devices : a. Nozzle and external work (Fig. 2.2.2). 5. The desired effect of a refrigerator is to
restored without any changes either in the system or surroundings. 2. by adjacent molecules inside the vessel. ii. As a result, when Diffuser : 1. The flow through a nozzle is characterized by following remove Q2 heat infiltrating into the cold space. 6. By using the
It can be observed from the consequence of second law of compared to an ideal gas, the pressure exerted by a real gas will be features : i. Shaft work is zero i.e., Ws = 0 ii. If the flow is reversible external work, it rejects Q1 heat to the high temperature reservoir
thermodynamics that all the natural processes are irreversible smaller by an amount equal to p. iii. This decrease in pressure is adiabatic manner, then Q = 0. iii. If the nozzle is horizontal, change in (surroundings). Therefore, Q2 is the heat infiltrating into the cold
because the available work energy either of the system or the directly proportional to the number of molecules affected by elevation, i.e., dZ will be zero  Z1 = Z2 . 2. Under these features, space of the refrigerator.
surroundings is converted into heat energy at the completion of molecular interaction or directly proportional to the square of SFEE for a nozzle / diffuser is reduced to h1+c21/2=h2+c22/2 3. For
reversed process which is an undesirable or outstanding change from density of gas as given by the expression. p = a 2 or 2 a V Where a a nozzle Enthalpy drop = Increase in kinetic energy 4. For a diffuser, second law of thermodynamics1. On the basis of limitations of first
the view point of second law of thermodynamics. a. Types of is the constant of proportionality which has a definite numerical Rise in enthalpy = decrease in kinetic energy. b. Boiler : 1. A boiler has law of thermodynamics, we have two statements of second law of
Irreversibility : Followings are various types of irreversibility : i. value for each gas. 2. When the above two corrections are following features : i. Shaft work is zero, Ws = 0. ii. Change in kinetic thermodynamics which are as follows : A. Kelvin-Planck Statement :
External Irreversibility : 1. It is due to dissipative effects like introduced, we get the Van der Waal’s equation of state 2 a p V     energy is negligible, 2 2 2 1 2 C C = 0. iii. Change in elevation According to this statement, “It is impossible to construct a heat
mechanical friction, viscosity, surface tension, magnetism etc. 2. It is    (V – b) = RT 3. In the above equation, a/V2 is known as force of between inlet and outlet point is negligible, Z1 = Z2 . 2. Therefore engine that operates in a cycle and produces no effect other than
due to finite temperature difference. ii. Internal Irreversibility : It is cohesion and ‘b’ is known as co-volume. SFEE is reduced to mh1 + Q = mh2 Q = m (h2 – h1 ) c. Turbine : 1. work output and exchange of heat with a single heat reservoir.” B.
related with dissipative effects within the working substance, e.g., A steam or gas turbine has following features : i. KE or d (kinetic Clausius Statement : According to this statement “It is impossible to
free expansion, throttling etc. A. Zeroth Law of Thermodynamics : 1. When a body A is in thermal energy) = 0 ii. PE or d (potential energy) = 0 iii. Q = 0 since walls are construct a device that operates in a cycle and produces no effect
equilibrium with a body B, and also separately with a body C, then B insulated. 2. Therefore, SFEE for a turbine is reduced to mh1 = mh2 + other than the transfer of heat from a region of low temperature to
A. Cyclic Process : 1. A thermodynamic cycle is defined as a series of and C will be in thermal equilibrium with each other. This is known as Ws or Ws = m(h1 – h2 ) Obviously the work is done by the turbine at another system at high temperature.” 2. The Clausius statement
state changes such that the final state is identical with the initial the zeroth law of thermodynamics. B. Temperature Measurement : 1. the expense of enthalpy. d. Compressor : 1. A compressor is implies that the heat itself cannot flow from a region of low
state. 2. The processes through which the system has passed can be Zeroth law forms the basis of temperature measurement. The characterized by following features : i. Shaft work is negative i.e., Ws temperature to a region of high temperature without the aid of
shown on a state diagram, but a complete section of the path temperature of a body can be determined by bringing another body = negative (Since work is done on the system and working fluid is external work. 3. Kelvin-Planck statement is applied to heat engines
requires in addition a statement of the heat and work crossing the (say a thermometer) in contact with the first body and allowing the compressed.) ii. Change in potential energy is negligible i.e., d(PE) = 0 while Clausius statement is concerned with heat pumps and
boundary of the system. 3. Fig. 1.7.1, shows such a cycle in which a thermal equilibrium to be attained. 2. The value of temperature is iii. Generally heat is lost to surroundings, Q is negative. 2. Therefore refrigerators. Both the statements of second law of thermodynamics
system commencing at condition ‘1’ changes in pressure and volume found out by measuring some temperature dependent property of SFEE for a compressor is reduced to m(h1+c21/2)-Q=m(h2+c22/2)-W are negative statements, they do not have any mathematical proofs.
