Download UNIT I PART B 1). (i). A spherical balloon of diameter

UNIT I
PART B
1). (i). A spherical balloon of diameter 0.5 m contains gas at 1 bar and 300 K. The gas is
heated and the balloon is allowed to expand. The pressure inside the balloon is directly
proportional to the square of the diameter. What would be the work done by the gas when
the pressure inside reaches 5 bar?
(ii) The turbines in a hydroelectric power plant are fed by water falling from a 50 m height.
Assuming 91% efficiency of for conversion of potential to kinetic energy and 8% loss of the
resulting power in transmission, what is the mass flow rate of water required to power a
200 W light bulb.
(Nov 2011).
2). Define the following terminologies of thermodynamics
(i). Reversible and irreversible processes
(ii). State functions and path functions.
(iii). Thermodynamic equilibrium. (Nov 2010)
Reversible Process:
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The process in which the system and surroundings can be restored to the initial state from the
final state without producing any changes in the thermodynamics properties of the universe is
called a reversible process. In the figure below, let us suppose that the system has undergone a
change from state A to state B. If the system can be restored from state B to state A, and there is
no change in the universe, then the process is said to be a reversible process. The reversible
process can be reversed completely and there is no trace left to show that the system had
undergone thermodynamic change.
For the system to undergo reversible change, it should occur infinitely slowly due to
infinitesimal gradient. During reversible process all the changes in state that occur in the system
are in thermodynamic equilibrium with each other.
Irreversible Process:
The irreversible process is also called the natural process because all the processes occurring in
nature are irreversible processes. The natural process occurs due to the finite gradient between
the two states of the system. For instance, heat flow between two bodies occurs due to the
temperature gradient between the two bodies; this is in fact the natural flow of heat. Similarly,
water flows from high level to low level, current moves from high potential to low potential, etc.
Here are some important points about the irreversible process:
1) In the irreversible process the initial state of the system and surroundings cannot be restored
from the final state.
2) During the irreversible process the various states of the system on the path of change from
initial state to final state are not in equilibrium with each other.
3) During the irreversible process the entropy of the system increases decisively and it cannot be
reduced back to its initial value.
4) The phenomenon of a system undergoing irreversible process is called as irreversibility.
STATE AND PATH FUNCTIONS:
There are a large number of properties of matter that we can express numerically. Examples of
these are mass, volume, temperature, pressure, number of moles, density, and so forth. All of
these properties have something in common: they can be determined by a single measurement at
a single moment in time, without any knowledge of the history of the system.
Properties that can be measured meaningfully at a single instant in time are called state functions.
Most physical and chemical properties that can be measured numerically are state functions.
Other examples (beyond those stated above) are distance, area, percent composition, velocity,
acceleration, and force.
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A second defining characteristic of a state function is that the change in a state function
can be determined by measuring the initial and final values of that function.
Work, on the other hand, is fundamentally different from any of the properties we’ve
listed above. One cannot speak meaningfully of the amount of “work” contained in an object,
because work is a process, requiring the passage of some amount of time. Furthermore, to
measure the amount of work that is done in a process, we must observe the entire process from
start to finish. As an example, suppose that you observe two people standing side by side at
12:00. You then leave for an hour. When you return at 1:00, the two people are still standing in
the same place. There is no apparent change in either person. Can you conclude that neither
person did any work? No, you cannot! Person A may have run up six flights of stairs, lifted
weights for 20 minutes, then eaten just enough food to counteract any weight loss (and fluid loss)
during the exercise, and finally returned to his original position; while person B just stood in
place for an hour.
Work is an example of a path function. A path function is a property whose numerical value
cannot be determined unless one observes the entire process. Path functions are not very
common in science, but they are common enough in everyday life. Here are a few other
examples of path functions:
• The distance you walked during some time period
• The number of times you changed your socks during some time period
• The amount of money you spent during some time period
In each case, an observer would need to watch you throughout the time period to measure the
property. Note that for a path function, a time period must be specified (as opposed to a state
function, where a single time must be specified).
