       HOME > Quantity requirements  DURANATE all Types & Grades   Application example   Resin compatibility   Quantity requirements   Test methods   Precautions   Technical data sheet  Catalog   Solvent properties   Contact   Formulation quantity The following equation and nomograph provide two ways to determine the appropriate quantity of Duranate™ (in weight parts) for a given polyol, from the Duranate™ NCO, the polyol quantity and OHV, and the desired NCO/OH ratio.
 Equation Example 1.
For Duranate™ TPA-100 (NCO = 23.1%), polyol having OHV=80, and NCO/OH = 1.0:

 Wt. parts Duranate™   = 80×42×100 561×23.1 ×(1.0)＝25.9
 Nomograph
To find the required Duranate™ quantity (in wt. parts per 100 wt. parts polyol): To find the required Duranate™ quantity (in wt. parts per 100 wt. parts polyol):Note: (1) For a Duranate™ grade having NCO<14%, use twice the actual NCO value in Step 1 and then multiply the value on Axis C in Step 2 by two, to obtain the required quantity for NCO/OH = 1. (2) For a polyol having an OHV value outside the Axis B range, multiply the actual OHV value by an appropriate factor (fraction or integer) to obtain an adjusted value within that range, use the adjusted value as the OHV value in Step 1, and multiply the value on Axis C in Step 2 by the same factor, to obtain the required quantity for NCO/OH = 1.

Example 2.
For the same materials as Example 1: Extend straight line from 23.1 on Axis A through 80 on Axis B to Axis C, and read the value of approx. 26 on Axis C as the Duranate™ quantity required for this application, in wt. parts per 100 wt. parts polyol.

Example 3.
For Duranate™ E402-90T (NCO = 8.5%), with all other materials as in Examples 1 and 2: Multiply NCO value by two, extend straight line from 17 on Axis A through 80 on Axis B to Axis C, read the value of approx. 35 on Axis C, and multiply by two to obtain the Duranate™ quantity of 70 wt. parts per 100 wt. parts polyol as the requirement for this application. TOP  ｜ Privacy Policy ｜ Terms and Conditions ｜   