Transcript Laws of Thermodynamics - Ohio Wesleyan University

Work in Thermodynamic Processes
• We are now ready to combine the energy transfer
mechanism of heat with another mechanism: work
– We will focus our discussion on work done on a gas
– Important in discussion of how system gets from one state
to another (described by state variables P, T, V, n, U)
– Will provide connection between microscopic and
macroscopic energy transfer mechanisms
• In Chap. 5 we spoke about work done by forces on
objects undergoing a displacement
• Now we will talk about the work done on a gas by
the environment W (or by gas on environment Wenv)
–
–
–
–
Work done on gas (W) can be positive, negative, or zero
W = 0 when no “mechanical action” taking place
W = –Wenv
Work influences internal energy
Work in Thermodynamic Processes
• Work done by a gas on a piston in a
cylinder: Wenv = FDy = PADy = PDV
= P(Vf – Vi)
– When the piston moves inward, Vf < Vi
and Wenv < 0
– When the piston moves outward, Vf > Vi
and Wenv > 0
– When piston doesn’t move, Wenv = 0
(from University Physics, 11th Ed.)
• Work done by the piston on the gas
= W = –Wenv
W   P DV   PV  V 
f
– When the piston
moves inward, Vf < Vi and W > 0
– When the piston moves outward, Vf > Vi
and W < 0
– When piston doesn’t move, W = 0
i
Work in Thermodynamic Processes
• This expression can only be used if the pressure
remains constant during the expansion or
compression
– This is called an isobaric (constant pressure) process (same
Greek root as “barometer”)
• If the pressure changes, the average pressure may be
used to estimate the work done
• If pressure and volume are known at each step of the
process, a PV diagram helps visualize process with
curves (paths) connecting initial and final states
• The area under the curve on a PV diagram is equal in
magnitude to the work done on a gas (W)
– True whether or not P remains constant
– W is positive (negative) when volume decreases (increases)
PV Diagrams
• For a gas being compressed at constant pressure:
(shaded area represents work done on
the gas; here W > 0)
• For a gas being compressed with varying pressure:
(shaded area, the area under the curve,
represents work done on the gas)
• The work done depends on the particular path
(same initial and final states, but
work done is different in each case)
Other Thermodynamic Processes
• Isovolumetric (or isochoric)
–
–
–
–
Volume stays constant
For example, maintain constant piston position
Vertical line on the PV diagram
No work done on gas
PV Processes Interactive
• Isothermal
– Temperature stays constant
– Heat flows between the system and a reservoir to keep
the system’s temperature constant
– For example, keep oven temperature constant
• Adiabatic
– No energy exchanged with the surroundings via heat
– For example, provide excellent insulation for system
– Or, process occurs so quickly that there is no time for
heat to flow in or out of system
Example Problem #12.5
A gas expands from I to F along the
three paths indicated in the
figure. Calculate the work done
on the gas along paths
(a) IAF
(b) IF, and
(c) IBF.
Solution (details given in class):
(a) – 810 J
(b) – 507 J
(c) – 203 J
First Law of Thermodynamics
• Both work and heat can change the internal energy
of a system
– Work can be done on a rubber ball by squeezing it,
stretching it, or throwing it onto a wall
– Energy can flow to the ball via heat by leaving it out in the
sun or putting it into a hot oven
• These 2 methods of increasing the internal energy of
a system lead to the first law of thermodynamics:
– The change in internal energy of a system is equal to the
heat flow into the system plus the work done on the system
DU  U f  U i  Q  W (really a statement of energy conservation!)
