Contributed by:
Work, Internal energy, Joule's law, Specific heat, Enthalpy
1.
Chap. 3
First Law of Thermodynamics
General Form
Atmospheric Science Applications
2.
Some general statements
• The energy of the universe is constant
• The First Law:
– defines internal energy
– states that heat is a form of energy.
– states that energy is conserved
• The First Law is the second fundamental
principle in (atmospheric) thermodynamics,
and is used extensively.
3.
Work of expansion
• Work is part of the First Law, but it can be
considered independently
• What is work?
– If a system (parcel) is not in mechanical
(pressure) equilibrium with its surroundings, it will
expand or contract. This involves “work”.
• Work is defined by the differential dw fds
4.
Cylinder
Piston
• dW = pAdx = pdV
(shaded region of the
p A
graph) e
r1
u
ss
• Specific work (work per e
r P
P
p
unit mass): Q
p2 B
dw = pd
• work is found by V1 dV V2
integrating this differential Volume
over the initial and final W area under the curve AB
volumes V1 and V2:
V2 This thermodynamic diagram (p-V)
W = ∫pdV represents the state of the system at every
V1
point along the line. (Recall that the p-V
diagram is a simple thermodynamic
diagram.)
5.
Example 3.1
Calculate the work done in compressing (isothermally) 2 kg
of dry air to one-tenth its initial volume at 15 ºC.
From the definition of work, W = pdV.
From the equation of state, p = dRdT = (m/V)RdT.
Then W = mRdT dlnV = mRdTln(V2/V1)
(remember the process is isothermal)
= (287 J K-1 kg-1)(288.15 K)(2 kg)(ln 0.1)
= -3.81 x 105 J.
The negative sign signifies that work is done on the volume
(parcel) by the surroundings.
6.
Work, cont.
• The quantity of work done depends on the path taken; work
is not an exact differential. If it was, work would depend only
on the beginning and end points (or initial and final
conditions.
• Reconsider Eq. (3.a) above, rewriting it as follows (noting
that the displacement dx = vdt, where v is the magnitude of
the velocity vector):
• dW = pA(dx) = pA(vdt)
• Since p = F/A (or F = pA), the above equation becomes
• dW = Fvdt or dW/dt = Fv
• Now from Newton’s Law, F = ma = mdv/dt. Substituting this
in the above yields
• or dW/dt = dK/dt (K = ½ mv2)
• Hmmmm, we just can’t escape dynamics!
7.
Example 3.2 (from problem 3.7 in
Tsonis): An ideal gas of p,V,T
undergoes the following successive
changes: (a) It is warmed under
constant pressure until its volume
doubles. (b) It is warmed under constant
volume until its pressure doubles. (c) It
expands isothermally until its pressure
returns to its original pressure p.
Calculate in each case the values of p,
V, T, and plot the three processes on a
(p,V) diagram.
The three processes are shown in the
graph on the right. In the first process,
the work is
2V
Wa = ∫pdV = p (2V −V ) = pV
v
In the second process, the work is zero since volume does not change.
In the third process, the value of work is similar to that done in Example 3.1. As
this process proceeds through steps a-c, the temperature increases such that
8.
More on thermodynamic work
9.
Internal energy and a mathematical
statement of the First Law
1. Consider a system which undergoes a change from some
heat input q per unit mass (q=Q/m).
2. The system responds through the work of expansion, w
(work per unit mass).
3. The excess energy beyond the work done by the system is
q-w.
4. If there is no change in macroscopic or bulk kinetic and
potential energy of the system, then it follows from
conservation of energy that the internal energy (u - per
unit mass) of the system must increase according to:
q - w = u (simple expression of the First Law) (3.2)
10.
General form of the First Law in terms of various energy terms:
We will sum the various forms of energy which have passed through the system-
surroundings boundary and set this new sum equal to the change in the system
internal energy, similar to what we did in the previous equation. This expression
differentiates thermal (LHS) and non-thermal (RHS) forms of energy:
u = q + ei, (a more general form of the First Law)
where q is again the net thermal energy (per unit mass) passing into the system
from the surroundings. [Thermal energy can be defined as the potential and kinetic
energies that change as a consequence of a temperature change.]
