Post on 17-Mar-2020
transcript
2.4, 2.10
Applications of the First Law to ideal gases
RTpv =
What we know
Equation of state
Internal energy is function only of temperature dTcdu v= dTcdh p=
Rcc vp =−Specific heats are related
pdvdw −="expansion work"
The fist law dwdqdu += vdpdqdh +=
pdvdqdTcv −=
vdpdqdTcp +=
dTcdu v= dTcdh p= Rcc vp =−
dwdqdu +=
vdpdqdh +=
pdvdw −=
The first law of thermodynamics for an ideal gas
Isothermal vs. Adiabatic
An isothermal process in one in which the initial and final temperatures are the same.
Isothermal processes are notnecessarily adiabatic.
An adiabatic process in one in which no heat is exchanged between the system and its surroundings.
0=dT
0=dq
Reversible Changes
• A reversible change is one that can be reversed by an infinitesimal modification of a variable.
gasenv pp =
• In a reversible expansion or compression
Lets consider an isothermal expansion 0=dT
dTcdu v=Because Internal energy is function only of temperature, the internal energy of the gas is unchanged
pdvdqdTcv −= pdvdq =
The first law of thermodynamics for an isothermal expansion
pdvdq =
isothermal expansion
If we want to integrate pdvdq =
pdvdw −=only expansion work" ∫ ≠ 0dw
What do we need???? Path!!!
v1 v2
AB
Lets consider the path of an isothermal reversible expansion
Equation of state is satisfied during all the stages of the expansion
RTpv =
isothermal expansion
pdvdq = RTpv =v
RTp =
vdvRTdq = ∫=Δ 2
1
v
v vdvRTq
⎟⎟⎠
⎞⎜⎜⎝
⎛=Δ
1
2lnvvRTq
pRTv = ⎟⎟
⎠
⎞⎜⎜⎝
⎛=Δ
2
1lnppRTq
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=Δ
2
1
1
2 lnlnppRT
vvRTq
If we want to integrate
isothermal expansion
Constant volume process 0=dv
dwdqdu += pdvdqdu −=
dqdu =
( )12 TTcq v −=Δ
dTcdu v=
The amount of heat required to raise the temperature of the gas from T1 to T2 at constant volume
dTcdq v=
dTcdu v= dTcdh p= Rcc vp =− pdvdw −=
Constant volume
vdpdqdh +=
Constant pressure process 0=dp
dqdh =
dTcdh p= dTcdq p=
The amount of heat required to raise the temperature of the gas from T1 to T2 at constant pressure
( )12 TTcq p −=Δ
dTcdu v= dTcdh p= Rcc vp =− pdvdw −=
Constant pressure
Adiabatic process Process in which NO HEAT is exchange between the system and its environment
0=dq
dwdqdu +=
dwdu =Fist law for a reversible adiabatic process
An adiabatic compression increases the internal energy of the system
q
q
w
pdvdu −=
Adiabatic process
pdvdqdTcv −= vdpdqdTcp +=
Adiabatic process 0=dq
pdvdTcv −= vdpdTcp =
Fist law for adiabatic expansion
The first law of thermodynamics for an ideal gas
Adiabatic process
Poisson’s equations
pdvdTcv −= RTpv =v
RTp =
dvv
RTdTcv −=vdvR
TdTcv −=
Lets assume constant vc There is a final and initial state
Lets consider a reversible adiabatic EXPANSION for an ideal gas
0=dq
∫∫ −= 2
1
2
1
v
v
T
Tv vdvR
TdTc
⎟⎟⎠
⎞⎜⎜⎝
⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛
1
2
1
2 lnlnvvR
TTcv
vcR
vv
TT
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛
2
1
1
2During an adiabatic expansion of a gas, the temperature decreases
∫∫ −= 2
1
2
1
v
v
T
Tv vdvR
TdTc
Poisson’s equations
Reverse process adiabatic Compression
Work is done on the gas and the temperature increases
vcR
vv
TT
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛
2
1
1
2
pcR
pp
TT
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛
1
2
1
2
vp
cc
vv
pp
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛
2
1
1
2
RTpv =
A given pressure decrease produces a smaller volume increase in the adiabatic case relative to the isothermal case
RTpv =
isothermal expansion Vs reversible adiabatic expansion
P, v diagram
Temperature also decreases during the adiabatic expansion
Dry Adiabatic Processes in the Atmosphere
For reversible adiabatic processes for an ideal gas
pcR
pp
TT
⎟⎟⎠
⎞⎜⎜⎝
⎛= 00
Orographic Lifting Frontal Lifting
Low-level convergence Vertical Mixing
Lifting Processes
p
T
0=dq
A dry adiabatic process as long CONDENSATION DOES NOT OCCUR!!
