1 metric ton of rice is to be dried from 25% moisture to 13% moisture by exposin
ID: 533733 • Letter: 1
Question
1 metric ton of rice is to be dried from 25% moisture to 13% moisture by exposing the batch through a stream of air flowing at 1.3 m^3/s. Air enters the dryer at 40 degree C, 20% RH, atmospheric pressure, and leaves the dryer at 22.78 degree C and 95% RH at the same pressure. Determine the absolute humidities of the inlet and outlet air. Determine how long would it take to dry the rice to the desired moisture level. If the inlet air is first preheated to 70 degree C before entering the dryer, what is the heat transfer rate required? If the preheated air is used to dry rice and the exit conditions remain the same as stated earlier, how long will the new drying time be?Explanation / Answer
To calculate the absolute humidities of the inlet and outlet air we first need to calculate the vapour pressure of inlet and outlet water vapour present in the air at the given temperatures.
We will calculate vapour pressure of water vapour using antoine equation, which is given by
ln Psat/KPa = A- [B/T/K+C] .................(1)
Where P is in KPa, and T is in kelvin
From literature we obtain the value of Antoine constans,
A= 16.3872, B=3885.70, C=-42.980
[The above data has been obtained from the literature : Introduction to Chemical Engineering Thermodynamics by Smith, Ven Ness and Abott ]
Now at At 40oC ,i.e,313 K, We calculate Vapour pressure of water vapour,
ln Psat/KPa = 16.3872- [3885.70/313-42.980 ]
From the above equation we get , Psat=7.365 kPA
Simlarly,
Now at At 22.78oC ,i.e,295.78 K, We calculate Vapour pressure of water vapour,
ln Psat/KPa = 16.3872- [3885.70/295.78-42.980 ]
From the above equation we get , Psat=2.7636 kPA
Now since realtive Humidity if given we will calculate the value Partial pressure of water vapour at the given temperature,
Relative Humidity (in %) =(Pa/Psat)*100
Now at 400 C,
Relative Humidity =20 %,
20=(Pa/7.365)*100
From above equation, Pa at 400 C=1.473 Kpa
Simlarly Partial pressure of water vapour at 22.780C is,
95=(Pa/2.763)*100
From the Abve equation , Pa at 22.780C =2.624
Now We can calculate Absolute Humidity,
Absolute Humidity (Y)= (Pa/P-Pa)*(18.02/28.97) Kg Water / Kg dry air
Where , Pa=Partial pressure of water vapour
P = System pressure (Here it will be atmospheric pressure)
For inlet air at 400C,
Y=1.473/(101.325-1.473)*(18.02/28.97) =0.009176 Kg water / Kg dry air
For outlet air at 22.780C ,
Y=(2.624/(101.325-2.624))*(18.02/28.97)=0.01654 Kg water/Kg dry air
Hence answer for 1st part ,
Absolute humidty of inlet air =0.009176 Kg water / Kg dry air
Absolute humidity of outlet air =0.01654 Kg water/Kg dry air
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