Chapter 14 Gas Vapor Gas Vapor Mixtures and Air Conditioning Air Conditioning Study Guide in PowerPoint to accompany Thermodynamics An Engineering Approach 8th edition by Yunus A engel and Michael A Boles 1 We will be concerned with the mixture of dry air and water vapor This mixture is often called atmospheric air The temperature of the atmospheric air in air conditioning applications ranges from about 10 to about 50oC Under these conditions we treat air as an ideal gas with constant specific heats Taking Cpa 1 005 kJ kg K the enthalpy of the dry air is given by assuming the reference state to be 0oC where the reference enthalpy is taken to be 0 kJ kga The assumption that the water vapor is an ideal gas is valid when the mixture temperature is below 50oC This means that the saturation pressure of the water vapor in the air vapor mixture is below 12 3 kPa For these conditions the enthalpy of the water vapor is approximated by hv T hg at mixture temperature T The following T s diagram for water illustrates the ideal gas behavior at low vapor pressures See Figure A 9 pressures A 9 for the actual T T ss diagram diagram 2 The saturated Th t t d vapor value l off the th enthalpy th l is i a ffunction ti off temperature t t and d can be b expressed as Note For the dry air water vapor mixture the partial pressure of the water vapor in the mixture is less that its saturation pressure at the temperature Pv Psat Tmix 3 Consider increasing the total pressure of an air water vapor mixture while the temperature of the mixture is held constant See if you can sketch the process on the P v diagram g relative to the saturation lines for the water alone g given below Assume that the water vapor is initially superheated P v When the mixture pressure is increased while keeping the mixture temperature constant the vapor partial pressure increases up to the vapor saturation pressure at the mixture temperature and condensation begins Therefore the partial pressure of the water vapor can never be greater than its saturation pressure corresponding to the temperature of the mixture 4 Definitions Dew Point Tdp p The dew point is the temperature at which vapor condenses or solidifies when cooled at constant pressure Consider cooling an air water vapor mixture while the mixture total pressure is held constant When the mixture is cooled to a temperature equal to the saturation temperature for the water vapor partial pressure condensation begins g When an atmospheric air vapor mixture is cooled at constant pressure such that the partial pressure of the water vapor is 1 491 kPa then the dew point temperature of that mixture is 12 12 95 95oC C Steam 375 325 275 T C 225 175 1 491 kPa 125 75 TDP 25 25 0 2 4 6 s kJ kg K 8 10 12 5 Relative Humidity Mass of vapor in air m v Mass off in saturated air mg Pv Pg Pv and Pg are shown on the following T s diagram for the water vapor alone Steam 125 T C 75 Tm 25 o Vapor State Pg 3 169 kPa Pv 1 491 kPa Tdp 25 0 2 4 6 8 10 12 s kJ kg K Since Pg Pv 1 or 100 Pv 1491 kPa 0 47 Pg 3169 kPa 6 Absolute humidity or specific humidity sometimes called humidity ratio Mass of water vapor in air mv M Mass off dry d air i ma PVM Pv M v v v Ru T PVM R T Pa M a a a u 0 622 Pv Pv 0 622 Pa P Pv Using the definition of the specific humidity humidity the relative humidity may be expressed as 0 622 Pg P and 0 622 Pg P Pg Volume of mixture per mass of dry air v v V m R T P m m m m ma ma After several steps we can show you should try this v V RT va a m ma Pa 7 So the volume of the mixture per unit mass of dry air is the specific volume of the dry air calculated at the mixture temperature and the partial pressure of the dry air Mass of mixture mv m ma mv ma 1 ma 1 ma m a Mass flow rate of dry air Based B d on th the volume l flflow rate t off mixture i t att a given i state t t the th mass flow fl rate t off dry d air is V m a v m3 s kga 3 m kga s Enthalpy of mixture per mass dry air h Hm Ha Hv ma ha mv hv h ma ma ma ha hv 8 Example 14 1 Atmospheric p air at 30oC 100 kPa has a dew point p of 21 3oC Find the relative humidity humidity ratio and h of the mixture per mass of dry air Pv 2 548 kPa 0 6 or 60 Pg 4 4 247 247 kPa 9 2 548 kPa kgv 0 622 0 01626 100 2 548 kPa kga h ha hv C p a T 25013 182 T kJ kgv kJ o o 1 005 30 C 0 01626 25013 182 30 C o kga C kga kgv kJ 71 71 kgga 10 Example 14 2 If the atmospheric p air in the last example p is conditioned to 20oC 40 percent p relative humidity what mass of water is added or removed per unit mass of dry air At 20oC Pg 2 339 kPa Pv Pg 0 4 2 339 kPa 0 936 kPa w 0 622 Pv 0 936 kPa 0 622 P Pv 100 0 936 kPa 0 00588 kgv kga The change in mass of water per mass of dry air is mv 2 mv 1 mv 2 mv 1 ma ma 2 1 0 00588 0 01626 kgv 0 01038 kga kgv kga 11 Or as the mixture changes from state 1 to state 2 0 01038 kg of water vapor is condensed for each kg of dry air Example 14 3 Atmospheric air is at 25oC 0 1 MPa 50 percent relative humidity If the mixture is cooled at constant pressure to 10oC C find the amount of water removed per mass of dry air Sketch the water vapor states relative to the saturation lines on the following g T s diagram T At 25oC Psat 3 170 kPa and with 1 50 s Pv 1 1 Pg 1 0 5 3 170 kPa 1 585 kPa Tdp 1 Tsat Pv 13 8o C 12 w1 0 622 Pv 1 P Pv 1 0 01001 0 622 15845 kPa 100 15845 kPa kgv kga Therefore when the mixture gets cooled to T2 10oC Tdp 1 the mixture is saturated and 2 100 100 Then Pv 2 Pg 2 1 228 1 228 kPa kPa w2 0 622 Pv 2 P Pv 2 0 00773 0 622 1 228 kPa 100 1 228 …