Thermodynamic Equilibrium
Whenever the system is in thermodynamic equilibrium, it tends to remain in this state infinitely
and will not change spontaneously. Thus when the system is in thermodynamic equilibrium there
won’t be any spontaneous change in its macroscopic properties.
Conditions for Thermodynamic Equilibrium
The system is said to be in thermodynamic equilibrium if the conditions for following three
equilibrium is satisfied:
1) Mechanical equilibrium
2) Chemical equilibrium
3) Thermal equilibrium
1) Mechanical equilibrium: When there are no unbalanced forces within the system and
between the system and the surrounding, the system is said to be under mechanical equilibrium.
The system is also said to be in mechanical equilibrium when the pressure throughout the system
and between the system and surrounding is same. Whenever some unbalance forces exist within
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the system, they will get neutralized to attain the condition of equilibrium. Two systems are said
to be in mechanical equilibrium with each other when their pressures are same.
2) Chemical equilibrium: The system is said to be in chemical equilibrium when there are no
chemical reactions going on within the system or there is no transfer of matter from one part of
the system to other due to diffusion. Two systems are said to be in chemical equilibrium with
each other when their chemical potentials are same.
3) Thermal equilibrium: When the system is in mechanical and chemical equilibrium and there
is no spontaneous change in any of its properties, the system is said to be in thermal equilibrium.
When the temperature of the system is uniform and not changing throughout the system and also
in the surroundings, the system is said to be thermal equilibrium. Two systems are said to be
thermal equilibrium with each other if their temperatures are same.
For the system to be thermodynamic equilibrium it is necessary that it should be under
mechanical, chemical and thermal equilibrium. If any one of the above condition are not
fulfilled, the system is said to be in non-equilibrium
3). (i). Distinguish between heat pump and heat engine. (Nov 2009).
Difference between a Heat Engine, Refrigerator and Heat Pump
Heat Engine
In heat engine the heat supplied to the engine is converted into useful work. If Q 2 is the heat
supplied to the engine and Q1 is the heat rejected from the engine, the net work done by the
engine is given by:
The performance of a heat engine is expressed by its efficiency. We know that the efficiency or
coefficient of performance of an engine,
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T1 >Ta
Refrigerator
Refrigerator is a reversed heat engine which either cool or maintain the temperature of a body
(T1) lower than the atmospheric temperature (Ta). This is done by extracting the Heat from a
cold body and delivering it to a hot body (Q 2). In doing so, work W R is required to be done on
the system. According to First law of thermodynamics,
WR = Q2 – Q1
The performance of a refrigerator is expressed by the ratio of amount of heat taken from the cold
body (Q1) to the amount of work required to be done on the system (W R). This ratio is called
coefficient of performance. Mathematically, coefficient of performance of a refrigerator,
T1 < Ta
Heat Pump
A refrigerator used for cooling in summer can be used as a heat pump for heating in winter. In
the similar way, as discussed for refrigerator, we have
Wp =- Q2 – Q1
The performance of a heat pump is expressed by the ratio of the amount of the heat delivered to
the hot body (Q2) to the amount of work required to be done on the system (Wp). This ratio is
called coefficient of performance or energy performance ratio (E.P.R.) of a heat pump.
Mathematically, coefficient of performance or energy performance ratio of a heat pump,
From above we see that the C.O.P. may be less than one or greater than one depending on the
type of refrigeration system used. But the C.O.P. of a heat pump is always greater than one.
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(ii) Discuss phase rule in detail with an example.
Gibbs Phase Rule is expressed by the simple formulation:
P + F = C + 2, where
P is the number of phases in the system
A phase is any physically separable material in the system. Every unique mineral is a
phase (including polymorphs); igneous melts, liquids (aqueous solutions), and vapor are
also considered unique phases. It is possible to have two or more phases in the same state
of matter (e.g. solid mineral assemblages, immiscible silicate and sulfide melts,
immiscible liquids such as water and hydrocarbons, etc.) Phases may either be pure
compounds or mixtures such as solid or aqueous solutions--but they must "behave" as a
coherent substance with fixed chemical and physical properties.
C is the minimum number of chemical components required to constitute all the phases in the
system
F is the number of degrees of freedom in the system (also referred to as the variance of the
system).