Note that this last expression deals with the classification of energy passing through
the system boundary -- it does not deal with a classification of energy within the
11.
What is thermal energy (q)?
• At this point, it is instructive to define
thermal energy.
• In the atmosphere, thermal energy can
include heating/cooling by
– Radiation (heating by absorption of SW or
LW; cooling by emission of LW)
– latent heating (associated with water phase
changes).
12.
Joule's Law: u depends only on T OK, prove it!
From statistical mechanics, for an ideal monatomic gas, the kinetic
energy of translations is given by (refer to Chap 2 of Knupp’s notes)
pV = (1/3)Nomu2 = (2/3)Ekin = RT,
where N0 is Avogadro's number (6.023x1023). Thus,
Ekin = (3/2)RT.
Since at constant temperature there are no energy changes in
electronic energy, rotational energy, etc., the internal energy of an
ideal gas is only a function of T. This is also true for polyatomic ideal
gases such as CO2 (and more generally for air).
u = f(T) (Is this a proof?)
13.
A general form of the First Law (in the differential form)
Dq = du + Dw = du + pD. (3.4)
[What has happened to the ei term?]
The operator "d" refers to an exact differential and "D" to inexact. One property
of the inexact differential (e.g., Dw) is that the closed integral is in general
nonzero, i.e., ∫Dw ≠0.
[see http://en.wikipedia.org/wiki/Inexact_differentials]
The first law requires that du be an exact differential -- one whose value
depends only on the initial and final states, and not on the specific path.
However, from here on, we will ignore (but not forget) this formal
distinction between exact and inexact differentials. Illustrate with p-V diagram
Aside: An exact differential can also be expressed as, for a function U = U(x,y)
(Tsonis, Section 2.1)
∂U ∂U
dU = dx + dy
∂x ∂y
14.
Review and applications
Applications of the equation of state, and connection with the First Law.
From http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/idgcon.html#c1
15.
Review and example problems
Review the information and example problems at the following web site:
16.
3.4 Specific heats
Consider the case where an incremental amount of heat dq is
added to a system.
The temperature of the system increases by an incremental
amount dT (assuming that a change of phase does not
occur).
The ratio dq/dT (or Dq/DT) is defined as the specific heat,
whose value is dependent on how the system changes as
heat is input.
c = dq/dT [note that C = mc = m(dq/dT)]
17.
Two atmospheric scenarios: constant volume and
constant pressure.
For constant volume: cv (dq/dT)=const
Since specific volume is constant, no work is done.
According to the First Law, Eq. (3.1), dq = du (since d =
0) and
cv = (du/dT)=const (3.5)
Then du=cvdT
and dq = cvdT + pd. (3.6)*
18.
For the isobaric process, the specific heat is
cp (dq/dT)p=const (3.7)
In this case, some of the heat added is used in the work of
expansion as the system expands against the constant
external pressure of the environment. The value of cp must
therefore be greater than that of cv.
To show this, we can write (3.6) as
dq = cvdT + d(p) - dp = d(u+p) - dp = dh - dp,
where h, the enthapy is defined as h u + p.
(Enthalpy is discussed further in the following section.)
19.
Since p = RT from the equation of state (for dry air), the
previous equation can be rewritten as
dq = (cv+R)dT - dp = cpdT - dp. (3.8)*
If pressure is constant then dp=0 and, using (3.6), we note
that cp exceeds cv by the amout R, the gas constant:
cp = cv + R. (3.9)*
20.
For dry air, the values are:
cv = 717 J K-1 kg-1
cp = 1005.7 J K-1 kg-1 [ = f(T,p); Bolton, 1980]
For ideal monatomic and diatomic (air) gases, it can be shown from statistical
mechanics theory that the ratios cp:cv:R are 5:3:2 and 7:5:2, respectively. (See
Tsonis, p. 32.) The variation of cp with T and p is presented in Table 3.1.
Table 3.1. Dependence of cpd (J K-1 kg-1) on T and p. From Iribarne and Godson
p (mb) T (C)
-80 -40 0 40
0 1003.3 1003.7 1004.0 1005.7
300 1004.4 1004.0 1004.4 1006.1
700 1006.5 1005.3 1005.3 1006.5
1000 1009.0 1006.5 1006.1 1007.4
Relative variation: (1007.4 – 1004.0) / 1007.4
Why is cp not constant? = 3.4/1007.4 = 0.0034
21.
