   # The comparison between the performances of the two units. ### Solid Waste Engineering

3rd Edition
Worrell + 1 other
Publisher: Cengage Learning,
ISBN: 9781305635203

#### Solutions

Chapter
Section ### Solid Waste Engineering

3rd Edition
Worrell + 1 other
Publisher: Cengage Learning,
ISBN: 9781305635203
Chapter 5, Problem 5.9P
Textbook Problem
3 views

##  To determine

The comparison between the performances of the two units.

### Explanation of Solution

Given:

Recovery of organics for manufacturer A is 80% .

Recovery of in organics for manufacturer A is 80% .

Overall recovery of manufacturer B is 60% .

Purity of the light fraction of manufacturer B is 60% .

Feed consist of 80% organics.

Calculations:

Assume that the total feed weighs 100lb .

Write the expression to calculate the recovery.

R1=x1x0   ...... (I)

Here, the recovery of the material is R1 , the extract quantity is x1 and the original quantity is x0 .

Calculate the organic recovered by manufacturer A.

Substitute 80% for R1 , 80lb for x0 and xorganic for x1 in Equation (I).

0.80=x organic80lbxorganic=64lb

Calculate the inorganics recovered by manufacturer A.

Substitute 80% for R1 , 20lb for x0 and xorganic for x1 in Equation (I).

0.80=x organic20lbxorganic=16lb

Write the expression to calculate the overall recovery rate.

RT=xorganic+xinorganicx0   ...... (II)

Substitute 64lb for xorganic , 16lb for xinorganic and 100lb for x0 in Equation (II).

RT=( 64+16)lb100lb=(0.8×100)%=80%

Write the expression to calculate the purity.

P1=x1x1+y1   ...... (III)

Here, the purity is P1 , the quantity of each material extract is x1 and y1 .

Calculate the purity of the light fraction.

Substitute 64lb for x1 , 16lb for y1 in Equation (III)

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