Consider a system in equilibrium containing P phases and C components. Assume that the
passage of a component from one phase to another does not contribute a chemical reaction. The
state of each phase of the system is completely specified by the two variables, temperature and
pressure and also by composition of each phase. So we have to describe the mole fraction of each
component. If C is the number of components required to describe the composition of a phase,
then the total number of composition variables for the P phases are PC. Besides, there are two
more variables, temperature and pressure, which have to be considered. The total number of
independent variables is PC + 2.
However, all the variables are not independent since in each phase, sum of the mole fractions
must equal to unity:
X1 + X2 + X3 +........XC = 1
∑ Xi = 1
Therefore the total numbers of independent variables to be specified are:
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CP + 2 - P
P(C- 1) + 2
When a heterogeneous system is in equilibrium at a constant temperature and pressure, the
chemical potential of a particular component must be the same in all the phases in which it
appears. For a one-component system having two phases α and β, the equality of chemical
potential implies
μ(α) = α (β)
Similarly, in a one component system having three phases a, p and y, we have,
μ (α) = μ(β)=μ (γ)
Similarly, for a system containing C number of components and P phases, the requirement that
the chemical potential of any component is same in all P phases at equilibrium leads to the
equations for the C components in P phases.
μ1 (α) = μ1 (β) = μ1 (γ) =………..= μ1 (P)
μ2 (α) = μ2 (β) = μ2 (γ) =………..= μ2 (P)
μ3 (α) = μ3 (β) = μ3 (γ) =………..= μ3 (P)
…………………………………………………………….
μC (α) = μc (β) = μc (γ) =………..= μc (P)
In general, for each component in P phases, (P - 1) relations are possible. Since there are C
components, the total number of these equations are C(P - 1).
F = total number of variables — total number of relations among the variables
= PC + 2 - P - C(P - 1)
=C -P+2
F+P=C+2
(iii). A special manometer fluid has a specific gravity of 3.65 and is used to measure a
pressure a pressure of 1.25 bar at a location where the barometric pressure is 760 mm Hg.
What height will the manometer fluid register?
4). Five kilograms of CO2 gas is contained in a piston cylinder assembly at a pressure of 7.5 bar
and a temperature of 300 K. The piston has a mass of 6000 Kg and a surface area of 1 m 2 . The
friction of the piston on the walls is insignificant. The atmospheric pressure is 1.0135 bar. The
latch holding the piston in position is suddenly removed and the gas is allowed to expand. The
expansion is arrested when the volume is double the original volume. Determine the work done
in the surroundings.
(Nov 2009)
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5). An elevator with a mass of 3 tons rests at a level 15 m above the base of an elevator shaft. It
is raised to 125 m above the base of the shaft. The elevator falls freely to the base of the shaft
and strikes a spring and comes to rest. Calculate.,
(i). the potential energy of the elevator in its initial and final position.
(ii) Work required to raise the elevator.
(iii). The velocity and kinetic energy of the elevator before it strikes the spring.
(iv). The potential energy of the compressed spring.
(v). If the elevator and spring is considered as a system, calculate the energy of the
systems at different conditions mentioned above. State your assumptions and explain the
interface. (Nov 2008).
6). Define reversible process. Summarize the characteristics of a reversible process.
Imagine a cylinder, with a perfectly smooth piston, which contains gas. If you push with a force
only just large enough to overcome the internal pressure, the volume will start to decrease
slowly. Then if you decrease the force only slightly, the volume will start to increase. This is the
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hallmark of a reversible process: an infinitesimal change in the external conditions reverses the
direction of the change. Heat flow is only reversible if the temperature difference between the
bodies is infinitesimally small.
Reversible processes require the absence of friction or other hysteresis effects. They must also be
carried out infinitesimally slowly. Otherwise pressure waves and finite temperature gradients
will be set up in the system, and irreversible dissipation and heat flow will occur.
Because reversible processes are very slow, the system is always very nearly in equilibrium at all
times. In that case all its state variables are well defined and uniform, and the state of the system
can be represented on a plot of, for instance, pressure versus volume. A finite reversible process
passes through an infinite set of such states, and so can be drawn as a solid line on the plot.