through a path 1-3-2 and returns to its initial condition ‘1’B. Quasi- the thermometer. Such a property of thermometer is known as f. Heat Exchanger : 1. A heat exchanger is characterized by the B. Equivalence of Kelvin-Plank and Clausius Statements : 1. Both
Static Process : 1. It is a succession of equilibrium states, “A process is thermometric property which can be volume of gases, pressure of following features : i. Shaft work is zero, Ws = 0 ii. Change in KE = 0 iii. Kelvin Planck and Clausius statements appear to be different but both
called quasi-static if it is carried out in such a way that at every gases, electrical resistance of solids, magnetic effects etc. 3. To give a Change in PE = 0 iv. It is a perfectly insulated system i.e., no external are interlinked and are complementary to each other. 2. Equivalence
instant the system departs only infinitesimally from previous numerical value to the thermal state of a body, it is imperative to heat interaction. 2. From energy balance equation, we can write, of these two statements can be proved by showing that violating one
thermodynamic equilibrium state”. 2. Only a quasi-static process can establish a temperature scale on which temperature of a body or Energy given by fluid A = Energy gained by fluid B statement leads to the violation of other statement and vice-versa. a.
be reversible and can be represented on a thermodynamic plane. 3. system can be read4. It needs selection of basic unit and a reference Violating Kelvin Planck Statement Leads to Violation of Clausius
Consider a gas contained in piston-cylinder assembly. System is state. For this purpose, generally two fixed points are used : i. Ice Statement : 1. Let’s consider a heat engine which violates Kelvin
initially in equilibrium state (p1 , V1 , T1 ). Case I : Whole weight is point : Ice point is the equilibrium temperature of ice with air Planck statement by absorbing heat from source at T1 and converts it
removed in one step. In this case, intermediate states passed through saturated water at standard atmospheric pressure. ii. Steam point : completely into work.  W = Q1 2. Now let’s introduce a refrigerator
by system are non-equilibrium statesCase II : Weight is removed in This is the equilibrium temperature of pure water with its own which gets work input from the engine. 3. The refrigerator extracts
steps. Now every state passed by system will be an equilibrium state vapour at standard atmospheric pressure. Q2 from the low temperature heat reservoir and rejects heat Q1 + Q2
as shown on p-V coordinate. “Such a process which is locus of all to the high temperature heat reservoir. 4. Combining the engine the
equilibrium points passed through by system is called quasi-state e first law of thermodynamics with its limitations1. First law of refrigerator into one system working between same temperature
process” thermodynamics is related with principle of conservation of energy limits, we observe that the sole effect of combined system is to
according to which the total energy of an isolated system is transfer Q2 from low temperature heat reservoir T2 to high
A. Work : 1. Work reflects the effect of a force on the system conserved. 2. It can be concluded that all forms of energies are temperature heat reservoir without any work input thus violating the
boundary. 2. When there occurs a physical displacement of a system equivalent and convertible. If one form of energy disappears, it must Clausius statement
boundary due to the action of an unbalanced force across the system appear in an equivalent amount of some other form of energy. 3. b. Violation of Clausius Statement Leads to Violation of Kelvin Planck
boundary, then work is done by or on the system. 3. If a part or Thus first law stipulates that when a thermodynamic process is Statement : 1. Let’s consider a refrigerator which violates Clausius
whole of a system boundary undergoes displacement under the carried out, energy is neither gained nor lost. 4. Energy is only statement as shown in the2. Refrigerator absorbs heat Q2 from low
action of an unbalanced force, then work done W = Force × transformed from one form into another and the energy balance is temperature heat reservoir and rejects the same to the high
Displacement 4. If work is done by a system on the surroundings, i.e., maintained. 5. First law fails to state the conditions under which temperature reservoir without the aid of any external work i.e., W =
a residual unbalanced force acting within the system pushes the energy conversion takes place. 6. The limitations of 1st law of 0. 3. Let’s introduce a heat engine which receives heat Q1 (Q1 > Q2 )
system boundary against the surroundings, the work is said to be thermodynamics can be explained with the help of following from the high temperature reservoir and rejects heat Q2 and
