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Example 2 - Solid Dryer

 

Problem 1

 

Material has the following conditions:

 

Type: solid particles

Feed moisture content = 0.03 kg/kg wet basis.

Feed temperature = 20 oC.

Product moisture content = 0.002 kg/kg wet basis.

Product temperature = 50 oC.

Specific heat of the absolute dry material = 1.26 kJ/kg.oC.

Mass flow rate wet basis = 2000 kg/hr.

 

Drying air has the following conditions:

 

Pressure at dryer inlet = 103.2 kPa.

Dry-bulb temperature at dryer inlet = 90 oC

Absolute Humidity at dryer inlet = 0.009 kg/kg

Dry-bulb temperature at dryer outlet = 48 oC

Pressure drop of air in dryer = 1.2 kPa

 

Calculate how much drying air is needed as well as what is the outlet humidity of the drying air.

 

Solution Steps:

 

1.      Start ProcesSimO. A new empty flowsheet is created with the default material of the software by default. 

2.      Go to Materials | Drying Materials to bring up the “Drying Materials” dialog. Click “Add” button to add a new drying material. In the “Name” field, change the name to “Generic Drying Material”. Input 1.26 in the “Specific Heat of Absolute Dry Material” field. Then click “OK” button and then close the “Drying Materials” dialog.

3.      Go to File | Close to close the current empty flowsheet.

4.      Go to Materials | New Flowsheet Settings to bring up the New Flowsheet Settings dialog. Select “Generic Drying Material” in the Select Drying Material list box. Then click “Set” button to set this material as the drying material of the new flowsheet to be created. Close this dialog.

5.      Edit | Application Preferences | Unit Systems tab to select SI-2 and click “Set Current” button to set SI-2 unit system as the current unit system. Close this dialog.

6.      Go to File | New to create a new empty flowsheet.

7.      Create a solid dryer on the flowsheet and double click the icon of the dryer to bring up the dryer’s editor.

8.      Input in “Mat 1” column 2000 kg/hr in the Mass Flow Rate (wet basis) field, 20oC in the Temperature field and 0.03 kg/kg in the Moisture Content (wet basis) field.

9.      Input in “Mat 2” column 50oC in the Temperature field and 0.002 kg/kg in the Moisture Content (wet basis) field.

10.  Input in “Gas 1” column 103.2 kPa in the Pressure filed, 90 oC in the Dry-bulb Temperature field and 0.009 kg/kg in the Absolute Humidity field.

11.  Input in “Gas 2” column 48 oC in the Dry-bulb Temperature field.

12.  Input in “Gas Pressure Drop” field of the Dryer group box 1.2 kPa.

13.  After the enter key of the last input is hit, the dryer is solved and all the calculated variable values are displayed as is shown in Figure 2.1.

 

 

Figure 2.1

 

Problem 2

 

All conditions are the same as those in Problem 1 except that the outlet gas relative humidity is known as 0.288 rather than the outlet air temperature is known.

 

Solution Steps:

 

All the steps are the same as in Problem 1 except that step 11 is replace by the following

 

11.  Input in “Gas 2” column 0.288 in the Relative Humidity field.

 

The calculated results are displayed in Figure 2.2.

 

 

Figure 2.2

 

Please note that the calculated gas outlet temperature and mass flow rates are slightly different from those in problem 1. This is caused by the truncation error of the outlet Relative Humidity. The more accurate value of the outlet Relative Humidity in problem 1   is 0.28787887 (To see more decimal digits of each field in the dryer’s editor you need to go to Edit | Application Preferences | Numeric Format tab’s Decimal Places field to increase decimal digits).

 

Problem 3

 

All conditions are the same as in problem 1 except that inlet gas mass flow rate (wet basis) is known as 5036.894 kg/hr rather than the outlet gas temperature is known.

 

Solution Steps:

 

All the steps are the same as in Problem 1 except that steps 10 and 11 are replace by the following one step

 

10.  Input in “Gas 1” column 5036.894 kg/hr in Mass Flow Rate (wet basis), 103.2 kPa in the Pressure filed, 90 oC in the Dry-bulb Temperature field and 0.009 kg/kg in the Absolute Humidity field.

 

The calculated results are displayed in Figure 2.3.

 

 

 

Figure 2.3

 

 

Problem 4

 

This problem is an extension of Problem 1. All material and drying air conditions are the same as in Problem 1.

 

Suppose this is a rotary dryer. The gas velocity in the dryer is 2.0 m/s. The length to diameter ratio is 8.0. Calculate the dimensions of the dryer using the simple scoping model.

 

Solution Steps:

 

1.      After the dryer balance calculation is solved, go to Calculation Type combo box on the dryer’s editor and select “Scoping” entry.

2.      Click “Scoping” menu on the dryer’s editor to bring up the scoping dialog.

3.      Select “Circle” for the cross section type because the cross section of a rotary dryer is circular.

4.      Input in Gas Velocity field 2.0 and in Length/Diameter Ratio field 8.0.

5.      After the enter key of the last input is hit, the dryer scoping is solved and the calculated variable values are displayed as is shown in Figure 2.4.

 

 

Figure 2.4

 

Please note that you can change the balance calculation specification and after the changed values are input into the dryer’s editor the calculated dimension variables in the scoping dialog is also automatically updated. 

 

 

Summary

 

1.      If the dryer has heat loss, heat input in addition to the heat from the drying gas, work input and heat loss by transport device for some specific dryers you can input a specified value in each of the corresponding field. The calculation will take them into account.

2.      In addition to the 3 cases displayed in the 3 examples, you can also know one of the outlet temperatures (either dry-bulb, or web-bulb or dew point) and one of the outlet humidities (either the absolute humidity or the relative humidity) rather than the corresponding inlet variables and satisfy the dryer’s balance calculation conditions and get the dryer solved.