HVAC MODULE 1.pdf

1.3 COURSE LEARNING OUTCOMES: By the end of this course, students are expected to: TLO1. understand the Principle of Ventilation and Air conditioning TLO2. identify the properties of moist air. TLO3. apply the use of Psychrometric equation and chart in determining the propertie...

1.3 COURSE LEARNING OUTCOMES: By the end of this course, students are expected to: TLO1. understand the Principle of Ventilation and Air conditioning TLO2. identify the properties of moist air. TLO3. apply the use of Psychrometric equation and chart in determining the properties of moist air. TLO4. understand the psychrometric process of air. 1.4 Air conditioning – process of simultaneous control of temperature, moisture content, air movement and quality of air in space. control of temperature – increase or decrease of temperature. moisture content – humidity or the amount of water vapor. air movement – air distribution or regulation of air velocity. quality of air – cleanliness or removal of particles. 1.4.1 Type of Air conditioning: A. Comfort air conditioning – process of treating air to control simultaneously its temperature, humidity, cleanliness and air distribution to meet the comfort requirement of the occupants of the conditioned space. B. Industrial air conditioning – providing at least a partial measure of comfort for the workers in hostile environment but also controlling air condition so that they are favorable to processing some object or materials. 1.5 Psychrometry – study of the properties of mixture of air and water vapor. 1.5.1Teminologies : A. Dry air – non condensing component of the mixture, mainly nitrogen and other inert gases. B. Moist air – binary mixture of dry air and water vapor. C. Vapor – condensable components of the mixture, the water vapor or steam which may exists in a saturated or superheated state. D. Saturated air – means that the vapor in the air is saturated. E. Unsaturated / superheated air – air containing superheated vapor. 1.5.2 Properties of Air a. Relative humidity, ɸ – ratio of the actual partial pressure of water vapor in the moist air to the saturated pressure corresponding to the temperature of dry air. ɸ = ( Ps/ Pd ) x 100% where: Ps = partial pressure of water vapor. Pd = saturation pressure corresponding to the dry bulb temperature of air. b. Humidity ratio, W – also called absolute humidity or specific humidity. – amount of water vapor per kilogram (or per pound ) of dry air. W = 0.622 (Ps/ Pa ) = 0.622 (Ps / PT - Ps ) where: Pa = partial pressure of dry air PT = total pressure Derivation: From Dalton’s Law: PT = Pa + Ps and from Ps Vs = ms Rs Ts.. P a Va = ma Ra Ta W = m s / ma W = [Ps Vs / Rs Ts ] / [ Pa Va / Ra Ta ] W = ( Ps / Rs ) / ( Pa / Ra ) and Vs = Va ; Ts = Ta where : Vs = volume of water vapor Va = volume of dry air Ts = temperature of water vapor Ta = temperature of dry air Rs = gas constant of water vapor = 0.4615 KJ / kg o K  Ra = gas constant of dry air = 0.287 KJ / kg o K W = ( Ps / 0.4615 KJ / kg o K) / ( Pa / 0.287 KJ / kg o K) W = ( 