Question

Wet stock containing 1.562 lb H2O/lb dry solids is to be dried to 0.099 lb H2Ob dry solid. For every lb of dry solid. 52.5 lb dry air passes through the dryer, the air leaves with a humidity of 0.0525 lb H2O/lb dry air. Some of the air leaving the dryer is recycled and mixed with the fresh air feed. The fresh air has a humidity of 0.0152 lb H2O.lb dry air. Find the fraction of air recycled.

Answer 1

Basis: 1 lb of dry solid Given data: ms(in) = mass flowrate of dry solid to dryer 1 lb mw(in) = mass flowrate of water to the dryre = 1.562 lb ma(1) = mass flowrate of air to the dryer = 52.5 lb = mass fraction of water in dry air from dryer = 0.0525 X = mass fraction of water in fresh dry air = 0.0152 mw (out) = mass flowrate of water in product = 0.099 lb Solution By converting given data into block diagram Xzma(out) ma(2) X2 ma(Rec) X2 ms(in) mw(in) ms(out) mw (out) Dryer ma(1) X X maſin)

By taking over all material balance over dryer ms(in) + mw(in) + ma(1) = Ms(out) + mw(out) + Ma(2) 1 + 1.562 + 52.5 = 1 +0.099 + ma(2) ma(2) 53.963 lb = Now take component balance for water mw(in) + ma(1) X1 = mw(out) + ma(2)X2 1.562 + 52.5 x X1 = 0.099 + 53.963 x 0.0525 1.562 + 52.5 x X1 0.099 + 53.963 X 0.0525 X1 = 0.02609 Taking material balance over recycle point ma(2) = Ma (Rec) + ma(out) (1) Taking material balance over mixing point ma(1) = Ma (Rec) + Maſin) (2) By subtracting 2 from 1 ma(1) - ma(2) ma(Rec) + malin) – ma(Rec) – Maſout) malin) – ma(out) = ma(1) – ma(2) malin) – ma(out) = -1.463 (3) Similarly for water malin) X – maout) X2 = ma(1)X1 – ma(2)X2 0.0152malin) - 0.0525ma(out) 52.5 x 0.02609-53.963 X 0.0525 0.0152malin) - 0.0525ma(out) = 1.3697 – 2.8333 0.0152malin) - 0.0525ma(out) = -1.3679

From equation 3 malin) ma(out) - 1.463 0.0152(ma(out) - 1.463) – 0.0525ma (out) = -1.3679 0.0152ma(out) -0.02223 – 0.0525malout) = -1.3679 -0.0373ma(out) = -1.34567 malout) = 36.077lb Material balance equation at recycle point = ma(2) -ma(out) ma(Rec) ma(Rec) 53.963 – 36.077 ma(Rec) 17.886 lb ma(Rec) fraction of recycled = ma(2) 17.886 53.963 0.3314 Conclusion Mass flowrate of recycle stream = 17.886 lb Fraction of dryer outlet air recycled = 0.3314