0.1 mole% caustic soda (NaOH) solution is to be concentrated in a continuous eva

Question

0.1 mole% caustic soda (NaOH) solution is to be concentrated in a continuous evaporator. The solution enters the unit at 25degreeC at a rate of 150 mol/min and is concentrated to 5 mol% at 50degree C. Hot dry air at 200degreeC and 1.1 bar absolute is bubbled through the evaporator and leaves saturated with water at 50degreeC and 1 atm. Calculate the required volumetric flow rate of the entering air and the rate at which heat must be transferred to or from the unit. Assume that the heat capacity per unit mass of all liquid solutions is that of pure liquid water.

Explanation / Answer

Cp of liquid water at 25 oC= 75.28 J mol-1 K-1

Heat absorbed by the liquid solution Qs= ms * Cp * (T2s-T1s) ............ (ms= molar flow rate of the solution)

= 150 * 75.28 * (50-25)

Qs = 282300 J/min = 282.3 kJ/min

Heat transfer rate to & from the unit = 282.3 kJ/min

Now, Heat absorbed by the liquid solution is equal to lost by the hot air.

QA= - MA * CpA * (T2A-T1A) = 282.3 kJ/min (MA= mass flow rate of air kg/min)

Cp of air at 200 oC is 1.026 kJ kg-1 K-1

Put this value in the above equation and solve for mA: mas flow rate of air in the entering stream

MA=mass flow rate of air= QA/ (CpA * (T1-T2)) = 282.30/ (1.026*150)= 1.8343 kg/min

Density of air at 200 oC is: 0.7461 kg/m3

Volumetric flow rate of air= 1.8343 /0.7461 = 2.45 m3/min

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