Solutions to questions from chapter 5 in GEF4310 - Cloud
Physics
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NB: THIS DOESN’T COVER ALL OF CHAPTER 5. WE WILL TAKE
THE LAST PART (WITH CAPE, MIXING AND SOME MORE ABOUT
CONDITIONS FOR STABILITY) IN THE NEXT GROUP SESSION!
Problem 1
What do we mean by...?
a) isobaric. Answer: constant pressure
b) isothermal. Answer: constant temperature
c) isopycnal. Answer: constant density
d) adiabatic. Answer: no exchange of heat with the surroundings. The heating/cooling
is only caused by the work done by compression/expansion
e) wet adiabatic. Answer: no exchange of heat with the surroundings. The heating/cooling is caused by the work done by compression/expansion AND by evaporation/condensation of liquid/vapor inside the air parcel.
Problem 2 - Some equations
Explain the different equations below
a)
p = pd + e = nRT
Answer: The ideal gas equation. It expressed how the pressure, p varies with the
temperature T and the molar density n. The molar density tells how many moles
we have per unit volume. n is just the sum of the molar density of water vapor nv
and the molar density of the rest of the air, nd . We can divide p into the partial
pressure excerted by the vapor molecules, e, and the partial pressure excerted by the
rest of the air, pd . R is the universal gas constant.
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b)
ε=
Mv
= 0.622
Md
Answer: The ratio of molar masses of water and dry air. We can see that the
molecular weight of water is lower than the molecular weight of dry air.
c)
−
dT
Mair g
= Γd
|dry =
dz
cp
Answer: This is the dry adiabatic lapse rate. Is says how the temperature decreases
with height when an unsaturated air parcel moves up adiabatically in the atmosphere.
From earlier courses, you’ve learned that it is given by g/c0p , where c0p is the mass
based specific heat for air (= 1004). This is the same as cp · Mair , since Mair is the
molecular mass of air, and cp is the molar heat capacity.
d)
−
dT
Mair g
1 d
|saturated =
+
(yv lv )
dz
cp
cp dz
Answer: This is the moist adiabatic lapse rate. The first term on the right side
expresses the same as the dry adiabatic lapse rate. The second term expresses the
additional heating/cooling we get when vapor/liquid inside the air parcel condeses/evaporates. yv is the vapor mole fraction and lv is the latent heat of vaporization.
We can see that when the vapor mole fraction decreases with height (as it does when
a saturated air parcel is lifted and vapor condenses into liquid), the second term at
the right side of the equation becomes negative. This means that the temperature
won’t decrease that much with height as it would if there where no condensation
(and this makes sense since condensation releases energy).
Problem 1 - Skew T - ln p chart
In this problem, you will get use of a skewT - lnp Chart. You find this at the web site of
the course under ”Schedule for the group sessions”.
We will look at an air parcel containing 1 g of water vapor and 400 g of dry air. The
temperature of the air parcel is 5.0 ◦ C, and it is located at 900 hPa. Give a definition
and find the following quantities
a) Mixing ratio, rv . Answer: The (mass) mixing ratio of water vapor, rv (sometimes
called w), is given by
rv =
ρv
nv M v
nv
qv
=
=ε
=
,
ρd
nd M d
nd
1 − qv
2
where ρv is the density of water vapor, ρd is the density of the dry air and qv is
the specific humidity, Md and Mv is the molar masses of dry air and water. We
calculate it for this case:
rv =
ρv
1 g/volume
1g
=
=
= 2.5 g/kg
ρd
400 g/volume
400 g
b) Specific humidity, qv . Answer: The specific humidity, qv , is given by
ρv
qv =
ρair
where ρv is the density of water vapor and ρair is the density of the air (including
the water vapor aswell). Since we don’t have that much water vaport inside our air
parcel, qv ≈ rv = 2, 5 g/kg
c) Saturation mixing ratio, rsf c . Answer: Saturation mixing ratio, rsf c is the mixing
ratio we would have needed at the given temperature and the given pressure for the
air parcel to be saturated. We find this by 1) draw the horizontal line for the given
pressure. 2) draw the strait, tilted line for the given temperature. 3) The intersection
between the line gives a point, P , that tells us the state of the air parcel. 4) Find
out which mixing-ratio-line (dashed line named w) that goes through this point. By
doing this for our parcel, we see that rsf c = 6 g/kg.
d) Dew point temperature, Td . Answer: The dew point temperature is the temperature
a moist air parcel needs to be cooled at constant pressure in order to bring the air
to saturation. We find this by 1) Draw the strait, tilted line that corresponds to the
mixing ratio of the air parcel (inthis case: 2.5 g/kg). 2) Start in the point P we
found in c). 3) Follow the dry adiabat from P until it intersects with the mixing
ratio line. 4) Follow the mixing ratio line down to the initial pressure. 5) Read
out the temperature by see which strait, tilted temperature line that intersects this
mixing ratio line at the initial pressure level. By doing this for our parcel, we see
that Td = −7 ◦ C.
e) Lifting condensation level, LCL. Answer: LCL is the height/pressure level we need
to lift a moist (but not saturated) air parcel to make the air parcel saturated. We
find this by 1) start in P . 2) Follow the dry adiabat until it intesects the mixing
ratio line you drew in d). 3) Read out the pressure level/height this corresponds to.
