Imagine a room, 3m x 4m x 2.5m say (volume 30m2).
Imagine that this room is perfectly sealed from the outside. ie
- No air can enter or leave
- No heat can enter or leave through the structure
The room is at 35oC and the air is completely dry.
Imagine a room, 3m x 4m x 2.5m say (volume 30m2). Imagine that this room is perfectly sealed from the outside. ie
The room is at 35oC and the air is completely dry. | | ||
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| Let 1 Tablespoon of water (15ml) be placed in the room. After some time all the water will have evaporated. The room temperature is now 34oC. | |||
The evaporation of the water requires heat. As no heat can pass through the walls the heat must be found from the air itself. The amount of heat released from 30m3 of air when it cools by 1oC is approximately the same as that required to evaporate 15g of water. |  | ||
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| More tablespoons of water can be added to the room, each reducing the temperature by 1oC upon evaporation. | |||
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| Similarly, at other starting temperatures, the result would be the same. | |||
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| Every point on a line represents an air/water vapour mixture with the same heat content Enthalpy. By coincidence the enthalpy of each line (in kJ/kg of dry air) has approximately the same value as the dry air temperature. | |||
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| The idea can be extended further and used for any size room by dividing the mass of air in the room. This ratio is called the Specific Humidity and represents the proportion of water vapour to air. | |||
Unit Mass of Air |
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| However, we can't continue to add water vapour forever and at some point the air will not hold any more. At this point it is saturated. The higher the air temperature the greater the amount of water which will vapourize. | |||
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| Thus, by saturating dry air at 35oC in a room completely 'heat sealed' the temperature will drop to 12.4oC. ie by 22.6oC. | |||
| Dry: 35oC Wet: 12.4oC Drop: 22.6oC Dry: 40oC Wet: 14.3oC Drop: 25.7oC |  | ||
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| Other saturation points may be added for other lines of constant enthalpy. | |||
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| The saturation temperatures are referred to as Wet Bulb Temperatures. Any air/water vapour mix with enthalpy 35kJ/kg has a Wet Bulb temperature of 12.4oC. | |||||
| Enthalpy: 35kJ/kg WBT: 12.4oC Enthalpy: 40kJ/kg WBT: 14.3oC | | ||||
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| Relative Humidity | |||
| For an air temperature of 11oC, saturation occurs when there is 8 grammes of water vapour to every kilogramme of air (approx 1m3). | |||
| Relative Humidity is effectively the ratio: mass of water vapour in the air: mass their would be when the air is saturated at that temperature. | |||
| At 11oC and SH=6g, 4g, 2g/kg RH=75, 50, 25% |
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| Continuing by marking in each quarter value of specific humidity. | |||
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| Joining up the points and we now have a, though incomplete, moderately useful Psychrometric Chart. | |||||
 | Remember the scale on the Vertical Lines IS linear. The Scale on the slanted lines IS NOT linear. | ||||
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| Determining RH | |||
| Given a Wet-Bulb Temperature of 15oC and Dry Bulb of 23oC, determine the relative humidity. | |||
| 1. Draw vertical line for 23oc. 2. Draw Line for 15oC. 3. At cross with Saturation Line Draw enthalpy line. 4. At cross with upper temperature, interpolate on Vertical Temperature Line. |  | ||
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| Determination of Dew Point Temperature. | |||
| The temperature for which the Specific Humidity represents a relative humidity of 100% ie Condensation. | |||
| At What temperature will air at 21oC, RH 36% be saturated? | |||
 | 1. Draw Vertical Line for 21oC. 2. Mark the point which represents 36% on this line. 3. Draw a horizontal Line from this point (constant Specific Humidity). 4. Draw Vertical Line where this joins Saturation Line and read off Dew Point Temperature. | ||
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| Dew Point Determination | |||
| It is important to assess where condensation could arise within the construction. | |||
| A steady state representation of Temperature Distribution in a wall section would look like this for a worst case of -5oC external temperature. It can be seen that condensation could arise within the insulation for a Dew Point Temperature of 6.5oC.For a breathing wall this would be a major problem. |  | ||
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| Evaporative Cooling | |||
| Two Main Categories of Evaporative Cooling | |||
| Direct | Evaporation Takes Place outside or inside the building reducing the temperature of the air entering or already inside the building |  | |
| Indirect | Evaporation Takes Place Outside the building away from vents and cools external surfaces |  | |
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| Climate Evaluation | |||
| Analysis of climate over entire cooling period is necessary to determine if evaporative cooling will be appropriate. | |||
| Example, Phoenix, Arizona Summer Solstice (21st June) Source: Meteonorm | |||
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| Using the Psychrometric Chart find the Wet Bulb Temperatures! | |||
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| Cooling Potential | |||
| A 100% efficient cooler could convert this air: | |||
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| Into this air: | |||
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| Direct Evaporative Cooling: Closed Facade incorporating Wetted Pads | |||
Double skin facade with outer constructed of an highly breathable material which is kept constantly wet. Windows incorporated in outer facade for daylighting/view. Exhaust Fan necessary to 'force' air through. |  | ||
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| Direct Evaporative Cooling: Cunningham/Thompson Tower | |||
Air enters through top of tower. Air takes on moisture and descends tower. Flow enhanced by addition of solar chimney. |  | ||
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| Indirect Evaporative Cooling: Principles Inside Cooler than Outside | |||
| Temperature Gradient across section of external structure in Hot Arid Climate: |  | ||
| Cooling External Surface, maintains gradients, reducing internal temperatures. |  | ||
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| Indirect Evaporative Cooling: Principles Outside Cooler than Inside | |||
| Temperature Gradient across section of external structure in Hot Arid Climate: | < | ||
| Cooling External Surface, maintains gradients, reducing internal temperatures. |  | ||
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