The air inside a room is at a temperature of 35C and has a mixing ratio of 7.6 g/kg. Answer the following in (S. I. UNITIS):
(a) What is the relative humidity?
(b) What is the dew point?
(c) If the room temperature decreases by 10°F per hour, how many hours will it take for the air to reach saturation?
(d) After reaching saturation, if the temperature of the room continues to decrease for one more hour, how many grams of water vapor (per kg of air) will have had to condense out of the air to maintain a relative humidity of 100%?

a. Relative Humidity = 20.57%

b. Dew point = 79.114° F

c. Time required for the air to reach Saturation point = 17 hours.

d. 7.6g/kg of water vapour

Explanation:

a. Relative Humidity

The formula for Relative Humidity is

(Mixing ratio ÷ Saturation mixing ratio) × 100

In the question

Mixing ratio = 7.6g/kg

Temperature = 35°C

Saturation mixing ratio at 35°C is given in Saturation mixing ratio Table = 36.94

Relative Humidity = (7.6g/kg ÷ 36.94) × 100

Relative Humidity = 20.57%

b. Dew point

Formula for Dew point =

T °F - (100 - Relative Humidity) ÷ 5

Temperature = 35°C

Conversion to Fahrenheit =

T°F = T°C × 1.8 + 32

35°C × 1.8 + 32

95°F

Dew point = 95°F - ( 100 - 20.57)÷5

95°F - 15.886

Dew point = 79.114° F

c. At Saturation Point, the Relative Humidity is always equal too 100%

In the question above, Room temperature decreases by 10°F per hour.

Time required for the air to reach Saturation point =

Time(hours) = (Temperature of the air in the room - Decrease in temperature) ÷ 5

Time(hours) = (95°F - 10°F) ÷ 5

= 85°F ÷ 5

= 17 hours

Therefore, the time required for the air to reach Saturation point = 17 hours.

d. It would take 7.6g/kg of water vapour will have to condense out of air to maintain the relative humidity at 100% .

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