– DU = internal energy change of system
– Q = energy transferred to system by heat from the outside
– W = work done on the system
Sign Conventions for First Law
Quantity
Definition
Meaning of + sign
Meaning of – sign
Q
Energy transferred
by heat flow
Heat flow into the
system
Heat flow out of
the system
W
Work
Work done on the
system
Work done by the
system
DU
Internal energy
change
Internal energy
increase
Internal energy
decrease
In–class example – Fill in the boxes with +, –, or 0:
Situation
System
(a) Rapidly pumping up bicycle tire
Air in the pump
(b) Pan of room-temperature water resting
on a hot stove
Water in the pan
(c) Air leaking quickly out of a balloon
Air rushing out of
balloon
Q
W DU
Applications of the First Law
• Isolated system: does not interact with its
surroundings
– No energy transfer takes place and no work is done
– Internal energy of the isolated system remains constant
– Q = W = DU = 0
• Cyclic process: originates and ends at the same state
– DU = 0 and Q = –W
– The net work done per cycle by the gas is equal to the area
enclosed by the path representing the process on a PV
diagram (important for describing heat engines)
 PV diagram for an ideal monatomic
gas confined in a cylinder by a movable
piston undergoing a cyclic process
Applications of the First Law
• Isothermal process: constanttemperature process
• Consider ideal monatomic gas
contained in cylinder with movable
piston
• The cylinder and gas are in thermal
contact with a large source of energy (reservoir)
• Allow energy to transfer into the gas by heat
• The gas expands (volume increases) and pressure
falls while maintaining a constant temperature
3
• Since DU  nRDT , DU = 0
2
– Therefore DU = 0 = Q + W
– So W = – Q < 0 (system supplies work to outside world)
Applications of the First Law
• Adiabatic process: no energy transferred by heat
– The work done is equal to the change in the internal energy
of the system (W = DU)
– One way to accomplish a process with no heat exchange is
to have it happen very quickly (mostly true in an internal
combustion engine)
– In an adiabatic expansion, the work done is negative and the
internal energy decreases
3
– For an ideal monatomic gas (see Chap. 10): DU  nRDT
2
• Isovolumetric process: no change in volume, so W = 0
– The energy added to the system goes into increasing the
internal energy of the system (DU = Q)
– Temperature will increase
• Isobaric process: no change in pressure
– W = – PDV, so DU = Q – PDV
The First Law and Human Metabolism
• The first law can be applied to living organisms
• The internal energy stored in humans goes into
other forms needed by the organs and into work and
heat
– When we eat, we replenish our supply of internal energy
• The metabolic rate (DU / Dt) is directly proportional
to the rate of oxygen consumption by volume
– Basal metabolic rate (to maintain and run organs, etc.) is
about 80 W (other rates shown in Table 12.4)
• The efficiency of the body is the ratio of the
mechanical power output to the metabolic rate
– What you get out divided by what you put in
– Efficiencies range from about 20% (cycling) to 3%
(shoveling)
Example Problem #12.16
A gas is taken through the cyclic
process described by the figure.
(a) Find the net energy transferred
to the system by heat during
one complete cycle.
(b) If the cycle is reversed  that
is, the process follows the path
ACBA  what is the net energy
transferred by heat per cycle?
Solution (details given in class):
(a) 12 kJ
(b) –12 kJ
Interactive Example Problem: Triangular
Cyclic Process on a PV Diagram
Simulation and solution details given in class.
(ActivPhysics Online Exercise #8.13, copyright Addison Wesley publishing)
Heat Engines
• A heat engine is a device that converts
internal energy to other useful forms,
such as electrical or mechanical energy
– Electrical power plants
– Internal combustion engine in cars
• A heat engine carries some “working
Animation
substance” through a cyclical process
• Energy is transferred from a source at a high
temperature (Qh)
• Work is done by the engine (Weng)
• Energy is expelled to a source at a lower
temperature (Qc)
• Example of piston steam engine (in-class movie):
Heat Engine
• Since it is a cyclical process, its initial and final
internal energies are the same
– So, DU = 0 = Q + W and Q = Qnet = – W = Weng
• The work done by the engine equals the net energy
absorbed by the engine
– Therefore, Weng  Qh  Qc
• The work is equal to the area enclosed by the curve
of the PV diagram (if working substance is a gas)
• Thermal efficiency is defined as the ratio of the work
done by the engine to the energy absorbed at the
higher temperature (what you get divided by what
you put in):
Weng Qh  Qc
Qc
e
Qh

Qh
 1
Qh
Second Law of Thermodynamics
• The efficiency of heat engines is addressed in the
second law of thermodynamics:
– No engine can have 100% efficiency
• Other statements and implications of the second
law:
– A system cannot convert heat solely into mechanical work
– Heat never flows spontaneously from a colder body to a
hotter body
– It is impossible for any process to have as its sole result
the transfer of heat from a cooler to a hotter body
– There is no such thing as a free lunch!
