Transcript Slide 1

* Reading Assignments:
2.2
2.2.1
2.2.2
2.3
2.4
2.4.1
2.4.2
2.5
3. The First Law
3.1 Internal Energy
Observations show that expansion work performed on an
adiabatic system only depends on the initial and final states.
So it behaves as a state variable. Further, changes in
expansion work are reflected by changes internal to the system,
i.e., the internal energy (u).
u  wad
Or for infinitesimal change,
du  wad
3.2 The First Law (Diabatic change)
If heat is exchanged between the system and the environment,
u  w
but
u  wad
So, the difference between
into the system,
q  w  wad
w and wad
is called the heat transfer
* For an incremental process,
du  q  w  q  pdv
(3.1)
This is the mathematical form of the first law.
It describes that the change of internal energy
between two states is the net effect between
the heat transfer into the system and
the work performed by the system on surroundings.
Four physical points can be deduced from the first law.
1) For an adiabatic process,
 q  0,
 du  w
For a process where no work is done,
 w  0,
 du  q
Because du is an exact differential and depends only on the
Initial and final states, two different processes can change the
internal energy of a system by the same amount.
2) Since u is a state variable,
Then,
 q   w  0,
 du  0
or
for a cyclic process.
 q   w
The net work performed by the system is balanced by the net heat
absorbed by the system during a cyclic process.
3)
q  0, and w  0
A closed system performs work through a conversion of heat absorbed
by it -- heat engine.
Example: Troposphere
4)
q  0, and w  0
A closed system rejects heat through a conversion of work performed
on it -- refrigerator.
Example: Stratosphere
* Another state variable, the specific enthalpy, can be defined
for the convenience,
h  u  pv
From this,
dh  du  pdv  vdp
Use the first law,
dh  (q  pdv)  pdv  vdp
Then,
dh  q  vdp
(3.2)
For constant pressure processes, the first law becomes
dh  q
The heat transferred into the system can be measured by
the change in enthalpy.
3.3 Heat Capacity
* How to determine the heat absorbed or released in
thermodynamic processes?
1) For a constant volume process, a homogeneous system,
no physical change of state and no chemical reaction,
The heat absorbed is proportional to its mass and temperature
variation,
Qv  cv m dT  Cv dT
or
qv  cv dT
 cv 
qv
dT
The constant of proportionality cv is called the specific heat capacity
at constant volume, and Cv ( mcv ) is called the heat capacity.
To relate
cv to state variables, consider u  u (T , v)
 u 
 u 
du    dT    dv
 T v
 v T
Use this, the first law can be written as
 u 
 u 
q    dT    dv  pdv
 T v
 v T
When the volume is held constant ( dv  0)
qv  u 
 u 
qv    dT or


dT  T  v
 T  v
 du  cv dT
2) Similarly, at constant pressure
Q p  c p m dT  C p dT
cp
So,
 cp 
q p
dT
is the specific heat capacity at constant pressure.
q p  c p dT
Changes of heat appear as changes of temperature for constant
pressure situations.
For
h  h(T , p) ,
we can relate c p to the enthalpy when p is held constant,
dh  c p dT
* What did the Joule experiment demonstrate?
Why is this result important?
 q  0 ,
 v1  v2
 du  w   pdv
is constant,
 dv  0,
 du  0
1) Consider
u  u (T , v) ,
 u 
 u 
 u 
du    dT    dv    dT
 T v
 v T
 T v
Experiment shows
dT  0
2) Consider
u  u (T , p) ,
 u 
 u 
du  
 dT    dp
 T  p
 p T
Since
dp  0 , u
can not depend on pressure.
Therefore, the internal energy (u ) of an ideal gas is a function
of temperature only, u  u (T ) .
An easy extension can be made to enthalpy,
 h  u (T )  pv  u (T )  RT
 h  h(T )
only.
* If we define u=0 and h=0 at T=0, integrate
dh  du  RdT ,
We get
T
T
T
0
0
0
c p  dT  cv  dT  R  dT
or
c pT  cvT  RT ,
 c p  cv  R
Based on simplified kinetic theory, in general,
1
du  jRdT
2
where j is the number of degree of freedom, R is the specific
gas constant.
So,
du 1
1
cv 
 j R , and c p  ( j  1) R
dT 2
2
For a monatomic gas, j=3, translation only.
For a diatomic gas, j=5, translation + rotation + vibration.
In dry air which mainly includes ( N2 , O2 )
,
we would expect j=5
5
Rd  717.5 J kg 1 K 1
2
7
c pd  Rd  1004.5 J kg 1 K 1
2
c vd 

cp
cv

7
 1.4
5
Rd 2

  0.286
c pd 7
* We will use the following forms of the first law frequently:
q  cv dT  pdv
(3.3)
q  c p dT  vdp
(3.4)
3.4 Adiabatic Processes
3.4.1 Poisson’s Equations
* If
q  0,
the other terms must balance in the first law.
Use
q  0  cv dT  pdv and
So
0  cv dT 

