the isotherm line of t,=20C with the line of constant relative humidity p,=50% on the psychrometric chart at the given total pressure p=101 325Pa. The enthalpy of the point 2. h=38.5 k/(kg dry air) Point 1 is the intersecting of the parallel line of f=5000 from point 2 with the isotherm line t,= 12, as shown in Fig 7-7. Other properties of the supply air at the state point I can be obtained from the psychrometric chart h, =23 k/(kg dry air) The specific humidity and the dew-point temperature, d,=4.2g//kg dry air),La=1.8C The relative humidity and the wet-bulb temperature, =48%, 4 1=7C The mass of the dry air enters the room per hour Q10 The mass of the moist air enters the room per hour m=m2(1+0.001d)=645.16(1+0.0042)=648.3kg/h 7. 4 Analysis and Application of Thermodynamic Process of Atmospheric Air Maintaining a living space or an industrial facility at a desired temperature and humidity requires some ng pr They incl ng process, SI process, humidifying process(adding moisture), and dehumidifying process(removing moisture),etc Sometimes two or more of these processes are needed so as to bring the air to a desired temperature and humidity level. Therefore, some processes commonly encountered in air-conditioning engineering are presented 7.4.1 Simple Heating Process When moist air flows over a heating coil in which hot water flows, the temperature and the enthalpy of the air increase, but the relative humidity decreases, the specific humidity remains constant, as shown in Fig. 7-8(a). That is, during the process 1<2,h<h2 at T Heat <d Fig.7-8 Sensible heati notice that the relative humidity of the air decreases during a heating process even if the specific moisture at a given temperature to the maximum moisture it can hold at the same temperature, and vith te l18
118 the isotherm line of 2 t = 20 ℃ with the line of constant relative humidity 2 = 50% on the psychrometric chart at the given total pressure p=101 325Pa. The enthalpy of the point 2, 2 h = 38.5 kJ/(kg dry air) Point 1 is the intersecting of the parallel line of = 5000 from point 2 with the isotherm line 1 t = 12 , as shown in Fig 7-7. Other properties of the supply air at the state point 1 can be obtained from the psychrometric chart, The enthalpy , 1 h = 23 kJ/(kg dry air） The specific humidity and the dew-point temperature, 1 1 4.2g/(kg , 1.8 d d t = = dry air） ℃ The relative humidity and the wet-bulb temperature, 1 1 48%, 7 w = = t ℃ The mass of the dry air enters the room per hour, 2 1 10 000 645.16 kg/h 38.5 23 a Q m h h = = = − − The mass of the moist air enters the room per hour, 1 (1 0.001 ) 645.16(1 0.004 2) 648.3 kg/h m m d = + = + = a 7.4 Analysis and Application of Thermodynamic Process of Atmospheric Air Maintaining a living space or an industrial facility at a desired temperature and humidity requires some processes called air-conditioning processes. They include simple heating process, simple cooling process, humidifying process (adding moisture), and dehumidifying process (removing moisture), etc. Sometimes two or more of these processes are needed so as to bring the air to a desired temperature and humidity level. Therefore, some processes commonly encountered in air-conditioning engineering are presented. 7.4.1 Simple Heating Process When moist air flows over a heating coil in which hot water flows, the temperature and the enthalpy of the air increase, but the relative humidity decreases, the specific humidity remains constant, as shown in Fig.7-8(a). That is, during the process, , d1 = d2 1 2 1 2 1 2 t t ,h h , Notice that the relative humidity of the air decreases during a heating process even if the specific humidity d remains constant. This is because the relative humidity is the ratio of the amount of moisture at a given temperature to the maximum moisture it can hold at the same temperature, and moisture capacity increases with temperature. (a) (b) Fig.7-8 Sensible heating process of moist air
