Shakshat Virtual Lab
INDIAN INSTITUTE OF TECHNOLOGY GUWAHATI
CALCULATIONS :
At Pressure = 1.5 cm of water
Weight of dry solid, S = 100 gm
Amount of initial moisture = 15 ml
Weight of solid, W= solid + water
Moisture content present in solid,
We have the experimental data as :
Time ,t (sec) |
Wt. of solid(solid + water ) W (gm) |
Moisture content present in solid X (gm water/gm of dry soild) |
---|---|---|
0 |
115
|
0.15
|
3 |
112
|
0.12
|
6 |
109
|
0.09
|
9 |
107
|
0.07
|
12 |
105
|
0.05
|
15 |
103
|
0.03
|
18 |
101
|
0.01
|
21 |
100
|
0
|
Now we plot a X vs t plot as shown
From this plot we calculate the slope = -dx/dt.
Now we have the drying rate as
For a constant area we hve the drying rate equation as :
Thus we have the table as :
Time,t (sec) |
Wt. of solid (solid + water ) W (gm) |
Moisture content present in solid X (gm water/gm of dry soild) |
Slope = -dx/dt |
N(Area const) gm/sec |
---|---|---|---|---|
0 |
115
|
0.15
|
0.0102
|
1.02
|
3 |
112
|
0.12
|
0.009324
|
0.9324
|
6 |
109
|
0.09
|
0.008448
|
0.8448
|
9 |
107
|
0.07
|
0.007572
|
0.7572
|
12 |
105
|
0.05
|
0.006696
|
0.6696
|
15 |
103
|
0.03
|
0.00582
|
0.582
|
18 |
101
|
0.01
|
0.004944
|
0.4944
|
Finally we plot a graph between X vs N as shown :
The same steps are repeated for other runs at different operating conditions.
COST CALCULATIONS :
Total time of drying = 21 min
Therefore No. of batches processed in a day = 24*60/21 = 68.57
Amount dried in one batch = 100 Kg
Therefore total weight = (100*68.57)Kg
Let us consider an orificemeter with the following diameters:
Diameter of the pipe , d1 = 0.039 m
Diameter of the orifice hole , d2 = 0.052 m
Co-effecient of discharge , Cd = 0.6
Therefore the areas are :
Area of the pipe , A1 =( 3.14*(0.039)2)/4 = 0.00119 m2
Area of the orifice hole , A2 = ( 3.14*(0.052)2)/4 = 0.00212 m2
Now we have the volumetric flow rate as :
where H is
= 1.5*((1000/1.2)-1)/100 m
=12.485 m
Therefore the volumetric flow rate = 0.013554 m^3/s
The mass flow rate G :
G = Q * Density of air
= 0.013554 * 1.2
= 0.016265 Kg/sec
We have the specific heat of air Cp= 1.005 kJ/kgK
Temperature at which drying takes place = 340 K
Therfore amount of heat removed Q1
=0.016265 *1.005 * (340-298)
= 0.683472 KW
= 3515.001 KWH(taking effeciency as 0.7)
Let us assume three industrial rate to be Rs 10 per unit
Therefore cost = 35150.01Rs
Cost per unit weight = 5.12Rs/Kg.