Document Sample

WARNING THIS MATERIAL IS COPYRIGHT PROTECTED PSYCHROMETRICS A. Psychrometric Information 1. Psychrometric psychro – meaning ‘cold’ metrics – meaning ‘measure of’ Actually, psychrometrics is more than the measurement of cold. It is a study of all the properties of moist air. AIR DRY WET 78% Nitrogen 78% Nitrogen 20.9% Oxygen 20.9% Oxygen 1% Argon 1% Argon .1% Other .1% Other Gases Gases PLUS Water Vapor 2. Atmospheric Air The weight of air pushing down on the earth is referred to as atmospheric pressure. At sea level, the pressure of 70o dry air is 14.696 lbs/in2(psi). 3. Standard Air At a barometric pressure of 29.921 inches of mercury (14.696 psi), one pound of 70o dry air will occupy 13.33 cubic feet. Air at these conditions is known as standard air. 4. Specific Density 1 Specific Density = ------------------------ Specific Volume 1 = --------- = .075 lbs/ft3 13.33 5. Relationship between Specific Volume and Specific Density 5. Relationship between Specific Volume and Specific Density Specific Volume Specific Density 6. Sensible Heat If we wish to calculate the Btu’s needed to raise the temperature of dry air, we would use the sensible heat formula. Sensible Heat Formula Btu/hr = Sp. Heat x Sp. Density x 60 min/hr x cfm x ΔT Btu/hr = .24 x .075 x cfm x ΔT Btu/hr = 1.08 x cfm x ΔT 7. BUT Moisture is almost always present in air and has a heat content of its own. This is known as latent heat. 8. Total Heat is Sensible Heat + Latent Heat 9. Enthalpy is the term used to indicate the total heat content of one pound of air. Enthalpy is measured with a wet bulb thermometer. 10. Total Heat Formula We use the total heat formula for changes in BOTH sensible and latent heat AND it is useful to determine the capacity of an air conditioning system. Total Heat Formula Btu/hr = Sp. Density x 60 min/hr x cfm x ΔH Btu/hr = .075 x 60 x cfm x ΔH Btu/hr = 4.5 x cfm x ΔH 11. Relative Humidity is a ratio of the amount of moisture present in the air to the amount it can hold at saturation. 12. Specific Humidity The amount of moisture present in the air expressed in grains of moisture per pound of dry air. 7,000 grains of moisture in one pound of water. 13. Dew Point The temperature at which the water vapor in the air becomes saturated and starts to condense into water droplets. In Summary: air has the following properties: • Density(dry or wet) • Volume • Sensible Heat • Latent Heat and the following measurements can be found: • Density • Volume • Temperature • Dry bulb • Wet bulb • Dew Point • Relative Humidity And now to the Psychrometric Chart The psychrometric chart is simply a tool that can be used to determine the properties of moist air. Construction of the Chart C. PROCESSES • Sensible Heat • Sensible Heat plus Humidification • Chemical Dehydration • Sensible Cooling • Cooling and Dehumidification • Evaporative Cooling 1. SENSIBLE HEAT PROCESS Entering Conditions: 69oF dry bulb o (return air) 55 F wet bulb Determine: grains ________(specific humidity) SENSIBLE HEAT PROCESS Entering Conditions: 69oF dry bulb o (return air) 55 F wet bulb Determine: grains ___42___(specific humidity) SENSIBLE HEAT PROCESS Entering Conditions: 69oF dry bulb o (return air) 55 F wet bulb Determine: grains ___42___(specific humidity) dew point ________oF SENSIBLE HEAT PROCESS Entering Conditions: 69oF dry bulb o (return air) 55 F wet bulb Determine: grains ___42___(specific humidity) dew point ___44___oF SENSIBLE HEAT PROCESS Entering Conditions: 