Phân tích giản đồ i d trong kỹ thuật sấy

THE PSYCHROMETRIC CHART: Theory and Application Perry Peralta NC State University PSYCHROMETRIC CHART Identify parts of the chart „ Determine moist air properties „ Use chart to analyze processes involving moist air „ Evaporative cooling: Example 10 (a) At state point 1: T1 = 65°C Tw1 = 57°C Tdp1 = 56.3°C RH1 = 66.9% HR1 = 123.1 g/kg of dry air v1 = 1.15 m3/kg of dry air h1 = 387.7 kJ/kg of dry air At state point 2: T2 = 62°C Tw2 = 57°C Tdp2 = 56.5°C RH2 = 77.3% HR2 = 124.5 g/kg of dry air v2 = 1.14 m3/kg of dry air h2 = 387.7 kJ/kg of dry air Evaporative cooling: Example 10 (b) Drying rate = ( ∆HR )( w a ) VF wa = v2 VF = ( A )( air flow rate ) Evaporative cooling: Example 10  V   Pl + Ss A= St Sw   Pl St − Ss  Pl Pw Bt   2.5 3.6 + 0.6    A= 0.019 * 0.038   3.6 * 0.019 − 0.6  3.6 *1.2 * 0.025   A = 1.47 m Evaporative cooling: Example 10 A = 1.47 m VF = ( A )( air flow rate ) m m   VF = (1.47 m3 )   = 2.9 s  s  Evaporative cooling: Example 10 m3 VF =2.9 s VF wa = v2 m3 2.9 kg dry air s = 2.6 wa = m s 1.14 kg dry air Evaporative cooling: Example 10 kg dry air w a = 2.6 s Drying rate = ( w a )( ∆HR )  kg dry air   g  Drying rate =  2.6  1.4  s kg dry air    g kg = 3.6 = 13.0 s h Adiabatic Mixing of Moist Air Stream „ A psychrometric process that involves no net heat loss or gain during the mixing of two air streams Adiabatic mixing: Example 11 T2=26.7ºC RH2=80% VF2=28 m3/min T3=43.3ºC Tw3=37.8ºC T1=43.3ºC Tw1=37.8ºC VF1=112 m3/min 37.8°C 80% 26.7°C 43.3°C 0.87 m3/kg d.a 0.95 m3/kg d.a Adiabatic mixing: Example 11 VF wa = v m3 112 kg dry air minute w a1 = = 117.9 m minute 0.95 kg dry air wa2 m3 28 kg dry air minute = = 32.2 m minute 0.87 kg dry air Adiabatic mixing: Example 11 line 1-3 w a2 32.2 = = = 0.21 line 1-2 w a2 +w a1 32.2 + 117.9 Therefore, length of line segment 1-3 is 0.21 times the length of line 1-2 37.8°C 35.6°C 80% 26.7°C 40°C 43.3°C 0.87 m3/kg d.a 0.95 m3/kg d.a Adiabatic mixing: Example 11 T3 = 40.0°C Tw3 = 35.6°C ... falls below the dewpoint temperature Cooling and dehumidifying: Example Moist air at 50°C dry bulb temperature and 32% relative humidity enters the cooling coil of a dehumidification kiln heat pump... If outside air is at 27°C dry bulb temperature and 80% relative humidity, how much outside air is needed per minute to carry away the evaporated moisture? 80 % 52°C 27°C 92 g/kg d. a 18 g/kg d. a... increase or decrease in the temperature of air without changing its humidity ratio „ Example: passing moist air over a room space heater and of kiln air over the heating coils „ Sensible heating: Example