Question

Question 1 A batch dryer removes water from a solid at a rate of 19.445 g/s during the constant rate period. The critical moisture content is 0.45 kg H20 /kg dry solid. The equilibrium moisture content of the material is 0.075 kg H20/ kg dry solid. The dryer is initially charged with 400 kg of dry material which contains 62.5% dry material and the remainder is moisture. The final product contains 0.085 kg H2O/kg dry solid and the falling rate period can be represented by a straight line. The total drying area is 1.3 m2 1.1. Draw the drying rate curve of the system. (4) 1.2. Determine the total drying time to produce the product (18)

Answer 1

Ne 20.0LS A gosto Perhep X=0.6 XA20.085 х* Xc=0.45 = 0.075 moisture content Ne= rate of drying area 19.4459 0.019445 kg moisture I see y 1.3m2 1.3 = 0.015 leg moisture m².s m²s as areas 1.3m2 = 250 We = mass of dry solid - 400 kg 80. 625 Xis - 1-0.625 = 0.6 kg moisture 0.625 kg dry solid kg Xc = 0,45 Xa= = 0,025 X = 0.085 As falling rate is rate is linear N = p Xtq where p = slope, q= intercept

WS t = W. (X; - Xc) an + In zilz ар

0 = pX® + 9 9 = -p8* NO NA = pX. +9 pX, +9 pXc - p8* PX; - pX* X - X* X, - X* 1 X - X; N. – N; X - X (pX. +9)-(pX, +9) Xc - - X* Nc p t = tc + tf = W,(X; - X), W, X.- X* a Nc Nc + In Xc - X* X, - X* a

- 250 (0.6 - 0.45) + 110.015) 25010.45-0.075) (1-2) (0.015) 0.45-0.075 xl 0.085-0.075 5.374 hrs tort ( १347, 743 Secs