During an irreversible process the system is not in an equilibrium state, and so cannot be
represented on the plot; an irreversible process is often drawn as a straight dotted line joining the
initial and final equilibrium states. The work done during the process is not then equal to the area
under the line, but will be greater than that for the corresponding reversible process.
7). Explain the following:
(i). Categorization of the System and the Process.
(ii). Classification of Energy.
System: The application of thermodynamics to any real problem starts with the identification of
a particular body of matter as the focus of attention which is call as system. A system is the
portion of the Universe under study which is separated from the remainder of the universe is real
or imaginary boundaries.
Surroundings: The rest of the Universe which may or may not interact by exchange of matter
and/or energy with the system is called the surroundings.
Types of Systems:
1). Isolated System: A system which can exchange neither energy nor matter with its
surroundings is called an Isolated system.
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2). Closed System: A system which can exchange energy but not matter with its surroundings is
called a closed system.
3). A system which can exchange matter as well as energy with its surroundings is said to be
open system.
Different Types of Processes:
Process: It is the operation by which a system changes from one state to another. This is
accompanied by energy.
Isothermal Process: When the temperature of the system remains constant during each stage of
operation it is said to be isothermal process.
Adiabatic Process: A process is said to be adiabatic if no, heat enters or leaves the system
during any step of the process. That is system is perfectly insulated.
Isobaric Process: When the pressure of the system, remains constant during each stage of
operation, it is said to be isobaric process.
Isochoric Process: When the volume of the system, remains constant during each stage of
operation, it is said to be isochoric process.
Reversible Process: A process is carried out infinitesimally slowly so that the driving force is
only infinitesimally greater than the opposing force is called a reversible process.
Irreversible Process: Any process which does not take place infinitesimally slowly is said to be
an irreversible process.
8). (i) Show that Work and Heat are Path functions, using a PV diagram.
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(ii) An auto mobile having a mass of 1400Kg is traveling at 30m/s. What is its
energy in KJ? How much work must be done to bring it to a stop?
kinetic
9). (i) Explain the concept “Heat and Work are Energy in Transition”.
In the strict sense both the forms of energy, viz., heat and work can not be stored at a
single point, and there is always a flow in the direction of high level to low level, is seen. Hence
heat and work energies are thermodynamically referred to as energies in transition.
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(ii) The potential energy of a body of mass 20Kg is 3.5KJ. What is the height of the
from the ground? If a body of mass 20Kg is moving at a velocity of 50m/s,
what
kinetic energy?
body
is its
10). (i) What is the scope and Limitations of Thermodynamics?
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(ii) Explain Macroscopic and Microscopic aspects of Thermodynamics.
Microscopic and macroscopic systems
It is useful, at this stage, to make a distinction between the different sizes of the systems that we
are going to examine. We shall call a system microscopic if it is roughly of atomic dimensions,
or smaller. On the other hand, we shall call a system macroscopic when it is large enough to be
visible in the ordinary sense. This is a rather inexact definition. The exact definition depends on
the number of particles in the system, which we shall call
. A system is macroscopic if
which means that statistical arguments can be applied to reasonable accuracy. For instance, if we
wish to keep the statistical error below one percent then a macroscopic system would have to
contain more than about ten thousand particles. Any system containing less than this number of
particles would be regarded as essentially microscopic, and, hence, statistical arguments could
not be applied to such a system without unacceptable error.
UNIT II
1). (i). Write short notes on internal energy, enthalpy, heat capacity and phase rule.
Internal Energy: This is the energy possessed by a substance because of motion and
configuration of its molecules, atoms and sub atomic particles which are referred as microscopic
modes of energy.
External energy: It is the product of pressure and volume and may be regarded as the energy of a
substance possessed by virtue of the space its occupies.
Apart from these, various other forms of energies can be mentioned like Surface energy,
Magnetic Energy, Electrical energy, Chemical Energy, thermal energy etc.
Enthalpy: Especially in flow processes, often we find terms like U + PV. To simplify this an
another thermodynamic property is defined which is called Enthalpy. It is the sum of internal and
external energy and is an important term in energy balance computations.
H = U + PV.
