An air conditioning system of an indoor ice skating rink processes V = 400.0 L/s of air. The air inlet is at T = 25 C, P = 1 atm, and has a relative humidity of rh = 93.0% (it is a very humid day). The air conditioner's outlet expels air at Tz = 0.01 C and P2 = 1 atm. You would like to determine how much water has condensed from the air and the rate of heat removed from the air. Follow these steps. VLAS n, mol air/s TC P, = 1 atm rh % mol water/mol air Via mol dry air/mol air Air conditioning in, mol air/s unit T, C P, = 1 atm y mol water/mol air y mol dry air/mol air r, mol liquid water/s T, = 0.01 C What is the flow rate of air into the AC unit? 4.08528 n = mol/s Incorrect What is the flow rate of water vapor into the AC unit? 0.1205 niw= mol H2O(g)/s Incorrect What is the flow rate of dry air into the AC unit? 3.96478 NIDA = mol D.A./S Incorrect What is the flow rate of dry air out of the AC unit? 3.96478 n3DA = mol D.A./s Incorrect What is the flow rate of water vapor out of the AC unit? 0.0158 n3w = mol H2O(g)/s Incorrect What is the flow rate of water condensing out of the AC unit? 0.1047 n2 = mol H, O(l)/ Incorrect In the remaining questions, use for air a reference state of air at 25 C and 1 atm and for water a reference state of liquid water at 0.01 C. Assume that changes in enthalpy due to pressure are insignificant. What is 1Da, the specific enthalpy of air at 25 C? DA = 0 kJ/mol What is the specific enthalpy of air at 0.01 C? 0 N ZDA = kJ/mol Incorrect What is the specific enthalpy of water vapor at 25 C? Note the units. 1.88 lw(g) (g) = kJ/mol Incorrect What is the specific enthalpy of water vapor at 0.01 C? 0 n zw(g) = kJ/mol Incorrect What is the specific enthalpy of liquid water at 0.01 C? 2wl) = 0 kJ/mol What is the rate of heat added to the AC unit? 0 o = = kJ/s Incorrect