Work in Thermodynamic Processes

flinkexistenceMechanics

Oct 27, 2013 (3 years and 9 months ago)

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Work in Thermodynamic Processes


Energy can be transferred to a
system by heat and/or work


The system will be a volume
of gas always in equilibrium


Consider a cylinder with a
movable piston


As piston is pressed a
distance
Δ
y, work is done on
the system reducing the
volume



W =
-
F
Δ
y =
-

P A
Δ
y

Work in Thermodynamic Processes


Cont.


Work done compressing
a system is defined to be
positive


Since
Δ
V is negative
(smaller final volume)
& A
Δ
y = V



W =
-

P
Δ
V


Gas compressed


W
on
gas
= pos.


Gas expands


W
on gas
=
neg.

Work in Thermodynamic Processes


Cont.


Can only be used if gas is
under constant pressure


An isobaric process (
iso

=
the same) P
1

= P
2


Represented on a pressure
vs. volume graph


a PV
diagram


Area under
any curve
=
work done on the gas


If volume decreases


work
is positive (work is done
on the system)

THERMODYNAMICS

First Law of Thermodynamics


Energy is conserved


Heat added to a system goes
into internal energy, work or
both


ΔU = Q + W


Heat
added to system


internal energy


Q is
positive


Work
done to the system


internal energy


W is
positive (again)


First Law


Cont.


A system will have a
certain amount of
internal energy (U)


It will
not have
certain
amounts of heat or work


These change the system


U depends only on state
of system, not what
brought it there


ΔU is independent of
process path (like
U
g
)



Isothermal Process


Temperature remains
constant


Since P = N
k
B

T / V

=
constant / V


An isotherm (line on
graph) is a hyperbola

Isothermal Process


Cont.


Moving from 1 to 2,
temperature is constant,
so P & V change


Work is done = area
under curve


Internal energy is
constant because
temperature is constant




Q =
-
W


Heat is converted into
mechanical work

Isometric (
isovolumetric
) Process


The volume is not
allowed to change


V
1

= V
2


Since no change in
volume, no work is done




Δ
U = Q


Heat added must go into
internal energy


it


Heat extracted is at the
expense

of internal
energy


it

Isometric Process


Cont.


The PV diagram
representation


No change in volume


Area under curve = 0


That is, no work done


The process moves from
one isotherm to another

Isobaric Process


As heat is added to system,
pressure is required to be
constant


The ratio of V / T = constant


Some of the heat does work
and the rest causes a change
in temperature


Thus moving to another
isotherm


Recall: changes in
temperature = changes in
internal energy



Δ
U = Q + W

Adiabatic Process


No heat is transferred
into or out of system


Q = 0




Δ
U = W


All work done to a system
goes into internal energy
increasing temperature


All work done by the
system comes from
internal energy & system
gets cooler

Adiabatic Process


Cont.


Either system is
insulated to not
allow heat
exchange or the
process happens
so fast there is
no time to
exchange heat

Adiabatic Process


Cont.


Since temperature
changes, we move
isotherms


The Second Law of
Thermodynamics & Heat Engines


Heat will not flow
spontaneously from a
colder body to a warmer


OR: Heat energy cannot
be transferred
completely into
mechanical work


OR: It is impossible to
construct an operational
perpetual motion
machine


Heat Engines


Any device that
converts heat energy
into work


Takes heat from a
high temperature
source (reservoir),
converts some into
work, then transfers
the rest to
surroundings (cold
reservoir) as waste
heat


Heat Engines


Cont.


Consider a cylinder and piston


Surround by water bath & allow
to expand along an isothermal


The heat flowing in (Q) along
AC equals the work done by the
gas as it expands (W) since
Δ
U =
0


To return to A along same
isothermal, work is done on the
gas and heat flows out


Work expanding = work
compressing

Heat Engines


Cont.


A cycle naturally can have
positive work done


In going from A to B work is
done by gas, temperature


(
Δ
U

) and heat enters
system


B to C


No work done, T

,
Δ
U

, &
heat leaves system


C to A


Δ
U = 0, heat leaving = work
done


The work out = the net heat
in (
Δ
U = 0)

Heat Engines


Cont.


Thermal Efficiency


Used to rate heat engines


efficiency = work out / heat in



e =
W
out

/ Q
in


Qin = heat into heat engine


Qout

= heat leaving heat engine


For one cycle, energy is
conserved




Q
in

= W +
Q
out


Since system returns to its
original state
Δ
U = 0


The Carnot Engine


Any cyclic heat engine
will always lose some
heat energy


What is the maximum
efficiency?


Solved by
Sadi

Carnot
(France) (Died at 36)


Must be reversible
adiabatic process

The Carnot Engine


Cont.


Carnot Cycle


A four stage reversible
process


2 isotherms & 2
adiabats


Consider a hypothetical
device


a cylinder & piston


Can alternately be brought
into contact with high or
low temperature
reservior


High temp


heat source


Low temp


heat sink


heat is exhausted


The Carnot Engine


Cont.


Step 1: an
isothermal
expansion, from A
to B


Cylinder receives
heat from source


Step 2: an adiabatic
expansion, from B
to C

The Carnot Engine


Cont.


Step 3: an isothermal
compression, C to D


Ejection of heat to sink at
low temp


Step 4: an adiabatic
compression, D to A


Represents the most
efficient (ideal) device


Sets the upper limit

Entropy


A measure of disorder


A messy room > neat
room


Pile of bricks > building
made from them


A puddle of water > ice
came from


All real processes
increase disorder


increase entropy




of entropy of one
system can be reduced at
the expense of another

Entropy


Cont.


Entropy of the universe
always increases


The universe only moves
in one direction


towards


entropy


This creates a “direction
of time flow”


Nature does not move
systems towards more
order


Entropy


Cont.


As entropy

, energy is
less able to do work


The “quality” of energy
has been reduced


Energy has “degraded”


Nature proceeds towards
what is most likely to
happen