Energy and the First Law of Thermodynamics Ch 2
Energy A property that must be transferred to an object
in order to
can be
Transfered
Stored
3 Always conserved
4 Units
are the same as work
English
Mechanical Concepts of Energy and Work
Fs
Ylogitsfitsbody
V2
path
recall from physics
of force
along the path of
component
the body
Fs
Newton's 2ⁿᵈ Law
ma
rearrange
integrate
says work performed by Fs as the bodymoves along the path
from s to s equals
Extending to IncludePotential Energy
resulting
II ythiaispy.cat
it in
forces
other
mg
rods
mg ds
Em v22 v
says Theworkdoneby force R onthe body is equal to
of
Work
Broadening Our Definition
Mechanical definition Js F as
of work
Work is done by a system on its
surroundings if thesoleeffect on everything external to the system
Thermodynamic definition
If
we definethe system as
1
Iii itdoti
2 The battery supplying
electrical current
Does it do work
Tio
Key Concepts
Work is not transferred it is
Work is dependent on the pathtaken in transferring this energy
for example
We could say
The change in volumefrom state 1 to state 2
Fat
But we couldnot say
The change in work fromstate 1 to state 2
Instead we couldsay
The workdone by or on thesystem in goingfrom state 1
to state 2
SW
Wiz
Sign ConventionofWork
Wnet
0
Workdoneby the system on thesurroundings
Wnit
0
Workdone on the systembythesurroundings
hence
The following are all equivalent statements
The network of the system is
20 KJ
The net energy consumed bythe system is 20 kJ
The net work into the system is 20 K J
Work andPower
Power
says
ffroÉnEIof
theforceandthevelocity at the point of
application of the force
Power
Units of power
SI
English
Othertypes of power
Rotating shaft
Electric power
Expansion
Compression Work
We have noted that mechanical work is W
consider how this applies to a spring moving
a mass
through some distance
Now consider an expanding gas moving a piston
through some distance
F ds
Path Dependence of Work
Both processes A B begin
at state 1 and end at
state 2 but the work
required for process B is
greater than that for A
Quasiequilibrium
An idealizedprocesses in which
departure from thermodynamic equilibrium is infinitesimal
The system is modeled as
A conceptual ideality similar to a
massless spring or frictionless surface
in mechanics
Special types
of processes usuallymodeled asquasiequilibrium
Adiabatic
No heat transfer
Isobaric Constant pressure
b
Isothermal
Constant temp
Isochoric
Constant volume
Polytropic
Process where PV^n is constant
*Will be told of this process is true
p
n can take on any value
if n = -1, p vs V is linear (is p=CV)
If n=0, p = C --> we would simply call this an
isobaric process
Everything that is not n = -1 is curved and concave up.
E Gas in a piston cylinder undergoes a process for which the relationship
between
pressure and volume is given by pH const
0.1 m
At state 1 p I bar
At state 2 p 9 bar
Find
a
b
Assumptions
Final volume in m
workforthe process in K
Quasiequilibruim process
Closed system
Schematic
Solution
a
p V^2 = const. ---> p1 V1^2 = p2 V2^2
V2 = SQRT( (p1 V1^2)/ p2) = SQRT( (1bar / 9bar)(0.1m^2)^2)
V2 = 0.033 m^3
b
W12 = integral(v1 --> v2) of p dV
pV^2 = C ---> p = C/V^2
Plug in p for the integral and just integrate. It's easy
For the actual answer, pick a initial condition.