positive. 5. Imagine a gas contained in a cylinder enclosed by a piston illustrations : a. Temperature of liquid contained in a vessel increases produces work, W = Q1 – Q2 . 4. Now combining the refrigerator and
expands by pushing the piston up in the same direction in which the when it is churned by paddle work. But paddle work can not be heat engine into one system. We observe that the combined system
residual unbalanced force acts. Hence, work of the system is restored on cooling the liquid to its initial state. b. When a block operates as a heat engine, which receives heat from a single high
positive : Work output = + W 6. In this case, work is done on a system slides down a rough place, it gets warmer. However, the reverse temperature reservoir as Q1 – Q2 and converts the same into equal
by the surroundings, e.g., when the piston compresses a gas, the process when the block slides up the plane and becomes cooler is amount of work energy without any heat rejection
work is said to negative, Work input to system = – W That is, all work not true even if the first law of thermodynamics still holds good. c. 8. Now let the reversible engine operates as a refrigerator and the
input to the system is negative. B. Heat : 1. Heat is thermal energy Electrical current flowing through a resistor produces heat according irreversible engine (EB ) still continues to act as an engine. Since the
that crosses a system boundary when there is a temperature gradient to equation, H = i 2Rt. Current once dissipated as heat cannot be engine EA is reversible, amount of heat and work interactions will
across the boundary, i.e., a net temperature difference between the converted back into electricity. d. Fuel (solid or liquid) burns with air remain the same but their directions will be reversed. Work input to
system and the surroundings is a must for heat transfer. If there is no and gets converted into products of combustion. Fuel once burnt drive the refrigerator may be received from the irreversible engine
temperature difference, then there is no heat transfer. This implies cannot be restored back to its original form. e. Work is easily through direct coupling between the two. 9. The Fig. 2.5.1(b)
that heat is a transient quantity. 2. For heat inflow into the system, Q converted into heat. However there is a maximum limit up to which represents heat and work interactions for the composite system
is positive. For heat outflow from the system to the surroundings, Q the conversion of heat is possible in a heat engine. Work is superior constituted by the reversible engine (which is now operating as
is negative.  Heat received by the system = + Q Heat rejected by the to heat, and a complete transformation of low grade energy (heat) refrigerator) and the irreversible engine. 10. The net effect is a. No
system = – Q into high grade energy (work) is not possible. 1. The expression (W) net interaction with the heat source at T1 , because it supplies and
cycle = ( Q) cycle applies only to systems undergoing cycles, and the recovers back the same magnitude of heat. b. The composite system
a. Avogadro’s Law : 1. Equal volumes of all gases, at a specified algebraic summation of all energy transfer across system boundaries withdraws (Q – WA ) – (Q – WB ) = WB – WA units of heat from the
temperature and pressure, contain equal numbers of molecules. 2. is zero. 2. But if a system undergoes a change of state during which heat sink at T2 and converts the same into an equal amount of work
Avogadro’s law states that the volume occupied by an ideal gas is both heat transfer and work transfer are involved, the net energy output. 11. Therefore the combination of the two constitutes a
directly proportional to the number of molecules of the gas present transfer will be stored for accumulated within the system. 3. If Q is perpetual motion machine of 2nd kind violating second law of
in the container. 3. The relation is given by 1 1 V n = 2 2 V n Where, n the amount of heat transferred to the system and W is the amount of thermodynamics. Thus the assumption that the irreversible engine is
is equal to number of molecules of gas. b. Boyle’s Law : 1. According work transferred from the system during the process (Fig. 1.27.1), the more efficient than a reversible engine is wrong. 12. Therefore an
to this law, volume of a given mass of a perfect gas varies inversely net energy transfer (Q – W) will be stored in the system. 4. Energy in irreversible engine cannot be more efficient than a reversible engine
with absolute pressure when temperature is kept constant. 2. storage is neither heat nor work, and is given the name internal operating between the same heat reservoirs.
