Environmental Engineering FOURTH EDITION

280 ENVIRONMENTAL. ENGINEERING

where (1 - Y) is the cumulative fraction larger than some size x. If

0.2 = exp [:In ,

then x = Lp, the screen size through which 80% of the product passes. Solving for
LP,

Lp = ~,(1.61)'/~.

Substituting this expression back into Eq. (14.4),

lOWi -- lOWi

W=

[xc( 1 .61)i/n]"2 - [LP]'/*

(14.5)

EXAMPLE 14.1. Assuming Wi = 400 and x, of the product is 1.62 cm, n = 1, and
Lp = 25 cm (about 10 in.; a realistic estimate for raw refuse), find the expected power
requirements for a shredder processing 10 tondh.
From Eq. (14.5)

(Note: and LF must be in Km.)

The power requirement is (16.7 kWh/ton)(lO tondh) = 167 kW.

General Expressions for Material Recovery

In separating any one material from a mixture, the separation is termed binary because

only two outputs are sought. When a device is to separate more than one material from

a mixture, the process is calledpolynary separation.

Figure 14-6 shows a binary separator receiving a mixed feed of xo and yo. The

objective is separation of the x fraction: the first exit stream is to contain the x com-

ponent, but the separation is not perfect and contains an amount of contamination y1.

This stream is called theproduct or extract, while the second stream, containing mostly

y but also some x, is the reject. The percent of x recovered in the first output stream,

R(xl), may be expressed as

R(q = (2) 100. (14.6)

R(xl) alone does not describe the performance of the binary separator adequately. If

the separator were turned 08 all of the feed would go to the first output, the extract

would be xo = XI, making = 100%. However, in this case there would have