Thermodynamics Chapter 4

1
The First Law ofThermodynamics
Meeting 6Section 4-1
2
So far weve studied two forms of energy transfer
Work Energy (W) Equivalent to raising a weight
Heat (Thermal) Energy (Q) Caused by a
temperature difference
3
A note about work and heat
and
Both Q and W are path dependent!
4

• Work and Heat are forms of energy transfers that
  happens at the boundary of a system.
• As Work and Heat cross the boundary, the system
  Energy changes.
• Work and Heat are not stored on the system but
  the Energy yes.

5
The Business of the First Law

• Energy is not destroyed but it is conserved.
• In fact during a thermodynamic process it is
  transformed in one type to another.
• The first law expresses a energy balance of the
  system.
• The energy fluxes in a system (Work and Heat) is
  equal to the change in the system Energy.

6
The system energy forms

• Prior to state the First Law, is necessary to
  define the system energy forms.

7
System energy consists of three components

• U is internal energy,
• KE is kinetic energy, and
• PE is potential energy.

NOTE All values are changes (deltas)
8
Internal energy..
Internal energy is the energy a molecule
possesses, mostly as a result of
Translation
Rotation
Vibration
All these are forms of kinetic energy. We will
neglect other forms of molecular energy which
exist on the atomic level.
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Translation

• Kinetic energy is possessed by a molecule as it
  moves through space. It transfers this energy to
  other systems by means of collisions in which its
  linear momentum changes. Collisions with such
  things as thermometers and thermocouples are the
  basis for temperature measurement.
• It is a characteristic of both polyatomic
  molecules and atoms.

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Vibration

• Molecules (not atoms) also vibrate along their
  intermolecular bonds.

The molecule has vibrational (kinetic) energy in
this mode.
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Rotation

• Molecules (and atoms) can also rotate and they
  possess kinetic energy in this rotational mode.
  They have angular momentum which can be changed
  to add or remove energy.

12
We will not worry about the microscopic details
of internal energy
Internal energy is a property of the system.
Often it shows up as a change in temperature or
pressure of the system but it can also show
up as a change in composition if its a mixture.
13
The kinetic energy is given by
14
The energy change in accelerating a mass of 10 kg
from Vi 0 to Vf 10 m/s is?
15
Gravity is another force acting on our system.
It shows up in the potential energy change.

• Work can be done by a change in elevation of the
  system

16
NOTE THE CONVERSION TO GET FROM m2/s2 to kJ/kg
REMEMBER IT! YOU WILL NEED IT.
17
TEAMPLAY
Lets say we have a 10 kg mass that we drop 100
m. We also have a device that will convert all
the potential energy into kinetic energy of an
object. If the objects mass is 1 kg and it is
initially at rest, what would be its final
velocity from absorbing the potential from a 100
m drop? Assume the object travels horizontally.
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Conservation of Energy
?E ?U ?PE ?KE Q ? W
Change in total energy in system during ?t
Changes during ?t in the amount of the various
forms that the energy of the system can take
Net amount of energy transferred to system as
heat.
Net amount of energy transferred out of system as
work.
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Some comments about this statement of the first
law
Q W ?E

• All terms on the left hand side are forms of
  energy that cross the boundary of the system
• Q in is positive, W out is positive
• Right hand side is a change in system energy
• Algebraic form of first law

20
The right hand side of the energy equation
consists of three terms
?E ?U ?KE ?PE

• ?KE - Motion of the system as a whole with
  respect to some fixed reference frame.
• ?PE - Position change of the system as a whole in
  the earths gravity field.
• ?U - Internal energy of the molecule--translation,
  rotation, vibration, and energy stored in
  electronic orbital states, nuclear spin, and
  others.

21
We previously had conservation of energy

• ?E ?U ?PE ?KE
• We can change the total energy E of a system by
• Changing the internal energy, perhaps best
  exemplified by heating.
• Changing the PE by raising or lowering.
• Changing the KE by accelerating or decelerating.

