Thermodynamics in Materials Engineering

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Thermodynamics in Materials Engineering Mat E
212 R. E. Napolitano Department of Materials
Science Engineering Iowa State
University More on entropy and the 1st and 2nd
laws
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Reminder Homework assignment 3 Gaskell
Problems 3.1, 3.2, 3.4 and 3.5 Due Friday, 1/26/07
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Reversible Processes
We consider a slow expansion followed by a slow
compression within an infinite constant-T heat
reservoir.
P1
PintPext
P1
P2
V1
P1
V2
Suppose we vary the external pressure very slowly
from P1 to P2, so that the internal pressure is
always equal to the external pressure. (i.e. an
infinitely slow process)
T
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Reversible Processes
We consider a slow expansion followed by a slow
compression within an infinite constant-T heat
reservoir.
Pext
P1
PintPext
P2
Pint
V1
V2
Suppose we vary the external pressure very slowly
from P1 to P2, so that the internal pressure is
always equal to the external pressure. (i.e. an
infinitely slow process)
T
5
Reversible Processes
We consider a slow expansion followed by a slow
compression within an infinite constant-T heat
reservoir.
Pext
P1
PintPext
P2
Pint
V1
V2
Suppose we vary the external pressure very slowly
from P1 to P2, so that the internal pressure is
always equal to the external pressure. (i.e. an
infinitely slow process)
T
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Reversible Processes
We consider a slow expansion followed by a slow
compression within an infinite constant-T heat
reservoir.
Pext
P1
PintPext
P2
Pint
V1
V2
Suppose we vary the external pressure very slowly
from P1 to P2, so that the internal pressure is
always equal to the external pressure. (i.e. an
infinitely slow process)
T
7
Reversible Processes
We consider a slow expansion followed by a slow
compression within an infinite constant-T heat
reservoir.
Pext
P1
PintPext
P2
Pint
V1
V2
Suppose we vary the external pressure very slowly
from P1 to P2, so that the internal pressure is
always equal to the external pressure. (i.e. an
infinitely slow process)
T
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Reversible Processes
We consider a slow expansion followed by a slow
compression within an infinite constant-T heat
reservoir.
P2
P1
w
P2
V1
P2
V2
The work done by the gas on the surroundings is
given by the area under the P-V curve.
T
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Reversible Processes
We consider a slow expansion followed by a slow
compression within an infinite constant-T heat
reservoir.
Because the process is isothermal, internal
energy is constant and the heat flow into the
cylinder is equal to the work done by the
internal gas.
P2
w
q
P2
T
TDS
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Irreversible Processes
We consider an expansion caused by a sudden
change in the external pressure.
We begin with a piston and cylinder in a
constant-T reservoir.
PextP1
-?P
We decrease the external pressure by a small
increment ?P.
The piston accelerates upward.
PintP1
Tres
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Irreversible Processes
We consider an expansion caused by a sudden
change in the external pressure.
We begin with a piston and cylinder in a
constant-T reservoir.
PextP1
-?P
We decrease the external pressure by a small
increment ?P.
The piston accelerates upward.
The gas cools upon the rapid expansion.
DV
Heat flows into the cylinder from the surrounding
reservoir.
TintltTres
After a very short time, PintPext, and the
piston stops.
Tres
The work done on the surroundings is given by
(P1-DP)?V.
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Irreversible Processes
We now consider an compression caused by a sudden
change in the external pressure.
We now increase the external pressure by ?P, thus
returning it to P1.
PextP1
-?P
The piston accelerates downward.
The gas heats upon the rapid compression, and
heat flows from the cylinder into the heat
reservoir.
After a very short time, PintPext and the piston
stops.
PintP1-DP
The work done on the gas is given by P1 ?V.
Tres
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Irreversible Processes
?V
T
Where does this work go?
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Degree of Irreversibility
P1
P
V1
V
Less work is done on the surroundings.
T
Some of the work done by the gas is degraded to
heat.
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Degree of Irreversibility
P1
P
V1
V
For a reversible process, no work is degraded to
heat.
T
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Degree of Irreversibility
P1
P
V1
V
T
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Degree of Irreversibility
P1
P
V1
V
T
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Degree of Irreversibility
For the reversible isothermal expansion
For an irreversible process
qd
T
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Example Problem
1/6
You start with a system of one mole of ideal gas,
where the starting conditions of the system are
P20 atm and V2 liters.The system undergoes the
following processes (a) A reversible
expansion to V10 L at constant internal energy
(b) A reversible adiabatic process, ending at
V15 L. (c) A reversible isothermal process,
ending at V3 L. (d) A reversible adiabatic
process ending at P20 atm.
1. Plot and label the processes (a,b,c,d) in P-V
space.
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Example Problem
2/6
(a) A reversible expansion to V10 L at constant
internal energy (b) A reversible adiabatic
process, ending at V15 L. (c) A reversible
isothermal process, ending at V3 L. (d) A
reversible adiabatic process ending at P20 atm.
4
2.04
10.2
2
488
488
372
372
488
Note State 5 State 1
Compute the unknown state variables
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Example Problem
3/6
(a) A reversible expansion to V10 L at constant
internal energy (b) A reversible adiabatic
process, ending at V15 L. (c) A reversible
isothermal process, ending at V3 L. (d) A
reversible adiabatic process ending at P20 atm.
Now we can plot the processes
2. Compute the total work for the cycle.
Recall The work for a cycle is given by the
area within the P-V loop.
The work done by the gas is positive for a
clockwise loop.
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Example Problem
4/6
Computing the work for each segment
Note wa and wb are positive. wc and wd are
negative.
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Example Problem
5/6
3. Sketch the cycle in Entropy Temperature
space.
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Example Problem
6/6
4. Compute the total heat flow into the system
for the cycle.
Note The heat we compute here is equal to the
work we computed previously. The area of the loop
in S-T space is equal to the area in P-V space.
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The Combined 1st and 2nd Law
We consider the combined 1st and 2nd laws
graphically by examining a differential element
of the internal energy surface, UU(S,V).
U
S
V
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The Combined 1st and 2nd Law
We consider the combined 1st and 2nd laws
graphically by examining a differential element
of the internal energy surface, UU(S,V).
U
S
V
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Thermodynamic Temperature Scale
1
For the cycle DU0
T2
2
4
3
T1
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Thermodynamic Temperature Scale
Lower cycle
Upper cycle
1
T3
Outer cycle
T2
2
4
3
T1
Because q1/q2 must be independent of T3
Therefore
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Thermodynamic Temperature Scale
1
T3
Employing the simplest possible function
T2
2
Now the efficiency is written as
4
3
T1
This defines a temperature scale, independent of
substance, where zero is a temperature which
makes a Carnot cycle 100 efficient.
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Exercise
Show that
A
T2
B
D
C
T1
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Illustrated for an ideal gas
A-B
A
B-C
C-D
T2
B
D-A
D
C
T1