A WARM WELCOME

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A WARM WELCOME TO ONE AND ALL FOR THE VTU -
EDUSAT PROGRAM -7
Dr. T.N. Shridhar, Professor, NIE, Mysore
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BASIC THERMODYNAMICS SUBJECT CODE
06ME33 SESSION 9 07.09.2007
Presented by Dr. T.N. Shridhar Professor, Dept.
of Mechanical Engg The National Institute of
Engineering, Mysore
Dr. T.N. Shridhar, Professor, NIE, Mysore
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OUTCOME OF SESSION - 1

• First law of thermodynamics for a closed system
  undergoing a cyclic process
• First law of thermodynamics for a closed system
  undergoing a non cyclic process
• Energy - a property of a system
• Illustrative Examples

Dr. T.N. Shridhar, Professor, NIE, Mysore
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UNIT 3 FIRST LAW OF THERMODYNAMICS
Dr. T.N. Shridhar, Professor, NIE, Mysore
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The first law of thermodynamics is often called
as the law of the conservation of energy, with
particular reference to heat energy and
mechanical energy i.e., work.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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First law of thermodynamics for a closed system
undergoing a cyclic process
The transfer of heat and the performance of work
may both cause the same effect in a system.
Energy which enters a system as heat may leave
the system as work, or energy which enters the
system as work may leave as heat.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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The first law of thermodynamics can therefore be
stated as follows When a system undergoes a
thermodynamic cyclic process, then the net heat
supplied to the system from the surroundings is
equal to the net work done by the system on its
surrounding.
?W
i.e.,
?Q
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Joules Experiment
Figure Joules Experiment
Figure Cycle completed by a system with two
energy interactions
Dr. T.N. Shridhar, Professor, NIE, Mysore
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The process 1-2 undergone by the system is shown
in figure i.e., W1-2. Let the insulation be
removed. The amount of heat transfer Q2-1 from
the system during this process 2-1 is shown in
figure. The system thus executes a cycle, which
consists of a definite amount of work input W1-2
to the system followed by the transfer of an
amount of heat Q2-1 from the system.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Based on this experimental evidence Joule stated
that, When a system (closed system) is
undergoing a cyclic process, the net heat
transfer to the system is directly proportional
to the net work done by the system. This
statement is referred to as the first law for a
closed system undergoing a cyclic process.
?W
?W
i.e.,
?Q ?
or
?Q
Dr. T.N. Shridhar, Professor, NIE, Mysore
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If the cycle involves many more heat and work
quantities as shown in figure, the same result
will be found.
Figure Cyclic Process on a Property Diagram
Dr. T.N. Shridhar, Professor, NIE, Mysore
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or 1Q2 2Q3 3Q4 4Q1 1W2 2W3 3W4 4W1
or (?Q)cycle (?W)cycle
Dr. T.N. Shridhar, Professor, NIE, Mysore
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First law for a closed system undergoing a
non-cyclic process (i.e., for a change of state)
If a system undergoes a change of state during
which both heat transfer and work transfer are
involved, the net energy transfer will be stored
or accumulated within the system.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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?Q-W ?E or Q ?E W If there are more
energy transfer quantities involved in the
process Therefore the first law is a particular
formulation of the principle of the conservation
of energy.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Energy A property of the system
For the process, 1-A-2-B-1,
Figure First law to a non cyclic process
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Considering the two separate processes, we have
--- (1)
Now consider another cycle, the system changing
from state 1 to state 2 by process A, as before
and returning to state 1 by process C. For this
cycle we can write
Dr. T.N. Shridhar, Professor, NIE, Mysore
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--- (2)
Subtracting (2) from (1), we get
Or, by rearranging,
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Since B and C represent arbitrary processes
between state 1 to state 2, we conclude that the
quantity (?Q - ?W) is the same for all processes
between state 1 and state 2. ? (?Q - ?W) depends
only on the initial and final states and not on
the path followed between the two states. ?This
is a point function and differential is a
property of the system. This property is called
the energy of the system, E.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Therefore, we can write ?Q - ?W dE Or ?Q dE
?W If it is integrated between initial and
final states, 1 and 2, we get Q1-2 E2 E1
W1-2 i.e., Q1-2 - W1-2 E2 E1
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Classification of Energy of the System
E Kinetic Energy (KE) Potential Energy (PE)
remaining forms of energy.
The remaining energies which can not be measured
directly and is the summation of all microscopic
energies is called internal energy of the system.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Internal Energy
It is the energy associated with internal
structure of matter. This energy can not be
determined in its absolute values. But it is
possible to determine the change in internal
energy of the system undergoing a process. By
first law of thermodynamics, ? Total energy E
KE PE IE
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Since the terms comprising E are point functions,
we can write dE d(KE) d (PE) dU The
first law of thermodynamics for a change of state
of a system may therefore be written as ?Q dU
d (KE) d (PE) ?W
?Q dU
d (mgZ) ?W
Dr. T.N. Shridhar, Professor, NIE, Mysore