3.5 Enthalpy
Many idealized and natural processes of interest in
atmospheric science occur at constant pressure. An
example is evaporation of rain. If heat is added isobarically
to a system such that both the internal energy u and
specific volume change, then the First Law (dq = du +
pd) can be integrated as
q = (u2-u1) + p(2 - 1) = (u2+p2) - (u1+p1)
= h2 - h 1,
where enthalpy h is defined as
h = u + p. (3.10)
22.
Upon differentiation, we obtain
dh = du + pd + dp [= dq + dp]
dq = dh - dp.
Comparing this with (3.8) we can redefine dh as
dh = cpdT. (3.11)
This can be integrated to give (assuming h=0 when T=0 K)
h = cpT.
Yet another form of the First Law is thus
dq = dh - dp = cpdT - dp. (3.12)*
Three useful forms of the first law: (3.6), (3.8) and (3.12).
23.
3.6 Example: an isothermal process and reversibility
Consider a fixed mass (m=const) of ideal gas confined in a cylinder with a
movable piston of variable weight. The piston weight and its cross-sectional area
determine the internal pressure. Assume that the entire assembly is maintained at
T=const (an isothermal process). Let the initial pressure be 10 atm and the initial
volume Vi be 1 liter (L).
Consider three different processes in going from A to B (below right)
24.
Process 1: The weight of the piston is reduced to change the cylinder pressure to 1
atm. The gas will expand until its pressure is 1 atm, and since pV=mRT=const, the
final volume Vf will be 10 L (see Fig. 3.1). The work of expansion is
W = pdV = psurr(Vf-Vi) = 1 atm * (10-1) L = 9 L-atm.
This is the work done on the surroundings.
Process 2: This will be a two-stage process: (i) Decrease (instantaneously) the
cylinder pressure to 2.5 atm; then the volume will be 4 L, since this is similar to
Process 1. (ii) Then further decrease the pressure (instantaneously) to 1 atm with a
volume of 10 L. The work is the sum of these two processes:
W = p1V1 + p2V2 = 2.5 atm * 3 L + 1 atm * 6 L = 13.5 L–atm
Process 3: The pressure is continuously reduced such that the pressure of the gas is
infinitesimally greater than that exerted by the piston at every instant during the
process (otherwise no expansion would occur). Then we must apply the integral form
of work to get
W = pdV = mRTln(V2/V1) = 23.03 L-atm.
Note that pV=mRT=10 L-atm = const in this example.
25.
This last process is reversible and represents the maximum work.
Definition: A reversible process is one in which the initial
conditions can be reproduced after a system goes through
at least one change in state.
In process 3 above, the initial state of the system can be realized by
increasing the pressure continuously until the initial volume is attained.
Note that the value of work for the
reversible process depends only on
the initial and final states, not on the
For the most part, idealized
atmospheric processes are reversible
since parcel pressure is assumed to
be equal to (i.e., differs infinitesimally
from) the ambient pressure.
26.
Wikipedia definition of reversibility
• In thermodynamics, a reversible process, or reversible cycle
if the process is cyclic, is a process that can be "reversed" by
means of infinitesimal changes in some property of the
system without loss or dissipation of energy. Due to these
infinitesimal changes, the system is at rest during the whole
process. Since it would take an infinite amount of time for the
process to finish, perfectly reversible processes are
impossible. However, if the system undergoing the changes
responds much faster than the applied change, the deviation
from reversibility may be negligible. In a reversible cycle, the
system and its surroundings will be exactly the same after
each cycle.[2]
• An alternative definition of a reversible process is a process
that, after it has taken place, can be reversed and causes no
change in either the system or its surroundings. In
thermodynamic terms, a process "taking place" would refer to
its transition from its initial state to its final state.
27.