By doing this for our parcel, we see that LCL = 760 hP a (≈ 2.2 km).
f) Potential temperature, θ. The potential temperature, θ is telling how large the temperature of an unsaturated air parcel would be if it was brought adiabatically down
to the reference pressure, p0 . It is given by
R/cp
p0
θ=T ·
p
We find this by 1) start in P. 2) Follow the dry adiabat down to the reference
pressure (p0 = 1000 hPa). By doing this, we find that θ = 14 ◦ C.
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g) Wet bulb temperature, Twb . Answer: The wet-bulb temperature is the temperature
we will get if we used energy from the air parcel to evaporate the amount of water
needed to make the air parcel saturated. We find this by 1) start in P. 2) Follow the
dry adiabat up to LCL (saturation). 3) Follow the moist adiabat down to the initial
pressure level. 4) Read out the temperature by see which strait, tilted temperature
line this point is at. By doing this for our parcel, we get that Twb = 0 ◦ C.
h) Equivalent temperature, Te . Answer: The equivalent temperature, Te , is the temperature we would get if all the water vapor inside a moist air parcel condensed and
rained out, but the released energy by the condensation is kept inside the parcel. We
find this by 1) start in P. 2) Follow the dry adiabat up to LCL. 3) Follow the moist
adiabat up until it is parallell to the dry adiabat. 4) Follow the dry adiabat down to
the initial pressure level. 5) Read out the temperature. By doing this for our parcel,
we see that Te = 11 ◦ C.
i) Equivalent potential temperature, θe . Answer: This is almost the same as Te , but
we need to go all the way down to the reference pressure (p0 = 1000 hPA). By doing
this for our parcel, we find that θe = 20 ◦ C.
j) Saturated equivalent potential temperature, θes . Answer: This is the equivalent
potential temperature the parcel would have if it was saturated. We find this by 1)
start in P. 2) Follow the moist adiabat up until it is parallell with the dry adiabat. 3)
Follow the dry adiabat down to the reference pressure. 4) Read out the temperature.
By doing this for our parcel, we find that θes = 31 ◦ C.
Problem 3 (related to Figure 5.2 in the book)
a) What is expressed by the equation below?
1 + rεv
Tv = T ·
1 + rv
Answer: The virtual temperature, Tv , of a moist air parcel tells us what the temperaure of an equally dense, but dry parcel needs to be at the same pressure. Since
drier air parcels are heavier (according to the equation of ε) above, we need the
temperature to be higher (since warmer air parcels are lighter) to make them equally dense. That’s why Tv always is larger than T . The facors in the equation are
explained in the other equations. We can see that the expression in the bracket is
larger than 1 since ε < 1. Maybe this explanation is easier to understand: you have
two parcels at the same pressure level: one moist and one dry. The moist parcel is
less dense because of the different molecular weight of water and dry air. To make
the dry parcel equally dense, it needs to be lighter. One way to make it lighter is
to increase the temperature (remember that warmer air parcels will rise because of
the lower density). The temperature it needs to be increased to is called the virtul
temperature.
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b) What is expressed by the equation below?
R/cpd
p0
θv = Tv
p
Answer: This is the virtual potential temperature. It is alomost the same as the
potential temperature. The difference is that we are using the virtual temperature
insted of the absolute temperature and the specific heat of dry air instead of the
molecular specific heat in the calculations. The reson for doing this is the increased
amount of water vapor as we move further down in the boundary layer. The water vapor contributes to a change in the heat capacity because of its difference in
molecular weight compared to that of dry air. This means that cp changes all the
way down in the boundary layer. By using the virtual temperature, one can use the
same heat capacity all the way down, which makes the calculations easier.
c) Why is the difference between T and Tv larger in the boundary layer than higher
up in the free troposphere? Answer: Because we have more water vapor in the
boundary layer. This gives us larger valus of rv , and thus larger values of Tv .
d) What is the advantage of using the virtual temperature? Answer: When using Tv
instead of T , we can use the ideal gas law for dry air when doing calculations on
moist air. It is also an advantage when calculating potential temperature (virtual
potential temperature) because we can use the same heat capacity for all the different
heights.
Figur 1: Figure 5.2 from the book
Problem 4 (related to Figure 5.1 in the book)
a) What is Tf , how is the value of Tf compared to Td , and why is that so? Answer:
Tf is the frost-point temperature. If we don’t get saturation before we lower the
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Figur 2: Figure 5.1 from the book
temperature below the ice point T0 = 0 ◦ C, the water vapor can depose on ice rather
than condense on water. We know that the saturation vapor pressure over ice is
lower than that over water, meaning that we don’t need to lower the temperature
that much to get saturation. Beacuse of this, Tf is always larger than Td .
Problem 5 - Buoyancy
a) In the book, you can see that the bouyancy is given by the net buoyancy force per
unit mass, and we can see this calculation:
FB
ρp V
(ρp − ρ)
= −g
ρp
Tvp − Tv
=g
Tv
B=
6
How do you get from the second last line to the last line? Answer: Use the ideal
gas law for moist air: p = ρair Rd Tv and get this:
p
p !
(ρp − ρ)
Rd Tvp − Rd Tv
= −g
−g
p
ρp
Rd Tvp
!
1
1
Tvp − Tv
= −g
1
Tvp

= −g 
Tvp
Tvp
−
Tvp
Tv


Tvp
Tvp
1−
= −g
Tvp
Tv
!
1
Tv − Tvp
= −g
Tv
Tvp − Tv
=g
Tv
b) Mark the right option: ”If the density of the parcel exceeds that of its environment,
that is ρp > ρ, Tvp is smaller than Tv , the buoyancy force is directed downward,
we have that B is negative and the parcel wil experience a downward acceleration.
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