Product of absolute pressure and volume of a given quantity of gas is energy or simply, the energy of the system. Therefore, Q – W = E
constant when the temperature is kept constant i.e., V  1 P or PV = Where E is the increase in the energy of the system or Q = E + W efficiency of an irreversible engine is always less than the efficiency of
Constantc. Charle’s Law : 1. According to this law, the volume of a Here Q, W, and E are all expressed in the same units (in joules). 5. If reversible engine. According to Carnot’s theorem, “No heat engine
given mass of a perfect gas varies directly with its absolute there are more energy transfer quantities involved in the process, as operating in a cycle between two given heat reservoirs, with fixed
temperature when pressure is kept constant i.e., V T = Constant. (At shown in Fig. 1.27.2, the first law gives (Q2 + Q3 – Q1 ) = E + (W2 + temperatures, can be more efficient than a reversible engine
constant pressure) 2. Charle’s law can also be defined as “The W3 – W1 – W4 ) 6. Energy is thus conserved in the operation. The operating between the same temperature limits (or same heat
absolute pressure of a perfect gas varies directly with absolute first law is a particular formulation of the principle of the reservoirs). 2. Let us consider a reversible engine EA and an
temperature if the volume of the gas is kept constant during the conservation of energy. 7. This definition does not give an absolute irreversible engine EB operating between the same heat reservoirs at
process.” P T = Constant (At constant volume value of energy E, but only the change of energy E for the process. temperatures T1 and T2 . 3. For the same quantity of heat withdrawn
from the high temperature reservoir, the work output from these
engines is WA and WB respectively4. Therefore heat rejected by
reversible engine will be Q – WA and by irreversible engine will be Q
– WB . 5. Let us assume that irreversible engine EB is more efficient
than the reversible engine EA . 6. Then it follows that W W B A Q Q 
or WB > WA . 7. That is to say that the output of the irreversible
engine is more than that of reversible engine.
 A. Irreversibility due to Finite Temperature Difference : 1. Consider a A. Refrigeration Effect : 1. It is defined as the amount of cooling A. Refrigerant : 1. A refrigerant is defined as any substance that
working of Carnot and reversed Carnot Cycle heat source at T1 and a body at T2 as shown in the adjoining2. Heat produced by a system. 2. This cooling is obtained at an expense of absorbs heat through expansion or vapourization and loses it
A. Carnot Cycle Operations (Processes) : 1. The system inside the Q can be transferred from heat source to the body by bringing them some energy. Hence, it is customary to define a term known as through condensation in a refrigeration system B. Function : 1. The
cylinder has an initial volume and initial pressure as indicated by in contact due to finite temperature difference. 3. If the process is coefficient of performance. 3. The ratio of heat extracted in the basic function of refrigerant is take heat from the evaporator and
the state point ‘1’ on p-V diagram. 2. Let Q1 be the heat supplied to to be reversed in order to restore the temperature of the body, it refrigerator to the work done on the refrigerant is known as loose the heat in the condenser with reasonable heat transfer rate
the system at T1 (source temperature). Since the heat supply requires a Carnot refrigerator (or heat pump) which would transfer coefficient of performance or theoretical coefficient of performance and establishes the effective heat exchange in the system. C.
(addition) takes place at constant temperature, the system volume heat Q from the body at the expense of external work W and reject of a refrigerator (COP). Mathematically, COP (theoretical) = Q W Desirable Properties of an Ideal Refrigerant : 1. Low boiling point. 2.
increases at constant temperature thus performing an isothermal (Q + W) to the heat source at T1 . 4. On the completion of reversed Where, Q = Amount of heat extracted in the refrigerator or capacity Low specific heat of liquid. 3. Low specific volume of vapour. 4. Low
expansion. a. Process 1-2 : (Isothermal Heat Addition Process) 1. process it is observed that work energy of surroundings has been of a refrigerator, and W = Amount of work done. B. Unit of cost. 5. High critical temperature. 6. High latent heat of
During this process, the working substance (air) expands transformed into heat energy. 5. This transformation of work Refrigeration : 1. The practical unit of refrigeration is expressed in vapourization. 7. Non-corrosive to metal. 8. Non-flammable and
isothermally from state ‘1’ to state ‘2’. 2. At point ‘2’ heat supply is energy into heat energy is undesirable from the view point of terms of ‘tonne of refrigeration’ or TR. 2. A tonne of refrigeration is non-explosive. 9. Non-toxic. 10. Easy to liquefy at moderate
cut off and cylinder head is brought in contact with an insulator or second law of thermodynamics. 6. Thus it can be concluded that all defined as the amount of refrigeration effect produced by the pressure and temperature. D. Classification of Refrigerants : The