3.7 Poisson’s Equations
An adiabatic process is defined as one in which dq=0. The two advanced
forms of the 1st Law (which are related by the equation of state) become
0 = dq = cvdT + pd
0 = dq = cpdT -dp
Using the equation of state in the form p=RdT (dry atmosphere), the above relations
can be manipulated to get the following differential equations:
0 = cvlnT + Rddln, [T, ]
0 = cpdlnT + Rddlnp, [T, p] (let’s look at this one)
0 = cvdlnp + cpdln, [p, ]
where the third expression was obtained using the equation of state. Integration yields
three forms of the so-called Poisson’s Equations:
T-1 = const (TcvRd = const)
Tp- = const (Tcpp-Rd = const
p = const (pcvcp = const)
= Rd/cp = 0.286 and = cp/cv = 1.403
28.
The last of the three above equations (p = const ) has a form similar to
that of the equation of state for an isothermal atmosphere (in which case
the exponent is 1). These relationships can be expressed in the more
general form
pn= const,
which are known as polytropic relations.
The exponent n can assume one of four values:
For n= 0, p = const isobaric process
For n = 1, p = const isothermal process
For n = , p = const adiabatic process
For n = isochoric process
Refer to Tsonis, pp. 34-36.
29.
3.8 Potential temperature and the adiabatic lapse rate
Potential temperature is defined as the temperature which an air parcel
attains upon rising (expansion) or sinking (compression) adiabatically to a
standard reference level of p0 = 100 kPa (1000 mb).
We use the (T,p) form of the First Law (3.5) , assuming an adiabatic process (dq=0)
dq = 0 = cpdT - dp.
Incorporate the equation of state, p=RdT, to eliminate , and rearrange to get
cp dT dp
− =0
Rd T p
Now integrate over the limits, in which a parcel has a temperature T at pressure p,
and then end with a (potential) temperature q at the reference pressure p0.
Although not strictly correct, we assume the cp and Rd are constant.
p
cp T
dT dp
∫ =∫
Rd θ T p0
p
30.
or cp T p
ln =ln
. Rd θ p0
Take the antilog of both sides and rearrange to isolate potential
temperature ():
⎛ p 0 ⎞κ (3.13a)
θ =T⎜ ⎟
⎝p ⎠
[This is also called Poisson's equation, since it a form of Poisson’s
equations, e.g., Tp = const] For dry air, = Rd/cp = 287/1005.7 =
0.286 [= 2/7 for a diatomic gas – from kinetic theory]. This value
changes somewhat for moist air because both cp and R (Rd) are
affected by water vapor (more so than by T,p), as we shall see in the
Bolton (1980) paper.
31.
Potential temperature has the property of being conserved for
unsaturated conditions (i.e., no condensation or evaporation),
assuming that the process is adiabatic (i.e., no mixing or
radiational heating/cooling of the parcel).
For a moist atmosphere, the exponent in Eq. (3.13a) is
multiplied by a correction factor involving the water vapor
mixing ratio rv, and q is expressed as (see Bolton 1980, eq. 7)
κ (1−0.28 rv )
⎛ p0 ⎞
θ =T ⎜⎜ ⎟⎟ (3.13b)*
⎝ p ⎠
where rv is the water vapor mixing ratio expressed in kg kg-1.
32.
Dry Adiabatic Lapse Rate
An associated quantity, the dry adiabatic lapse rate, which is
used to evaluate static stability.
The term "lapse rate" refers to a rate of temperature change with
height (or vertical temperature gradient), i.e., T/z.
[Aside: It is important to differentiate the static stability of the
atmosphere, as given the the vertical gradient of temperature,
T/z, from the Lagrangian temperature change that results
when a parcel moves adiabically in the vertical direction. The
parcel change of temperature would be dT/dt = (dT/dz)(dz/dt) =
33.
Our starting point is once again the First Law (3.5) with dq = 0
(adiabatic process).
dq = cpdT - dp = 0.
For a hydrostatic atmosphere (hydrostatic implies no vertical
acceleration, and will be defined more fully later) the vertical
pressure gradient is
dp/dz = p/z = -g = -g/ (hydrostatic equation)
Solving the above for and substituting into the First Law, we
cpdT + gdz = 0.
34.
Thus, the value of the dry adiabatic lapse rate (d) is
(dT/dz)d = -g/cp = d = -9.81 m s-2 /1005.7 J K-1 kg-1 [J = kg m2 s-2]
= -9.75 K km-1. (3.14)
Again, one should be aware that this value changes slightly for a moist
(subscript m) atmosphere (one with water vapor), since the addition of water
vapor effectively yields a modified value of the specific heat at const
pressure, which has the following dependence on water vapor:
cpm = cpd(1+0.887rv),
where rv is in units of kg kg-1 (Bolton, 1980). (We will see this
difference in graphical form later.)
Specifically, for moist air,
m = d / (1 + 0.887rv) d (1-0.887rv).
35.
Some uses of
• Atmospheric structure; conserved for
subsaturated motion; atmospheric bores
• Vertical cross sections
• Analysis of static stability
36.
3.9 Heat capacities of moist air; effects on constants
The exponent of Poisson's equation ( = Rd/cp) requires adjustment when water
vapor is present. Why?
The water vapor molecule (H2O) is a triatomic and nonlinear molecule, whose
position can be described by 3 translational and 3 rotational coordinates.
Dry air is very closely approximated as a diatomic molecule (N2, O2)
(See web site
The specific heats for water vapor are therefore quite different from (much
larger than) that of dry air:
cwv = 1463 J K-1 kg-1 (w subscript designates the water phase)
cwp = 1952 J K-1 kg-1,
For Poisson's eq. (3.10a) the exponent Rd/cp is adjusted using the correction
term (Rd/cp)(1-0.28rv) (Bolton 1980), where the water vapor mixing ratio rv is
expressed in kg kg-1. Also the "constants" Rd and cp can be corrected for moist
air as follows:
c = c (1+0.887r ),
37.
3.10 Diabatic processes, Latent Heats and Kirchoff's equation
diabatic process dq 0.
Two examples of diabatic heating/cooling
1. absorption/emission of radiation;
2. heating/cooling associated with water phase changes
In the moist atmosphere, there are cases where heat supplied to a parcel without a
corresponding change in temperature.
Under such conditions, the water substance is changing phase, and the change in
internal energy is associated with a change in the molecular configuration of the
water molecule, i.e., a change of phase. (latent heating)
Heat gain
Deposition
Freezing Condensation
Lil Liquid Water
Ice Lvl
Water vapor
Melting Evaporation
Liv
Sublimation
Heat loss
38.
Notation on the latent heating terms: the two subscripts define the change in phase of
the water substance.
Lvl is the latent of condensation with the subscript vl denoting a change in phase from
vapor (v) to liquid (l). Thus, condensation is implied from the order of subscripts:
vapor to liquid.
For the sake of simplicity we can define the latent heats as follows:
Lvl = 2.50 x 106 J kg-1 (0 C) latent heat of condensation (function of T)
Lvl = 2.25 x 106 J kg-1 (100 C) (yes, Lvl does vary by 10% between 0-100 C)
Lil = 0.334 x 106 J kg-1 latent heat of melting
Lvi = 2.83 x 106 J kg-1 (0 C) latent heat of deposition
Note: Lvi = Lvl + Lil
In general, these terms are defined for p=const . . .
but Lvl does vary with temperature!
39.
Why is L a function of temperature?
From the First Law: L is related to an enthalpy change, i.e., L = h.
[Proof: Since dp=0, the First Law can be written as dq = cpdT = dh. Also recall that
enthalpy can be derived from the First Law.]
To examine this temperature dependence, expand the differential dh, based on the
definition of the total derivative (note that h = f(T,p):
dh = (h/T)p dT + (h/p)T dp (definition of the exact differential),
and apply this to two states a and b (h=L=hb-ha):
d(h) = (h/T)p dT + (h/p)T dp.
For an isobaric process, only the second term vanishes and we have
d(h)p ≡ dL = (h/T)p dT = (hb/T)pdT - (ha/T)pdT
= (cpb - cpa)dT.
This latter equivalence is based on the definition of specific heat (see Section
3.4), cp dq/dT = dh/dT for an isobaric process.
40.
From the previous equation, we can write Kirchoff's equation
(L/T)p = cp, (3.15)*
Thus, the temperature dependence of L is related to the temperature
dependence of cp.
Bolton (1980) provides an empirical equation that has a linear form for the
temperature correction of Lvl:
Lvl = (2.501 - aTc) x 106 J kg-1, (3.16)
where a = 0.00237 C-1 and Tc is the dry bulb (actual air) temperature in C.