69oF dry bulb o (return air) 55 F wet bulb Determine: grains ___42___(specific humidity) dew point ___44___oF enthalpy ________Btu/lb SENSIBLE HEAT PROCESS Entering Conditions: 69oF dry bulb o (return air) 55 F wet bulb Determine: grains ___42___(specific humidity) dew point ___44___oF enthalpy __23.22__Btu/lb SENSIBLE HEAT PROCESS Entering Conditions: 69oF dry bulb o (return air) 55 F wet bulb Determine: grains ___42___(specific humidity) dew point ___44___oF enthalpy __23.22__Btu/lb relative humidity ______% SENSIBLE HEAT PROCESS Entering Conditions: 69oF dry bulb o (return air) 55 F wet bulb Determine: grains ___42___(specific humidity) dew point ___44___oF enthalpy __23.22__Btu/lb relative humidity __40__% SENSIBLE HEAT PROCESS Entering Conditions: 69oF dry bulb o (return air) 55 F wet bulb Determine: grains ___42___(specific humidity) dew point ___44___oF enthalpy __23.22__Btu/lb relative humidity __40__% specific volume ________Ft3/lb. SENSIBLE HEAT PROCESS Entering Conditions: 69oF dry bulb o (return air) 55 F wet bulb Determine: grains ___42___(specific humidity) dew point ___44___oF enthalpy __23.22__Btu/lb relative humidity __40__% specific volume __13.45__Ft3/lb. SENSIBLE HEAT PROCESS Entering Conditions: 69oF dry bulb o (return air) 55 F wet bulb Determine: grains ___42___(specific humidity) dew point ___44___oF enthalpy __23.22__Btu/lb relative humidity __40__% specific volume __13.45__Ft3/lb. specific density _______lbs/ft3 SENSIBLE HEAT PROCESS Entering Conditions: 69oF dry bulb o (return air) 55 F wet bulb Determine: grains ___42___(specific humidity) dew point ___44___oF enthalpy __23.22__Btu/lb relative humidity __40__% specific volume __13.45__Ft3/lb. specific density __.074__lbs/ft3 SENSIBLE HEAT PROCESS Leaving Conditions: 95oF dry bulb (supply air) 64.5oF wet bulb Determine: grains _______(specific humidity) SENSIBLE HEAT PROCESS Leaving Conditions: 95oF dry bulb (supply air) 64.5oF wet bulb Determine: grains __42__(specific humidity) SENSIBLE HEAT PROCESS Leaving Conditions: 95oF dry bulb (supply air) 64.5oF wet bulb Determine: grains __42__(specific humidity) dew point ______oF SENSIBLE HEAT PROCESS Leaving Conditions: 95oF dry bulb (supply air) 64.5oF wet bulb Determine: grains __42__(specific humidity) dew point __44__oF SENSIBLE HEAT PROCESS Leaving Conditions: 95oF dry bulb (supply air) 64.5oF wet bulb Determine: grains __42__(specific humidity) dew point __44__oF enthalpy ______Btu/lb SENSIBLE HEAT PROCESS Leaving Conditions: 95oF dry bulb (supply air) 64.5oF wet bulb Determine: grains __42__(specific humidity) dew point __44__oF enthalpy __29.68__Btu/lb SENSIBLE HEAT PROCESS Leaving Conditions: 95oF dry bulb (supply air) 64.5oF wet bulb Determine: grains __42__(specific humidity) dew point __44__oF enthalpy __29.68__Btu/lb relative humidity ______% SENSIBLE HEAT PROCESS Leaving Conditions: 95oF dry bulb (supply air) 64.5oF wet bulb Determine: grains __42__(specific humidity) dew point __44__oF enthalpy __29.68__Btu/lb relative humidity __18__% SENSIBLE HEAT PROCESS Leaving Conditions: 95oF dry bulb (supply air) 64.5oF wet bulb Determine: grains __42__(specific humidity) dew point __44__oF enthalpy __29.68__Btu/lb relative humidity __18__% specific volume ______Ft3/lb. SENSIBLE HEAT PROCESS Leaving Conditions: 95oF dry bulb (supply air) 64.5oF wet bulb Determine: grains __42__(specific humidity) dew point __44__oF enthalpy __29.68__Btu/lb relative humidity __18__% specific volume __14.1__Ft3/lb. SENSIBLE HEAT PROCESS Leaving Conditions: 95oF dry bulb (supply air) 64.5oF wet bulb Determine: grains __42__(specific humidity) dew point __44__oF enthalpy __29.68__Btu/lb relative humidity __18__% specific volume __14.1__Ft3/lb. specific density ______lbs/ft3 SENSIBLE HEAT PROCESS Leaving Conditions: 95oF dry bulb (supply air) 64.5oF wet bulb Determine: grains __42__(specific humidity) dew point __44__oF enthalpy __29.68__Btu/lb relative humidity __18__% specific volume __14.1__Ft3/lb. specific density __.071__lbs/ft3 SENSIBLE HEAT PROCESS Leaving Conditions: 95oF dry bulb (supply air) 64.5oF wet bulb Determine: grains __42__(specific humidity) dew point __44__oF enthalpy __29.68__Btu/lb relative humidity __18__% specific volume __14.1__Ft3/lb. specific density __.071__lbs/ft3 sensible heat factor ______ SENSIBLE HEAT PROCESS Leaving Conditions: 95oF dry bulb (supply air) 64.5oF wet bulb Determine: grains __42__(specific humidity) dew point __44__oF enthalpy __29.68__Btu/lb relative humidity __18__% specific volume __14.1__Ft3/lb. specific density __.071__lbs/ft3 sensible heat factor __1.00__ Sensible Heat Added Btu/hr = 1.08 x cfm x ΔT Btu/hr = 1.08 x 1000 x (95 – 69) Btu/hr = 1.08 x 1000 x 26 Btu/hr = 28,080 2. COOLING and DEHUMIDIFICATION Entering Conditions: 80oF dry bulb o (return air) 63.5 F wet bulb Determine: grains _______(specific humidity) COOLING and DEHUMIDIFICATION Entering Conditions: 80oF dry bulb o (return air) 63.5 F wet bulb Determine: grains __61___(specific humidity) COOLING and DEHUMIDIFICATION Entering Conditions: 80oF dry bulb o (return air) 63.5 F wet bulb Determine: grains __61___(specific humidity) dew point ______oF COOLING and DEHUMIDIFICATION Entering Conditions: 80oF dry bulb o (return air) 63.5 F wet bulb Determine: grains __61___(specific humidity) dew point __53.5__oF COOLING and DEHUMIDIFICATION Entering Conditions: 80oF dry bulb o (return air) 63.5 F wet bulb Determine: grains __61___(specific humidity) dew point __53.5__oF enthalpy ______Btu/lb COOLING and DEHUMIDIFICATION Entering Conditions: 80oF dry bulb o (return air) 63.5 F wet bulb Determine: grains __61___(specific humidity) dew point __53.5__oF enthalpy __28.94__Btu/lb COOLING and DEHUMIDIFICATION Entering Conditions: 80oF dry bulb o (return air) 63.5 F wet bulb Determine: grains __61___(specific humidity) dew point __53.5__oF enthalpy __28.94__Btu/lb relative humidity ______% COOLING and DEHUMIDIFICATION Entering Conditions: 80oF dry bulb o (return air) 63.5 F wet bulb Determine: grains __61___(specific humidity) dew point __53.5__oF enthalpy __28.94__Btu/lb relative humidity __40__% COOLING and DEHUMIDIFICATION Entering Conditions: 80oF dry bulb o (return air) 63.5 F wet bulb Determine: grains __61___(specific humidity) dew point __53.5__oF enthalpy __28.94__Btu/lb relative humidity __40__% specific volume ______ Ft3/lb. COOLING and DEHUMIDIFICATION Entering Conditions: 80oF dry bulb o (return air) 63.5 F wet bulb Determine: grains __61___(specific humidity) dew point __53.5__oF enthalpy __28.94__Btu/lb relative humidity __40__% specific volume __13.78__ Ft3/lb. COOLING and DEHUMIDIFICATION Entering Conditions: 80oF dry bulb o (return air) 63.5 F wet bulb Determine: grains __61___(specific humidity) dew point __53.5__oF enthalpy __28.94__Btu/lb relative humidity __40__% specific volume __13.78__ Ft3/lb. specific density ______ lbs/ft3 COOLING and DEHUMIDIFICATION Entering Conditions: 80oF dry bulb o (return air) 63.5 F wet