Question

If the cooling water (cp = 4180 J/kg·ºC) from the above problem is to be provided evenly by 10 cooling towers. Humid air enters each tower at 1 atm, 15 ºC and 20% relative humidity, and leaves at 20 ºC and 0.014 kg of water/kg of dry air. Determine the required volume flow rate of air through each cooling tower.

Air exit 20 °C, 0 = 0.014 kg of water/kg of air Warm water 40 °C Air inlet 1 atm 15 °C, o = 20%, 30 °C Cool water Makeup water

Answer 1

+ hater til wate 4 1 makeup water 6.3 204 Ti = 150 0.014 ( ks legal Ta 20° Ww= T3 = 46° T4= 30° for. Ta ise 0, 20% 0.0021 lagu lesa) Wya water man .W, + siz my + mi az. We (mis, na l wa-wi) (mis -mu) L makeupwater [ima, masina day ain ]

Energy balance: & mihi Emehe ry.hu na hot riz h3= m'achat mia. hat maihi + 2 mizhz = [miz - mal wsuwo ]" m'a [( he-he) - (W) - m) by] n's I ha.hu) mis (cp) ( T3-Tu) [che-he) - (hy-wo,) bu mia --- Note ma (mis) lcp) C T3 - Tu) b) -- -- Mg not given.. from psychometric chart: Ø 2014 @ 7,- ppc 20-42 (t5|23.0) hit (ksel aga) 0.014 20°C, @ Tz 55.64 ( 15 dega) ha = 88 pse ) Steam-tables @ TB4= 30°C, h4 = 12574 (kJ/kg)

ra a) [ 4.16) (40-20) (55-64 20:42) - (0.014-0.002) (125.71) (mig) [ 1.24) - mia 11.24 ) riz mall flow rate of water thir) at inlet. mia. Vi volumetric How rate of air at inlet. from psychometric chalt: @T, = 15°C $ -20%. Via 0.819 (m 3 /kg) via wax & (1. 24). m3 X 0.819 miz ( 1.01556) are Vi (!.0/2.) 13 G) lu * volumetric flow rate of at inlet for each looling tower