The simple heating process of moist air proceeds in the direction of increasing the dry-bulb temperature, following a line of constant specific humidity as shown in Fig 7-8(b), which appears to be a vertical line. During heating process, the moisture capacity of the air enhances. The heat added to the air can be calculated by (7-14) where h, and h, are the enthalpies per unit mass of dry air at the inlet and the exit of the heating 7.4.2 Simple Cooling Process A cooling process at constant specific humidity can be accomplished by making the air flow over some coils in which refrigerant or chilled water flows. And the external surface temperature of the coil is gher than the dew point temperature of the air, as shown in Fig. 7-9(a). After the air flows through the cooling coil, its dry-bulb temperature drops, and thus the enthalpy decreases, but the specific humidity remains constant and the relative humidity increases. However, the air temperature at the exit is still higher than the dew point temperature of the inlet air, that is d1=d2,1>h2>1,>h2,< ling coils Air T2 T,o1,bicol 2=01 7-9 Simple The simple cooling process is shown in Fig. 7-9(b) by the process line 1-2. The heat released during this cooling process is q=h-h k/(kg dry air where h, and h, are the enthalpies per unit mass of dry air at the inlet and exit of the cooling section, respectively Notice that simple heating and cooling processes appear as vertical lines on this chart since the humidity ratio remains constant (d=constant)during these processes 7.4.3 Cooling with Dehumidification As shown in Fig. 7-10(a), hot, moist air enters a cooling section at state 1. As it passes through the cooling coils, its temperature decreases and the relative humidity increases at constant specific humidity. If the cooling section is long enough, air reaches its dew point(state A, saturated air) Further cooling of air will result in the condensation of the moisture. Air remains saturated during the entire condensation process. And the process follows the saturated air line to proceed until it reaches the final state(state 2), as illustrated in Fig. 7-10(b) This process can achieve the purposes of cooling the air with dehumidification. It can only occur when the surface temperature of the cooling coil is below the dew point temperature of the moist air 119
119 The simple heating process of moist air proceeds in the direction of increasing the dry-bulb temperature, following a line of constant specific humidity as shown in Fig. 7-8（b）, which appears to be a vertical line. During heating process, the moisture capacity of the air enhances. The heat added to the air can be calculated by 2 1 ( ) Q m h h = − a （7-14） where 1h and h2 are the enthalpies per unit mass of dry air at the inlet and the exit of the heating section, respectively. 7.4.2 Simple Cooling Process A cooling process at constant specific humidity can be accomplished by making the air flow over some coils in which refrigerant or chilled water flows. And the external surface temperature of the coil is higher than the dew point temperature of the air, as shown in Fig. 7-9（a）. After the air flows through the cooling coil, its dry-bulb temperature drops, and thus the enthalpy decreases, but the specific humidity remains constant and the relative humidity increases. However, the air temperature at the exit is still higher than the dew point temperature of the inlet air, that is, 1 2 d d = , 1 2 1, 1 2 1 2 , , dew t t t h h The simple cooling process is shown in Fig.7-9 (b) by the process line 1-2. The heat released during this cooling process is 2 1 q h h = − kJ/(kg dry air） (7-15) where 1h and h2 are the enthalpies per unit mass of dry air at the inlet and exit of the cooling section, respectively. Notice that simple heating and cooling processes appear as vertical lines on this chart since the humidity ratio remains constant ( d =constant) during these processes. 7.4.3 Cooling with Dehumidification As shown in Fig. 7-10 (a), hot, moist air enters a cooling section at state 1. As it passes through the cooling coils, its temperature decreases and the relative humidity increases at constant specific humidity. If the cooling section is long enough, air reaches its dew point (state A, saturated air). Further cooling of air will result in the condensation of the moisture. Air remains saturated during the entire condensation process. And the process follows the saturated air line to proceed until it reaches the final state (state 2), as illustrated in Fig. 7-10 (b), This process can achieve the purposes of cooling the air with dehumidification. It can only occur when the surface temperature of the cooling coil is below the dew point temperature of the moist air. (a) (b) Fig.7-9 Simple cooling process of moist air
