HVAC Systems: Air Conditioning Dr. Harjit Singh Room 116 Howell Building harjit. singh@brunel. ac. uk Fridays 09. 30 – 11. 30am Unless otherwise specified For students enquiries: 1 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Introduction to the module
• Credit: 15
• Teaching materials: – Provided for the MSc students. – Undergrads: Buy them from the stores.
• Assignment: to be handed two weeks before Xmas, deadline-28 Jan 2013
• Students enquiries: – Fridays 09. 30 – 11. 0am Unless otherwise specified – Outside these hours: email me for appointment: harjit. singh@brunel. ac. uk
• Is the teaching materials enough: yes, but further reading is recommended – Eastop, T.

D and Watson. E. (1992). Mechanical Services for Buildings, Longman Scientific and Technical – CIBSE Guide A, Section 1: Environmental criteria for design, CIBSE, 2006 – Jones, W. P. (2003). Air Conditioning Engineering (5th edition), Arnold Publishing. 2 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel UniversityModule Content
• Part 1 – Basic principles of air conditioning system design – Psychrometry – Basic Air Conditioning Processes (heating, cooling, humidification/dehumidification) – Thermal Comfort – Design Conditions – Basic Air Conditioning System Design
• Part 2 – Air Conditioning Systems (Design and Applications) – Air Conditioning System Classification – All-air systems (constant volume and variable volume) – Air and water systems (induction and fan coil systems – panel type systems) – All water systems (non ventilation fan coils and heat pumps) 3Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Basic Requirements
• For revising the basics (Chapter 1 ‘Building Air Conditioning’ book) – – – – – Thermodynamic properties and how to obtain them from tables and diagrams Open and closed systems First law of thermodynamics (SFEE equation).

Equation of state for pure substances Second law of thermodynamics » Thermodynamic cycles » Thermal efficiency » Entropy
• Only after a good understanding of this chapter (more like “remembering the information! ”, you can comfortably proceed to later chapters eg.

Get quality help now
writer-Charlotte
Verified

Proficient in: Chemistry

4.7 (348)

“ Amazing as always, gave her a week to finish a big assignment and came through way ahead of time. ”

+84 relevant experts are online
Hire writer

Chapter 2 – ‘Psychrometry’ 4Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Heat transfer Basics “Fundamentals of Thermal-Fluid Sciences” by Cengel & Turner, McGraw-Hill, ISBN 0-07-239054-9 5 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Psychrometery (Chapter 2: ‘Building Air Conditioning’ book) ? This chapter explains the principles of psychrometry as applied to the study of modern air conditioning systems ?Learning Objectives After covering this chapter and completing the personal feedback questions, you should be able to: ? Understand the properties of humid air Given two properties of humid air, calculate other properties of interest for the same state point ?Be able to use psychrometric tables and the psychrometric chart as practical tools for obtaining quickly the properties of humid air. 6 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Basics- Definition of Air Conditioning Air conditioning may be defined as the simultaneous control of all (or at least the first three) of those factors affecting both the physical & the chemical conditions of the surrounding atmosphere within a structure. These factors are: 1. 2. . 4. 5. 6. 7. Temperature Humidity Air movement Dust Bacteria Odours Toxic gases 7 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Basics- Air conditioned air Air conditioned air: is air that has undergone one or a combination of the following processes: 1. Heating 2. Cooling 3. Humidification 4. Dehumidification 5. Circulation 6. Cleaning and filtering The ultimate goal of air conditioning process is achieving the comfort conditions for the occupants of a building 8 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel UniversityPsychrometry (Chapter 2: ‘Building Air Conditioning’ book) Psychrometry is the branch of science covering the properties of humid air Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University 9 Introduction ? Air is the main career (working fluid) of heating or cooling energy in air conditioning systems ? Before it is supplied to the indoor spaces of a building, the air is conditioned by passing it through the various system components. These may include heaters, coolers humidifiers, etc. ?The study of the properties of humid air is called psychrometry 10Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Humid Air
• Humid air is a mixture of dry air and water vapour.