adiabatic cover. b. Process 2-3 : (Reversible Adiabatic Expansion) the processes in which heat transfer is due to finite temperature uniform melting of one tonne of ice at 0 °C in 24 hours. Since the refrigerants are basically classified as follow : a. Primary
1. Adiabatic cover is brought in contact with the cylinder head and difference are irreversible. B. Free Expansion Process : 1. Let us latent heat of ice is 335 kJ / kg, therefore one tonne of refrigeration Refrigerants : 1. Those refrigerants which directly take part in the
during this process, the working substance is allowed to expand consider an insulated container divided in two compartments A and is given by, 1 TR = 1000 × 335 kJ in 24 hours = 1000 335 24 60   = refrigeration system and cool the substance by the absorption of
adiabatically so that its temperature becomes T2 c. Process 3-4 : B separated by means of a diaphragm as shown in th2. 232.6 kJ / min Note : In actual practice, one tonne of refrigeration is latent heat. Example : Ammonia, Caron dioxide, Sulphur dioxide,
(Reversible Isothermal Heat Rejection) 1. Adiabatic cover is Compartment A has a gas system maintained at p, V and T. In taken as equivalent to 210 kJ / min or 3.5 kW (i.e., 3.5 kJ / s) Methyl chloride, Freon group etc. 2. The primary refrigerants are
removed and heat sink is brought in contact with the cylinder head. compartment B, vacuum is maintained. 3. If the diaphragm is further classified into the following four groups : i. Halo-carbon
2. The working substance is compressed isothermally thus removed, gas in the compartment A expands into B until an A. Bell-Coleman or Reversed Brayton or Joule Cycle : 1. This cycle refrigerants, ii. Azeotrope refrigerants, iii. Inorganic refrigerants,
transferring heat Q2 to the heat sink at T2 . d. Process 4-1 : equilibrium state is established. 4. During this expansion process; consists of a compressor, a cooler, an expander and a refrigerator 2. and iv. Hydro-carbon refrigerant. i. Halo-Carbon Refrigerants : 1.
(Reversible Adiabatic Compression) 1. The adiabatic cover is again heat Q = 0 (since the system is insulated) and work W = 0 (i.e., no The Bell-Coleman cycle is a modification of reversed Carnot cycle. 3. The halocarbon compounds are obtained after replacing one or
brought in contact with the cylinder head and the system is external work transfer) 5. From the first law of thermodynamics, The four processes of this cycle are shown on p-v and T-s diagram more of hydrogen atoms of hydrocarbon methane or ethane by one
compressed adiabatically. 2. During this process the temperature of For a closed system, Q = dU + W Here Q and W are zero. 6. below : 4. The four process of the cycle are as follows : a. Isentropic or more of the three halogens : chlorine, fluorine and bromine. 2.
system is raised to T1 from T2 and it is brought to its initial state. Change in internal energy, dU = 0 U1 = U2 7. Only pressure and Compression Process (1-2) : 1. During the compression stroke, the Some of the commonly used halo-carbon compounds are given in
volume of the gas are changed. Now if the process has to be cold air from the refrigerator is drawn into the compressor cylinder the following table R-11 Trichloromonofluoromethane CCl3 F R-12
B. Reversed Carnot Cycle : 1. A reversed Carnot cycle, using air as reversed so that it may attain its initial state, it requires an where it is compressed isentropically in the compressor. 2. In this Dichlorodifluoromethane CCl2 F2 R-13
working medium (or refrigerant) is shown on p-v and T-s diagrams isothermal compression process in which work ‘W’ is supplied to process, both the pressure and temperature increases and the Monochlorotrifluoromethane CClF3 R-14 Carbontetrafluoride CF4
in Fig. 2.12.3(a) and 2.12.3(b) respectively. 2. At point 1, let p1 , v1 , the system from the surroundings with equivalent amount of heat specific volume of air at delivery from compressor reduces from v1 R-21 Dichloromonofluoromethane CHCl2 F R-22
T1 be the pressure, volume and temperature of air respectively. 3. to be rejected from the system to the surroundings even if the to v2 . b. Constant Pressure Cooling Process (2-3) : 1. During this Monochlorodifluoromethane CHClF2 R-30 Methylene chloride
The four processes of the cycle are as follows : a. Isentropic process is carried out in the absence of friction. 8. The sole effect of process, the warm air from the compressor is passed into the CH2Cl2 R-40 Methyl chloride CH3Cl R-100 Ethyl chloride C2H5Cl R-
Compression Process : 1. The air is compressed isentropically as the process is the transformation of work energy of the cooler where it is cooled at constant pressure p3 (= p2 ), reducing 113 Trichlorotrifluoroethane CCl2 FCClF2 R-114
shown by the curve 1-2 on p-v and T-s diagrams. 2. During this surroundings into the heat energy. 9. That’s why free expansion the temperature from T2 to T3 (temperature of cooling water). The Dichlorotetrafluoroethane CClF2CClF2 R-115