The amount of moisture removed △d=d2-d1g/ kg dry air) The amount of heat removed by the cooling water is q=(h2-h1)-(d1-d2 (7-17) Cooling coils 2=14c T1=30°C 中2=100%14°C Condensate ⅵ=10m/min removal (a) Fig 7-10 Dehumidifying(Removing Moisture)Process where h, is the specific enthalpy of the water condensed 7.4.4 Humidification Process (1)Adiabatic humidification When moist air passes through a humidifying section(process 1-2), in which water is sprayed into the airstream, as shown in Fig 7-11(a). Tiny water droplets absorb heat from the air and evaporate. As a result, the airstream is cooled during the process. And the temperature of the airstream decreases and its humidity increases(state 2). In the limiting case, the air leaves the humidification section as saturated air at state 3. This is the lowest temperature that can be achieved in this process essentially identical to the adiabatic saturation process since the latent heat the water absorbed comes from the moist air itself when water droplets evaporate. Therefore, the evaporative cooling process follows a line of constant wet-bulb temperature on the psychrometric chart. Note that this will not exactly be the case if the liquid water is supplied at a temperature different from the exit temperature of the airstream. ) Since the constant wet-bulb temperature lines almost coincide with the constant-enthalpy lines, the enthalpy of the airstream can also be assumed to remain constant In fact, the process is as the same as the adiabatic saturator process, but the outlet temperature does not reach saturated temperature of moist air state COOL HOT hn=hy MOIST Fig 7-11 Adiabatic Humidifying As the amount of the moisture added is very small, the magnitude ge in enthalpy be neglected. Thus, the enthalpy remains constant during adiabatic humidification process. On the h-d
120 The amount of moisture removed is 2 1 = − d d d g /(kg dry air) The amount of heat removed by the cooling water is 2 1 1 2 ( ) ( ) l q h h d d h = − − − （7-17） where l h is the specific enthalpy of the water condensed. 7.4.4 Humidification Process ⑴ Adiabatic humidification When moist air passes through a humidifying section (process1-2), in which water is sprayed into the airstream, as shown in Fig. 7–11(a). Tiny water droplets absorb heat from the air and evaporate. As a result, the airstream is cooled during the process. And the temperature of the airstream decreases and its humidity increases (state 2). In the limiting case, the air leaves the humidification section as saturated air at state 3. This is the lowest temperature that can be achieved in this process. The process is essentially identical to the adiabatic saturation process since the latent heat the water absorbed comes from the moist air itself when water droplets evaporate. Therefore, the evaporative cooling process follows a line of constant wet-bulb temperature on the psychrometric chart. (Note that this will not exactly be the case if the liquid water is supplied at a temperature different from the exit temperature of the airstream.) Since the constant wet-bulb temperature lines almost coincide with the constant-enthalpy lines, the enthalpy of the airstream can also be assumed to remain constant. In fact, the process is as the same as the adiabatic saturator process, but the outlet temperature does not reach saturated temperature of moist air state. As the amount of the moisture added is very small, the magnitude of the change in enthalpy can be neglected. Thus, the enthalpy remains constant during adiabatic humidification process. On the h-d (a) (b) Fig.7-11 Adiabatic Humidifying (a) (b) Fig.7-10 Dehumidifying (Removing Moisture) Process
diagram, this process proceeds along the constant enthalpy line in the direction of increasing the amount of moisture, as shown in Figure 7-11(b)(the process 1-2) In terms of mass conservation, the amount of water added equals to the increase in the amount of the moisture m2=0001mn(d2-d1)or=0.001(d2-d1) where subscript I represents liquid water. According to energy conservation, for adiabatic steady flow without external work, q=0 and w=0, there is mh1+0.001(d2-41mh=m2h2 h2+0.00(d2-d1m=h2 Enthalpy of water is relatively small, the difference of d,-d, is also very small. So the enthalpy of water added is negligible, that is, 0.001(d,,)h,0. Thus, h2 (7-18) As shown in Fig. 7-11(b), the process 1-2 will proceed along constant enthalpy line in