T (C) Liv (106 J kg-1) Lil (106 J kg-1) Llv (106 J kg-1)
-100 2.8240
-80 2.8320
-60 2.8370
-40 2.8387 0.2357 2.6030
-20 2.8383 0.2889 2.5494
0 2.8345 0.3337 2.5008
20 2.4535
40 2.4062
41.
3.11 Equivalent potential temperature and the saturated adiabatic lapse rate
3.11.1 Equivalent potential temperature (approximate form)
Saturation condensation latent heating by condensation (Lvl, multiplied by the
mass of water vapor condensed)
Expressed by a change in the saturation mixing ratio, rvs
[Question: Is this an adiabatic process since dq ≠ 0?]
Starting point – First Law (dq is now nonzero due to latent heating)
dq = -Lvldrvs = cpdT - dp. (3.17)
42.
Substitute the equation of state p=RT for , and rearrange
drvs dT dp
−L vl =c p −R d
T T p
Take the log differential of Poisson's eq. (3.10a):
Rd
d ln θ =d ln T − d ln p
cp
dθ dT dp
cp =c p −R d
θ T p
43.
Combine the preceding with eq (3.14):
L vl dθ
− drvs =
c pT θ
Physical interpretation: the latent heating changes the potential temperature of the
parcel, such that a reduction in rvs (drvs < 0) corresponds to a positive d.
The LHS of the preceding equation is cumbersome to integrate, as it currently
stands, because Lvl = Lvl (T). With the use of a graphical (thermodynamic)
diagram, it will later be shown that
Lvl ⎡Lvl rvs ⎤ ,
drvs ≈d ⎢ ⎥
c pT ⎣ c pT ⎥
⎢ ⎦
(this is an approximation, but it provides an exact differential)
Then we can integrate the following
⎡ L r ⎤ dθ
vl vs
d⎢ ⎥=
⎣ c pT ⎦ θ
44.
assuming that -> e as rvs/T -> 0, to get
⎡Lvl rvs ⎤ ⎛θ ⎞
−⎢ ⎥ =ln ⎜⎜ ⎟⎟
⎣ c pT ⎥
⎢ ⎦ ⎝θ e ⎠
o
r ⎛L r ⎞
vl vs
θe = θ exp⎜ ⎟ (approximate form) (rvs in kg kg-1) (3.18)
c T
⎝ p sp ⎠
which is an approximation for equivalent potential temperature, e.
Note the approximations used here:
(1) assumed that cp and Lvl are independent of rv and/or T and p;
(2) made the approximation in the differential (Lvl/cpT)drvs d(Lrvs/cpT).
In essence, we have assumed that Lvl is independent of temperature, which
sacrifices precision in the integrated form. Tsp in Eq (3.18) is the
temperature of the parcel's saturation point (SP), traditionally called the
lifting condensation level or LCL, and rvs is the mixing ratio at the LCL (or
45.
A semi-empirical formula for e, superior to Eq. (3.15) and
accurate
to within ~0.5 K, is
⎛ 2675rvs ⎞
θ e =θ exp⎜ ⎟ (within 0.5 K) (rvs in kg kg-1) (3.19)*
⎜ T ⎟
⎝ sp ⎠
Better!
(I don’t recall the source of this, but it is given as Eq. (2.36) in Rogers and Yau
1989. Note that the numerical value 2675 replaces the ratio Lvl/cp, so this implies
some constant values for cp, and especially Lvl. This form is good for quick,
relatively accurate calculation of e. An accurate calculation of e requires an
accurate determination of Tsp, , cp, and an integrated form that does not assume
that Lvl is consant. These steps are detailed in the paper by Bolton (1980).
Note: L/cp = 2.5 x106 / 1005 = 2488
L/cp = 2675; L = 2675 cp = 2675 x 1005.7
= 2.69 x 106
46.
3.11.2 Equivalent potential temperature (accurate form)
Because e is conserved for moist adiabatic processes, it is widely used,
and its accurate calculation has received much effort. An analytic solution is not
possible. The approximate form derived in the previous section may produce
errors of 3-4 K under very humid conditions (i.e., large rv).
