Humidity Problem Solving Practice Solving Problems using the PRS.

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Humidity Problem Solving Practice Solving Problems using the PRS

You will need… 1) A calculator 2) Your RH & Capacity Charts 3) Your clicker

Scenario #1 On a January morning you wake up for school and take a warm shower. The warm water quickly adds water vapor into the air. Your home’s thermostat is set to 68°F (20°C), but the outside air temperature is only 16°F (-9°C) —which means that the inside surface of the window in your bathroom is very cool— 38°F (3°C).

Your home’s thermostat is set to 68°F (20°C), but the outside air temperature is only 16°F (- 9°C) —which means that the inside surface of the window in your bathroom is very cool—38°F (3°C). Given this information, how much water vapor could possibly be in the air in your bathroom? (give your answer in g/kg)

Solution Capacity is dictated by the air temperature. Plug the air temperature into the capacity chart to find the answer. Temp = 68°F so… (68-32) *5/9 = 20°C 20°C has a capacity of 15g/kg

Now, suppose that the moisture from the shower saturates the air in the bathroom… What will happen when this air contacts the surface of the window in the bathroom? 1)Nothing4) Sublimation 2)Evaporation 5) Condensation 3)Deposition6) Freezing

Now, suppose that the moisture from the shower saturates the air in the bathroom… What will happen when this air contacts the surface of the window in the bathroom? 1)Nothing4) Sublimation 2)Evaporation 5) Condensation 3)Deposition6) Freezing

Given that your home’s thermostat is set to 68°F (20°C), but the outside air temperature is only 16°F (-9°C) —which means that the inside surface of the window in your bathroom is very cool—38°F (3°C). And that the shower has saturated the air in the room… How much condensation will take place on the window? (give your answer in g/kg)

Solution Find the specific humidity of the air and compare it to the capacity of the window’s inside surface. Window surface temp = (38-32)*5/9 = 3.3°C Since SH = Cap. ; the SH = 15 g/kg Air at 3°C has a Cap. Of 4.8 g/kg Thus the amount of condensation will be… 15 – 4.8 = 10.2g/kg

When you get out of shower, you realize you left your towel in your bedroom. As you leave the bathroom to get it you feel very cold since… 1)The air temp. is only 68°F 2)Evaporation takes energy from you to break hydrogen bonds 3)Evaporation takes energy from you to break covalent bonds 4)Evaporation takes energy from you to form hydrogen bonds 5)Evaporation takes energy from you to form covalent bonds

When you get out of shower, you realize you left your towel in your bedroom. As you leave the bathroom to get it you feel very cold since… 1)The air temp. is only 68°F 2)Evaporation takes energy from you to break hydrogen bonds 3)Evaporation takes energy from you to break covalent bonds 4)Evaporation takes energy from you to form hydrogen bonds 5)Evaporation takes energy from you to form covalent bonds

Scenario #2 One October day the air temperature gets up to a high temperature of 62°F (17°C) and cools down overnight so that the air is 35°F (2°C) the next morning. You happen to live right next to the Farmington River which according to a local website has a fairly consistent temperature of 50°F (10°C) at this time of year.

Given this situation, what do you expect would happen? One October day the air temperature gets up to a high temperature of 62°F. After cooling overnight, the temperature early the next morning reads 35°F. You happen to live right next to the Farmington River which according to a local website has a fairly consistent temperature of 50°F at this time of year. 1.Fog will be seen over the land during the middle of the day 2.Fog will be seen over the river during the middle of the day 3.Fog will be seen over the land early the next morning 4.Fog will be seen over the river early the next morning

Given this situation, what do you expect would happen? One October day the air temperature gets up to a high temperature of 62°F. After cooling overnight, the temperature early the next morning reads 35°F. You happen to live right next to the Farmington River which according to a local website has a fairly consistent temperature of 50°F at this time of year. 1.Fog will be seen over the land during the middle of the day 2.Fog will be seen over the river during the middle of the day 3.Fog will be seen over the land early the next morning 4.Fog will be seen over the river early the next morning

WHY? The warmer water will evaporate into the air at a relatively fast rate. However, the air above the river is not warm enough to keep this moisture in the vapor phase—so it condenses back into a liquid in the air.

One October day the air temperature gets up to a high temperature of 62°F (17°C). After cooling overnight, the temperature early the next morning reads 35°F (2°C). You happen to live right next to the Farmington River which according to a local website has a fairly consistent temperature of 50°F (10°C) at this time of year. If the river evaporates H 2 O such that it puts 6.8g/kg of water vapor in the air, how much water vapor will condense by the next morning? (enter your answer in g/kg)

Solution: Daytime temp = (62-32)*5/9 = 16.67°C (17°C) Morning temp = (35-32)*5/9 = 1.67°C (2°C) SH = 6.8g/kg Capacity of daytime air = ~ 12.3 g/kg so there is no condensation during the day Capacity of morning air = ~ 4.4 g/kg, so 2.4 g/kg of moisture condense by morning.

Scenario 3 This morning you watched the weather forecast and the weatherman said that today the air would reach a high temperature of 90°F (32°C) and have a dew point of 73°F (23°C). What is the specific humidity of this air? (enter your answer in g/kg)

Solution: If given a dew point, you have a direct indicator of SH. This is true since… Dew pt is the temp. when air is saturated When air is saturated, SH = Capacity Thus plug DP temp. into the capacity chart and you will find the SH of that air. DP = (73-32)*5/9 = 22.78°C (23°C) This means that the SH = 18.1 g/kg

This morning you watched the weather forecast and the weatherman said that today the air would reach a high temperature of 90°F (32°C) and have a dew point of 73°F (23°C). What is the relative humidity of this air? (enter your answer in percent)

Solution: RH = SH/C Air temp = (90-32)*5/9= 32.2 °C (32°C) Capacity of 32°C= 31.2g/kg SH = 18.1 g/kg RH = 18.1/31.2 = 58%