All the above three terms has the units of energy and are state functions.
Heat Capacity
The heat capacity of a substance is the quantity of heat to be supplied to effect a temperature rise
of one degree.
dQ = C dT
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Where C is know as heat capacity of the substance.
Heat capacity of unit mass of substance is known as specific heat of the substance.
The heat capacity depends on the way in which heat is supplied. When heat is supplied
to a system at constant volume the system is unable to do any work and the quantity of heat
required is given by
dQ = Cv dT, where Cv is known as the heat capacity at constant volume.
Cv = dQ/dT. We know that a at constant volume dQ = dU
Therefore Cv = dU/dT
dU = Cv dT
Similarly if heat is supplied at constant pressure, it is free to expand doing work against the
constant pressure applied.
dQ = Cp dT where Cp is known as specific heat at constant pressure.
Cp = dQ/dT
We know from first law of TD dU = dQ – dW = dQ – PdV
dQ = dU + PdV = dH.
That is for process at constant pressure, dQ = dH
Hence dH = Cp dT
Obviously for a given temperature change, the heat required is more in a constant
pressure process than that is a constant volume process, Cp is always greater than Cv.
We know dH = dU + d(PV)
For ideal gases PV = RT. Hence for ideal gases dH = dU + d(RT) = dU + R dT
But we know that dH = Cp dT and dU = Cv dT
Hence substituting the values of dH and dU interms of specific heats,
CpdT = CvdT + RdT , Dividing through out by dT
Cp = Cv + R
Cp – Cv = R vallied only for ideal gases.
(ii). A system undergoes a process 1-2 in which it absorbs 100 KJ energy as heat and does
40 KJ work. Then it flows to another process 2-3 in which 50 KJ of work is done on it while
it rejects 30 KJ as heat of it is desired to restore the system to the initial state by an
adiabatic path. Calculate the work and heat interaction during this process. Also
determine the network and heat interactions. (April 2011).
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2). (i). One Kg mole of an ideal gas at 1 atmosphere and 300 K is compressed isothermally
and reversibly to 10 atm. Calculate Q, W, ∆U and ∆H. Assume the value of R.
(ii). A system consisting of a gas confined in a cylinder is undergoing the series of process
before it is brought back to the initial conditions.
Step 1: A constant pressure process when it receives 500 J of work and gives up 25 J of
heat.
Step 2: A constant volume process when it receives 75 J of heat.
Step 3: An adiabatic process. Determine the change in internal energy during each step
and the work done during the adiabatic process. (April 2011)
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3). (i). Derive the expression for first law of thermodynamics as applied to steady state flow
processes. State the assumptions.
First law of Thermodynamics for a flow Process:
Consider a fluid flowing thorough the apparatus from section (1) to section (2). The
internal energy, velocity, specific volume, Pressure and Height are represented by U, V, v, P and
Z with subscripts 1 and 2 correspondingly for sections (1) and (2), as shown in the figure. Heat Q
is added per unit mass of the fluid by means of the heat exchanger and shaft work W s is extracted
per unit mass by means of a turbine.
According to I law of Thermodynamics the total energy with which the fluid is entering
at section (1), plus the total energy imparted to the fluid inside the system must be equal to the
total energy with which the fluid is leaving the system and the work extracted from the system.
Net energy content of fluid at section (1) :Internal energy + Potential energy + Kinetic energy +
external energy
Net energy content of fluid at section (1) : mU1 + mgz1 + ½ mv12 + mP1 V1
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Similarly Net energy content of fluid at section (2) = mU2 + mgz2 + ½ mv22 + mP2 V2
Total energy imparted to the fluid = mQ
Work that is obtained form the system = mWs
As per I law of TD Energy in = Energy out for a steady state system.
