![]() Countercurrent-induced draft towers are the most prevalent in the process industries, capable of cooling water within just over 1 ☌ (2 ☏) of the wet-bulb air temperature. Most packaged cooling towers use fans to force or draw air through the tower. $W =\dfrac) \times 58.2 = 1.7625 \times 1.2 \times 58.2 = 123. Stephen Hall, in Branans Rules of Thumb for Chemical Engineers (Fifth Edition), 2012. Determine the cooling, in kJ/kg dry air, required for this process. $p_v = 0.5 \times p_s = 0.5 \times 6.2795 = 3.13975$ kPa Humid air at 1 atm, 30C, and 60 percent relative humidity is cooled at constant pressure to the dew-point temperature. Since the relative humidity is 50%, the vapour pressure of water in air (p$_v$) is: Atmospheric air enters the cooling tower at 30 C, with. Makeup water is added in a separate stream at 20 C. A stream of cooled water is returned to the condenser at the same flowrate. Water exiting the condenser of a power plant at 45 C enters a cooling tower with a mas flow rate of 15,000 kg/s. An induced draft fan draws the air across the wetted fill and expels it through the top of the structure. Solve the following using the psychrometric chart. Find the humidity ratio, dew point temperature and enthalpy of moist air on this day.Īt 37$^o$C the saturation pressure (p$_s$) of water vapour is obtained from steam tables as 6.2795 kPa. air, however, is introduced at the side either on one side (single-flow tower) or opposite sides (double-flow tower). ![]() On a particular day the weather forecast states that the dry bulb temperature is 37$^o$C, while the relative humidity is 50% and the barometric pressure is 101.325 kPa.
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