• Summary of 1st and 2nd laws:
– 1st law: We cannot get more energy out than we put in
– 2nd law: We cannot break even
Internal Combustion Engines
(from College Physics, Giambattista)
(from University Physics, 11th Ed.)
• An idealized model of the thermodynamic processes of a gas
engine is called the Otto cycle
– Intake stroke: Gas-air mixture drawn into cylinder (a  b)
– Compression stroke: Intake valve closes, mixture compressed (adiabatic
compression; b  c)
– Ignition: Spark plug ignites mixture (heating at constant volume; c  d)
– Power stroke: Hot burned mixture pushes mixture down, doing work
(adiabatic expansion; d  e)
– Cooling: Energy released via heat at constant volume (e  b)
– Exhaust stroke: Exhaust valve opens and burned mixture is pushed out
of cylinder (b  a)
Heat Pumps: Refrigerators, Air conditioners
• Refrigerators
(from University Physics, 11th Ed.)
– Energy is extracted from the cold reservoir (food cabin)
and transferred to hot reservoir (kitchen)
• Air conditioner works on the same principle
(from University Physics, 11th Ed.)
• Both are examples of a heat pump (reverse heat
engines)
Heat Pumps Interactive
Carnot Engine
• The most efficient heat engine possible is the Carnot
engine
– A theoretical heat engine operating in an ideal, reversible
cycle (a cycle in which system can be returned to its initial
state along the same path) called the Carnot cycle
– Note that a truly reversible process is an idealization, and
real processes are irreversible (although some are close
to being reversible)
• PV diagram of the cycle:
– Consists of 2 adiabatic and 2
isothermal processes, all reversible
– Net work done W is net energy
received in one cycle: W = Qh – Qc
Carnot Cycle
• Assume the substance that changes
temperature is an ideal gas
• Gas is contained in cylinder
with movable piston at one end
• Steps of the cycle:
– A  B: Gas expands isothermally
while in contact with reservoir at Th
– B  C: Gas expands adiabatically (Q = 0)
– C  D: Gas compressed
isothermally while in contact with reservoir
at Tc < Th
– D  A: Gas compressed adiabatically
– Note that upward (downward) red arrows on
piston indicate removal (addition) of sand
Carnot Engine
• Carnot showed that the efficiency of the engine
depends on the temperatures of the reservoirs:
–
–
–
–
–
–
–
Tc
ec  1 
Th
Temperatures must be in Kelvin
All Carnot engines operating between the same two
temperatures will have the same efficiency
The efficiency increases as Tc is lowered and as Th is raised
Efficiency is 0 if Th = Tc
Efficiency is 100% only if Tc = 0 K (not possible from third
law of thermodynamics)
In most practical cases Tc is near room temperature (300 K)
so generally Th is raised to increase efficiency
All real engines are less efficient than the Carnot engine
(friction, cycle completed in brief time period)
Example Problem #12.27
One of the most efficient engines ever built is a coalfired steam turbine engine in the Ohio River valley,
driving an electric generator as it operates
between 1870°C and 430°C.
(a)What is its maximum theoretical efficiency?
(b)Its actual efficiency is 42.0%. How much
mechanical power does the engine deliver if it
absorbs 1.40  105 J of energy each second from
the hot reservoir?
Solution (details given in class):
(a) 0.672 (or 67.2%)
(b) 58.8 kW