• It is reasonably accurate to assume that both dry air and water vapour
• • obey the ideal gas law. Hence, it is rational to assume that humid air behaves as an ideal gas mixture. If we consider that a closed system consists of a gaseous mixture of two or more ideal gases, then the composition of the mixture can be described either by giving the mass or the number of moles of each component present. The total mass of the mixture, m, is the sum of the masses of its components: ? m1 ? m2 ? m3 ? …. ? mn m ? ? mi i ? 1 i ? n or, 11 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Humid Air
• Total number of moles in the mixture, n, is the sum of the number of moles of each of its components: or, n = n1 + n 2 + n 3 + … + n n n = ? ni i ? 1 i ? n
• The apparent or average molecular weight of the mixture, M, is defined as the ratio of the total mass of the mixture, m, to the total number of moles of the mixture, n. n1 M 1 + n2 M 2 + … + nn M n M = n 12 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel UniversityDalton’s law of Partial Pressures
• Pressure exerted by a mixture of ideal gases is the sum of the pressures exerted by each gas if it were alone at the same temperature and volume of the mixture P1 P2 P = P1+P2 X X X X X X O X O O O O O O O T X X X X V X X X X X X + T O O O O V O = X O X X O X O X O X O O X O O X O X O T O V O GAS A MASS M1 GAS B MASS M2 P ? P ? P2 ? P3 ? …. ? Pn 1 P ? ? Pi i ? 1 i ? n MIXTURE A + B MASS M = M1 + M2
• Pressure exerted by each gas in the mixture is called its Partial Pressure 13 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel UniversityDry Air
• Atmospheric air contains several gaseous components plus water vapour and contaminants such as dust and pollutants.
• The term dry air refers only to the gaseous components when all water vapour and contaminants have been removed.
• The molar analysis of a typical sample of dry air is given in the Table. Gaseous Component Nitrogen N2 Oxygen O2 Carbon Dioxide CO2 Hydrogen H2 Argon Ar Total Molar Fraction % 78. 03 20. 99 0. 03 0. 01 0. 94 100. 00 Molecular Weight 28. 02 32. 00 44. 00 2. 02 39. 91
• The apparent molecular weight of dry air is: Ma ? ? (Molar Fraction ? Mol. Weight) i ? 1 i ? n
• Ma = (28. 2 x 0. 7803) + (32 x 0. 2099) + (44 x 0. 0003) + (2. 02 x 0. 0001) + (39. 91 x 0. 0094) = 28. 97
• The apparent molecular weight of dry air = 28. 97 kg/kmol 14 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Water Vapour WATER VAPOUR WATER VAPOUR LIQUID WATER (A) LIQUID WATER (B) (C) Consider the vessel shown in the Fig. A above which contains a unit mass of saturated liquid water (1 kg) at a pressure of 1. 01325 bar (1 atm). This state is represented by point 1 on the T-v diagram in Figure D. Now suppose the water is slowly heated while its pressure is kept constant.This will result in the formation of vapour with considerable increase in the specific volume. As shown in Figure B, the system would now consist of a two-phase liquid vapour mixture. If the contents of the vessel are at an equilibrium condition, that is the temperatures of the vapour and liquid are equal, the liquid phase is saturated liquid and the vapour phase is saturated vapour. The pressure that the vapour exerts at this condition is known as the saturation vapour pressure (SVP). 15 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Water Vapour SVP Vapour pressure TemperatureIf more heat is added to the vessel, more water will be evaporated and when all liquid has evaporated the system would reach point 2, the saturated vapour state. Further addition of heat at constant pressure will result in an increase in both temperature and specific volume. The condition of the system would now be represented by point 3 in Figure D. The state of the system represented by point 3 is often referred to as a superheated vapour state. The Saturated Vapour Pressure varies with temperature as shown in the above figure, and values are published in property Tables for Saturated Water and Steam (Rogers and Mayhew) or Steam Tables. 6 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Molecular Weight of Water Vapour The molecular weight of water vapour can be obtained from the molecular weights of its chemical composition, H2O. Ms = 2 x 1. 01 + 1 x 16 = 18. 02 kg/kmol 17 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Relative Humidity Relative humidity is defined as the ratio of the vapour pressure of water vapour in the air, Ps, to the saturated vapour pressure at the same air temperature, Pss. This ratio is expressed as a percentage.Relative humidity is denoted by % RH or ?. Ps ? = x 100 P ss A Pss Ps Temperature oC 100% RH corresponds to the saturated vapour line. Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University 18 Relative Humidity Air at 25 °C has a vapour pressure of 17 mbar. Using the Thermodynamic Property Tables calculate the relative humidity. 19 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Relative Humidity Air at 25 °C has a vapour pressure of 17 mbar. Using the Thermodynamic Property Tables calculate the relative humidity. ? =Ps x 100 P ss 20 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Relative Humidity 21 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Moisture Content
• The moisture content of humid air, g, is defined as the mass of water vapour contained in 1 kg of dry air. The moisture content is sometimes referred to as Specific Humidity, Absolute Humidity or Humidity Ratio.