mU1 + mgz1 + ½ mv12 + mP1 V1 + mQ = mU2 + mgz2 + ½ mv22 + mP2 V2 + mWs
Dividing though out by m, and rearranging,
(U2 - U1) + g(Z1 –Z2) + ½ (v22 – v12 ) + (P2 V2 - P1 V1 ) = Q – Ws
ΔU + Δ(PV) + g(ΔZ) + 1/2Δv2 = Q – Ws
But we know that ΔU + Δ(PV) = Δ H
Hence ΔH + g(ΔZ) + 1/2Δv2 = Q – Ws
This is the mathematical statement of first law of thermodynamics for flow process and can be
used to determine power required for pumps and compressors etc. For most applications in
thermodynamics the kinetic energy and potential energy values are small compared to other
terms and hence
ΔH = Q – Ws
(ii). A Tank containing 20 kg of water at 20˚C is fitted with a stirrer that delivers work to
water at the rate of 0.25 kW. How long does it take for the temperature of water to rise to
30˚C if no heat is lost from the water? Cp for water may be taken as 4.184 KJ/Kg˚C
(Nov 2011).
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4). (i). A domestic refrigerator transfers energy in the form of heat from the cold space to
the ambient atmosphere at higher temperature. Does it violate the Clausius statement of
the II law of thermodynamics? Explain in detail.
This does not violate Clasusius statement, because on its own without any external energy supply
heat transfer from lower level to higher level is impossible. In a domestic refrigerator, transfer of
energy in the form of heat from the cold space to the ambient atmosphere at high temperature is
made possible with the addition of work to the system in the form of compressor.
(ii). An inventor claims to have developed a cyclically working device which absorbs 500
KJ as heat from a reservoir at 800K and 300KJ from a reservoir at 400k and rejects 100
KJ as heat to a reservoir at 600K and 50KJ to a reservoir at 300K, while it delivers 650 KJ
work. Would you agree with his claim? Justify your answer on thermodynamic grounds.
(Nov 2010).
5). (i) Explain in detail the Carnot cycle. Derive the equation for the efficiency of Carnot
engine.
Carnot Engine: A heat engine operating in a completely reversible manner (with 100%
efficiency) was first described by Carnot and is called after him as Carnot engine.
A Carnot engine performs a cycle of operation as follows.
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(1). A system initially in thermal equilibrium with a cold reservoir at temperature Tc
undergoes, a reversible adiabatic process that causes its temperature to rise to that of a hot
reservoir T H
(2). The system maintains contact the hot reservoir at T H and undergoes a reversible
isothermal process during which heat |Q H| is absorbed from the hot reservoir.
(3). The system undergoes a reversible adiabatic process in the opposite direction of step (1),
that brings its temperature back to that of cold reservoir at Tc.
(4). The system maintains contact with the reservoir at Tc and undergoes a reversible
isothermal process in the opposite direction of step (2) that returns it to its initial state with
rejection of heat |Qc| to the cold reservoir.
Since Carnot engine is reversible it may be operated in reverse that is in the opposite
direction and it becomes a reversible refrigeration cycle for which the quantities |Q H|, |Qc| and
|W| are the same as for the engine cycle but are reversed in the direction.
CARNOT THEOREM:
Statement: For two given heat reservoirs no engine can have a higher efficiency than a Carnot
engine.
Proof: Consider an engine operating a Carnot refrigerator C. Let a Carnot engine absorbs heat
|Q H| from reservoir, produces work |W| and discards
|Q H| - |Qc| to a cold reservoir. Assume a second engine E with a greater thermal efficiency
operating between the same heat reservoirs absorbing heat |Q H*|, producing the same work
and discarding |Q H*| - |W|., then if its efficiency is higher than that of Carnot’s then,
Now let engine E drive the Carnot engine backward as a heat pump that is Carnot refrigerator
as shown in the figure.
Heat extracted from the cold reservoir = [|Q H| - |W| - { |Q H*| - |W|}]
= |Q H| - |Q H*|
Heat delivered to the hot reservoir = |Q H| - |Q H*|
This suggests that heat is being transferred from lower temp source to high temperature
source, which violates II law of thermodynamics. Hence no other engine can have efficiency
greater than that of Carnot’s engine.
Carnot’s II theorem:
All reversible heat engines operating between the same two given thermal reservoirs
with fixed temperatures have the same efficiency. The efficiency of a heat engine does not
depend on the nature of the working medium but depends only on the temperatures of the
reservoirs between with it operates.
