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Forced draft tray dryer
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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, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathvariant=¨bold¨»X«/mi»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mrow»«mo»(«/mo»«mi mathvariant=¨bold¨»W«/mi»«mo»-«/mo»«mi mathvariant=¨bold¨»S«/mi»«mo»)«/mo»«/mrow»«mi mathvariant=¨bold¨»S«/mi»«/mfrac»«/math»

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

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathvariant=¨bold¨»N«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mo»-«/mo»«mi mathvariant=¨bold¨»S«/mi»«mfrac»«mrow»«mi mathvariant=¨bold¨»d«/mi»«mi mathvariant=¨bold¨»x«/mi»«/mrow»«mrow»«mi mathvariant=¨bold¨»d«/mi»«mi mathvariant=¨bold¨»t«/mi»«/mrow»«/mfrac»«mfrac»«mn»1«/mn»«mi mathvariant=¨bold¨»A«/mi»«/mfrac»«/math»

For a constant area we hve the drying rate equation as :

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathvariant=¨bold¨»N«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mo»-«/mo»«mi mathvariant=¨bold¨»S«/mi»«mfrac»«mrow»«mi mathvariant=¨bold¨»d«/mi»«mi mathvariant=¨bold¨»x«/mi»«/mrow»«mrow»«mi mathvariant=¨bold¨»d«/mi»«mi mathvariant=¨bold¨»t«/mi»«/mrow»«/mfrac»«/math»

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 :

 

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathvariant=¨bold¨»Q«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mrow»«msub»«mi mathvariant=¨bold¨»A«/mi»«mn»1«/mn»«/msub»«msub»«mi mathvariant=¨bold¨»A«/mi»«mn»2«/mn»«/msub»«msub»«mi mathvariant=¨bold¨»C«/mi»«mi mathvariant=¨bold¨»d«/mi»«/msub»«/mrow»«msqrt»«mrow»«msup»«msub»«mi mathvariant=¨bold¨»A«/mi»«mn»2«/mn»«/msub»«mn»2«/mn»«/msup»«mo»-«/mo»«msup»«msub»«mi mathvariant=¨bold¨»A«/mi»«mn»1«/mn»«/msub»«mn»2«/mn»«/msup»«mo»§nbsp;«/mo»«/mrow»«/msqrt»«/mfrac»«msqrt»«mrow»«mn»2«/mn»«mi mathvariant=¨bold¨»g«/mi»«mi mathvariant=¨bold¨»h«/mi»«/mrow»«/msqrt»«/math»

 

Where H is

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathvariant=¨bold¨»H«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mi mathvariant=¨bold¨»R«/mi»«mn»100«/mn»«/mfrac»«mfenced»«mrow»«mfrac»«msub»«mi mathvariant=¨bold¨»§#929;«/mi»«mi mathvariant=¨bold¨»w«/mi»«/msub»«msub»«mi mathvariant=¨bold¨»§#929;«/mi»«mi mathvariant=¨bold¨»a«/mi»«/msub»«/mfrac»«mo»§nbsp;«/mo»«mo»-«/mo»«mo»§nbsp;«/mo»«mn»1«/mn»«/mrow»«/mfenced»«/math»

= 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

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi mathvariant=¨bold¨»Q«/mi»«mn»1«/mn»«/msub»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mi mathvariant=¨bold¨»G«/mi»«mo»§#215;«/mo»«msub»«mi mathvariant=¨bold¨»C«/mi»«mi mathvariant=¨bold¨»p«/mi»«/msub»«mo»§#215;«/mo»«mo»(«/mo»«mi mathvariant=¨bold¨»T«/mi»«mo»-«/mo»«mn»298«/mn»«mo»)«/mo»«/math»

=0.016265 *1.005 * (340-298)

= 0.683472 KW

= 3515.001 KWH(taking effeciency as 0.7)

 

Let us assume thre industrial rate to be Rs 10 per unit

Therfore cost = 35150.01Rs

Cost per unit weight = 5.12Rs/Kg.

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