• Note: some properties of humid air are based on 1 kg of dry air
• • From the definition of moisture content: g = ms ma kg of water vapour/ kg of dry air From the ideal gas law,
• for dry air: P a V a = ma R a T a
• for water vapour: ? Pa V a ma = Ra T a ? ms = Ps V s Rs T s P s V s = ms R s T s V a =V s But for the mixture and T a = Ts Note: g and gs are interchangeably used for the moisture content in air Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University 22 Moisture Content substituting equations (2. 3) and (2. 4) in equation (2. 2) gives: 0. 2869 P s g = Ra Ps = = 0. 622 P s R s P a 0. 4615 P a Pa Also 18. 02 P s g ? M s Ps = = 0. 622 P s M a P a 28. 7 P a Pa – But, P a = P at – P s – then, Ps g = 0. 622 kg/ kg of dry air P at – P s 23 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Moisture Content Determine the moisture content of air at atmospheric pressure of 1013 mbar and vapour pressure 20 mbar 24 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Moisture Content Determine the moisture content of air at atmospheric pressure of 1013 mbar and vapour pressure 20 mbar R a P s = 0. 2869 P s = 0. 622 P s g= R s P a 0. 615 P a Pa Also M s P s = 18. 02 P s = 0. 622 P s g? M a P a 28. 97 P a Pa 25 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Moisture Content at Saturation
• The moisture content corresponding to the saturation vapour pressure is known as the saturation moisture content, gss
• It can be calculated from the equation for g by substituting Pss for Ps. The saturation moisture content can be plotted against temperature, as shown in the Figure below. g ss = 0. 622 P ss Pat – P ss Variation of saturation moisture content 26Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Percentage Saturation
• Percentage saturation, µ , is defined as the ratio of the moisture content in the air to the moisture content at saturation at the same temperature, expressed as a percentage. The percentage saturation of the humid air at condition A in Figure below is therefore given by: gs ? = x 100 g ss A gss gs Temperature oC Note: g and gs are interchangeably used for moisture content in air Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University 27Percentage Saturation Rearranging, taking into account the equations for gs ; gss Ps g P at – P s x 100 ? = s x 100 = g ss 0. 622 P ss P at – P ss 0. 622 – simplifying: ( ) ( ) ? = Ps ? Pat Pss ? 100 ? ? = ? ? Pat Pss P ss ( Pat – P s ) ( Pat – P s ) – At a given dry bulb temperature this expression will vary with the vapour pressure but as Ps and Pss are small compared with Pat, the variation between %RH and µ will also be small. – The quantity ( P at – P ss ) ( P at – P s ) is nearly 1 – So for all practical purposes, the percentage saturation is interchangeable with percentage relative humidity. 8 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Percentage Saturation
• Explain the difference between percentage saturation and relative humidity.
• For air at atmospheric pressure of 1013 mbar, 20 °C dry bulb temperature and 13. 67 mb vapour pressure, determine: a) the relative humidity b) the difference between the values of relative humidity and percentage saturation. 29 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Specific Volume The specific volume, v, of humid air is defined as the volume of humid air m3) per unit mass of dry air (kg) and the water vapour associated with it. It is defined as: v= From Dalton’s law, V = Va = Vs and from the ideal gas law: V ma 3 m /kg of dry air V =V a = but, ma R a T a Pa Pa = Pat – Ps hence, v = R a Ta (Pat – Ps) 30 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Specific Volume In relation to the 100% saturation line, lines of constant specific volume can be drawn on a temperature versus moisture content diagram, as shown in Figure below 100% saturation Lines of constant Specific volume gss gs Temperature oC 1 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Relationship between air density and specific volume Air density, ? , is defined as the mass of humid air per unit volume. ?= m ma + ms 1 + g = = V V V ma v= V ma but, hence, ?= (1 + g) v Although the difference between density and the reciprocal of specific volume is relatively small, it is important to note that, for air conditioning calculations, the two quantities are different. Specific volume is defined in terms of unit mass of dry air; density is defined in terms of unit mass of the mixture. 2 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Dry-Bulb Temperature The dry bulb temperature is the temperature indicated by a dry buld thermometer shielded from radiation. The dry bulb temperature indicates the degree of the sensible heat content of the air but it does not give any information about the latent heat content. The dry bulb temperature is denoted by t and is measured in °C. The thermometer used to measure t 33 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel UniversityWet-Bulb Temperature The wet bulb temperature is the temperature indicated by a thermometer whose bulb is covered by a muslin sleeve which is kept moist with water, is freely exposed to the air and is free from radiation. The reading obtained is affected by air movement and for this reason there are two wet bulb temperatures — sling and screen. 34 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Wet-Bulb Temperature A. The sling wet bulb temperature is obtained in a moving air stream, which has a velocity of about 2 ms-1.The Sling WBT is denoted by t? , and is measured in °C B. The screen wet bulb temperature is taken in still air. The wet bulb thermometer is usually installed in a Stevenson screen. The Stevenson screen is a method more favoured by meteorological and weather stations and is essentially a louvred box in which the thermometers are housed. The louvres allow the free passage of air whilst eliminating radiation effects. The screen WBT is denoted by tsc, and is measured in °C B A As air conditioning involves air movement the sling WBT is preferred for air conditioning calculations. 5 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Wet-Bulb Temperature
• The numerical difference between the DBT and the WBT is known as the wet-bulb depression.
