How to Calculate Dyeing Recipe in Textile Wet Processing?

Dyeing Recipe Calculation:

Dyeing recipe calculation is not a tough task that we have normally seemed. The person who is engaged in the textile dyeing industry has to know the dyeing recipe calculation method to dye the fabric perfectly. This article has presented an easy method for calculating dyeing recipe calculation in the textile wet processing sector.
Dyeing recipe calculation method
Fig: Dyeing recipe calculation method
Lab Dip Calculation Method Followed in Textile Dyeing Industry:
The below equipment has needed to calculate the dyeing recipe in the textile wet processing industry.

  1. Digital weighting meter with glass box (Explorer, USA),
  2. Digital pipette,
  3. Microprocessor pH meter.

Dyeing Recipe Calculation Formula:

The below formulas have needed to calculate lab dip in the textile dyeing industry.
 
1. Amount of dye calculation formula:
The required amount of dye,
    (Weight of the fabric in gram × Shade percentage
=…………………………………………………………………………………………
                   Percentage of Stock solution

Or,
The required amount of solution,
     Fw × Sp
=……………………
         Cs

Where,
Fw indicates the weight of yarn, or fiber, or fabric.
Sp indicates the shade percentage,
Cs indicates the concentration of the stock solution.
 
2. Auxiliaries or Chemicals Calculation Formula:
The below formula has to follow to calculate the required amount of auxiliaries or chemicals.

The required amount of solution (mls),
     (Weight of substrate x Gram per liquor required x Liquor ratio)
=………..…………………………………………………………………………………………….……
               (Concentration percentage of stock solution× 10)
 
3. Additional Auxiliaries Calculation Formula:
The formula for the addition of auxiliaries in solids form such as salt, soda is:

Salt in gram per liquor,
    Sample weight x Liquor Ratio× required amount (%)
=…………………………………………………………………………………….……
                                     1000
4. Percentage to Gram Conversion Calculation Formula:
Conversion formula from percentage to gram/ liquor is in the below:

Gram per liquor,
= Required amount in percentage x 10

If alkali concentration is given then the below formula has to follow for calculating this in gram per liquor:

The required amount of solution (mls),
     (Weight of substrate x Gram per liquor required x Liquor ratio)
=………………………………………………………………………………………………..……………….
             (10 x Concentration percentage of stock solution)

Or, 
The required amount of solution (mls),
     Weight of substrate x Liquor ratio× required amount (%)
= ……………………………………………………………………………………………………….
              Concentration percentage of stock solution

 
Lab Deep Calculation Example: 
Recipe:
Dyes: 
  1. Rema Blue RR = 1.122%
  2. React Red KHW = 2.014%
  3. React Yellow KHW = 1.486%
Chemicals:
Soda Ash (concentration.20%) = 5 gram/liter
Sample Weight. = 5 gram
Caustic Soda = 1.32%
Salt = 70%
Stock Solution (%) = 1
M: L = 1:8

Now, calculate the required recipe for dyes and auxiliaries or chemicals in gram per liter.

Solution:

In the case of dyes,

We know,
The required amount of dye,
   (Weight of the fabric in gram × Shade percentage
=…………………………………………………………………………………………
                  Percentage of Stock solution

Calculation of dyestuff will be,
For,
  1. Rema Blue RR = (1.12 2x 5)/1=5.61 g/l
  2. React Yellow KHW = (1.486 x 5)/1= 7.43 g/l,
  3. React Red KHW = (2.014 x 5)/1= 10.07 g/l
In the case of auxiliaries or chemicals,

We know,
Salt in gram per liquor,
     Sample weight x Liquor Ratio× required amount (%)
=……………………………………………………………………………………..…………
                                      1000
     5×8×70
=……………………….
      1000
=   2.8gm

So, the required amount of salt is 2.8gm

 
In the case of Soda Ash (Concentration 20%):
The required amount of solution (mls),
    (Weight of substrate x Gram per liquor required x Liquor ratio)
=………………………………………………………………………………………………………………….
             (10 x Concentration percentage of stock solution)
    5×5×8
=……………………
    10×20
= 1ml

So, the required quantity of soda ash in c.c stands at 1.0

Extra Amount of Water Required:
= M: L – (Required amount of water to make solution of auxiliaries and dyes)
= [(5 x 8) – {(5.61+10.07+7.43) + (1.0+0.12)}]
= 40 – 24.112
= 15.77

So, 15.77 amount of salt is added in solid form

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