W12 = -20 kJ
Broadening Our Understanding of Energy
So far we have seen energy in theforms of
Macroscopic kinetic energy
Gravitational potential
energy
1/2mv^2
---> Mechanical Energy
mgh
But we havealsodiscussed other forms of energy
Chemical thermal electrical potential etc
We can broadly statethis as
Total
Energy
Macro. kinetic
Energy
Gravitational
potential energy
All other forms
of energy
Internal Energy
Kinetic energy and potenical engery
All forms of energy otherthan
U = extensive internal energy
units kJ or Btn
U = intensive ----> specific internal energy
units kJ/kg or Btn/lb
Microscopically has 3 components
1 Etherm Energyassociated with
the translation, vibration, and rotation of molecules and atoms
2 Uchem Energy associatedwith chemical bonds and electron orbital states
3 Unus
Energyassociated with
nuclear binding forces
You cannot solve for U only AU Internal energyvalues listed in
reference
look
point
up tables chap 3 are always with respect to a
TotalEnergy
As we've said there a 3 ways of storingenergy in a system
1 Kinetic energy
2 Potential energy
3 Internalenergy
Therefore thechange in totalenergyof a system is
deltaE = deltaKE + deltaPE + deltaU
or
if going say from state 1 to state 2
which is
im not typing ts. Just add a subscript to each 12
We have 3 ways to transfer energy to move between system states
1
2
3
Work
---> only work if there is a closed system
Heat
Mass flow
---> (all three) works for open system
1ˢᵗ Law of Thermodynamics for a Closed System
deltaE = Q (heat transfer into the system) - W (net work done by the system on the suroundings)
deltaKE + deltaPE + deltaU = Q - W
Q - W b/c you are looking at the system, not around
the system. +W taking away energy from system, but +Q
is taking away energy. So they must be opposite
Notethat the sign conventionforQ is oppositethatforW
0
Q
0
Energy transfer into the system by heat
0
Energy transfer out of the system by heat
In differential form
dE =
As a rate of change
At steady state properties of the system do not change with time
Energy Transfer by Heat
Heat
Energy transfer due to a temp difference
Like work heat
Not a property of the system (like work, its the process)
Depends on thedetails
on the
of theprocess notjustthe end states i.e depends
path taken between states
Is represented by an inexactdifferential IQ hence
d It is not technically correct to say an object hasheat the correct term
A
theÉimÉñéaÉÉransfer is more precisely stated as
Energy transferby heat
Heat transfer always travels in the directionof
decreasing temperature
Energy may be enteringandexiting a system by heat via multiple pathways
Qnet = Qin - Qout
is the net transferofenergy tothesystem by heatand work that we
consider in our 1ˢᵗ Law formulation
It
ΔE
Qnet Wnet
Units
Terminology
Heat transfer Q
Heat transfer rate
units
kJ or Btn
Q
heat transfer per unit time
Q_dot = dQ / dt
units
kJ/s = kw or Btn / hr
of your texbook calls
this
it at fTxq
heat transfer per unit time per unit
area perpendicular to thedirection of heat transfer
units
Adiabatic
kw/m^2 or Btn / hr*ft^2
No heat is transferred systems and surroundings
Can be achieved in 2 ways
1 "Insulated" system
2
The system and surroundings are in equilibrium (T_sys = T_sur)
The Three Modes of HeatTransfer
1 Conduction just random particles bumping into each other and making them bump into others
Heat transfer from more energetic to adjacent less energetic particles due to their random motion
Transport Law
Fourier's Law
Area of which heat transfer occurs
T_hot
Q_x = -kA (dt/dx)
T_cold
gradient derivative
Q_x
withrespect to a
spacial dimension
rate of conductive heat transfer in the x-direction
Negative sign because heat transfer is in
the direction of decreasing temp
K is thermal conductivity, a material property
(look it up)
units
large k
small k
w/(m*k)
Good conductor of heat
Good insulator of heat
Important simplifiedcase
1 Dimensional
Steady State
Plane wall
e.g Q in x direction only
system temp not a functionoftime
simplest conductiongeometry
K is constant throughout wall
No energy generation withinthewall
If these 5 conditions are met andtheyoftenare
T Thot
Q_x = (-kA(T_2-T_1))/L
Linear
T2 Tcold
L
or