• Dry-bulb and Wet-bulb temperatures measured together are the most popular method for obtaining the air properties of humid air for air conditioning calculations. 36 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Dew Point Temperature If air is cooled at constant moisture content, see Figure below, 100% saturationA gs Temperature oC tdb t the temperature will eventually reach the SVP curve and at this point condensation of the water vapour will begin to take place. This temperature is known as the dew point temperature (DPT). Dew point temperature may also be defined as the temperature of saturated air which has the same vapour pressure as the moist air under consideration. 37 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Dew Point Temperature When reaching the dew point temperature, dew, clouds or fog can begin to form 38Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Specific Enthalpy The specific enthalpy of humid air is equal to the sum of the individual partial enthalpies of its constituents. Consider 1 kg of dry air and associated moisture content, g, at a dry bulb temperature t °C. Using a temperature datum of 0°C, the sensible heat content of the air, h1, is given by: h1 = 1 ? c pa ? (t – 0) = 1. 005 ? t where: c pa = specific heat of dry air = 1. 005 kJ/kg ? C and the sensible heat of water vapour, h2, is: h2 = g ? c ps ? (t – 0) = 1. 89 g t where: ps = specific heat of water vapour = 1. 89 kJ/kg ? C If it is considered that all the water vapour in the air has been formed by evaporation of water at 0o C, the latent heat associated with the moisture content, h3, is given by: where: h3 = g ? h fg h fg = the latent heat of evaporation of water at 0o C = 2501 kJ/ kg 39 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Specific Enthalpy
• The specific heat of humid air, h, is thus given by: or, h = h1 + h 2 + h 3 rearranging gives: h = 1. 005 t + 1. 89 g t + 2501 g h = (1. 005 + 1. 89 g) t + 2501 g the term (1. 005 + 1. 9 g)is known as the humid specific heat. 40 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Psychrometric Equation Consider the wet-bulb thermometer where air flows across the bulb, which is covered by a muslin sleeve. The muslin sleeve is kept moist by a reservoir of water. At equilibrium conditions, the heat lost due to moisture evaporation will equal the sensible heat gained; the air close to the thermometer bulb is saturated and its moisture content is gss 41 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Psychrometric EquationThe latent heat lost is proportional to the moisture content difference between the air close to the thermometer bulb and the ambient air ( g ss? – g s ) The sensible heat gained is proportional to the difference in temperature between the ambient air and the thermometer bulb i. e. (t – t? ) At equilibrium conditions: C1 ( g ss? – g s ) = C 2 (t – t ? ) Where C1 and C2 are constants related to the surface area, specific heat and latent heat of evaporation Ps p s? s But, and g s = 0. 622 g ss? = 0. 622 P at – P s Pat – P s? s 42 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel UniversityPsychrometric Equation Ps and Pss are small compared to Pat and so the above equations can be written as: g s = 0. 622 Ps , P ? g ss? = 0. 622 ss Pat Pat Subsituting the above equations in equation 2. 10 gives: 0. 622 C1 Pat ( P ss? – P s ) = C 2 (t – t ? ) or, Ps = Pss? – Pat A (t – t ? ) The above equation is known as the psychrometric equation The psychrometric constant depends on the air velocity across the bulb. it will therefore be different for sling and screen psychrometers. 43 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel UniversityPsychrometric constants Sling A = 6. 66 x 10 -4 A = 5. 94 x 10 -4 K -1 K -1 for t ? ? 0 o C for t ? < 0 o C Screen A = 7. 99 x 10 -4 A = 7. 20 x 10 -4 K -1 K -1 for t ? ? 0 o C for t ? < 0 o C 44 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Adiabatic Saturation INSULATION t 1 g 1 t 2 g 2 In thermodynamics, an adiabatic process is one in which no external heat is allowed to enter or leave the system. Consider the above figure in which air flows through a duct at the bottom of which there is an open water tank.The duct casing is insulated so that there is no heat flow into or out of the duct from or to the surroundings. 45 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Adiabatic Saturation
• Humid air enters the duct at dry-bulb temperature t1, pressure P and moisture content g1 and as it passes through the duct it comes into contact with the pool of water. If the entering air is not saturated, some of the water would evaporate causing an increase in the moisture content of the air leaving the duct.The energy required for evaporation of the water would come from the humid air and its temperature would decrease. At steady state, the sensible heat lost by the humid air would be equal to the latent heat gained from the increase in its moisture content. If the air leaves with a dry-bulb temperature t2 and moisture content g2, then applying the steady flow energy equation to the control volume shown in Figure gives: dividing through by ,and substituting for h in terms of the specific heat and temperature gives:
• • ( ma h a1 + ms1 h s1 ) – mw h fg = ( ma h a 2 + ms2 h s2 ) ? ? ? ? ? 1 ( c pa + g 1 c ps ) – ( g 2 – g 1 ) h fg = t 2 ( c pa + g 2 c ps ) 46 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Adiabatic Saturation
• where: g 2 – g1 = ? mw ? ma
• but for humid air, c p = c pa + g c p s
• hence, c p ( t 1 – t 2 ) = ( g 2 – g 1 ) h fg