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In the integral form this equation is, assuming
g as a constant
mg (Z2 Z1) W1-2
Q1-2 U2 U1
??Q dU ?W or Q1-2 U2 U1 W1-2
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Law of conservation of energy (2nd corollary of
first law of thermodynamics)
From first law of thermodynamics Q1-2 E2 E1
W1-2 This equation in effect, a statement of
the conservation of energy.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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For an isolated system, Q 0, W 0 ?E2 E1
0 ?For an isolated system, the energy of the
system remains constant.
Therefore, the first law of thermodynamics. may
also be stated as follows, Heat and work are
mutually convertible but since energy can neither
be created nor destroyed, the total energy
associated with an energy conversion remains
constant.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Perpetual Machine of first kind (3rd Corollary)
Any system which violates the first law of
thermodynamics is called the Perpetual Motion
machine of first kind. i.e., It is impossible to
construct a perpetual motion machine of first
kind. A perpetual machine is one which can do
continuous work without receiving energy from
other systems or surroundings.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Numerical / Illustrative Examples 1) In a cyclic
process, heat transfers are 14.7 kJ, -25.2 kJ,
-3.56 kJ and 31.5 kJ. What is the net work for
this cyclic process? Solution 1st law of T.D.
for a cyclic process is i.e.,Net work 14.7
25.2 -3.56 31.5 17.44 kJ
Dr. T.N. Shridhar, Professor, NIE, Mysore
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2) A domestic refrigerator is loaded with food
and the door closed. During a certain period the
machine consumes 1 kWhr of energy and the
internal energy of the system drops by 5000 kJ.
Find the net heat transfer for the
system. Solution W1-2 1kWhr 1(kJ/s)hr -1
x3600 kJ U2 U1 -5000 kJ From 1st law,
Q1-2 (U2-U1) W1-2 -5000 -3600
-8600 kJ -8.6 MJ
Dr. T.N. Shridhar, Professor, NIE, Mysore
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3. On a warm summer day, a housewife decides to
beat the heat by closing the windows and doors in
the kitchen and opening the refrigerator door. At
first she feels cool and refreshed, but after a
while the effect begins to wear off. Evaluate the
situation as it relates to first law, considering
the room including the refrigerator as the
system.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Solution
At first the temperature of air in the room falls
since it communicates with the cool refrigerator.
This makes the housewife feel cool.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Considering the room and its contents as a
system, and assuming walls, windows and doors
non-conducting, Q 0. For the operation of
refrigerator, electricity is supplied from
outside and hence electrical work We is done on
the system.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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?From first law of TD Q ?E We 0 ?E -
We ??E We Positive sign of energy indicates
the increase in energy of the system with time.
As the energy is increasing, the temperature of
air increases and hence effect of coolness
gradually begins to wear off.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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4) A household refrigerator is loaded with fresh
food and closed. Consider the whole refrigerator
and the contents as a system. The machine uses 1
kWhr of electrical energy in cooling the food and
the internal energy of the food (system)
decreases by 5250 kJ, as the temperature drops.
Find the magnitude and direction of heat transfer
during the process.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Solution
Given Electrical work We 1 kWhr
1x(kW/s)hr 1 x 60 x 60 kJ ?We - 3600
kJ Given, ?U - 5250 kJ
From first law of TD Q ?U We - 5250
3600.8 - 8850 kJ Negative sign indicates
heat flows from the refrigerator to the
surroundings
Dr. T.N. Shridhar, Professor, NIE, Mysore
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5) The average heat transfer from a person to the
surroundings when he is not actively working is
about 950 kJ/hr. Suppose that in the auditorium
containing 1000 people the ventilation system
fails. a) How much does the internal energy of
air in the auditorium increase during the first
15 minutes after the ventilation fails? b)
Considering the auditorium and all the people as
system and assuming no heat transfer. to
surroundings, how much does the int. energy of
the system change? How do you account for the
fact that the temperature of air increases?
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Solution a) Average heat transfer per
person, 950 kJ/hr 950 / 60 15.83 kJ
/min ?Average heat transfer / person for 15 min
237.5 kJ ?Average heat transfer for 15 min
in the auditorium containing 1000 people Q
237.5 x 1000 237500 kJ/min
Dr. T.N. Shridhar, Professor, NIE, Mysore
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From first law of TD, we have Q ?E
W 237500 ?E 0 ??E 237.5 MJ
b) Considering the auditorium and all the people
as system, Q 0 W 0 ?Q ?E W 0
?E 0 ??E 0 Increase in internal energy of
the air due to increase in its temperature is
compensated by the decrease in internal energy of
the people.
Dr. T.N. Shridhar, Professor, NIE, Mysore
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6) For the following process in a closed system
find the missing data (all in kJ)
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Process (a) Q ?U W U2 U1
W1-2 But U2-U1 15 ?U2 5 ? Q 15 20
35 kJ Process (b) Q U2 U1 W 15 -6 -
U1 - 6 i.e., 27 - U1 ?U1
-27 kJ ??U U2 U1 - 6 27 21 kJ
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Process (c) - 7 U2 20 10 ?U2 3 kJ
??U 3 - 20 -17 kJ Process
(d) ?U U2 U1 - 20 8 U1 ?U1
28 kJ We have, Q 8 28 7
- 27 kJ
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Please mail to tns_nie_at_yahoo.com
Dr. T.N. Shridhar, Professor, NIE, Mysore
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Thank you
Dr. T.N. Shridhar, Professor, NIE, Mysore