Refer to the paper of Bolton (1980) for a presentation of the accurate calculation of
e. We will consider this in some detail.
[Assignment: Read the paper by Bolton (1980).]
Bolton’s curve-fitted form of e is (more clumsy with the calculator, but is easily
⎡⎛ 3.376 ⎞ 3
⎤
θ e =θ exp ⎢⎜ −0.00254 ⎟ × rv (1 + 0.81×10 rv )⎥
−
(3.20)
⎢⎣⎜⎝ Tsp ⎟
⎠ ⎥
⎦
(within 0.04 K) (rv in g kg-1)
47.
3.11.3 Saturated (psuedo) adiabatic lapse rate – preliminary form
The derivation of the saturated adiabatic lapse rate is complicated and requires
advanced concepts/tools that are developed in later chapters. The derivation below
is first order only. A complete derivation will be presented in Chapter 6 (my notes).
When water droplets condense within an ascending parcel, two limiting situations can be
assumed:
1) the condensed water immediately leaves the parcel (the irreversible pseudo-
adiabatic process), or
2) all condensed water remains within the parcel (the reversible saturated-
adiabatic process).
In reality, the processes acting within clouds lie somewhere in between. Here we
consider the pseudo-adiabatic process.
With the addition of latent heating, one may anticipate that a rising parcel undergoing
condensation will cool less rapidly that an unsaturated parcel. Our starting point is
(3.14), but with the hydrostatic equation term gdz substituted for dp in (3.14). We
can then write
-Lvldrvs = cpdT + gdz. (remember dp/dz = -g)
48.
Ignore the effects of water vapor being heated along with the dry air and
write the above as
dT −L vl drvs g
= −
dz c p dz c p
Applying the chain rule to drvs/dz
dT −L vl drvs dT g
= −
dz c p dT dz c p
The second term g/cp defines the dry adiabatic lapse rate. The first term is new, and
is the somewhat messy water vapor term (first term on the RHS). Solving the above
for dT/dz, the approximate saturated adiabatic lapse rate is given as
⎛ dT ⎞ Γd
⎜ ⎟ =Γs = (3.21)
⎝ dz ⎠s Lvl drvs
1+
c p dT
A functional relationship for rvs is obtained from the Clausius-Clapeyron
equation, to be considered Chapter 5 (my notes). We also note that the magnitude
of s is not constant, but decreases (nonlinearly) as T increases. This is not obvious
from Eq. (3.21), but will become more apparent when we examine and analyze the
Clausius-Clapeyron equation.
49.
More on the saturated adiabatic
lapse rate
• Pseudo adiabatic process – all condensed
water leaves the parcel; irreversible
• Saturated adiabatic process – all
condensed water is carried with the parcel;
reversible
• More on this in Chap. 6
50.
More on e
• Recall the approximate form (for illustrative
puroses only):
⎛L r ⎞
vl vs
θe = θ exp⎜ ⎟
⎝ c p Tsp ⎠
• rvs is the initial parcel water vapor mixing ratio (r v)
• Tsp is the so-called “saturation point” temperature.
Bolton defines this is TL.
51.
How is Tsp determined?
• Tsp will be derived in Chap. 6
• For now, use Bolton’s formula:
2840
Tsp = + 55.
3.5 ln T −ln e −4.805
e is water vapor pressure; T is temperature in K
1
Tsp = + 55
1 ln( f / 100)
−
T −55 2840
f is relative humidity
52.
3.12 Atmospheric Thermodynamic Diagrams
Refer to Chapter 9 in Tsonis.
thermodynamic diagram - a valuable tool to illustrate the relationship between dry
adiabats, saturated adiabats and other thermodynamic variables
A number of thermodynamic diagrams used for atmospheric applications have been
constructed. We have considered only the simple p-V diagram so far (e.g., Fig. 3.1).
Any diagram is based on two thermodynamic coordinates. With two thermodynamic
variables being defined, other variables and processes can be determined based on
the Equation of State, First Law, and ancilliary relations that we have considered, or
will consider.
For example, the dry and saturated adiabatic lapse rates can (and must) be drawn on an
atmospheric thermodynamic diagram to illustrate static stability.
