Dyeing of 100% polyester woven fabric with Disperse dye (High Temperature Method).

 

Experiment Name: Dyeing of 100% polyester woven fabric with Disperse dye (High Temperature Method).

Theory:  The term “disperse dye” have been applied to the organic colouring substances which are free from ionizing groups, are of low water solubility and are suitable for dyeing hydrophobic fibres. Disperse dyes have substantivity for one or more hydrophobic fibres e.g. cellulose acetate, nylon, polyester, acrylic and other synthetic fibres. The negative charge on the surface of hydrophobic fibres like polyester cannot be reduced by any means, so non-ionic dyes like disperse dyes are used which are not influenced by that surface charge.

Objective:

1.       To learn about dyeing process of polyester fabric by disperse dye.

2.       To dye polyester fabric by disperse dye.

Nature of sample:

Pre-treated 100%  polyester woven fabric.

Apparatus required:

1.Beakers

2.Glass Rod.

3.Pipette.

4. Measuring Cylinder.

5. Digital Balance.

6. Tri-pod stand.

7. Gas Burner.

8. Thermometer.

9. Pot.

Process Sequence:

• Collection of sample
• Set water level
• Add Acetic acid solution
• check pH
• Add dispersing agent solution
• Add disperse dye solution
• Add fabric sample
• Raise temperature to 90deg C
• Run time for 20 minutes
• Bath Drop
• Rinsing

Table1: Dyeing Recipe

SL	Process Parameter	Unit	Dossing	Stock soln:
01	Disperse Dye	%	1	1%
02	Dispersing Agent	g/L	1	1%
03	Aceic Acid	ml	0.5	-
04	M : L	---	1:30	-
05	Fabric Weight	gm	5	-
06	Temperature	deg C	130	-
07	Time	Min	20	-

Calculation:

Total Liquor = Fabric weight X L  [M:L = 01:30]

                                = 5 X 30

                                = 150mL

Dye 1%  = (Fabric weight X x%) / stock solution%

                = (5X1%) / 1%

                =5mL

Dispersing Agent = (Total liquor X 'x'g/L) / (stock solution%  X 1000)

                                = (150 X 1) / 1% X 1000

                                = 15mL

Acetic Acid : 1ml

Initial Water = total liquor - (chemical 1 + chemical 2 + chemical 3 ...)

                = 150- (5+15+1) mL

                = 150 -21 mL

                = 129mL

Process Curve:

Sample Attachment & Observation: 

Conclusion: >>>>>>>>>>