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Proof: The cycle traversed by Carnot’s engine having an ideal gas serving as the working fluid
consists of four steps
(1). Adiabatic compression, with no heat flow in and out, during which temperature increases
from Tc to T H
(2). Isothermal expansion with the input of energy |Q H|, maintaining same temperature T H
(3). Adiabatic expansion with no heat flow in and out, during with temperature drops from T H
to Tc.
(4). Isothermal compression with the liberation of energy |Qc|, during which temperature is
maintained at Tc
For isothermal process dT = 0 and dU = 0, from I law TD dU = dQ – dW
dQ = dW = PdV
Consider isothermal step b to c, dQ = PdV For ideal gas PV = RT, P = RT/V
dQ = RT H dV/V , integrating between the limits, Vc to Vb
Q = RTH ln Vc/Vb --------(1)
Consider isothermal step d to a, dQ = PdV For ideal gas PV = RT, P = RT/V
dQ = RTc dV/V , integrating between the limits, Vd to Va - Q = RTc ln Va/Vd
( - because heat is liberated from the system)
or Q = RTc ln Vd/Va -----(2)
For an adiabatic process dQ = 0, hence from I law dU = - dW = - PdV
Cv dT = - PdV = RT dV/v
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(ii) A Carnot engine operating between 800 oC and 25oC is used to run a Carnot
refrigerator operating between -20oC and 25oC. If the engine absorbs 10 KJ/s from the
reservoir at 800oC, determine the capacity of the refrigerator.
6). (i) Derive the first law of thermodynamics for a steady-state steady-non flow process.
The first law of Thermodynamics requires that the change in the total energy of the system be
compensated by an equal but opposite change in the total energy of the surroundings.
Δ (Energy of the system) + Δ (Energy of the surroundings) = 0
Energy from a closed system can get transferred as heat and work.
Sign convention: Work done by a system on its surroundings is assigned a positive sign and
work done by the surroundings is assigned a negative sign. The energy transfer as heat from
surroundings to system is assigned a positive sign, and energy transfer as heat from system to its
surroundings is assigned a negative value.
If Q is the heat transferred to the system and w is the work extracted fro it, during process,
(ΔE) Surroundings = - Q + W
For a closed system undergoing only changes in the Kinetic, Potential and internal energies the
total energy change of the system is given by,
(ΔE) system = Δ(PE) + Δ(KE) + Δ(U)
As per first law,
Δ(PE) + Δ(KE) + Δ(U) - Q + W = 0
For a steady state non flow process Δ(PE) = 0 and Δ(KE) = 0
Hence Δ(U) = Q - W = 0
For a differential changes in the thermodynamic state of a closed system,
dU = dQ - dW
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(ii) A rigid and insulated container of 2m3 capacity is divided into two equal compartments
by a membrane. One compartment contains Helium at 200KPa and 127oC while the
second compartment contains Nitrogen at 400KPa and 227 oC. The membrane is punctured
and the gases are allowed to mix. Determine the temperature
and
pressure
after
equilibrium has been established. Consider Helium and Nitrogen as ideal gases with their
CV as 1.5R and 2.5R respectively.
7). (i) Explain the second law of thermodynamics with suitable example.
All efforts to device a process for the continuous conversion of heat completely into work or
into mechanical or electrical energy have failed. Regardless of improvements to the devices
conversion efficiencies never exceeded 40%. To explain these shortcomings second law of
thermodynamics came to existence.
Statement: There are several statements of the second law of thermodynamics and the
essence of all these are to deny the possibility of a spontaneous process reverting on its own.
Kelvin plank statement: No apparatus can operate in such a way that its own effect (in
system and surroundings) is to convert heat absorbed by a system completely into work.
Clausius statement: No process is possible, which consists of solely in the transfer of heat
from one temperature level to a higher one.
Heat engine: A cyclically operating device for the continuous conversion of internal energy
(transferred as heat) of matter into work is called heat engine. It is an energy conversion
device which continuously converts energy absorbed as heat into work.
Ex: An automobile engine. Steam power plant etc.
Steam Power Plant:
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(i). Liquid water at ambient temperature is pumped into a boiler
(ii). Heat from fuel or from nuclear reaction is transferred
in the boiler to the water,
converting it to the stream at high pressure and temperature.
