Sample exam 2
Part 2 •
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Exercise 8 Thermodynamic Diagram
Given the thermodynamic diagram of Lindenberg (Germany), taken July 26, 2008 (Appendix).
10p. 8a.
Has the weather balloon been launched at night or during the day? How is this visible
in the diagram?
10p. 8b. What was the depth of the atmospheric boundary layer in the previous day? Motivate
your answer.
8c.
Motivate the answer of the following questions using either a construction or a
calculation. Considering the air parcel at 400hPa, what is its:
5p.
1. Dew point temperature
5p.
2. Relative humidity
5p.
3. Virtual temperature
5p.
4. Potential wet bulb temperature
5p.
5. Lifting Condensation Level
5p. 8d. Determine the stability of the layer between 850 hPa and 780 hPa?
10p. 8e.
How many °C must the surface temperature rise in order to form a Cumulonimbus
cloud. Motivate your answer by a construction.
a) At night, since the temperature increases with height going up from the surface. This
indicates the radiation inversion.
b) 780 hPa. At this pressure the actual temperature increases with height more quickly than
the dry-adiabates. Also from that level the dew point temperature (red dashed line) start to
drop, indicating the dry free atmosphere.
c1) -45 ◦C. The dew point temperature is indicated by the red dashed line. At the 400 hPa line,
this point is exactly between the -50◦C and the -40◦C, thus Tdew= -45◦C.
c2) RH=q/qs= 0.18/1.2 = 15%. q is the actual mixing ratio and can be read from the isohume
that intersects the dew point temperature (red dashed line) at 400 hPa and amounts to 0.18
g/kg. qs is the saturated mixing ratio and can be read from the isohume that intersects the
temperature (full red line) at 400 hPa and amounts to 1.2 g/kg.
c3) Tv=T(1+0.61*q) = (-25+273.15)*(1+0.61*0.18.10-3)= 248 K of 25 ◦C. Note that Tv cannot be
read from the diagram directly. Note that T should be denoted in Kelvin and q in kg/kg.
c4) +18◦C. See the construction below. From the dewpoint at 400 hPa you follow the isohume
upward. Also, from the temperature you follow the dry adiabat upward until both intersect.
This is the LCL (see question c5). From the LCL you follow the moist adiabat downward to the
surface.
c5) 280 hPa (see construction and c4)
d) Conditionally stable. The observed temperature gradient (full red line) at 850 hPa has a
slope between the dry and moist adiabat. Therefore the layer is conditionally stable.
e) To 29.6 ◦C, so temperature change is about 8 K. See the construction: take the dew point at
the surface, and follow its isohume until it intersects the observed temperature (full red line).
From that point go dry adiabatically down to the surface.
Exercise 9 Hole in the cabin
On July 25, 2008 a hole in the cabin of a Quantas plane of type Boeing 747-400ER was formed
during a flight between Hong Kong to Melbourne (See photo below). During the accident, the
plane flew at an altitude of about 11 km, with an outside air temperature of −32.3ºC and a dew
Sample exam 2
Part 2 • Open Book
point temperature of −66.3ºC. Within the plane the temperature was 18ºC with a relative
humidity of 60%. Usually, the cabin pressure is kept constant at 850 hPa.
Also some technical specifications of the B747-400ER are given.
Basic Dimensions
Wing Span
Overall Length
Tail Height
Interior Cabin Width
5p.
9a.
5p.
5p.
9b.
9c.
211 ft 5 in (64.4 m)
231 ft 10 in (70.6 m)
63 ft 8 in (19.4 m)
20 ft (6.1 m)
Calculate the outside pressure at cruise height. Assume an atmospheric scale height of
7400 m.
Calculate the specific humidity at cruise height.
Idem for the air within the cabin.
Due to the hole in the cabin, the pressure in the cabin dropped rapidly, and oxygen masks
were provided to the passengers.
10p. 9d.
Calculate or determine the percentage available oxygen compared to the normal
situation.
