Thermodynamics of Atmospheres and Oceans
Worksheet 13 (6.1)
1. The enthalpy of a water cloud is
dH = md c pd + mv cpv + m l c l dT + Llv dm v
where the subscripts d, v, and l refer to dry air, water vapor and liquid water,
respectively. Which terms in the parentheses are usually neglected and why?
mv c pv
because both m’s are significantly smaller than md
ml cl
3. Write the entropy equation for dry air
d  c pd d (ln T )  Rd d (ln pd )
4. Consider the entropy equation for a cloud
d = c pd + wt cl d(ln T)  R d d(ln pd ) + d
L lvwv
A lv
+ wv d
T
T
Describe the terms that are different from #3.
L w 
d  lv v  accounts for change in latent heat of vaporization times water vapor mixing
 T 
A 
ratio for a given temperature change, wv d  lv  applies the water vapor mixing ratio
 T 
times the change in affinity for vaporization for a given temperature change, wt cl gives
the total water vapor mixing ration times the specific heat of liquid and in general alters
the general specific heat term
Isobaric cooling
5. During the isobaric cooling of unsaturated air, the relative humidity (increases,
decreases, remains the same).
6. During the isobaric cooling of saturated air, the relative humidity (increases,
decreases, remains the same).
7. As the dew point depression decreases, relative humidity (increases, decreases,
remains the same)
8. The frost point temperature is (greater than, less than, equal to) the dew point
temperature
9. During the isobaric cooling of saturated air the temperature decrease of the air is
proportional to the cooling (true, false)
10. During reversible condensation (i.e. no water falls out), which of the following is
true
a) dwl < -dws
b) dwl = -dws
c) dwl > -dws
For a pressure of 1000 hPa, determine the following. You may use the es Table in the
Appendix D. Given:
11. ws = 5 g kg-1, find T
ws  
es 
es
p
ws p


1000  5 103
 8.039
0.622
T  4C
12. T = 25°C, find ws
es (25C ) 31.668

0.622  0.02
p
1000
ws  20 g / kg
ws  
13. T = 30°C and w = 15 g/kg, find H
w
15
0.015


 0.568
ws es (30C )  42.421 0.622
1000
p
H  56.8%
H
14. T = 20°C and TD = 15, find H
H
es (TD ) 17.042

 0.729
es (T ) 23.371
H  72.9%
15. T = 15°C and H = 0.8, find TD
es (TD )  H  es (T )  0.8 17.042  13.6336
TD  11.5C
16. w = 20 g/kg, find TD
es (TD )
p
wp .02 1000
es (TD ) 

 32.15

0.622
TD  25.5C
w  ws (TD )  
17. TF = –10°C, find TD
e  esi (TF )  2.597
2.597  es (TD )
TD  11.5C
Cooling and Moistening by Evaporation of Water
18. Write the first law of thermodynamics, enthalpy form, intensive, for an adiabatic
isobaric process that includes evaporation of water
dh  c p dT (2.16)
dh  c p dT  Llv dws
c p dT  dq  vdp (2.18b)
c p dT  Llv dws  dq  vdp dq  0
c p dT  Llv dws  vdp
19. The wet bulb temperature is (greater, less) the (physical) temperature
20. The wet bulb temperature is (greater, less) than the dew point temperature.
21. What is the temperature of a ran drop falling through an unsaturated layer
a) T, the temperature of the air
b) Tw, wet bulb temperature
c) Td, dew point temperature
d) none of the above
Saturation by Adiabatic Isobaric Mixing
Describe how the mixing of two unsaturated parcels can result in saturation.
The non linear nature of the vapor pressure curve and the linear nature of parcel
mixing allows for cases where we have two parcels with high pressure and temperature
gradients to mix in a position that slips behind the curve allowing for condensation to
occur.