• •
• If the duct is sufficiently long, at the end of the process the air will be completely saturated. The temperature at which this occurs is known as the adiabatic saturation temperature, tas, and the corresponding moisture content is gas. Therefore, c p ( t 1 – t as ) = ( g as – g 1 ) h fg 7 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Adiabatic Saturation
• The equation can be rearranged as follows: ( g as – g 1 ) cp = = Constant ( t 1 – t as ) h fg
• This equation indicates that, on a dry-bulb temperature and moisture content co-ordinate system, there is a line of constant adiabatic saturation temperature joining the state points (tas,gas) and (t1,g1). This line can be represented on the T-v diagram as shown in Figure . 48 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel UniversityAdiabatic Saturation For dry air and water vapour mixtures, the adiabatic saturation temperature is the same as the sling wet-bulb temperature. At the saturated condition, the dry-bulb temperature, t2, the wet-bulb temperature, t2′, and the adiabatic saturation temperature, tas, are equal. 49 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Example The dry and wet bulb temperature readings given by a sling psychrometer in a room at an atmospheric pressure of 1013 mbar are 20°C dry bulb and 15 °C wet bulb respectively.Determine from first principles: 1) vapour pressure 2) relative humidity 3) moisture content 4) enthalpy 5) dew point temperature 50 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Problem PF2-4
• A sample of air at a total pressure of 900 mbar has a measured condition of 28 °C dry bulb and 20°C wet bulb. Using data from Rogers ; Mayhew tables, calculate for this sample of air: 1) vapour pressure 2) relative humidity 3) dew point temperature 4) T he moisture content 5) the humid volume 6) the enthalpy 51Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Psychrometric Tables ? The properties of humid air are published in tables which for a given dry-bulb temperature and atmospheric pressure, usually 1013. 25 mbar, list property values, as shown in Table 2. 2 at the end of this chapter. It is important to note that the moisture content, the specific enthalpy and the specific volume are given per kg of dry air. Also, at 100% saturation the relative humidity is also 100% and all the temperatures are equal. The vapour pressure at this condition is the saturation vapour pressure. Property values which are not specified in the tables can be determined with reasonable accuracy by performing linear interpolation between the closest values specified in the Tables. 52 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Psychrometric Chart The psychrometric tables are useful in accurately determining the properties of humid air at a given state point but provide little information on how the properties change during a process. This information is provided by the psychrometric chart which is a graphical representation of the important properties of humid air.The chart provides a picture of the way in which the state of moist air alters as an air conditioning process takes place or a physical process occurs. A psychrometric chart is shown schematically in Figure 2-14. A full scale chart published by CIBSE is given at the end of this volume. 53 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Psychrometric Chart 0. 9 90% 80 % g kg/kg 0. 018 40 % 70 % 0. 85 60 % 20 % 50 0. 015 15 0. 8 10 5 % 30 0. 010 % 20 0. 005 10% 5 10 15 20 25 30 35 40 t C 54 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel UniversityPsychrometric Chart What are the common types of psychrometric charts, which are in use by Engineers? 55 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Psychrometric Chart 56 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Psychrometric Chart 57 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University The Psychrometric Chart We Are Going To Use! Percentage Saturation ~(Relative humedity) Based on Parametric Pressure of 101 325 kPa Dry Bulb Temperature °C Specific Enthalpy Moisture Content 58Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Problem PF2-5
• For air at atmospheric pressure of 1013 mbar, dry bulb temperature of 30 °C and relative humidity of 60%, determine without using the psychrometric chart:
• 1) moisture content 2) enthalpy 3) percentage saturation 4) specific volume 5) percentage saturation 6) dew point temperature
• For the same conditions but now using the chart, determine: 1) the wet bulb temperature 2) enthalpy 3) moisture content 4) dew point temperature 59 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel UniversityProblem PF2-6 Complete the following table, using the psychrometric chart: Dry bulb °C 30 40 20 0. 012 20 20 40 0. 9 Wet bulb °C 24 80 Dew point Moisture content kg/kg Relative humidity % Enthalpy kJ/kg Specific volume m3/kg 60 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Problem PF2-6 Complete the following table, using the psychrometric chart: Dry bulb °C 30 40 Wet bulb °C 24 Dew point Moisture content kg/kg Relative humidity % Enthalpy kJ/kg Specific volume m3/kg 21. 5 20. 7 20 0. 0162 35 38. 6 20 26. 2 24. 4 23. 6 20 16. 9 20 59. 5 0. 155 31. 6 0. 0148 40 27 0. 012 0. 0148 100 71. 8 80 73. 1 69. 6 57. 5 0. 882 0. 909 0. 893 0. 9 0. 85 61 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Chapter 3: Typical Air Conditioning Processes Learning Objectives of This Chapter
• Represent psychrometric processes on the psychrometric chart.