bulb Determine: grains __61___(specific humidity) dew point __53.5__oF enthalpy __28.94__Btu/lb relative humidity __40__% specific volume __13.78__ Ft3/lb. specific density __.073__ lbs/ft3 COOLING and DEHUMIDIFICATION Leaving Conditions: 60oF dry bulb o (supply air) 53.5 F wet bulb Determine: grains _______(specific humidity) COOLING and DEHUMIDIFICATION Leaving Conditions: 60oF dry bulb o (supply air) 53.5 F wet bulb Determine: grains __51__(specific humidity) COOLING and DEHUMIDIFICATION Leaving Conditions: 60oF dry bulb o (supply air) 53.5 F wet bulb Determine: grains __51__(specific humidity) dew point ______oF COOLING and DEHUMIDIFICATION Leaving Conditions: 60oF dry bulb o (supply air) 53.5 F wet bulb Determine: grains __51__(specific humidity) dew point __48__oF COOLING and DEHUMIDIFICATION Leaving Conditions: 60oF dry bulb o (supply air) 53.5 F wet bulb Determine: grains __51__(specific humidity) dew point __48__oF enthalpy ______Btu/lb COOLING and DEHUMIDIFICATION Leaving Conditions: 60oF dry bulb o (supply air) 53.5 F wet bulb Determine: grains __51__(specific humidity) dew point __48__oF enthalpy __22.30__Btu/lb COOLING and DEHUMIDIFICATION Leaving Conditions: 60oF dry bulb o (supply air) 53.5 F wet bulb Determine: grains __51__(specific humidity) dew point __48__oF enthalpy __22.30__Btu/lb relative humidity ______% COOLING and DEHUMIDIFICATION Leaving Conditions: 60oF dry bulb o (supply air) 53.5 F wet bulb Determine: grains __51__(specific humidity) dew point __48__oF enthalpy __22.30__Btu/lb relative humidity __67__% COOLING and DEHUMIDIFICATION Leaving Conditions: 60oF dry bulb o (supply air) 53.5 F wet bulb Determine: grains __51__(specific humidity) dew point __48__oF enthalpy __22.30__Btu/lb relative humidity __67__% specific volume ______Ft3/lb. COOLING and DEHUMIDIFICATION Leaving Conditions: 60oF dry bulb o (supply air) 53.5 F wet bulb Determine: grains __51__(specific humidity) dew point __48__oF enthalpy __22.30__Btu/lb relative humidity __67__% specific volume __13.24__Ft3/lb. COOLING and DEHUMIDIFICATION Leaving Conditions: 60oF dry bulb o (supply air) 53.5 F wet bulb Determine: grains __51__(specific humidity) dew point __48__oF enthalpy __22.30__Btu/lb relative humidity __67__% specific volume __13.24__Ft3/lb. specific density ______lbs/ft3 COOLING and DEHUMIDIFICATION Leaving Conditions: 60oF dry bulb o (supply air) 53.5 F wet bulb Determine: grains __51__(specific humidity) dew point __48__oF enthalpy __22.30__Btu/lb relative humidity __67__% specific volume __13.24__Ft3/lb. specific density __.076__lbs/ft3 COOLING and DEHUMIDIFICATION Leaving Conditions: 60oF dry bulb o (supply air) 53.5 F wet bulb Determine: grains __51__(specific humidity) dew point __48__oF enthalpy __22.30__Btu/lb relative humidity __67__% specific volume __13.24__Ft3/lb. specific density __.076__lbs/ft3 sensible heat factor ______ COOLING and DEHUMIDIFICATION Leaving Conditions: 60oF dry bulb o (supply air) 53.5 F wet bulb Determine: grains __51__(specific humidity) dew point __48__oF enthalpy __22.30__Btu/lb relative humidity __67__% specific volume __13.24__Ft3/lb. specific density __.076__lbs/ft3 sensible heat factor __.75__ Sensible Heat Removed Btu/hr = 1.08 x cfm x ΔT Btu/hr = 1.08 x 1600 x (80 – 60) Btu/hr = 1.08 x 1600 x 20 Btu/hr = 34,560 Btu/hr TOTAL Heat Removed Btu/hr = 4.5 x cfm x ΔH Btu/hr = 4.5 x 1600 x (28.94 – 22.30) Btu/hr = 4.5 x 1600 x 6.64 Btu/hr = 47,808 qs SHR = -------- QT 34,560 SHR = ----------- 47,808 •SHR = .73 IF 1. You extend the process line through the 100% saturation line, then 2. the air would be completely saturated as it leaves the coil, thus 3. the air temperature would be equal to the coil temperature. What is that temperature? o 43 F a. APPARATUS DEW POINT The temperature of the air at which it leaves the coil saturated. 