Page 135 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

90 Fundamentals of Water Treatment Unit Processes: Physical, Chemical, and Biological

r ¼ specific mass of suspended matter as deposited on Step 8: Substitute Equation Ex5.3.7 in Equation Ex5.3.5
3
screen (kg=m )
3
2
HLR ¼ hydraulic loading rate (m =m =s) dX(mat) h L (mat)
M ¼ mass of suspended matter deposited on screen at dt r ¼ k(mat) X(mat) C r (Ex5:3:10)
any given time (kg)
V ¼ volume of suspended matter deposited on screen at Step 9: Separate the variables and integrate
3
any given time (m )
v ¼ velocity of flow of water through screen (m=s) Q
2
A ¼ area of screen (m ) ð k(mat)h L (mat)C r ð M
r ¼ radius of screen (m) X(mat)dX(mat) ¼ rv dQ (Ex5:3:11)
t ¼ time for rotation of screen from initial water line (s) 0
k(screen) ¼ coefficient of hydraulic conductivity for
screen (m=s) giving a ‘‘final’’ equation:
k(mat) ¼ coefficient of hydraulic conductivity for mat of
suspended matter (m=s) 1 2 k(mat)h L (mat)C r (Ex5:3:12)
2 X(mat) ¼ rv Q M
Step 4: State materials balance for screen
To simplify, substitute Equation Ex5.3.9 for h L (mat) to give
Rate of mass retention on screen
1 k(mat)h L C r
¼ (mass rate of suspended solids to screen) X(mat) ¼ Q M (Ex5:3:13)
2
2 rv
(mass rate of suspended solids leaving screen)
(Ex5:3:1) Step 10: Illustrate graphically selected relations in
Equation Ex5.3.13
Step 5: Express materials balance mathematically
Discussion: Figure 5.11 says that the mat builds up
dM quickly and then declines toward an asymptote. Since
¼ QC o QC (Ex5:3:2)
dt Q is fixed within narrow limits a rapid buildup of mat is
inevitable. At the same time, the mat thickness declines
Substitute: M ¼ V r exponentially with v and then declines toward a lower
asymptote. In other words, a slight increase in v can do
d(Vr)
¼ QC o QC (Ex5:3:3) much to reduce the mat thickness. The mat thickness is
dt that at Q M .
Substitute: V ¼ A X Step 11: Find relation for Q

d(XAr)
¼ QC o QC (Ex5:3:4) ð A
dt
Q ¼ v dA (Ex5:3:14)
Note that v ¼ Q=A
ð A
d(X) h L (mat)
r ¼ vC r (Ex5:3:5) Q ¼ k(mat) dA (Ex5:3:15)
dt X(mat)
Step 6: Apply Darcy’s law for flow across screen Q ð M
h L (mat)
Lr dQ
h L (screen) Q ¼ k(mat) (2k(mat)h L (mat)C r )=(rv) Q
(Ex5:3:6)
v ¼ k(screen) 0
X(screen)
(Ex5:3:16)
Step7:AgainapplyDarcy’slaw,thistimeforflowacrossmat
1=2
2rvk(mat)h L (mat)Q M
h L (mat) Q ¼ Lr (Ex5:3:17)
(Ex5:3:7) C r
X(mat)
v ¼ k(mat)
Note: By continuity, the velocity across the screen equals
the velocity across the mat, thus Equations Ex5.3.6 and
Ex5.3.7 are equal.
Also, for later reference recall, x x
h L ¼ h L (screen) þ h L (mat) (Ex5:3:8)
(a) Θ (b) ω
Since, for most of the screening duration, h L (mat)
h L (screen), then we can neglect h L (screen) in Equation
Ex5.3.8 to give FIGURE 5.11 Mat thickness, calculated as function of submer-
gence angle and rotational velocity. (a) Relation x versus Q.
h L h L (mat) (Ex5:3:9) (b) Relation x versus v.