• Apply the psychrometric principles developed in Chapter 2 to psychrometric processes, using psychrometric parameters obtained from both first principles and the psychrometric chart.
• Analyze both water and steam injection humidification processes. Distinguish between sensible heating and cooling, humidification and dehumidification.
• Distinguish between mechanical and chemical dehumidification.
• Analyze spray water processes. 62 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Introduction
• We can represent complete air conditioning systems on the psychrometric chart by combining psychrometric processes in each component of the system. The heat and mass flows across each component are represented by the state points of the air at entry and exit from the component. The most common processes which take place as the air flows through a single component or a combination of components in air conditioning systems are: – mixing of two air streams – sensible heating and cooling – humidification and dehumidification
• In this chapter we will consider the basic air conditioning processes, illustrate their representation on the psychrometric chart and show how important information on each process can be derived from the chart. We will concentrate on the psychrometrics of each process rather than the components of the air conditioning system. 3 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Adiabatic mixing Why Air Mixing?
• To keep Air Fresh
• Fresh air mixed with recycled air
• More efficient process than heating or cooling fresh air (except in special circumstances where no recycled air is allowed) 64 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Adiabatic mixing Mass and energy balance equations for the process: ? ? ? ma1 + ma2 = ma3 (dry air) ? ? ? ms1 + ms2 = ms3 (water vapour) ? ? ms = g ma ? ? Q =W = 0 1 ma1 + g 2 ma2 = g 3 ma3 ? ? ? and ignoring kinetic and potential energy terms ? ? ? ma1 ( ha1 + g1 hs1 ) + ma2 ( ha2 + g 2 hs2 ) = ma3 ( ha3 + g 3 hs3 ) ? ? ? ma1 h1 + ma2 h2 = ma3 h3 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University 65 Adiabatic mixing on the Chart ? ? ? ma1 h1 + ma2 h2 = ma3 h3 1 3 ? ? ? ma1 h1 + ma2 h2 = ma3 h3 2 Dry Bulb Temperature °C 66 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Moisture Content ? ? ? ma1 h1 + ma2 h2 = ma3 h3 Adiabatic mixingThe three state points lie on a straight line on the chart with the mixture state point dividing the line in two segments whose ratio is inversely proportional to the ratio of the masses of dry air such that: ? ma 2 h1 ? h3 g1 ? g 3 ? ? ? ma1 h3 ? h2 g 3 ? g 2 In design calculations we can assume with reasonable accuracy that the mass flow of dry air is equal to the mass flow of humid air. This arises from the fact that the mass of water vapour in one kg of moist air is very small compared with the mass of dry air in the mixture. 67 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel UniversityAdiabatic mixing: Example A stream of 2 kgs-1 of moist air at a dry-bulb temperature of 5 oC and moisture content of 0. 002 kg/kgof_dry_air, mixes adiabatically with a second stream of 3 kgs-1 at a dry-bulb temperature of 20 oC and 50% relative humidity. The pressure is constant at 1 atmosphere. Determine the enthalpy, moisture content, and dry-bulb temperature of the mixture. Assuming the mass flow of dry air is equal to the mass flow of moist air: Stream 1: ma1=m1=2kgs-1, ta1=5oC and g=0. 002kg/kgdry-air mixes adiabatically with: Stream 2: ma2= m2=3 kgs-1 ta1=20 oC and ? 2=50%. h3=? , g3=? , and t3=? 68Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Adiabatic mixing: Example Solution:
• Using the psychrometric chart we get: h1 = 10. 05 kJ/kg, g2 = 0. 0074 kg/kg, h2 = 38. 84 kJ/kg