100% RH What is the temperature at which the air left our coil? o 60 F WHY? b. BYPASS AIR Because some of the air was bypassed and unaffected by the coil temperature. This is known as Bypass Factor c. Conditions which affect the BYPASS FACTOR 1. Fin Spacing 2. Number of Rows & Depth of Coil 3. Type of Fin 4. Velocity of Air 5. If Coil is Wet or Dry 6. Conditions of System D. NOW application of PSYCHROMETRICS APPLICATION 1 new unit installed in existing building 3 TON Day 1 Conditions o o • Entering Air - 80 DB, 73 WB, 72%RH o o • Leaving Air - 68 DB, 65 WB, 85%RH • Determine: • Sensible heat • Latent heat • Sensible Heat Ratio Locate these two conditions on the Psychrometric Chart o o Entering Air - 80 DB, 73 WB o o Leaving Air - 68 DB, 65 WB Draw a line connecting the two points. Draw a vertical line down from the entering conditions. Draw a line horizontally to the right from the leaving conditions. At the intersection of these two lines, draw a line upwards following the wet bulb line until it crosses the line connecting the two points. • NOTE that the VERTICAL line represents the latent load, and • NOTE that the HORIZONTAL line represents the sensible load. Where does this point of crossing occur, in terms of distance from either point? SO Use the SWAG method to determine the approximate amount of sensible load and latent load you have. Another method to determine the amount of sensible heat to latent heat is: 1. Locate the 80DB, 67WB reference dot. 2. Place your pencil point on the dot. 3. Lay a straight edge against the pencil point and use the dot as a pivot point. 4. Rotate the straight edge until it is parallel to your original line. 5. Read the sensible heat percentage on the far right of the chart. NOW Let us do it again for the ‘Day 2’ conditions. Day 2 Conditions o o • Entering Air - 80 DB, 67 WB • Leaving Air - 63oDB, 58oWB • Determine: • Sensible heat • Latent heat • Sensible Heat Ratio Locate these two conditions on the Psychrometric Chart o o Entering Air - 80 DB, 67 WB o o Leaving Air - 63 DB, 58 WB Draw a line connecting the two points. Draw a vertical line down from the entering conditions. Draw a line horizontally to the right from the leaving conditions. At the intersection of these two lines, draw a line upwards following the wet bulb line until it crosses the line connecting the two points. • NOTE that the VERTICAL line represents the latent load, and • NOTE that the HORIZONTAL line represents the sensible load. Where does this point of crossing occur, in terms of distance from either point? SO Use the SWAG method to determine the approximate amount of sensible load and latent load you have. Another method to determine the amount of sensible heat to latent heat is: 1. Locate the 80DB, 67WB reference dot. 2. Place your pencil point on the dot. 3. Lay a straight edge against the pencil point and use the dot as a pivot point. 