# Chapter 28 : Environmental management¶

## Example 28.3 Page no 430¶

In [1]:
from __future__ import division
print "Example 28.3 page no 430\n\n"
#we have to determine the minimum distance downstream from a cement dust emitting  source that will be free of cement deposit
#the souce is equipped with a cyclone located 150 ft above ground level
#neglect meteorological aspects
h=150#cyclone height from ground level,ft
v_w=3/3600#wind velocity,miles/second
SG=1.96#specific gravity of cement dust
rho_w=62.4#density of water,lb/ft**3
rho_p=SG*rho_w#/density cement particles
#applying ideal gas law for density of air
P=1#pressure,atm
M= 29#mol. weight of air
R=0.73#gas constant
T=520#temperature,Rankine
rho_a=P*M/(R*T)#density of air
meu=1.22e-5#viscosity of air,lb/ft.s
g=32.174#grav. acc.
d_p=2.5/(25400*12)#particle diameter,ft
K = d_p*(g*rho_p*rho_a/(meu**2))**(1/3)#settling factor
print "\n settling factor K=%.3f "%(K)#
#since K=0.103<3.3,sokes law rane applies
v= g*d_p**2*rho_p/(18*meu)#terminal settling velocity)
print "\nsettling velocity v=%.3e ft/s"%(v)#
t=h/v#time for desent
print "\n desent time t=%.2f sec"%(t)#
x=v_w*t#horizontal distance travelled in miles
print "\n minimum horizontal distance x=%.2f miles"%(x)##printing mistake in book

Example 28.3 page no 430

settling factor K=0.104

settling velocity v=1.205e-03 ft/s

desent time t=124430.77 sec

minimum horizontal distance x=103.69 miles


## Example 28.4 Page no 432¶

In [2]:
from __future__ import division
from math import pi,sqrt
print "Example 28.4 page  no 432\n\n"
#it is proposed to install a pulse jet fabric filter system to clean an air stream containing particulate pollutants
#we have to select the most apporpriate filter beg fabric
q_scfm=10000#volumetric flow rate of polluted air stream at 60 deg F ,1 atm
T=520#temperature,R
T_o=710#operating temparature ,R
q_acfm=q_scfm*(T_o/T)#flow rate in acfm
v_f=2.5#filteration velocity,ft/min
S_c=q_acfm/v_f#filtering beg area
print "\n filtering beg area S_c=%.2f ft**2"%(S_c)#
#(1) for bag A ,the area and N number of bags are
D_a=8/12#diamter,ft
H_a=16#height,ft
S_a =pi*D_a*H_a#area
N_a= round(S_c/S_a)#no. of bags
print "\n area S_a=%.2f ft**2\n number og bags N_a=%.2f "%(S_a,N_a)#
#(2) for bag B
D_b=10/12#diameter,ft
H_b=16#height,ft
S_b=pi*D_b*H_b#area
N_b=round(S_c/S_b)#no. of bags
print "\n area S_b=%.2f ft**2\n no. of bags N_b=%.2f "%(S_b,N_b)#
#total cost for each bag
#for bag A
c_a=26#cost per bag
TC_a=round(N_a*c_a)#total cost for A bag
print "\n total cost TC_a=%.2f $"%(TC_a)# #for bag B c_b=38#cost per bag TC_b=N_b*c_b#total cost for bag B print "\n total cost TC_b=%.2f$"%(TC_b)#
#since the total cost for bag A is less than bag B,select bag A

Example 28.4 page  no 432

filtering beg area S_c=5461.54 ft**2

area S_a=33.51 ft**2
number og bags N_a=163.00

area S_b=41.89 ft**2
no. of bags N_b=130.00

total cost TC_a=4238.00 $total cost TC_b=4940.00$


## Example 28.5 Page no 433¶

In [3]:
from __future__ import division
from math import pi,sqrt
print "\n Example 28.5 page no 433\n\n"
#we have to determine the number if filtering bags required and cleaning frequency for a plant equipped with a fabric system
q=50000#volumetric flow rate of gas stream,acfm
v_f=10#filteration velocity,ft/min
D=1#diameter of filtering bag,ft
L=15#length of filtering bag,ft
S_c=q/v_f#filtering area,ft**2
S=pi*D*L#area per bag,ft**2
N=S_c/S#no. of bags
print "\n no. of bags N=%.2f "%(N)#
c=0.0007143#dust concentration ,lb/ft**2
P_drop=8#pressure drop ,in H20
t=(P_drop-(0.2*v_f))/(5*c*v_f**2)#time sic ethe bags were cleaned
print "\n time t=%.2f min"%(t)#

 Example 28.5 page no 433

no. of bags N=106.10

time t=16.80 min


## Example 28.6 Page no 434¶

In [4]:
from __future__ import division
from math import pi,sqrt
print "Example 28.6 page no 434\n\n"
#comparison between flow in pipes and open channel flow
#water is passing through a trapezodial channel
l_b=20#length of bottom base,ft
l_t=50#length of top base,ft
h=7.5#height of channel,ft
A = (l_b+ l_t)*(h/2)#cross sectional area
P = l_b +sqrt(h**2+ (2*h)**2)#perimeter of trapezoid
S=0.0008#coeff. in manning equation
n=0.02#coeff. in manning eq.
q = 1.486*A*r_h**(2/3)*S**(1/2)/n#manning equation to determine flow rate
print "\n volumetric flow rate q=%.2f ft**3/s"%(q)#