• from mass balance ? ? ? m3 = m2 + m1 = 2 + 3 = 5 kg/s ? ma 2 h1 ? h3 g1 ? g 3 ? ? ? ma1 h3 ? h2 g 3 ? g 2 ? ? m1 h1 + m2 h2 2 ? 10. 05 + 3 ? 38. 84 = = 27. 5 kJ/kg dry air h3 = 5 ? m3 ? ? m1 g 1 + m2 g 2 2 ? 0. 002 + 3 ? 0. 0074 = = 0. 00524 kg/kg dry air g3 = 5 ? m3 69 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Adiabatic mixing: Example 70Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University PF3-1 A stream of recalculated air at a state of 25 oC db and 40 KJkg-1 of dry air specific enthalpy, is to be mixed with a fresh air stream at 20 oC db and 12 oC wb. If the mixture by mass of recalculated air to fresh air is 75% to 25%, calculate the supply air conditions. 71 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Sensible heating and cooling In sensible heating and cooling, the temperature of the air increases or decreases but the moisture content remains constant.Hence, these processes may be defined by horizontal lines on the psychrometric chart. Can you think of an example sensible heating system? 72 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Sensible heating An example sensible heating system is: The hair dryer! Heating element 1 Air in 2 Air out 1 2 g1=g2 t2 t1 Sensible heating can be achieved by passing the air through any sort of heater. ? Q = I ? V ? ma ( h2 – h1 ) ? Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University 73 Sensible cooling Cooling fluid in 1 Air in Cooling fluid out 2 Air out 2 1 g1=g2 2 t1 Sensible cooling, can be achieved by passing the air through cooler coils. The final temperature must always be above the dew point temperature otherwise dehumidification will occur as water condenses on the coil, introducing mass transfer. 74 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Example Moist air enters a duct at 10 oC dry-bulb temperature and 80% relative humidity, with a volumetric flow rate of 120 m3/min. The air is sensibly heated at constant pressure of 1 atmosphere to 25 oC dry-bulb temperature. Determine the rate of heat transfer during the process in kW. ? Q = ma ( h2 – h1 ) 75 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Example Solution o o ? t 1 = 10 C , t 2 = 25 C ,V a = Given: 120 = 2 m3 /s , ? 1 = 80% 60 From the psychrometric chart: h1 = 25. 5 kJ/kg dry air , g 1 = 0. 0061 kg/kg dry air, 3 v1 = 0. 81 m /kg dry air mass flow of dry air: ? V 2 = = = 2. 47 kg/s ? ma v 0. 81 g 1 = g 2 hence state 2 of the process can be plotted on the psychrometric chart using the dry-bulb temperature and the moisture content. 76 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel UniversityExample Heating fluid in 1 Air in Heating fluid out 2 Air out 40. 8 kJ/kg 25. 5 kJ/kg 80% 1 2 0. 0061 kg/kg From chart h2 = 40. 6 kJ/kg dry air Hence, ? ? Q = ma ( h2 – h1 ) = 2. 47 ? (40. 6 – 25. 5) = 37. 3 kW 10 C 25 C 77 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Examples PF3-2
• Calculate the load on a cooler coil which sensibly cools 1. 5 m3s-1 of moist air initially at 21 oC db, 15 oC wb and 1013. 25 mb, by 5 oC. PF3-3
• Calculate the load on a heater battery which heats 1. 5 m3s-1 of moist air initially at 21 oC db, 15 oC wb and 1013. 25 mb by 20 oC.If low pressure hot water at 85 oC flow and 75 oC return is used, calculate the flow rate of water in kgs-1. Take the specific heat of water as 4. 2 kJkg-1K-1. 78 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Humidification
• Can be defined as: “The addition of moisture to the Air”
• In air conditioning systems, it is often necessary to increase the moisture content of the air supplied to the conditioned spaces.
• It may be accomplished either by the addition of water or steam. Each process is entirely different in terms of psychrometry and equipment required to produce the change. 9 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Humidification The most common methods of using water as a humidifying agent are:
• •
• The injection of water drops of aerosol size into the air duct or directly into the space being conditioned. The passage of air through a spray chamber containing a very large number of small water droplets (ie. air washer). The passage of air over a large wetted surface or pool of water (ie. pan humidifier). 80 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Humidification The humidification equipment does not operate at 100% efficiency. The efficiency definition is usually based on the overall change of state that the air undergoes.
