# 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 , d_{1} = 0.039 m

Diameter of the orifice hole , d_{2} = 0.052 m

Co-effecient of discharge , C_{d} = 0.6

Therefore the areas are :

Area of the pipe , A_{1} =( 3.14*(0.039)^{2})/4 = 0.00119 m^{2}

Area of the orifice hole , A_{2} = ( 3.14*(0.052)^{2})/4 = 0.00212 m^{2}

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 C_{p}= 1.005 kJ/kgK

Temperature at which drying takes place = 340 K

Therfore amount of heat removed Q_{1}

=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.