Question

1. (relative humidity) The partial pressure of water vapor in air is usually less than the equilibrium vapor pressure (or saturation pressure) at the ambient temperature; this is why a cup of water will spontaneously evaporate. The ratio of the partial pressure of water vapor to the equilibrium vapor pressure (at the same temperature) is called the relative humidity. Saturated air, at 100% relative humidity, would be in equilibrium with a cup of liquid water. The saturation pressure depends strongly (in the exponent) on temperature: T(°C) Ps (kPa) 0 0.61 5 0.87 10 1.23 15 1.71 20 2.34 25 3.17 30 4.25 35 5.63 Answer the following short questions; try to be brief and to the point. (a) The dew point is the temperature at which the relative humidity would be 100%, for a given partial pressure of water vapor. The temperature in a certain Fall day is 25°C. What is the dew point if the relative humidity is 80%? 50%? (Interpolate if needed.) (b) Suppose the dew point on a given night is 10°C, and the temperature drops from 10°C to 5°C. Compare the energy released by water condensation with that released by the cooling air. (Relevant numbers are AHvap = 44.6kJ/mol for water and Cp = 1kJ/kg-K for dry air.) How would that affect the cooling rate? (c) Can there be humidity (presence of H2O molecules) in air at -10°C? Justify using the phase diagram of water. If yes, can you say roughly how much?

Answer 1

1.a) Dew point and relative humidity are related to each other by the following expression:

$Td = T- \frac{(100 - RH)}{5}$

where, Td is the dew point temperature (^oC) , T is the known temperature (^oC), and RH is the relative humidity

So, from the given values, at RH= 80%

$Td = 25- \frac{(100 - 80)}{5}$

= 21^oC

Similarly at RH= 50%

$Td = 25- \frac{(100 - 50)}{5}$

=15^oC

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