• For a 100% efficient process, the final state point of the air would be on the 100% saturation line. 81 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Humidification 1 Air in pump 2 Air out h2 h1 h3 3 2 1 t2 g3 g2 g1 humidifier t1 The effectiveness of the humidification system can be defined as: E= ( h2 – h1 ) ( h3 – h1 ) or ( g 2 – g1 ) E= ( g 3 – g1 ) ? = 100 ? E 82 and the humidifying efficiency as:Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Water injection humidification In cases where all the water injected into the air is evaporated, the humidification process is illustrated in Figure. 1 Air in pump humidifier Energy balance 2 Air out h2 h1 h3 3 2 1 t2 t1 g3 g2 g1 ? ? ? ? ? ma1 ha1 + ms1 hs1 + m? h? = ma2 ha2 + ms2 hs2 but ? ? ? ma1 = ma2 = ma 83 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Water injection humidification Dividing above equation by ? magives ? mw ha1 + g 1 hs1 + h w = h a2 + g 2 h s 2 ? a but hence ha + g hs = h ? mw h1 + h w = h2 ? ma ? ? ? ? ? ma1 + ms1 + mw = ma2 + ms2 dividing through by ? ma gives: ? mw = g2 g1 + ? ma Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University 84 Example In a water injection humidification process, water at a temperature of 0 oC is injected into the air stream at a rate of 0. 002 kgs-1. If the air entering the process is at 20 oC dry-bulb and 15 oC wet-bulb, and the mass flow of dry air is 1. 5 kgs-1, determine the enthalpy and moisture content of the moist air at the end of the process and draw the process on the sychrometric chart. Assume that all the water injected is evaporated. 85 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Effect of water temperature on water injection humidification 1 Air in pump humidifier 2,3,4 Air out tw =0 C tw =100 C 23 4 7o 1 g2 g1 We can observe from this figure that the final condition of any water injection humidification process, where all the water is evaporated, will lie between the limits of the limits of the 0oC and 100oC processes and will be very close to the wet bulb temperature temperature process.It is therefore reasonable to assume for all practical purposes that a purposes that a water injection humidification process, where all the injected water evaporates, follows a constant wet bulb temperature line on the psychrometric psychrometric chart. 86 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Steam Injection Humidification 1 Air in steam Steam humidifier 2 Air out 2 3 ts =100 C ts =234 C 4o g2 g1 1
• Steam injection humidification is achieved by injecting steam through a number of fine nozzles into the air stream.If none of the injected steam condenses during the process, we can analyze the steam injection process by applying the mass and energy balance equations in a similar manner to that of the water injection process.
• For all practical purposes, we can therefore assume that in a steam injection humidification process, where none of the steam condenses, the change of state takes place along a constant dry bulb temperature line. 87 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University ExampleIn a steam injection humidification process, 0. 02 kg/s of dry saturated steam is injected into an air stream initially at 25 oC dry bulb and 12 oC wet bulb temperature. If none of the steam condenses and the mass flow rate of dry air is 2 kgs-1 determine the final enthalpy and moisture content of the air stream leaving the process. 88 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Dehumidification Although sensible heating reduces the percentage saturation of moist air, it does not reduce its moisture moisture content.In air conditioning applications, the the principal methods employed to reduce the moisture moisture content of moist air are: •cooling to a temperature below the dew point •chemical methods (sorbent dehumidifiers) 89 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Cooling to a temperature below the dew point Mechanical dehumidification Cooling fluid in Cooling coil d 2 Air out 1 1 3 g1 Cooling fluid out In the case of using cooler coils, tADP is known as the mean coil surface temperature. 3= tADP t1 ADP: Apparatus Dew Temperature 1-d-3: Assuming the process involves 100% of the air passing through: 90 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Cooling to a temperature below the dew point In practise, some of the air will escape the process (contact with the coil surface) and hence will keep its moisture. After the coil, this air will mix with the dehumidified air . Therefore, the actual process does not follow 1-3, but 1-2.In air conditioning applications, dehumidification process is represented by a straight line between 1 and 3, where 2 is on this line. 3 2 t3 t2 d 1 g1 g2 g3 t1 91 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Cooling to a temperature below the dew point h1 Cooling fluid in h2 1 2 h3 d 1 g1 g2 g3 Air out Cooling fluid out 3 2 t1 ? t3= tADP t2 The effectiveness of the dehumidification process is defined by the terms contact factor (1 – by-pass factor or the ? ) The contact factor is defined as: Similarly, by-pass factor: g 1 – g 2 ) ( h1 – h2 ) ? = = ( g 1 – g 3 ) ( h1 – h3 ) ( g 2 – g 3 ) ( h 2 – h3 ) (1 – ? ) = = ( g 1 – g 3 ) ( h1 – h3 ) 92 Building Services Engineering: Building Air Conditioning Module (ME5508),Brunel University h1 h2 h3 3 2 1 g1 g2 g3 t3= tADP t2 t1 93 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Cooling to a temperature below the dew point For all practical purposes, the above equations can be written in terms of the dry bulb temperature as follows: ( g 1 – g 2 ) ( h1 – h2 ) ? = ( g 1 – g 3 ) ( h1 – h3 ) ( – ) ? = t1 t 2 ( t1 – t 3 ) and, ( g 2 – g 3 ) ( h 2 – h3 ) (1 – ? ) = = ( g 1 – g 3 ) ( h1 – h3 ) ( t2 – t3 ) (1 – ? ) = ( t1 – t 3 ) For accurate calculations, we will not use this approximation! 94 Building Services Engineering: Building Air Conditioning Module (ME5508), Brunel University Example A cooling coil has an apparatus dew point of 6 oC and a contact factor of 80%. If moist air at 29 oC dry bulb temperature, 50% saturation enters the coil, determine the leaving dry bulb temperature and moisture content of the air.

Cite this page

Air Conditioning. (2019, Mar 19). Retrieved from https://paperap.com/paper-on-essay-air-conditioning/

Air Conditioning
Let’s chat?  We're online 24/7