8-538.74 (cont’d) The outside air is first cooled to a temperature at which the required amount of water is condensed, and the cold air is then reheated to 55°F. Since haremains constant in the second step, the condition of the air following the cooling step must lie at the intersection of the ha=0 0075.line and the saturation curve ⇒=°T49 F References:Same as Fig. 8.4-2 [including H OF2l,32°bg] substance ±min²Hin±mout²HoutAir HOF2l,49°76.5 — 45.5— 76.51.2 21.417.0±mairin lbmD.A./min ²Hairin Btu/lbmD.A. ±m2in lbm/min, ²H2in Btu/lbmQH==−−=Δ765 21445512 0009...,+1.2(17.0) (Btu)60 min1 ton coolingmin1 hBtu h.1 tons coolingb.6m7m2mom(76.5 lb DA/min)
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This note was uploaded on 01/07/2012 for the course CH E 210 at Pennsylvania State University, University Park.