process, the pressure of air increases from p1 to p2 , specific process is highly irreversible specific volume also reduces from v2 to v3 . 2. Heat rejected by the Monochloropentafluoroethane CClF2CF3 ii. Azeotrope Refrigerant :
volume decreases from v1 to v2 and temperature increases from T1 air during constant pressure per kg of air is given by, Q2–3 = c p (T2 1. The term ‘azeotrope’ refers to a stable mixture of refrigerants
to T2 . 3. We know that during isentropic compression, no heat is – T3 ) c. Isentropic Expansion Process (3-4) : 1. During this process, whose vapour and liquid phases retain identical compositions over
absorbed or rejected by the air. b. Isothermal Heat Rejection 1. According to the third law of thermodynamics, “Entropy of all the air from the cooler is now drawn into the expander cylinder a wide range of temperatures. 2. Some of the azeotropes are given
Process : 1. The air is now compressed isothermally (i.e., at homogeneous crystalline substances in equilibrium state is zero at where it is expanded isentropically from pressure p3 to the in the following R-500 73.8 % R-12 and 26.2 % R-152 CCl2 F2
constant temperature, T2 = T3 ) as shown by the curve 2-3 on p-v absolute zero temperature.” 2. The degree of atomic or molecular refrigerator pressure p4 which is equal to the atmospheric pressure /CH3CHF2 R-502 48.8 % R-22 and 51.2 % R-115 CHClF2 /CClF2CF3
and T-s diagrams. 2. During this process, the pressure of air activity of a substance depends on its temperature. As the absolute and the temperature of air during expansion falls from T3 to T4 . R-503 40.1 % R-23 and 59.9 % R-13 CHF3 /CClF3 R-504 48.2 % R-32
increases from p2 to p3 and specific volume decreases from v2 to zero temperature is approached, the randomness of molecules The specific volume of air at entry to the refrigerator increases from and 51.8 % R-115 CH2 F2 /CClF2CF3 iii. Inorganic Refrigerant : 1.
v3 . 3. We know that the heat rejected by the air during isothermal tends to decrease and at absolute zero temperature the entropy v3 to v4 . d. Constant Pressure Expansion Process (4-1) : 1. During The inorganic refrigerants were most commonly used before the
compression per kg of air, qR = q2 – 3 = Area 2–3–3–2 = T3 (s 2 – s becomes zero. 3. It suggests that the entropy ceases to be a this process, the cold air from expander is passed to the refrigerator introduction of hydro-carbon group for all purposes. 2. These
3 ) = T2 (s 2 – s 3 ) c. Isentropic Expansion Process : 1. The air is function of state at absolute zero temperature Mathematically, 0 where it is expanded at constant pressure p4 (= p1 ) and the refrigerants are still in use due to their inhereand physical
now expanded isentropically as shown by the curve 3-4 on p-v and limT S  = 0 4. But for many substances like alloys, amorphous temperature of air increases from T4 to T1 . Due to heat from the properties. 3. The important inorganic refrigerants are given in the
T-s diagrams. 2. The pressure of air decreases from p3 to p4 , bodies, chemical compounds such as carbon mono oxide, NO, etc; refrigerator, the specific volume of the air changes from v4 to v1 following table R-717 Ammonia NH3 R-729 Air — R-744 Carbon
specific volume increases from v3 to v4 and the temperature entropy does not tend to zero as T  0 but takes some finite dioxide CO2 R-764 Sulphur dioxide SO2 R-118 Water H2O iv. Hydro-
decreases from T3 to T4 . 3. We know that during isentropic positive value at absolute zero temperature. The reason being that A. Vapour Compression Refrigeration System : 1. It is an improved Carbon Refrigerants : 1. Most of the refrigerants of this group are
expansion, no heat is absorbed or rejected by the air. d. these substances are not found in equilibrium state. 5. The third type of air refrigeration system in which a suitable working organic compounds and these are successfully used in industrial
Isothermal Heat Addition Process : 1. The air is now expanded law of thermodynamics provides absolute base which helps in substance termed as refrigerant is used. 2. The refrigerant does not and commercial installations. 2. Some of the important refrigerants
isothermally (i.e., at constant temperature, T4 = T1 ) as shown by measuring the entropy of each substance. The third law is also leave the system. Condensed and evaporated alternately and is of this group are given in the table below : R-170 Ethane C2H6 R-
the curve 4-1 on p-v and T-s diagrams. 2. The pressure of air helpful in measuring the chemical affinity (i.e., action of chemical circulated throughout the system. 3. During evaporation, the 290 Propane C3H8 R-600 Butane C4H10 R-600a Isobutane CH(CH3 )
decreases from p4 to p1 , and specific volume increases from v4 to forces of reacting substances), explaining behaviour of solids at low refrigerant absorbs its latent heat from the brine (salt water) which 3 R-1120 Trichloroethylene C2HCl3 R-1130 Dichloroethylene
v1 . 3. We know that the heat absorbed by the air (or heat temperature and analyzing the chemical and phase equilibrium is used for circulating it around the cold chamber. 4. While C2H2Cl2 R-1150 Ethylene C2H4 R-1270 Propylene C3H6 b.