direction of d o increasing Utilizing evaporative cooling in regions with hot and dry weather and abundant in natural water resources to replace chilled water for air conditioning is an energy saving technique (2) Humidification at constant temperature(Humidification through spraying water vapor) This is accomplished by passing the air through a humidifying section(process 2-3)into which steam is introduced as shown in Fig. 7-1l(a), this will result in humidification When dealing with moist air, a little amount of steam is sprayed into the air so as to humidify the air. On the psychrometric chart, this process will follow the route of 1-2, as shown in Fig. 7-11(b) Based on the mass and energy conservation, we can get d2=d1+×103 (7-19) h2-h1=0001(d2-4 mo After spraying steam into the air, the enthalpy, the humidity ratio and the relative humidity of the During this process, the ratio of the enthalpy change to the specific humidity change is equal to h,=2501+1.86f. It is the same as the slope of the isotherm line. Thus, the humidifying fulfilled by spraying steam into the air is also called humidifying process at constant temperatupocess If steam is available as a heat source, for example in textile mills, humidification at constant temperature can be adopted in this case 7. 4.5Adia batic Mixing of Air Streams In air-conditioning engineering many applications require the mixing of two airstreams, particularly for large buildings. Most manufactory and process plants, and hospitals require that the conditioned air be mixed with a certain fraction of fresh outside air before it is routed into the living space. The mixing is accomplished by simply merging the two airstreams, as shown in Fig. 7-12 The heat transfer with the surroundings is usually small, and thus the mixing processes can be assumed to be adiabatic Mixing processes normally involve no work interactions, and the changes in kinetic and potential energies, if any, are negligible. Then the mass and energy balances for the diabatic mixing of two airstreams reduce to
121 diagram, this process proceeds along the constant enthalpy line in the direction of increasing the amount of moisture, as shown in Figure 7-11 (b) (the process 1-2). In terms of mass conservation, the amount of water added equals to the increase in the amount of the moisture. 2 1 2 1 0.001 ( ) or 0.001( ) l l a a m m m d d d d m = − = − where subscript l represents liquid water. According to energy conservation, for adiabatic steady flow without external work, q w = = 0 and 0 , there is 1 2 1 2 0.001( ) m h d d m h m h a a l a + − = 1 2 1 2 0.001( ) l h d d h h + − = Enthalpy of water is relatively small, the difference of 2 1 d d − is also very small. So the enthalpy of water added is negligible, that is, 2 1 0.001( ) 0 l d d h − . Thus, 1 2 h h （7-18） As shown in Fig. 7-11(b), the process 1-2 will proceed along constant enthalpy line in the direction of d、 increasing. Utilizing evaporative cooling in regions with hot and dry weather and abundant in natural water resources to replace chilled water for air conditioning is an energy saving technique. ⑵ Humidification at constant temperature (Humidification through spraying water vapor) This is accomplished by passing the air through a humidifying section (process 2-3) into which steam is introduced as shown in Fig. 7–11(a), this will result in humidification When dealing with moist air, a little amount of steam is sprayed into the air so as to humidify the air. On the psychrometric chart, this process will follow the route of ' 1 2 − , as shown in Fig. 7-11 (b). Based on the mass and energy conservation, we can get, 3 2 1 10 v a m d d m = + （7-19） 2 1 2 1 0.001( ) v h h d d h − = − （7-20） After spraying steam into the air, the enthalpy, the humidity ratio and the relative humidity of the moist air increase. During this process, the ratio of the enthalpy change to the specific humidity change is equal to 2501 1.86 v h t = + . It is the same as the slope of the isotherm line. Thus, the humidifying process fulfilled by spraying steam into the air is also called humidifying process at constant temperature. If steam is available as a heat source, for example in textile mills, humidification at constant temperature can be adopted in this case. 7.4.5 Adiabatic Mixing of Air Streams In air-conditioning engineering, many applications require the mixing of two airstreams, particularly for large buildings. Most manufactory and process plants, and hospitals require that the conditioned air be mixed with a certain fraction of fresh outside air before it is routed into the living space. The mixing is accomplished by simply merging the two airstreams, as shown in Fig. 7–12. The heat transfer with the surroundings is usually small, and thus the mixing processes can be assumed to be adiabatic. Mixing processes normally involve no work interactions, and the changes in kinetic and potential energies, if any, are negligible. Then the mass and energy balances for the adiabatic mixing of two airstreams reduce to
