The following data were obtained in a resin burn-off test of an E-glass-polyeste

Question

The following data were obtained in a resin burn-off test of an E-glass-polyester sample: Weight of an crucible = 10.1528g, Weight of crucible + sample before burn off = 10.5219g, Weight of crucible + sample after burn off = 10.3221g.  

Calculate the fiber weight fraction, the fiber volume fraction, and the density of the composite sample.  Assume Pf = 2.54g/ml, and Pm = 1.25g/ml.  Do you expect the calculated value to be higher or lower than the actual value?

tips: Vf = ((Wf/Pf))/((Wf/Pf) + (1-Wf)/Pm)

         Pc = (1)/((Wf/Pf) + (1-Wf)/Pm)

Explanation / Answer

weight of crucible = 10.5219 - 10.1528 = 0.3691 g

weight of fibre = 10.3221 - 0.3691 = 9.953 g

fibre weight fraction (Wf) = 9.953/10.1528 = 0.9803

Vf = [(Wf/Pf)]/[(Wf/Pf) + (1-Wf)/Pm]

     = [0.9803/2.54]/[(0.9803/2.54) + (1 - 0.9803)/1.25]

    = (0.386)/[(0.386) + 0.0158]

    = 0.961

Pc = (1)/[(Wf/Pf) + (1-Wf)/Pm]

     = (1)/[(Wf/Pf) + (1-Wf)/Pm]

     = 1/[(0.386) + 0.0158]

     = 2.489

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