Just after the hole was formed, fog was observed in the cabin. Student Gert-Jan suggests that
this was due to decompression, but student Bert claims fog occurs due to mixing of
atmospheric air and cabin air.
15p. 9e.
15p. 9f.
Calculate whether the fog occurs due to mixing of atmospheric air and cabin air.
Assume air mixes in a ratio as follows outside air: cabin air is 1:4.
Calculate whether the fog could occur due to decompression.
a) Use the barometric height equation (eq. 1.8 and eq. 3.27 in WH06).
p(11km)=1013.25*exp(-11000/(29.3*253)) = 229 hPa.
b) q= εe/p. ε is a constant 0.622. The vapor pressure can be calculated with Tetens formula by
applying the dew point temperature (-66.3 ºC) as input, i.e. e=611*exp(b(T-T1)/(T-T2)) =
611*exp(b(-66.3+273.15-T1)/(-66.3+273.15-T2)). Consequently, q= εe/p = q=
0.622*611*exp(b(T-T1)/(T-T2))/22900= 2.0.10-5 kg/kg.
c) First calculate the saturated specific humidity based on the temperature in the cabin. qs=
ε*e/p=0.622*611*exp(b(18+273.15-T1)/(18+273.15-T2))/85000 . Then continue with
q=qs*RH = 0.00906 kg/kg
d) Normally pO2 = 0.2* 850 hPa. Due to the event, the O2 concentration of the outdoor air
will also apply inside. The new pO2 is 0.2*229 hPa. Thus the relative contribution amounts
to 229/850 = 0.269 = 27 %
e) Tmix = (Toutside+4Tcabin)/5 => Tmix= 7.94 ºC.
qmix = (qoutside+4qcabin)/5= > qmix= 0.00725 kg/kg. qoutside and qcabin are obtained from b) and c)
qsat(Tmix) = 0.031489 kg/kg. Use the Tetens formula and apply it to Tmix to determine the
saturated mixing ratio of the mixture.
qmix < qsat(Tmix), this no fog.
Sample exam 2
f)
Part 2 • Open Book
Before the pressure drop p = 850hPa, T =18ºC, e = 1238 Pa, ρ = 1.017 kg/m3. After the
pressure drop p =229 hPa. Since the process goes very fast, you can assume an adiabatic
Rd
0.286
 p2  C p
 229 

process, and eq. 3.8 applies. T 2 = T 1
= (18 + 273.)
= 200 K =-73.2ºC.

 850 
 p1 
Then calculate the qsat for this new temperature using Tetens: qsat(-73.2ºC) = 1.0.10-3 kg/kg.
It appears that qsat(-73.2ºC) << qcabine. Hence fog will appear.
Exercise 10: Convective boundary layer growth
The growth of the atmospheric boundary layer (ABL) during the daytime is very important to
estimate air pollution concentrations at the end of the afternoon. In this exercise, we estimate
the ABL depth, for a case with a constant surface sensible heat flux, H(t) = 50 W/m2, and for a
case with a diurnal cycle:
100
. Assume ρ =1 kg/m3, and Cp =1000 J/kg/K.
Figure 5: Diurnal cycle of sensible heat flux; grey: constant in time black: varying in time.
15p. 10a. Calculate the accumulated heat (in J/m2) for the constant H case (H=50 W/m2).
20p. 10b. Calculate the accumulated heat (in J/m2) for the diurnal cycle case.
Hint: sin2(x)= 0.5*(1-cos(2x)).
The early morning sounding reads as follows the vertical profile
10
20p. 10c. Calculate the PBL height at the end of the day, assuming no entrainment, assuming the
heat input as calculated in 10b. If you did not find the answer of 10b, use 1000 J/m2 for
the accumulated heat.
10a: Qak=50*40000=2 MJ/m2.
10b:
100
100
50 &
1
1
2
4"
$
80000
'
= 2 MJ/m2
Sample exam 2
10c: Part 2 • Open Book
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Exam February 2012
Name:...........................................................................
Registration number:..................................................
LCL
Appendix 1