Answer:
a) steam used = 8440 kg/hr
b) Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection
Explanation:
Saturated steam pressure = 42KPag=1.42 bar
From steam table, Steam temperature = 110 C
Latent heat of this steam = 2230 KJ/kg
Process side pressure = 20 Kpaa = .2 bar
Water latent heat at this Pressue = 2360 KJ/kg
Water boiling Point at this Pressure =60 C
Feed Inlet temperature = 15.6 C
Total Heat required = Heat required to rise feed temperature from 15.6 to 60 degree C + Evaporation of water to concentrate the feed
Total Water evaporation required = 9072*(100-10)/100-9072*.1/.5*.5=7257.6 Kg/hr
Specific heat of feed assumed = 4.2 KJ/kg/K
=> Total heat required = 9072*4.2*(60-15.6)+7257.6*2360=18819682 KJ/hr
LMTD = ((110-60)-(110-15.6))/LN((110-60)/(110-15.6))=69.8 C
U, Overall Heat transfer coefficient = 1988 W/m2/K
Total heat required = U*A*LMTD
=> Area required of evaporator, A=18819682 *1000/3600/1988/59.8=44 m2
Steam used = 18819682/2230=8440 kg/hr
Energy required to condense vaporised feed = 7257.6*2360=17127936 KJ/hr
=> Steam efficiency = (18819682-17127936)/18819682=9%
b) Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection
