Question 30.3: A rotary drum filter with 30 percent submergence is to be us......

Equation (30.34) wi11 be used. The quantities needed for substitution are

{\frac{\dot{m}_{\mathrm{c}}}{A_{T}}}=\left({\frac{2c\,\Delta p^{1-s}\,g_{c}f n}{\alpha_{0}\,\mu}}\right)^{1/2}   (30.34)

\Delta p=20{\frac{14.69}{29.92}}\times144=1414 \ {\mathrm{lb}}_{f}/{\mathrm{ft}}^{2}

f=0.30\qquad t_{c}=5\times60=300\,{\mathrm{s}}\qquad n=\frac{1}{300}\,{\mathrm{s}}^{-1}

From Example 30.2

\alpha_{0}=2.{9}0\times10^{10}\,{\mathrm{ft}}/{\mathrm{lb}}\qquad s=0.26

Also

\mu=1\,{\mathrm{cP}}=6.72\times10^{-4} \ {\mathrm{lb/ft-s}}\qquad\rho=62.3\,{\mathrm{lb/ft}}^{3}

Quantity c is found from Eq. (30.19). The slurry concentration c_F is 14.7 lb/ft³. Since the cake contains 50 percent moisture, m_{F}/m_{c}=2. Substitution of these quantities in Eq. (30.19) gives

c={\frac{c_{F}}{1-[(m_{F}/m_{\mathrm{c}})-1]c_s{/{\rho}}}}    (30.19)

c={\frac{14.7}{1-(2-1)(14.7/62.3)}}=19.24 \ {\mathrm{lb/ft}}^{3}

Solving Eq. (30.34) for A_T gives

A_{T}=\dot{m}_{\mathrm{c}}\Bigl(\frac{\alpha_{0}\mu}{2c\,\Delta p^{1-s}\,g_{c}f n}\Bigr)^{1/2}   (30.35)

The solids production rate \dot{m}_{\mathrm{c}} equals the slurry flow rate times its concentration c_F. Thus since the density of CaCO_3 is 168.8 lb³/ft³,

{\dot{m}}_{c}={\frac{10}{60}}{\frac{1}{7.48}}\left({\frac{1}{(14.7/168.8)+1}}\right)14.7=0.302 \ {\mathrm{lb}}/s

Substitution in Eq. (30.35) gives

A_{T}=0.302\biggl({\frac{2.90\times10^{10}\times6.72\times10^{-4}}{2\times19.24\times1414^{0.74}\times32.17\times0.30\times(1/300)}}\biggr)^{1/2}

=81.7\,\mathrm{ft}^{2}\left(7.59\,\mathrm{m}^{2}\right)