(iii). Energy is transferred as shaft work from the steam to surroundings by a device such as
turbine.
(iv). Exhaust steam is condensed by the transfer of heat to cooling water thus completing the
cycle.
In total here energy is absorbed as heat from a high temperature body (source) and energy
is rejected as heat to a lower temperature body (sink). Network is delivered.
In the theoretical treatment of heat engines, the two temperature levels that characterize
their operation are maintained by heat reservoirs, bodies imagined capable of absorbing or
rejecting an infinite quantity of heat without any temperature change. The fluid used in the
device to absorb and reject energy in the form of heat and which undergoes a cyclic change is
called the working medium. In operation, the working fluid of a heat engine absorbs heat |Q H|
from a heat reservoir, produces a net work w, discards |Qc| to a cold reservoir and returns to its
initial state.
According to first law ΔU = 0 (being cyclic process)
Hence Q – W = 0 or Q = W.
The effective heat utilized in the process is Q H – Qc
W = QH – Qc
(b) A reversible heat engine absorbs 1000Kcal of heat at 327 oC and produces work and
discards heat at 29oC. What is the Entropy change of heat source
and heat sink?
What is the total Entropy change?
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8). State and prove Carnot’s theorem.
CARNOT THEOREM:
Statement: For two given heat reservoirs no engine can have a higher efficiency than a Carnot
engine.
Proof: Consider an engine operating a Carnot refrigerator C. Let a Carnot engine absorbs heat
|Q H| from reservoir, produces work |W| and discards
|Q H| - |Qc| to a cold reservoir. Assume a second engine E with a greater thermal efficiency
operating between the same heat reservoirs absorbing heat |Q H*|, producing the same work
and discarding |Q H*| - |W|., then if its efficiency is higher than that of Carnot’s then,
Now let engine E drive the Carnot engine backward as a heat pump that is Carnot refrigerator
as shown in the figure.
Heat extracted from the cold reservoir = [|Q H| - |W| - { |Q H*| - |W|}]
= |Q H| - |Q H*|
Heat delivered to the hot reservoir = |Q H| - |Q H*|
This suggests that heat is being transferred from lower temp source to high temperature
source, which violates II law of thermodynamics. Hence no other engine can have efficiency
greater than that of Carnot’s engine.
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9). (a) A central power plant, rated at 900MW, generates stem at 600K and discards heat
to a river at 300K.If the thermal efficiency of the plant is 80% of the maximum possible
value, how much heat is discarded to the river at rated power?
(b) Calculate the minimum work required to produce 10Kg ice cubes from water initially
at 273K. Assume that the surroundings are at 300K. The latent heat of fusion of water at
273K is 333.5KJ/Kg.
10). (a) Obtain an expression for finding the entropy changes of an ideal gas from the first
law of thermodynamics.
ENTROPY:
For Carnot’s engine |QH|/TH = |Qc |/Tc
If the heat quantities, refer to the engine taking heat into the system Q H is positive, giving out
heat Qc is negative,
then, |Q H|/TH + |Qc |/Tc = 0
Thus for a complete cycle of a Carnot engine, two quantities Q/T associated with the
absorption and rejection of heat by the working fluid of the engine sum to zero. As in Carnot
engine, engine returns to its original state properties such as temperature, pressure and internal
energy return to their initial value even though they vary from one point of the cycle to
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another. The principle characteristic of the property is that sum of its changes is zero, for any
complete cycle. Thus the above equation suggests the existence of a property whose value
changes are given by the quantity Q/T.
For a complete cycle ∑ Qrev / T = 0
This equation suggests Q/T has the qualities of state function, Clausius called this function as
Entropy change. It differential changes are given by,
ds = dQrev / T or dQrev = TdS
Entropy chanes of an Ideal gas:
(b) A lump of steel weighing 30Kg at a temperature of 427 oC is dropped in 150Kg of oil at
27oC. The specific heats of the steel and oil are 0.5 and 2.5KJ/Kg K
respectively.
Estimate the entropy change of the steel, the oil and the system consisting of oil and the
lump of steel.
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