22
Conservation of Energyfor Stationary System

• Stationary means not moving -so ?PE and ?KE are
  zero and the first law becomes

23
First Law Forms for Stationary Systems

• Differential Form
• Rate Form
• Integrated Form

24
We can also write the first law in differential
terms
dE ?Q ?W, and
dU dPE dKE ?Q ?W
Differential amount of energy transferred out ()
or in (-) by work interaction.
Differential amount of energy trans-ferred in ()
or out (-) by heat transfer.
Change in the amount of energy of the system
during some time interval.
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If we are analyzing a transient process, well
need the rate form of the first law

• Where

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Rate form will allow us to calculate

• Changes in temperature with time
• Changes in pressure with time
• Changes in speed with time
• Changes in altitude with time

27
Hints to set up a problem

1. Define the system carefully indicating clearly
   its boundaries.
2. Enroll all the simplifying hypothesis to the
   case.
3. Draw the heat and work fluxes at the boundaries
   including their signals.
4. Sketch a process representation on a
   thermodynamic diagram Pv or Tv.

28
System Energy Change
?E Q ? W (Qin ? Qout) ? (Wout ?
Win)
?E (15 - 3) 6 18 kJ
29
Example 4-1
0.01 kg of air is compressed in a
piston-cylinder. Find the rate of temperature
rise at an instant of time when T 400K. Work is
being done at a rate of 8.165 KW and Heat is
being removed at a rate of 1.0 KW. Solution on
the black board

30
Example 4-2Isothermal Process
An ideal gas is compressed reversibly and
isothermally from a volume of 0.01 m3 and a
pressure of 0.1 MPa to a pressure of 1 MPa. How
much heat is transferred during this
process? Solution on the black board

31
Example 4-3Isobaric Process
The volume below a weighted piston contains 0.01
kg of water. The piston area is of 0.01 m2 and
the piston mass is of 102 kg. The top face of the
piston is at atmospheric pressure, 0.1 MPa.
Initially the water is at 25 oC and the final
state is saturated vapor (x1). How much heat and
work are done on or by the water? Solution on the
black board

32
Isobaric Process
For a constant-pressure process, Wb ?U P?V
?U ?(UPV) ?H Thus,
Q - Wother ? H ? KE ? PE (kJ)
Example Boil water at constant pressure
33
Example
An insulated tank is divided into two parts by a
partition. One part of the tank contains 2.5 kg
of compressed liquid water at 60oC and 600 kPa
while the other part is evacuated. The partition
is now removed, and the water expands to fill the
entire tank. Determine the final temperature of
the water and the volume of the tank for a final
pressure of 10 kPa.

