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 DURANATETM 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 DURANATETM 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 content 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 outside the Axis B range, multiply the actual OHV by an appropriate factor (fraction or integer) to obtain an adjusted value within that range, use the adjusted value as the OHV 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 DURANATETM 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 content 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 DURANATETM quantity of 70 wt. parts per 100 wt. parts polyol as the requirement for this application.
