Question

fluid

12 6 Problem 5.12 An evaporative cooling tower (see the figure below) is used to cool water from 110 to 80 °F . Water enters the tower at a rate of 250,000 lbm/hr. Dry air (no water vapor) flows into the tower at a rate of 151,000 lbm/hr. If the rate of wet airflow out of the tower is 156,900 Ibm/hr. determine the rate of water evaporation in lbm/hr and the rate of cooled water flow in lbm/hr. Wet Water in Air in Water out Solution 5.12 For steady flow of dry air

Answer 1

cooling tower is as Schematic Diagram of follows . ↑ Wet Air out Water - In ① Air In 5 Water ou Consider the flow of day air is steady let mz= mass flow rate of day air at inlet M2. Day alx = mass flow rate of day air at Section © . ma, water = mass flow rate of cuater of section ® now we can write miza miz, say air So that mass flow rate at section ② is given as me = M2, D24 aux + m2, water = mg + m2, water Ma, water = ma - miz o

- We have given. mg = 156,900 obm/hr and miz= so that from equation ®, the evaporation is given as : ma, water = 156,900 – 151,000 riz, water = 5,900 lbm/hr 151,000 (bm/hr rate of water det m = mass flow rate of warm water at section 0 We have given , mi = 250,000 lbm/ho Now rate of cooled water flow (ra) is given as . mia = mi - M2, water = 250,000 - 5,900 → ma= 244,100 Sbm/ ho