Once the coordinates are defined, then other isolines (e.g., saturation mixing ratio,
dry/saturated adiabats, isotherms, isobars) can be defined and drawn on the diagram
to graphically represent atmospheric processes and evaluate static stability.
53.
The most commonly used diagrams are the skew-T and the tephigram. Interestingly,
it seems as though the "tropical" meteorologists favor the tephigram (with the
exception of Rogers and Yau, who are Canadians). The skew-T is most widely used
in the research and operational sectors in the U.S.
An ideal atmospheric thermodynamic diagram has the following features:
• area equivalence: the area traced out by some process, e.g., the Carnot cycle, is
proportional to energy;
• a maximum number of straight lines;
• coordinate variables that are easily mearured in the atmosphere;
• a large angle between adiabats and isotherms;
• a vertical coordinate that is approximately linear with height.
The tephigram and skew-T closely satisfy nearly all these criteria. The table below
summarizes the important aspects for some diagrams. Note that the skew-T, which
we will use in this class, exhibits most of the ideal properties.
See Irabarne and Godson (1973, pp. 79-90; handout) and Tsonis (Chapter 9) for a more
complete discussion.
54.
Table 3.3. Summary of thermodynamic diagram properties.
Diagram Abscissa Ordinate Straight line characteristics Angle between adiabats and
isotherms
isobars adiabats isotherms
Skew-T, ln p T ln p Yes no Yes nearly 90 (variable)
Tephigram T ln no Yes Yes 90
Stuve T p Yes Yes Yes
Psueo- T -pd Yes Yes Yes ~45
Clapeyron -p Yes No No small
Emagram T -ln p yes No Yes ~45
Refsdal ln T -Tlnp no No yes ~45
55.
Various isolines are drawn on thermodynamic diagrams, including isobars, isotherms,
lines of constant saturation mixing ratio, dry adiabats and saturated adiabats.
On-line references:
Skew-T diagram:
A good tutorial with bad graphics:
Another one:
%20the%20skew-t%22
An excellent resource to access recent and archived soundings (U. Wyoming):
Available commercially:
57.
Applications
• representation of vertical profiles of temperature and moisture
(i.e., soundings)
• evaluation of static stability and potential thunderstorm
intensity
• quick estimation of derived thermodynamic quantities such as:
relative humidity, given the temperature (T) and dewpoint
temperature (Td)
mixing ratio, given Td and pressure p,
potential temperature and equivalent potential
temperature, given p, T and Td
others
• determination of cloud/environment mixing processes
• determination of thickness (1000-500 mb thickness)
• mixing processes (advanced application)
58.
Sources of skew-T diagrams (real-time and historical)
1) NCAR/RAP – the best Skew-T on the web:
2) University of Wyoming – flexible site, data, skew-T or Stuve diagram;
historical data
3) Unisys
Other valuable information:
GOES satellite sounding page – good information on skew-T’s and their
applications. We will examine many of these during this course.
RAOB program:
59.
Soundings plotted on a skew-T, ln p diagram
60.
Variations in Tv vertical profiles around an
atmospheric bore
63.
A more detailed analysis on the skew-T
64.
Animation of parcel ascent
• Buoyancy and CAPE
65.
Chap. 3 HW
• Problems 1-7 in notes, plus Petty 5.4, 5.7,
5.10
– ATS 441 students may waive number 1
66.
Examples
3.39 (W&H) A person perspires. How much liquid water (as a percentage of the
mass of the person) must evaporate to lower the temperature of the person by 5
°C? (Assume that the latent heat of evaporation of water is 2.5 x 106 J kg-1, and
the specific heat of the human body is 4.2 x 103 J K-1 kg-1.)
3.38 (W&H) Consider a parcel of dry air moving with the speed of sound, cs =
(RdT)½, where = cpcv = 1.40, Rd is the gas constant for a unit mass of dry air,
and T is the temperature of the air in degrees kelvin.
(a) Derive a relationship between the macroscopic kinetic energy of the air parcel
Km and its enthalpy H.
(b) Derive an expression for the fractional change in the speed of sound per
degree Kelvin change in temperature in terms of cv, Rd, and T.
+1-315-636-4485
+91 98151-74050 