4. Rotate the straight edge until it is parallel to your original line. 5. Read the sensible heat percentage on the far right of the chart. NOW Let us do it again for the ‘One Week Later’ conditions. One Week Later Conditions o o • Entering Air - 78 DB, 62 WB o o • Leaving Air - 53 DB, 51 WB • Determine: • Sensible heat • Latent heat • Sensible Heat Ratio Locate these two conditions on the Psychrometric Chart o o Entering Air - 78 DB, 62 WB o o Leaving Air - 53 DB, 51 WB Draw a line connecting the two points. Draw a vertical line down from the entering conditions. Draw a line horizontally to the right from the leaving conditions. At the intersection of these two lines, draw a line upwards following the wet bulb line until it crosses the line connecting the two points. • NOTE that the VERTICAL line represents the latent load, and • NOTE that the HORIZONTAL line represents the sensible load. Where does this point of crossing occur, in terms of distance from either point? SO Use the SWAG method to determine the approximate amount of sensible load and latent load you have. Another method to determine the amount of sensible heat to latent heat is: 1. Locate the 78DB, 62WB reference dot. 2. Place your pencil point on the dot. 3. Lay a straight edge against the pencil point and use the dot as a pivot point. 4. Rotate the straight edge until it is parallel to your original line. 5. Read the sensible heat percentage on the far right of the chart. Comparison of Critical Data DAY 1 DAY 2 ONE WEEK Temperature Temperature LATER Difference Difference Temperature Difference 12 17 25 Comparison of Critical Data DAY 1 DAY 2 ONE WEEK Sensible Sensible LATER Heat Ratio Heat Ratio Sensible Heat Ratio .43 .62 .86 Comparison of Critical Data DAY 1 DAY 2 ONE WEEK Temperature Temperature LATER Difference Difference Temperature Difference 12 17 25 CFM REQUIREMENTS Temperature DROP For COOLING 1. Temperature DROP o o 18 – 22 o Minimum = 15 o Maximum = 25 2. Application Building with Sensible Load HIGH Should the temperature drop be closer to 15 or 25? 2. Application Building with Latent Load HIGH Should the temperature drop be closer to 15 or 25? 3. Cooling Temperature Splits (temperature drops) Outdoor Indoor Indoor DB Indoor DB Indoor DB DB WB oF 75oF 78oF 80oF 59 22 24 25 85oF 63 19 21 23 67 15 17 19 59 21 23 24 95oF 63 18 20 22 67 15 17 19 63 17 20 21 105oF 67 14 17 18 71 11 13 15 63 17 19 21 115oF 67 13 16 17 A17 71 10 13 14 APPLICATION 2 MIXTURE TEMPERATURES Mix outdoor air (OA) with Return Air (RA) THEN The Mixture Air (MA) passes over the coil QUESTION: What should be the temperature of the mixed air? Problem Outdoor Ambient Temperature = 95 DB Return Air Temperature = 78 DB Required to have: 25% OA 75% RA Two Methods 1. Formula 2. Psychrometric Chart Formula TEMPMA = (%OA x TEMPOA) + (%RA x TEMPRA) TEMPMA = (.25 x 95) + (.75 x 78) TEMPMA = 23.75 + 58.5 TEMPMA = 82.25oF Psychrometric Chart 1. Plot the following two points on the chart. OUTDOOR AIR 95DB, 83WB RETURN AIR 78DB, 65WB 2. Draw a line between the two points. 3. Locate a point approximately 25% from the condition which has the MOST air. Use the SWAG method. 4. That will be the mixed air temperature. Question: What if you want to check the %OA on an existing job? Problem Outdoor Ambient Temperature = 95 DB Return Air Temperature = 78 DB Mixed Air Temperature = 82.25 DB Formula TMA - TRA %OA = ----------------- TOA - TRA TMA - TRA %OA = ----------------- TOA - TRA 82.25 - 78 = ----------------- = .25 or 25% 95 - 78

DOCUMENT INFO

Shared By:

Categories:

Tags:

Stats:

views: | 1 |

posted: | 3/29/2013 |

language: | English |

pages: | 180 |

OTHER DOCS BY xiaopangnv

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.