RT
dv
v
0  cv d ln T  Rd ln v
Integrate
0  cv  d ln T  R  d ln v

cv ln T  R ln v  const

ln T cv  ln v R  const

Finally,
pv  RT
ln T cv v R  const
T cv v R  new const
(3.5)
* Similarly, use
q  0  c p dT  vdp and
We have
pv  RT
RT
0  c p dT 
dp
p
and
T p p  R  const
c
(3.6)
* To get a relationship in p & v, use log derivatives on idea gas law
pv  RT 
So,
ln p  ln v  ln R  ln T
d ln p  d ln v  d ln T
From,
RT
0  c p dT 
dp  0  c p d ln T  Rd ln p
p

0  c p d ln p  d ln v   Rd ln p

0  (c p  R)d ln p  c p d ln v

0  cv d ln p  c p d ln v

0  d ln p cv  d ln v
cp
Integrate to get
ln p cv  ln v
or
Since
p cv v
cp
cp
 const
 new const
  c p / cv ,   R / c p ,
from (3.5)-(3.7),
(3.7)
We can get Poisson’s equations
Tv  1  const
Tp   const
pv   const
(3.8)
3.4.2 Potential Temperature
From the relationship between pressure and temperature for
an adiabatic process, we can derive another state variable,
i.e., potential temperature ( ) .
Consider
q  0  c p dT  vdp
Integrate it from
(T , p ) to ( , p0 ) , p0  1000 mb
We have

cp 
T

p0 dp
dT
 R
 0
p
T
p

p0
T
p
c p  d ln T  R  d ln p  0



c p ln
ln
ln

T

T
 R ln

p0
 0
p
R p0
ln
 0
cp
p

p 
 ln  0 
T
 p
Finally,
 p0 

 p
  T 
R
cp
 0
R
cp
(3.9)
The potential temperature of an air parcel is the temperature that it
would take if we compressed or expanded it adiabatically to a
reference pressure of 1000 mb.  is conserved during an adiabatic
process.
q  cdT
q  cdT
Tp    const

R
cp  c
q  cdT
3.4.3 Vertical Motions and Adiabatic Constraints
To describe the change of temperature with height under adiabatic
condition, we need to combine two equations:
1) q  0  c p dT '  v ' dp '
2)
dp   gdz
primes are value for an air parcel
for the environment
'
Note that the parcel pressure p also obeys 2) since horizontal
pressure difference between the parcel and environment adjust
at sonic speeds. So
dp '  dp   gdz
Use this in 1)
c p dT '  v'  g dz  0

c p dT  ' g dz  0

'
Rearrange it, we have
dT '
g 
   ' 
dz
cp   
Since
p

RT
p'
and  
RT '
'
 T'
 '  1
 T
Finally
dT ' g


 d
dz c p
d is called the dry adiabatic lapse rate.
3.5
Diabatic Processes
We can intuitively predict that
diabatic processes occur.
How do changes in
 is not constant when significant
q (q ) relate to changes in  (d ) ?
Use log derivatives on
p 
We have

  T  0 
 p
ln   ln T   ln p0   ln p
d ln   d ln T  d ln p
(3.10)
We can also put the first law in log derivative form,
q  c p dT  vdp
RT
dp
p
q
R
 d ln T  d ln p
c pT
cp
q  c p dT 
(3.11)
Compare (3.10) and (3.11), we have
q
c pT
 d ln 
(3.12)
This is an alternative form of the first law. It shows that the heat
transferred into the system is directly related to the change in
potential temperature.
Homework (2)
1. A sample of dry air has an initial pressure p1=1000 hPa and temperature
T1=300K. It undergoes a process that takes it to a new pressure p2=500hPa with
unchanged temperature T2=T1. Compute the mechanical work per unit mass performed
by the sample under the following scenarios:
a) Isochoric pressure reduction to p2 followed by isobaric expansion to final state.
b) Isobaric expansion to final specific volume v2 followed by isochoric pressure reduction
to final state.
c) Isothermal expansion to final state.
2. A unit mass of dry air goes through the following steps: a) adiabatic compression from
60 kPa and 0oC to a temperature of 25oC, and b) isothermal expansion to a pressure of
70 kPa. Calculate the work done by the air during each step.
3. Through sloping convection, dry air initially at 20oC ascends from sea level to 700 mb.
Calculate a) its initial and final specific volume, b) its final temperature, c) the specific
work performed, and d) changes in its specific energy and enthalpy.
4. A plume of heated air leaves the cooling tower of a power plant at 1000 mb with a
temperature of 30oC. If the air may be treated as dry, to what level the plume will ascend
if the ambient temperature varies with altitude as:
a) T(z)=20-8z (oC) and b) T(z)=20+z (oC), with z in kilometers.