Mass balance of dry air m,+m=m Mass balance of water vapor md, +mo,d =m d h Figure 7-12 Adiabatic mixing of two streams Energy balance of the moist air ma + ma2=mahe Eliminating m from the three equations above we obtain h,-h h-h h,-h (7-21) Thus we can conclude from the two equation above that the state of the mixture( state c)lies on the straight line connecting states I and 2 on the h-d chart when two airstreams at two different states(states I and 2)are mixed adiabatically, and the ratio of the distances 2-c and c-I is equal to the ratio of mass flow mo, to m, The mixing process of air streams can help to save energy consumption during air conditioning 【 Example7-5】 Moist air enters into a heater at41=25℃,tn=20℃. It is heated to 2=90 C, then it enters a drying chamber(Fig. 7-14). When it leaves the drying chamber, the emperature rises to 1=40 C. The local atmospheric pressure is p=0. 1013 MPa. Determine (1) d,, h, ta and p,(2)the amount of moisture which l kg dry air absorbs in the drying chamber, and (3) the amount of dry air needed to absorb l kg water and the heat absorbed in the heater Solution I Drying equipment is shown in Fig. 7-14 (1)In Fig 7-15, the intersection point of the isotherms line t,,=20 C and the relative humidity p=100% line is point a; the intersection point of the constant enthalpy line through point a with the isothermal line t=25C is point 1. We can read h,=h=56.5 kJ/(kg dry air) The intersection point of the constant specific humidity line through point I with the saturated line is point A, which is the dew point. We can read ta=17C The intersection point of the constant specific humidity line through point I with the vapor pressure conversion line is point B. We can obtain P,=1.9 kPa
122 Mass balance of dry air m m m a a ac 1 2 + = Mass balance of water vapor m d m d m d a a ac c 1 1 2 2 + = Energy balance of the moist air m h m h m h a a ac c 1 1 2 2 + = Eliminating mac from the three equations above, we obtain 2 1 1 1 2 2 a c c a c c m h h d d m h h d d − − = = − − 1 2 1 2 c c c c h h h h d d d d − − = − − （7-21） Thus we can conclude from the two equation above that the state of the mixture (state c) lies on the straight line connecting states 1 and 2 on the h d − chart when two airstreams at two different states (states 1 and 2) are mixed adiabatically, and the ratio of the distances 2-c and c-1 is equal to the ratio of mass flow ma1 to ma2 . The mixing process of air streams can help to save energy consumption during air conditioning process. 【 Example 7-5】 Moist air enters into a heater at 1 t = 25 ℃, 1 20 w t = ℃.It is heated to 2 t = 90 ℃,then it enters a drying chamber(Fig.7-14). When it leaves the drying chamber, the temperature rises to 3 t = 40 ℃. The local atmospheric pressure is p=0.1013 MPa. Determine (1) 1 d , 1 h , d1 t and v1 p , (2) the amount of moisture which 1 kg dry air absorbs in the drying chamber, and （3）the amount of dry air needed to absorb 1 kg water and the heat absorbed in the heater. 【Solution】Drying equipment is shown in Fig. 7-14. (1) In Fig. 7-15, the intersection point of the isotherms line 1 20 w t = ℃ and the relative humidity =100% line is point a; the intersection point of the constant enthalpy line through point a with the isothermal line t = 25 ℃ is point 1. We can read 1 56.5 kJ/(kg dry air a h h = = ）. The intersection point of the constant specific humidity line through point 1 with the saturated air line is point A, which is the dew point. We can read 1 17 d t = ℃. The intersection point of the constant specific humidity line through point 1 with the vapor pressure conversion line is point B. We can obtain v1 p =1.9 kPa. (a) (b) Figure 7-12 Adiabatic mixing of two streams (Adding Moisture) Process