34
Example
m 2.5 kg T1 60oC P1 600 kPa P2 10 kPa

?E Q - W
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Solution - page 1
First Law Q - W ?E Q W ?KE ?PE 0 ?E
?U m(u2 - u1) 0 u1 u2
No Work and no Heat therefore the internal
Energy is kept constant
36
Solution - page 2
State 1 compressed liquid P1 600 kPa, T1
60oC vf vf_at_60oC 0.001017 m3/kg uf uf_at_60oC
251.11 kJ/kg State 2 saturated liquid-vapor
mixture P2 10 kPa, u2 u1 251.11
kJ/kg uf 191.82 kJ/Kg, ufg 2246.1 kJ/kg
37
Solution - page 3
Thus, T2 Tsat_at_10 kPa 45.81oC v2 vf x2vg
0.001010.0264(14.67 - 0.00101) m3/kg
0.388 m3/kg V2 mv2 (2.5 kg)(0.388m3/kg)
0.97 m3
38
Example
One kilogram of water is contained in a
piston-cylinder device at 100oC. The piston
rests on lower stops such that the volume
occupied by the water is 0.835 m3. The cylinder
is fitted with an upper set of stops. The volume
enclosed by the piston-cylinder device is 0.841
m3 when the piston rests against the upper stops.
A pressure of 200 kPa is required to support the
piston. Heat is added to the water until the
water exists as a saturated vapor. How much work
does the water do on the piston?
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Example Work
m 1 kg T1 100oC V1 0.835 m3 V2 0.841 m3
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T-v Diagram
T
211.3 kPa
200 kPa
101.3 kPa
v
v1
v2
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Solution - page 1
State 1 saturated liquid-vapor mixture
T1 100oC, vf0.001044 m3/kg ,
vg1.6729 m3/kg vf lt v lt vg gt saturation
P1101.35 kPa
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Solution - page 2
Process 1-2 The volume stay constant until the
pressure increases to 200 kPa. Then the piston
will move. Process 2-3 Piston lifts off the
bottom stop while the pressure stays constant.
State 2 saturated liquid-vapor mixture
P2 200 kPa , v2 v1 0.835 m3
Does the piston hit upper stops before or after
reaching the saturated vapor state?
43
Solution - page 3
State 3 Saturated liquid-vapor mixture
P3 P2 200 kPa vf 0.001061 m3/kg
, vg 0.8857 m3/kg vf lt v3 lt vg gt piston hit
the upper stops before water reaches the
saturated vapor state.
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Solution - page 4
Process 3-4 With the piston against the upper
stops, the volume remains constant during the
final heating to the saturated vapor state and
the pressure increases.
State 4 Saturated vapor state v4
v3 0.841 m3/kg vg P4 211.3
kPa , T4 122oC
45
Solution - page 5
(gt 0, done by the system)
46
Example Heat Transfer
Find the require heat transfer for the water in
previous example.
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Solution - page 1
First Law Conservation of Energy Q - W ?E
?U ?KE ?PE Q14 Wb,14 ?U14 ?U14 m(u4 -
u1)
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Solution - page 2
State 1 saturated liquid-vapor mixture
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Solution - page 3
State 4 saturated vapor state
v4 0.841 m3/kg vg u4
2531.48 kJ/kg (interpolation)
(gt 0, added to the water)
50
TEAMPLAY EX. 4-6
A pressure cooker with volume of 2 liters
operates at 0.2 MPa with water at x 0.5. After
operation the pressure cooker is left aside
allowing its contents to cool. The heat loss is
50 watts, how long does it take for the pressure
drop to 0.1 MPa? What is the state of the water
at this point? Indicate the process on a T-v
diagram.
51
Ex4.6)
52
P10,2MPa Tsat120ºC v1L0,001m3/kg v1G0,8919m3/k
g u1L503,5KJ/kg u1G2025,8KJ/kg x0,5
P20,1MPa Tsat100ºC v2L0,001m3/kg v2G1,6729m3/k
g u2L418,9KJ/kg u2G2087,5KJ/kg
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Massa de Água vV/M?MV/v M210-3/0,446 M0,004kg
Aplicando na 1ºLei
54
TEAMPLAY EX. 4-7
A powerful 847 W blender is used to raise 1.36kg
of water from a temperature of 20oC to 70oC. If
the water loses heat to the surroundings at the
rate of 0.176 W, how much time will the process
take?
55
Ex4.7)
1HP745W 1lbm0,453kg ºC(ºF-32)/1,8 1Btu1,055J
Wmec1,2HP894W 68ºF20ºC (água no estado
líquido) 158ºF70ºC (água no estado
líquido) Q10Btu/min0,176W 3lbm1,359kg cvcp4,1
80KJ/kgºC
56
TEAMPLAY EX. 4-10
Air, assumed to be ideal gas with constant
specific heats, is compressed in a closed
piston-cylinder device in a reversible polytropic
process with n 1.27. The air temperature before
compression is 30oC and after compression is
130oC. Compute the heat transferred on the
compression process.
57
Ex4.10)
Para comprimir do estado 1 ao 2 é necessário
transferir 38,3 KJ por kg de ar comprimido.