Método de Kynch

Qual o diâmetro de um decantador capaz de tratar 1.200 t/d se uma suspensão com a concentração de 125 g/L e de obter uma lama de concentração 720 g/l

Os dados de um ensaio proveta estão na folha de excel des.xls na coluna B e C nas linhas 17 a 36.

import xlrd
import math
from sage.plot.colors import rainbow
workbook=xlrd.open_workbook('des.xls')
she=workbook.sheet_names()
xl_sheet=workbook.sheet_by_name(she[0])

t=[]
z=[]
for i in range(16,35):
    t.append(xl_sheet.cell_value(i,1))
    z.append(xl_sheet.cell_value(i,2))

O ensaio proveta

G=list_plot(zip(t,z),size=18)
S=spline(zip(t,z))
G+=plot(S,t[0],t[-1],color='gray',axes_labels=['$t\ (min.)$','$h\ (cm)$'])
print ('Ensaio de proveta')
pretty_print(table(columns=[(t[i],z[i]) for i in range(len(t))],
      header_column=['t (min)','h (cm)'], frame=True))
G.show(figsize=4)

Ensaio de proveta

t (min)	$0.0$	$2.0$	$4.0$	$6.0$	$8.0$	$10.0$	$13.0$	$16.0$	$20.0$	$25.0$	$30.0$	$37.0$	$44.0$	$50.0$	$58.0$	$65.0$	$74.0$	$86.0$	$95.0$
h (cm)	$116.3$	$103.5$	$92.6$	$83.4$	$76.0$	$69.6$	$61.2$	$54.3$	$47.2$	$40.2$	$34.5$	$28.3$	$24.0$	$21.2$	$17.8$	$15.4$	$13.3$	$11.0$	$9.6$

Determinação da área do decantador pelo método de Kynch

• Carga 1200 t/d
• Concentração de entrada 125 g/L
• Concentração da lama 720 g/L $$u_i=\frac{z_i-z}{t}$$ $$C_i=\frac{C_0\cdot z_0}{z_i}$$

Gg=G
zi=[]
ui=[float('nan')]
ci=[125]
cu=720
colors= rainbow(10)
for i in range(len(t)):
    m=S.derivative(t[i])      # derivada no ponto t
    zi.append(z[i]-m*t[i])    # zi - ordenada na origem da tangente em t
    if t[i]!=0:
        ui.append((zi[i]-z[i])/t[i])  # ui=(zi-z)/t
        ci.append(ci[0]*z[0]/zi[i])   # ci=co.zo/zi
    if i%2:
        Gg+=point((t[i],z[i]),color=colors[int(i/2)],size=40)
        Gg+=line([(0,zi[i]),(1.2*t[i],m*1.2*t[i]+zi[i])],color=colors[int(i/2)],linestyle='--')
Gg.show()
pretty_print(table(columns=[(t[i],z[i],round(zi[i],1),
    round(ui[i],2),round(ci[i],1)) for i in range(len(t))],
    header_column=[' t    ( min )  ','   h     (   cm    )    ','$z_i$(cm)','$u_i$(cm/min)','$c_i(g/L)$'],
    frame=True))

t ( min )	$0.0$	$2.0$	$4.0$	$6.0$	$8.0$	$10.0$	$13.0$	$16.0$	$20.0$	$25.0$	$30.0$	$37.0$	$44.0$	$50.0$	$58.0$	$65.0$	$74.0$	$86.0$	$95.0$
h ( cm )	$116.3$	$103.5$	$92.6$	$83.4$	$76.0$	$69.6$	$61.2$	$54.3$	$47.2$	$40.2$	$34.5$	$28.3$	$24.0$	$21.2$	$17.8$	$15.4$	$13.3$	$11.0$	$9.6$
$z_i$(cm)	$116.3$	$115.5$	$112.6$	$108.2$	$103.1$	$99.9$	$94.4$	$87.1$	$78.5$	$71.6$	$65.4$	$55.7$	$46.5$	$43.3$	$40.5$	$34.1$	$28.4$	$25.7$	$23.7$
$u_i$(cm/min)	$NaN$	$5.98$	$5.01$	$4.13$	$3.39$	$3.03$	$2.55$	$2.05$	$1.56$	$1.26$	$1.03$	$0.74$	$0.51$	$0.44$	$0.39$	$0.29$	$0.2$	$0.17$	$0.15$
$c_i(g/L)$	$125.0$	$125.9$	$129.1$	$134.4$	$141.0$	$145.5$	$154.0$	$167.0$	$185.2$	$203.0$	$222.2$	$261.2$	$312.3$	$335.4$	$358.7$	$426.5$	$511.2$	$566.7$	$614.1$

coqo=1200
A=[]
for i in range(1,len(t)):
# cm^2 -> m^2     /1e4
# g/L -> g/cm^3   1/1000
# t/d -> g/min    /60/24/1e6
    A.append(coqo*(1/ci[i]-1/cu)/ui[i]*1e3/60/24*1e6/1e4)
print u'Área de %.0f m^2'%max(A)
print u'diâmetro de %.1f m'%(sqrt(max(A)*4/pi()))

Área de 300 m^2
diâmetro de 19.5 m