0.287 / 0.4615 ) ( Ps / Pa ) W = 0.622 ( Ps / Pa ) c. Enthalpy, h – heat content of mixed air. h = ha + hs where: ha = specific enthalpy of dry air = Cp ( T ) hs = specific enthalpy of water vapor = W hg h = Cp ( T ) + W hg where: Cp = specific heat of dry air. T = dry bulb temperature of dry air. W = humidity ratio of moist air. hg = enthalpy of saturated vapor (steam) at temperature of air-water vapor mixture. Note: Approximate formula for hg ( based for 00 C reference temperature ) hg = 2,501 + 1.863 ( T ) where: hg at 0 0C so that, h = Cp ( T ) + W [ 2,501 + 1.863 ( T ) ]  d. Specific volume, v – volume of mixed air per kilogram of dry air. v = R T / Pa = R T / PT – Ps e. Dry bulb temperature, T db – is the temperature of air as measured by an ordinary thermometer. f. Wet bulb temperature, T wb – is the temperature of air as registered by a thermometer whose bulb is covered by a wetted gauze and exposed to a current rapidly moving air. g. Dew point temperature, Tdp – is the saturation temperature corresponding to the actual partial pressure of the water vapor. – temperature at which condensation of moisture begins when air is cooled at constant pressure.0 Tdp = TSAT ] Ps 1.5.3 Psychrometric Chart – a plotted data used for determining the condition of air such as the amount of water, relative humidity and other useful information on any point at a particular state.  Fig. 1 Psychrometric Chart Example 1 Moist air at dry bulb temperature of 250C has a relative humidity of 50% when the barometric pressure is 101.325 KPa. Determine : a.) partial pressure of water vapor and dry air b.) dew point temperature c.) specific humidity or humidity ratio d.) enthalpy e.) specific volume Solution: a. Ps  ɸ = ( Ps / Pd ) x 100% Pd = PSAT] Tdb = PSAT] 25C = 3.169 KPa Ps = ɸ Pd = 0.50 (3. 169 ) Ps = 1.58 KPa Ans. Pa = 101.325 – 1.58 = 99.745 KPa Ans. b. Tdp Tdp = TSAT ] Ps = Tdp = TSAT ] 1.58 KPa Tdp = 13.87 0 C Ans. c. W W = 0.622 ( Ps / Pa ) = 0.622 (Ps / PT - Ps ) W = 0.622 [( 1.58 ) / ( 101.325 – 1.58 ) ] W = 0.00985 kgwv / kg da Ans. d. h h = Cp ( T ) + W [ 2,501 + 1.863 ( T ) ] h = 1.0 KJ/kg 0K ( 25 ) 0K + 0.00985 kgwv / kg da [ 2,501KJ/kg + 1.863 KJ/kg 0K ( 250K ) Note: Cp ( T ) is actually Cp ( ΔT ) = Cp ( T – 00C ) 0K , for a reference temperature of 00C. But for convenience it can be written Cp ( T ) and unit is 0 K , since it is ΔT. Same for 1.863 (T), it is actually 1.863 KJ/kg 0K ( ΔT ) = 1.863 KJ/kg 0K ( T - 00C ) hence, h = 50.09 KJ / kg da Ans.  e. v v = R T / Pa = R T / PT – Ps v = (0.287 KN – m) /kg 0K) (25 + 273) 0K / (101.325 – 1.58) KN/m v = 0.857 m3 / kg Ans. Using Psychrometric chart: Fig.2 Example 1 Example 2. A certain air has a dry bulb temperature of 340C and a wet bulb temperature of 250C. Determine: a.) relative humidity b.) dew point temperature c.) humidity ratio d.) enthalpy e.) specific volume. Solution: a. ɸ = ( Ps / Pd ) x 100% From Carrier’s equation: Ps = PS’ – [ ( PT - PS’ ) ( Tdb – Twb ) / 2830 – 1.44 (Twb ) ] → psi Tdb = 340C = 93.20F ; Twb = 250C = 77 0F PT = atmospheric