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 5 103 8.039 0.622 T 4C 12. T = 25°C, find ws es (25C ) 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 (30C ) 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 11.5C 16. w = 20 g/kg, find TD es (TD ) p wp .02 1000 es (TD ) 32.15 0.622 TD 25.5C w ws (TD ) 17. TF = –10°C, find TD e esi (TF ) 2.597 2.597 es (TD ) TD 11.5C 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.