Lifting:
pcR
pp
T ⎟⎟⎠
⎞⎜⎜⎝
⎛= 0θ
If for mb 10000 =p the temperature is , θ
pcR
pp
TT
⎟⎟⎠
⎞⎜⎜⎝
⎛= 00
pcRWhere for dry air is 286.0
72
25
==+
=+
=RR
RRc
RcR
vp
θ Potential Temperature
Temperature a parcel of gas (e.g. dry air) would have if compressed (or expanded) in an reversible adiabatic process from a state, and , to a a pressure of mb 10000 =pp T
θ Also a state variable, invariant during a reversible adiabatic process: Conservative quantity!!!!
pcR
pp
T ⎟⎟⎠
⎞⎜⎜⎝
⎛= 0θ
-1km C6Ο=Γ
Consider a Temperature profile with
For mb 1000<p T>θ⇒ Adiabatic compression to lower the parcel
mb 1000>p T<θ⇒
mb 1000=p T=θ⇒
T2 T1
Adiabatic expansion to rise the parcel
dzdT
−=Γ
Lapse rate
pcR
pp
T ⎟⎟⎠
⎞⎜⎜⎝
⎛= 0θ
Considering Water vapor
( ) ( )vpdpvvpdvp qccqcqc 87.011 +≈+−=Specific heat for moist air
vq Specific humidityd For dry air v For water vapor
( )vd qRR 608.01+=
And knowing(from Ch. 1) ( )v
pd
d
v
v
pd
d
p
qcR
cR
cR 26.01
87.01608.01
−≈⎟⎟⎠
⎞⎜⎜⎝
⎛++
=
pcR
?? Considering water vapor
( )pd
vdc
qR
pp
T26.01
0
−
⎟⎟⎠
⎞⎜⎜⎝
⎛=θ
Potential Temperature for Moist Air
The difference between dry-air and moist air potential temperature is generally
less than 0.1 degree.
Important: is not conserved if a change of
phase occurs
θ
Adiabatic expansion or compression of moisture air can be treated as if
it were dry air
Liquid to vapor
solid to Liquid
Virtual Potential Temperature
Virtual Temperaturethe temperature a parcel which contains no moisture would have to equal the density of a parcel at a specific temperature and humidity.
temperature a parcel at a specific pressure level and virtual temperature would have if it were lowered or raised to 1000 mb.
vθ
vT
pdd
cR
vv ppT ⎟⎟⎠
⎞⎜⎜⎝
⎛= 0θ
Virtual Potential Temperature
Neglect the dependence on water vapor from the exponent, and replace temperature by virtual temperature
( )pd
vdc
qR
pp
T26.01
0
−
⎟⎟⎠
⎞⎜⎜⎝
⎛=θ
dzdT
−=Γ 1-km C8.9 Ο≈=Γpd
d cg
( )vpd qcg
87.01+=Γ
Adiabatic Ascent of a Parcel
T θ
vdpdTc p =
gdzdTc p −=
constant
Rate of decrease of temperature with height (first law, enthalpy form)
If there are no large vertical accelerations (hydrostatic relation applies)
vdpgdz =−
Lapse rate for dry air
For moist air
First law for adiabatic process