1
sign
we could distribute the negative
incorporate that heat hot cold
Q_x = (kA(T_hot - T_cold))/L
2
heat transfer between a solid surface and an adjacent moving fluid
Convection
Transport Law
I
Newton's Law of Cooling
fluid temp
Q_c = hA(T_hot - T_cold)
body temp
A is the area that is IN CONTACT
with the fluid
If That Tb heattransfer
from the body to the fluid
If That Tf heattransfer
is
from the fluid to the body
h is the convective heat transfer coeffecient
emperical
function
i e experimentallydetermined
of fluid velocity
properties surface geometry etc
can't simply look up in a table
units w/(M^2 *K)
in this class it will begivento you
3 Radiation
Energy transmited by electromagnetic waves
Transport Law
Stefan-Boltzmanns law
Ts
Absolute temperature of surroundings
Absolute surface temp a body [K or R
A is the surface area of the body
sigma is the stefan_boltzmann
constant = 5.67*10^-8 (w/(m^2*k^4))
epsilon is the surface emissiviy
how well the surface radiates
0 and 1
value is between 1 is perfect emitter
unitless
from surface to surroundings
If That Ts heattransfer
is surroundings to the surface
property of the surface
indicates
If That Tb heattransfer
0 doesn't emit anything at all
Assumptions
Schematic
Solution
Conduction is negligible
Energy Analysis of Cycles
Thermodynamic Cycle
Sequence of process that begins and ends at the same state
power generation, refrigeration, and
Are the essence of most methods of vehicle propulsion
Because beginning andendstates are identical
AFaya
it follows that
E_end - E_beginning = 0
change in the amount ofenergystored inthe
system fromthe beginning to the end of the cycle
So
for example for a 4 step cycle
deltaE = 0 = all of the states = 0
and since
ΔE
Q W
W_cyc = Q_cyc
D These are the net amounts
of energy transfer by heat
and work for the cycle
Measuring the performance
of a cycle
Goal: What we hope to achieve with the cycle
Cost: What energy we need to put in to reach goal
Goal/Cost
Cycle Performance
cycle whose goal is to produce some net power out
Power Cycle Thermodynamic
eg. Engine, anything that produces work out
For a power cycle
positive
Wnet is always
T_hot
system (engine)
e
P
g Consider this 3 step powercycle
2
T_cold
s
here these are
W_cyc = Q_in - Q_out
Qin
magnitudes
with direction indicated by
subscripts
Qout for a power cycle
The performance of a power cycle is measured by its
thermmal efficiency
= Goal / Cost = W_cyc / Q_in =
y
is always < 1 for any real power cycle
the only way it = 1 is if Q_out = 0 or Q_in = infinity, (both impossible)
Refrigeration and Heat Pump Cycles
Thermodynamic cycles that transfer energy by heat from a lower temperature space to a higher temperature space
For a refrigeration heat pump cycle
negative
Wnet is always
T_hot
W_in > W_out
system
W_cyc = Q_out - Q_in
6
T_cold
again these are
magnitudes
with direction indicated by
Qout
Qin for refrigeration heat pump cycles
subcriptes
The performance of a refrigeration or heat pump cycle is measured
coefficient of performance (COP)
its
by
cycle performance
Refrigeration
from a cold space
8
Goal is to remove energy by heat transfer
into the system
t must trans er this heat
so the goal is
Q_in
COP_in = Q_in / W_cyc = Q_in / (Q_out - Q_in)
Heat Pump Goal is to add energy by heat transfer
to a hot space
so the goal
is
must transfer this heat
from the system
Q_out
For the same reasons that y must be
1
Assumptions
W_out = 0
Solution
a
b
W_cyc = W_in - W_out
W_cyc = W_in = 0.125 kW
W_cyc = Q_out - Q_in
0.125 = 0.61 - Q_in
Q_in = 0.485kW
B = Q_in / W_cyc = 0.485kW / 0.125kW
B = 3.88
Given Gas in a piston cylinder undergoes the following cycle
1 2
Compression with
pH const
2 3 Constant volume heat exchange with surroundings
3 1
Constant pressure expansion
known properties at eachstate
p 1bar
V2
Us
2m
U
240 KJ
0 2m
120 KJ
In process 2 73 the heatexchange occurs over an area
of 1.5 m over an 8 sec duration with Tsys To 60 C
Tsar Tf 20 C and h 150Km K
Find a
Waye following
sign conventionfor work
b
Power or refrigerationcycle
c
Magnitudes
of Qout and Qin
d cycle efficiency or coefficient of performance
KE = PE = 0
Assumptions Q_23 by convection only
Schematic
P bar
m
Solution
a
b
c
d