extracted from the cold body) during isothermal expansion per kg condensing, it gives out its latent heat to the circulating water of Secondary Refrigerants : 1. Those refrigerants which are first cooled
of air, 4. We know that work done during the cycle per kg of air5. the cooler. 5. Therefore, vapour compression refrigeration system is with the help of the primary refrigerants and are then employed for
Coefficient of performance of the refrigeration system working on a latent heat pump, as it pumps its latent heat from the brine and cooling purposes are known as secondary refrigerant. 2. These
reversed Carnot cycle, (COP)R = Heat absorbed Work done qa/qr- A. Available and Unavailable Energy : 1. The part of heat energy delivers it to the cooler. B. Advantages of Vapour Compression refrigerants cool substances by absorption of their sensible heat. 3.
qa= q4-1/q2-3 – q4-1 == T1(s2-s3)/(t2-t1)(s2-s3)= T1/T2-T16. input in a cyclic heat engine which gets converted into mechanical Refrigeration System over Air Refrigeration System : 1. It has smaller The commonly used secondary refrigerants are as follows : i. Water,
Though the reversed Carnot cycle is the most efficient between the work is known as available energy. 2. That part of heat energy size for the given capacity of refrigeration. 2. It has less running ii. Sodium chloride brine, iii. Calcium chloride brine, iv. Ethylene
fixed temperature limits, yet no refrigerator has been made using which is not utilizable and is to be rejected to the surroundings is cost. 3. It can be employed over a large range of temperatures. 4. glycol, and v. Propylene glycol etcnt thermodynamic
this cycle. 7. This is due to the reason that the isentropic processes known as ‘unavailable energy’. 3. The term ‘exergy’ is synonymous The coefficient of performance is quite high.
of the cycle require high speed while the isothermal processes with available energy and the term ‘anergy’ is synonymous with
require an extremely low speed. 8. This variation in speed of air is unavailable energy. Therefore, Energy = exergy + anergy. 4. The
not practicable. concept of availability is related to the maximum amount of a. Compressor : 1. The low pressure and temperature vapour
theoretical work (without dissipative effects) which can be obtained refrigerant from evaporator is drawn into the compressor through
from a system at a given state upto in dead state. 5. Maximum the inlet or suction valve A, where it is compressed to a high
A. Carnot’s Theorem : 1. It states that of all heat engines operating useful work obtained under such ideal conditions is known as pressure and temperature. 2. This high pressure and temperature
between a given constant temperature source and a given constant available energy of the system and the part of energy rejected is vapour refrigerant is discharged into the condenser through the
temperature sink, none has a higher efficiency than a reversible known as unavailable energy 6. It is worthwhile to note that when delivery or discharge valve B. b. Condenser : 1. The condenser or
engineThe cyclic heat engines and operating between the same the system reaches its dead state, the transfer of energy ceases, cooler consists of coils of pipe in which the high pressure and
source and sink of which is reversible2. Let two heat engines EA and though the system contains internal energy but this energy cannot temperature vapour refrigerant is cooled and condensed. 2. The
EB operate between the given source at temperature T1 and given be referred to as available energy. refrigerant, while passing through the condenser, gives up its latent
sink at temperature T2 as shown in Fig. 2.14.1. 3. Let EA be any heat heat to the surrounding condensing medium which is normally air
engine and EB be any reversible heat engine. We have to prove that A. Exergy Destruction : 1. In thermodynamics, the exergy of a or water. c. Receiver : 1. The condensed liquid refrigerant from the
the efficiency of EB is more than that of EA . 4. Let us assume that system is the maximum useful work possible during a process that condenser is stored in a vessel known as receiver from where it is
this is not true and A > B . 5. Let the rates of working of the brings the system into equilibrium with a heat reservoir. 2. When supplied to the evaporator through the expansion valve or
engines be such that Q1A = Q1B = Q1 Since A > BW Q > 1 B B W the surroundings are the reservoir, exergy is the potential of a refrigerant control valve d. Expansion Valve : 1. It is also called
