Question

Answer 1

= 0.4m/ TOBT= 16°C RH = 604. o v=0.2m/ TOBT: 28c RH = 50%. RH sp humidity -0.0076 kg dA @ state point I from psychometric a o specific volume = 0.83 m3/kg, w ke specific enthalpy 33 kJ/kg = und kg of dA in = 0.482 kg/ @ state pointe 2 From psychometric chart o 0.87 m/kg, we 0.012 in kg . ha 58 kJ/kg - in der 0.2 = M2 x 0.87 in = 0.2298 kg/s Now Applying Energy balance mxt, & ita = inzxt 3 0:482* 16 + 0.2298x28-(0:482+0.22999 tz = 19.874°C DBT of mixture Now, applying "mass of vapour consenation mw, & mior a moz mw → mass of 3 x 3 Inge mass flow rate of mixture 182 +0.2298) 3 = DBT of mixture Water vapour)

Now we know that w mo, mai w sp. humidity mw = mass of water rapour ma = mass of dry air mw, ma, xw, Now, mwit mer = mwe maxw,+ na xwe = maz* wz So = 0.48220.007 + 0.2298 0.012 z wz 0.482 +0.2298) W3 = 0.00 Mixture -0.0086 kg/kg of dA [Meeste hemidity] Now, from psychometric chart corresponding to W, = 0.0086 and T3 = 19.87 c Relative humidity & 59.. at state point 3 or the mixture relative humidity - LID and wer Bull temperature - 14°C and specific enthalpy ~ 41 kJ Now. mixture is cooled sensibly to igoc at state point 3 Po 3=41 kg - &1kg at state point - 4 h4 = 39 ks (Froment) son cooling load in [he-in] [0.482 +0.2298) *[41-39] 14236 KW 18 19.8 To Br