pressure / total pressure = 14.7 psi PS’ = PSAT ] Twb = 25 C = 3.169 KPa x 14.7 psi / 101.325 KPa = 0.4597 psi hence, Ps = 0.4597 lb/in2 – [ ( 14.7 – 0.4597) lb/ in2 ( 93.2 – 77) 0R / 2830 Btu/lb – 1.44Btu/lb 0R (77 0R) ] Ps = 0.375 lb/in2 x 101.325 KP / 14.7 lb/in2 = 2.58 KPa Pd = PSAT] T=34C = 5.324 KPa so that, ɸ = ( 2.58 / 5.324 ) x 100% = 48.46 % Ans. b. Tdp = TSAT] Ps = 2.58 KPa = 21.56 0C Ans. c. W = 0.622 ( Ps / Pa ) = 0.622 (Ps / PT - Ps ) = 0.622 [( 2.58 ) / ( 101.325 – 2.58)] W = 0.01625 kg wv / kg da Ans. d. h = Cp ( T ) + W [ 2,501 + 1.863 ( T ) ] h = 1.0 ( 34 ) + 0.01625 [ 2,501 + 1.863 ( 34) ] h = 75. 67 KJ/kg Ans.  e. v = R T / Pa = R T / PT – Ps v = (0.287) ( 34 + 273) / (101.325 – 2.58) v = 0. 892 m3 / kg Ans. Using Psychrometric chart: Fig.3 Example 2 Example 3. A mixture of air and water vapor at 101.3 KPa and 160C has a dew point temperature of 5 0C. Determine a.) relative humidity b.) specific volume of dry air and water vapor c.) enthalpy. Solution: a. ɸ = (Ps / Pd) x 100% Ps = PSAT] Tdp = 5C = 0.8721 KPa Pd = PSAT] Tdp = 16C = 1.8181 Kpa  ɸ = ( Ps / Pd ) x 100% = (0.8721 / 1.8181) x 100% = 47.97% Ans. b. v = R T / Pa = R T / PT – Ps = (0.287)(16 + 273) / (101.3 – 0.8721) v = 0.826 m3 / kg Ans. c. h = Cp ( T ) + W [ 2,501 + 1.863 ( T ) ] and W = 0.622 ( Ps / Pa ) = 0.622 (0.8721) / (101.3 – 0.8721) W = 0.00540 kg wv / kg da so that, h = 1.0 ( 16) + (0.00540)[ 2,501 + 1.863(16)] h = 29.66 KJ/kg Ans. Using Psychrometric chart: Fig.4 Example 3 Example 4. An air conditioned room is maintained at a room temperature of 210C and a relative humidity of 55% when barometric pressure is 740 mm Hg. A. Calculate the humidity ratio of the air-water vapor mixture. B. Calculate also the temperature of the inside of the window if moisture is beginning to form in the inside window. C. What mass of water vapor per kg dry air in the room must be removed from the mixture in order to prevent condensation on the window when the temperature drops to 40C? D. Calculate also the relative humidity to satisfy the condition if the temperature remain at 210C. The barometric pressure remains constant. Solution: A. W = 0.622 ( Ps / Pa ) and ɸ = ( Ps / Pd ) x 100% PT = 740 mmHg ( 101.325 KPa / 760 mm Hg) = 98.650 KPa Pd = PSAT] 21 C = 2.487 KPa hence, Ps = 0.55 (2.487) = 1.37 KPa therefore, W = 0.622 [1.37 / ( 98.650 – 1.37 ) ] = 0.00875 kg wv / kg da Ans. B. Tdp = TSAT] Ps = 1.37 KPa = 11.62 0C Ans. C. mass of water to be removed = W 11.62 C – W 4 C W 4C = 0.622 ( Ps 4 C / PT - Ps 4 C ) Ps 4 C = PSAT ] T = 4C = 0.8131 KPa hence, W 4C = 0.622 [( 0.8131 / 98.650 – 0.8131 )] = 0.005168 kg wv / kg da so that , mass of water to be removed = 0.00875 – 0.005168 mass of water to be remove = 0.00358 kg wv / kg da Ans. D. ɸ = ( Ps / Pd ) x 100% = (0.8131 / 2.487) x 100% ɸ = 32.69% Ans. 1.6 Basic Psychrometric Processes: Fig. 5. Basic Psychrometric Processes Process : 0 → 1 : Sensible Heating 0 → 2 : Sensible Cooling 0 → 3 : Humidifying 0 → 4 : Dehumidifying 0→ 5 : Heating And Humidifying 0→6: Cooling And Dehumidifying  0→7: Cooling And Humidifying 0→8: Heating And Dehumidifying 1.6.1 Sensible Heating – increase in temperature at constant humidity ratio. Fig.6 by energy balance: ma = ma1 = ma2 Q = ma ( h2 – h1 ) Q = ma Cpm ( T2 – T1 ) where: Q = heat ma = mass flow rate of air h1 and h2 = incoming and leaving enthalpy of air respectively T1 and T2 = incoming and leaving temperature of air respectively Cpm = specific heat of moist air at constant temperature  and Cpm = 1 + W (1.863) → KJ / kg 0K or Cpm = 1 + W (0.444) → Btu / lb 0R where: 1.863 KJ / kg 0K and 0.444 Btu / lb 0R are specific of steam or water vapor in SI and English unit respectively. 1.6.2 Sensible Cooling – decrease in temperature @ constant humidity ratio. Fig 7 by energy balance: ma = ma1 = ma2 Q = ma ( h1 – h2 ) Q = ma Cpm ( T1 – T2 ) where: Q = heat  ma = mass flow rate of air h1 and h2 = incoming and leaving enthalpy of air respectively T1 and T2 = incoming and leaving temperature of air respectively Cpm = specific heat of moist air at constant temperature and Cpm = 1 + W (1.863) → KJ / kg 0K or Cpm = 1 + W (0.444) → Btu / lb 0R Example 1. Determine the quantity of heat needed to raise 14 m 3/ min of air to 35 0C and 40% relative humidity from its saturation condition. Atmospheric pressure is 101.325. Solution: Using Psychrometric equation: Q = ma Cpm ( T2 – T1 ) and Cpm = 1 + W (1.863) but W2 = 0.622 [ ( Ps2 / (PT – Ps2 )] ɸ2 = ( Ps2 / Pd2) x 100% Pd2 = PSAT] 35 C = 5.628 Kpa Ps2 = (0.40) (5.628) = 2.2512 Kpa hence, W2 = 0.622 [ ( 2.2512 / (101.325 – 2.2512 )] = 0.01413 kg wv / kg da and W 1 = W2 therefore , Cpm = 1 + (0.01413) (1.863) = 1.0263 KJ / kg 0K for ma,  v1 = V1 / ma m a = V 1 / v1 v1 = R T1 / Pa1 = R T1 / (PT – Ps1 ) Note: For W1 = W2 , Ps1 = Ps2 = 2.2512 Kpa since the initial condition of air is saturated, T 1 = Tdp1 = TSAT] PS1= 2.2512 Kpa = 19.38 0C so that , v = R T1 / (PT – Ps1 ) = (0.287)(19.38 + 273) / ( 101.325 – 2.2512 ) v = 0.847 m3/kg hence, ma = V1 / v1 = 14 m3/min / 0.847 m3/kg = 16.529 kg/min therefore, Q = 16.529 kg/min (1.0263 KJ / kg 0K ( 35 – 19.38 ) 0K Q = 264.97 KJ/min Ans. Psychrometric representation: Fig.8 Example 2. Find the heat transfer required to cool 42.5 m3/min of air at 35 0C and 50% relative humidity to 26 0C. What is the final relative humidity? Barometric pressure is 760 mm Hg. Solution: A. Q = ma Cpm ( T1 – T2 ) and Cpm = 1 + W (1.863) W1 = 0.622 [ ( Ps1 / (PT – Ps1 )] ɸ1 = ( Ps1 / Pd1) x 100% Pd1 = PSAT] 35 C = 5.628 KPa Ps1 = ɸ1 Pd1 = 0.50 (5.628 ) = 2.814 KPa so that, W1 = 0.622 [ ( 2.814 / ( 101.325 – 2.814 )] = 0.01776 kg wv / kg da Cpm = 1 + (0.01776 ) (1.863) = 1.033 KJ / kg 0K and m a = V 1 / v1 and v1 = R T1 / Pa1 = R T1 / (PT - Ps1 ) v1 = (0.287) ( 35 + 273 ) / (101.325 – 2.814 ) v1 = 0.897 m3/kg ma = 42.5 / 0.897 = 47.38 kg/min therefore, Q = 47.38 kg/min (1.033 KJ / kg 0K) ( 35 – 26) 0K Q = 440.49 KJ/ min Ans. B. ɸ2 = ( Ps2 / Pd2) x 100% Pd2 = PSAT] 26 C = 3.363 KPa since, W1 = W2 , therefore Ps1 = Ps2 = 2.814 Kpa ɸ2 = ( 2.814 / 3.363 ) x 100% = 83.5% Ans. Psychrometric representation: Fig.9