Calculation Basis for Chemicals used in Textile Processing

For any textile process we often need to calculate the amount of chemicals to be added in the process liquor and many a times we need to convert or dilute their concentrations in order to suit the required concentration.

Basis of Chemicals to be taken for processing
Basis of Chemicals to be taken for processing

NOTE: Both ‘P’ & ‘C’ should be in same concentration units either ‘%’ or ‘gpl’. If different, for example Required conc. is in gpl and available conc. is in %, then we have to convert the unit first as shown below.

For converting ‘%’ into ‘gpl’ the % is multiplied by 10. For example to convert 5% into gpl, multiply it with 10,

i.e.,                        5% X 10 = 50 gpl

For 10%,               10% X 10 = 100 gpl

For 40%,               40% X 10 = 400 gpl

Now, for converting ‘gpl’ into ‘%’ divide it by 10. For example, to convert 10gpl into %, divide it by 10,

i.e.,                       10gpl / 10 = 1%

For 5gpl                5gpl / 10 = 0.5%

For 25gpl              25gpl / 10 = 2.5%


DERIVATION #1 for Calculations based on Weight of Material

Let the Required conc. = P %

Weight of material = W

Available conc. = C %

Total Volume of Liquor = V

Now,

for 100gms of material –> ‘P’ gms of chemical is required,

Therefore for ‘W’ gms of material –> W X P / 100 gms of chemical is required,

      ………….(by cross multiplication)

Similarly,

‘C’ gms of chemical is present in –> 100 ml of solution

Therefore, (W X P / 100) gms of chemical is present in –> {(W X P / 100) * 100} / C ml of solution

…………….(by cross multiplication)

= (W X P / C) ml of solution

Hence Proved, for ‘W’ gms of Material, (W X P / C) ml of solution is taken

 


DERIVATION #2 for Calculations based on Volume of Liquor

Let the Required Conc. = P gpl

Weight of Material = W

Available conc. = C gpl

Total Volume of Liquor = V

Now,

‘P’ gpl means, P gms of chemical is present in –> 1000 ml of solution

i.e., for 1000 ml of solution –> ‘P’ gms of chemical is required

Therefore, for ‘V’ ml of solution –> (V X P / 1000) gm of chemical is required.

                ………….(by cross multiplication)

Similarly,

‘C’ gpl means, C gm of chemical is present in –> 1000 ml of solution

Therefore, (V X P / 1000) gms of chemical is present in –> {(V X P / 1000) * 1000 / C}

………….(by cross multiplication)

= (V X P / C) ml of solution

Hence Proved, for ‘V’ ml of Liquor, (V X P / C) ml of solution is taken

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