Example 28.6 page no 434

volumetric flow rate q=2045.26 ft**3/s


## Example 28.7 Page no 435¶

In [5]:
from __future__ import division
print "\n Example 28.7 page no 435\n\n"
#waste water treatment plant
#we have  to compare the total nitrogen discharge from the watershed with that of the city 's sewage treatment plant
q_w=10#flow rate from waste water treatment plant
c=35#nitoren concentration,mg/l
m_dot_w=c*q_w*8.34#discharge   from the treatment plant
print "\n fdischarge from the treatment plant m_dot_w=%.2f lb/day"%(m_dot_w)#
S=8#area of watershed,mi**2
r=0.06#rate of rainfall,ml/day
n=.5#50% rain reaches the sewers
q=n*r*S*(5280**2/(3600*12))#volumetric flow rate of the runoff
c_r=9#tota# nitrogen conentration in runoff,mg/l
rho=62.4#/density of water
m_r=q*c_r*1e-6*(3600*24)*rho#total nitrogen discharge  from runoff
print "\n total nitrogen discharge m_r=%.2f lb/day "%(m_r)#
#since the durinf rain ,the runoff is over 2.5 times that for the tratment plant

 Example 28.7 page no 435

fdischarge from the treatment plant m_dot_w=2919.00 lb/day

total nitrogen discharge m_r=7515.12 lb/day


## Example 28.8 Page no 436¶

In [6]:
from __future__ import division
print "Example 28.8 page no 436\n\n"
#we have to determine the siaze an aerobic digester to treat the solids
m=1000#mass of solid that is generate by municipality,lb
VS=.78#volatile solids
t_h=20#detention time hydraulic, days
TS=0.044#percentage solids enterning digester
print "\n volume based on hyraulic load V_hl=%.2f ft**3"%(V_hl)#
#since V_hl >V_ol,the hdraulic time controls and the design volume is V_hl

Example 28.8 page no 436

volume based on hyraulic load V_hl=7295.09 ft**3


## Example 28.9 Page no 437¶

In [7]:
from __future__ import division
print "Example 28.9 page no 437\n\n"
#a large deep cavern has been proposed as an ultimate disposal site for both solid hazardous and municipal wastes
V_c=0.78#approximate total volume of cavern,mi**2
V_s=.75#% volume availiable for solid waste depositry
V=V_c*V_s*(5280)**3#volume of the cavern availible for the solid waste ,factor 5280 to convert mi**3 into ft**3
print "\n volume of cavern availible for solid waste V=%.2e ft**3"%(V)
r=20000#proposed maximum waste feed rate to cavern ,lb/day
rho=30#average bulk density,lb/ft**3
q=(r/rho)*(6*52)#volume rate of solid deposited within the cavern in ft**3/year
print "\n q=%.2f "%(q)#
t=V/q#time to fill the cavern
print "\n time to fill the cavern t=%.2e year"%(t)#

Example 28.9 page no 437

volume of cavern availible for solid waste V=8.61e+10 ft**3

q=208000.00

time to fill the cavern t=4.14e+05 year


## Example 28.10 Page no 438¶

In [8]:
from __future__ import division
from math import sqrt,pi
print "Example 28.10 page no 438\n\n"
# a compliance stack test on a facility yields the results ,we have to determine whether the incineratormeets the state particulate standard of 0.05 gr/dscf
g=9.807#grav. acc
rho_l=1000#density of manometer fluid,kg/m**3
rho=1.084#density of flue gas,kg/m**3
C=0.85#pitot tube constant
h=0.3772#mean pitot tube reading ,in H2O
m=0.16#mass of particulate collected ,g
V=35#volume sampled,dscf
C_p=m*15.43/V#partculate concentration,gr/dscf
print "\n particulate con. C_p=%.2f gr/dscf"%(C_p)#
#since this does not exceed the particulate standard of 0.05 gr/dscf,the facility is not in compliance
#the stack flow rate is calculated from the velocity measurement
v=C*sqrt(2*g*(rho_l/rho)* 0.0254*h)/.3048#velocity
print "\n velocity v=%.2f fps"%(v)#
D=2#diameter of stack,ft
v_s=(v*pi*D**2/4)*60#stack flow rate
print "\n stack flow rate v_s=%.2f acfm"%(v_s)#
w_mo=0.07#% moisture in stack gas
v_dry=(1-w_mo)*v_s#dry volumetric flow rate
#correct to standard conditions of  70 deg F and 1 atm
T_s=530# standard temprature deg R
P_s=29.9#standard pressure,psi
P_g=29.6#pressure of stack gas,psi
T_g=600#temprature of standard gas,deg R
q_s=v_dry*(T_s/T_g)*(P_g/P_s)#standard volumetric flow rate
print "\n standard volumetric flow rate q_s=%.2f dscfm"%(q_s)
R_e=C_p*q_s*(1440/7000)#particulate emission rate
print "\n particulate emmision rate R_e=%.2f lb/day"%(R_e)#
w_co2=0.14#percentage of co2 by volume
w_N2=0.79#percentage of N2 by volume
mw_o=32#molecular weight of oxygen
mw_co2=44#molecular weight of co2
mw_N2=28#molecular weight of N2
MW_d=w_mo*mw_o + w_co2*mw_co2 +w_N2*mw_N2#molecular weight of flue gas on dry basis
print "\n mol. weight of flue gas on dry basis MW_d=%.2f lb/lbmol"%(MW_d)#


Example 28.10 page no 438

particulate con. C_p=0.07 gr/dscf

velocity v=36.72 fps

stack flow rate v_s=6921.12 acfm

standard volumetric flow rate q_s=5628.65 dscfm

particulate emmision rate R_e=81.67 lb/day

mol. weight of flue gas on dry basis MW_d=30.52 lb/lbmol