Prob #3-Spray drying is a process in which a liquid containing dissolved or susp
ID: 703636 • Letter: P
Question
Prob #3-Spray drying is a process in which a liquid containing dissolved or suspended solids is injected into a chamber through a spray nozzle or centrifugal disk atomizer. The resulting mist is contacted with hot air, which evaporates most or all of the liquid, leaving the dried solids to fall to a conveyor belt at the bottom of the chamber. Milk (70% H2O) Wet air 11 m/min @83°C,1 atm Dried milk 167°C,-40 cm H,O a Powdered milk is produced in a spray dryer 6 m in diameter by 6 m high. Air enters at 167°C and a -40 cm H2O gauge pressure. The milk fed to the atomizer contains 70% water by mass, all of which evaporates. The outlet gas contains 12 mole% water and leaves the chamber at 83°C and 1 atm (absolute) at a rate of 311 m/min. Calculate the production rate of dried milk and the volumetric flow rate of the inlet air. Estimate the upward velocity of air (m/s) at the bottom of the dryer.Explanation / Answer
For outlet wet air
Pressure P = 1 atm
Temperature T = 83 + 273.15= 356.15 K
Volume V = 311 m3/min
Gas constant R = 0.0821 m3-atm/kmol-K
From the ideal gas equation
Moles of wet air = PV/RT
= (1 atm x 311 m3/min) /(0.0821 m3-atm/kmol-K x 356.15K)
= 10.64 kmol/min
Water balance
Water in liquid inlet = water in wet air at oulet
0.70 x mass flow of liquid inlet = 10.64 kmol/min x 18 kg/kmol x 0.12
mass flow of liquid inlet = 32.9 kg/min
Solid balance
mass flow of liquid inlet x 0.30 = mass flow of dried milk at outlet
mass flow of dried milk at outlet = 0.30 x 32.9 = 9.8 kg/min
Moles of dry air inlet = (100 - 12)% x Moles of wet air
= 0.88 x 10.64
= 9.36 kmol/min
Volumetric flow rate of dry inlet air
= moles x Gas constant x temperature / pressure
= [9.36 kmol/min x 0.0821 m3-atm/kmol-K x 440 K] / [ (1033-40)cmH2O x 1 atm/1033cmH2O]
= 352 m3/min
Upward velocity of air
= Volumetric flow rate of dry inlet air / area of cross section
=( 352 m3/min x 1min/60s) / (3.14 x 6 x 6 m2 / 4)
= 0.21 m/s
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