Q  WA > WB 6. Now, let EB be reversed. Since EB is a reversible system to cause a change as it achieves equilibrium with its throttle or refrigerant control valve. 2. The function of the
heat engine, the magnitude of heat and work transfer quantities environment. 3. Exergy is the energy that is available to be used. expansion valve is to allow the liquid refrigerant under high
will remain the same, but their directions will be reversed, 7. Since After the system and surroundings reach equilibrium, the exergy is pressure and temperature to pass at a controlled rate after reducing
WA > WB , some part of WA (equal to WB ) may be fed to drive the zero. 4. Determining exergy was also the first goal of its pressure and temperature. e. Evaporator : 1. An evaporator
reversed heat engine B . 8. Since Q1A = Q1B = Q1 , the heat thermodynamics. 5. Exergy is never destroyed during a process, it consists of coils of pipe in which the liquid-vapour refrigerant at low
discharged by B may be supplied to EA . 9. The source may, changes from one form to another. In contrast, exergy accounts for pressure and temperature is evaporated and changed into vapour
therefore, be eliminated (Fig. 2.14.3). The net result is that EA and the irreversibility of a process due to increase in entropy. 6. Exergy refrigerant at low pressure and temperature. 2. During evaporation,
B together constitute a heat engine which, operating in a cycle, is always destroyed when a process involves a temperature change. the liquid vapour refrigerant absorbs its latent heat of vapourization
produces net work WA – WB , while exchanging heat with a single This destruction is proportional to the entropy increase of the from the medium (air, water or brine) which is to be cooled.
reservoir at T2 . Violates the Kelvin-Planck statement of the second system together with its surroundings. The destroyed exergy has
law. Hence the assumption that A > B is wrong. Therefore B  been called anergy
B. Corollary of Carnot’ Theorem : 1. The efficiency of all reversible
heat engines operating between the same temperature levels is the e importance of availability, effectiveness and irreversibility 1. First
same. 2. Let both the heat engines EA and EB (Fig. 2.14.2) be law of thermodynamics has given the concept of efficiency which
reversible. 3. Let us assume A > B . 4. Similar if EB is reversed to can be applied to cycles only, while the concept of availability,
run, say as a heat pump using some part of the work output (WA ) irreversibility and effectiveness have been derived from second law
of engine EA , we see that the combined system of heat pump EB of thermodynamics and these concepts are applicable to both
and engine EA , becomes a PMM2. 5. So A cannot be greater than processes and cycles. 2. These concepts also help in analyzing the
B . 6. Similarly, if B > A and reverse the engine EA , then B can processes since they show the deviation of actual processes from
not be greater than A . 7. Therefore A = 8. Since the efficiencies ideal processes; therefore, these concepts suggest the
of all reversible heat engines operating between the same heat improvement in thermodynamic cycles. 3. The derived concepts of
reservoirs are the same, the efficiency of a reversible engine is availability, irreversibility and effectiveness are particularly useful in
independent of the nature or amount of the working substance heat transfer processes between two fluids while first law of
undergoing the cycle. thermodynamics does not show any irreversibility during the
process.

A. Dead State : 1. Dead state refers to the state at which system and
environment are at chemical, thermal and mechanical equilibrium.
2. Thus neither there can be any spontaneous change within the
system or within the environment, nor any spontaneous interaction
between the two. Dead state being a limiting state is also called
‘restricted dead state’. 3. At dead state the system is at same
temperature and pressure as that of its surroundings and shall have
no kinetic energy or potential energy relative to surroundings. 4.
Thus, a system shall have zero availability at dead state and yield
maximum possible work only when it follows a reversible process
from its state to the state of its surroundings (dead state). B.
Second Law Efficiency : 1. The second law efficiency ‘II’of a process
is defined as the ratio of the minimum available energy which must
be consumed to do a work divided by the actual amount of
available energy consumed in performing the work. II = Minimum
available energy to do the work Actual available energy consumed
or II = Amin A Where, A is the availability or exergy