Answer:
(a). The relative humidity is 20.57%
(b). The dew point is 79.114°F
(c). It take for the air to reach saturation is 17 hours.
(d). The water is 7.6g/kg.
Explanation:
Given that,
Temperature = 35°
Mixing ratio = 7.6 g/kg
(a). We need to calculate the relative humidity
Using formula of relative humidity
[tex]relative humidity=\dfrac{mixing\ ratio}{36}\times100[/tex]
[tex]relative humidity=\dfrac{7.6}{36.94}\times100[/tex]
[tex]relative humidity=20.57\%[/tex]
(b). We need to calculate the dew point
Using formula of dew point
[tex]dew\ point=T^{\circ}F-\dfrac{(100-relative\ humidity)}{5}[/tex]
Put the value into the formula
[tex]dew\ point=95-\dfrac{100-20.57}{5}[/tex]
[tex]dew\ point=79.114^{\circ}F[/tex]
(c). If the room temperature decreases by 5°C per hour
We need to calculate the time
Using formula of time
[tex]t=\dfrac{T_{a}-increase\ temperature}{5}[/tex]
Put the value into the formula
[tex]t=\dfrac{95-10}{5}[/tex]
[tex]t=17\ hours[/tex]
(d). If the temperature of the room continues to decrease for one more hour,
We need to calculate the how many grams of water vapor
Using given data
The water of 7.6g/kg vapour will have to condense out of air for the maintain relative humidity at 100%
Hence, (a). The relative humidity is 20.57%
(b). The dew point is 79.114°F
(c). It take for the air to reach saturation is 17 hours.
(d). The water is 7.6g/kg.
Answer:
temp: 35° c (95°f)
MR:7.6
SMR: 36.5
RH: 21% (20.82)
Dew Point: 10° C
5°c per hour: 5 hours for the 35°c air temp to reach the 10°c dew point
1 more hour: Temperature will decrease to 5°c, half the water molecules dewpoint "size" I view it as, and 3.8g/kg of water vapour has condensed out
so
1. 21%
2. 10° C
3. 5 hours
4. 3.8g/kg
