Chapter 20 : Ventilation¶

Example 20.2 Page no 269¶

In [1]:
from __future__ import division
print "\n Example 20.2 page no 269\n\n"
#ventilation required in an indoor work area where a toluene containing adhesive in a nanotechnology process is used.
#equation  for estimate the dilution air requirement
C_a=80e-6#concentration of toluene
q=3/8#volumatric flow rate, gal/h
v=0.4#adhesive contains 4 volume % toluene
S_g=0.87#specific gravity
print "\n C_a concentration of toluene=%.2f \n q  volumatric flow rate q=%.2f gal/h \n S_g specific gravity=%.2f "%(C_a,q,S_g)#
#mass flow rate of toluene
m_dot_tol=q*v*S_g*(8.34)#factor 8.34 for lb
print "\n mass flow rate m_dot-tol=%.2f lb/h"%(m_dot_tol)#
m_dot_g=m_dot_tol*(454/60)#unit conversion of mass flow rate in g/min
print "\n mass flow rate in g/min m_dot_g=%.2f g/min"%(m_dot_g)#
M_w=92#molecular weight of toluene
n_dot_tol=m_dot_g/M_w#no. of gm moles of toluene/min
print "\n no. of moles n_dot_tol=%.2f gmol/min"%(n_dot_tol)#
#resultant toluene vapor volumatric flow rate q_tol is directly calculated from th eidal gas law
#applying ideal gas law
R=0.08206#gas constant
P=1#standard pressure
T=293#standard temperature
print "\n R gas constant=%.2f atm.L/(gmol.K)\n T temperature=%.2f K\n P pressure =%.2f atm"%(R,T,P)#
q_tol=n_dot_tol*R*T/P#toluene vapor volumatric flow rate
print "\n toluene vapor vol. flow rate q_tol=%.2f L/min"%(q_tol)#
q_tol=2.15#round off value
#the required diluent volumatric flow rtae
K=5#dimensionless mixing factor
q_dil=K*q_tol/(C_a)#diluent vol. flow rate
print "\n diluent vol. flow rate q_dil=%.2f L/min"%(q_dil)#
Example 20.2 page no 269

C_a concentration of toluene=0.00
q  volumatric flow rate q=0.38 gal/h
S_g specific gravity=0.87

mass flow rate m_dot-tol=1.09 lb/h

mass flow rate in g/min m_dot_g=8.24 g/min

no. of moles n_dot_tol=0.09 gmol/min

R gas constant=0.08 atm.L/(gmol.K)
T temperature=293.00 K
P pressure =1.00 atm

toluene vapor vol. flow rate q_tol=2.15 L/min

diluent vol. flow rate q_dil=134375.00 L/min

Example 20.3 Page no 270¶

In [2]:
from __future__ import division
print "Example 20.3 page no 270 \n\n"
# a certain poorly ventilated room chemical stroage room has a ceiling fan
#inside this room bottle of iron(3) sulfide sits next to a bottle sulfuric acid  containg 1 lb H2SO4 in water
# an earthquake sends the botlles on the shelf crashing to the floor where bottles break and their contant mix and react to form iron(3) sulfate and hydrogen sulfide
#we have to calculate maximum H2S concentration that could be reached in the room
Mw_Fe2S3=208#mol. weight of Fe2S3
Mw_H2SO4=98#mol. weight of H2SO4
Mw_H2S=34#mol. weight of H2S
Mw_air=29#mol. weight of air
#balancing chemical reaction
# from the stiochiometric of the reaction ,sulfuric acid is the limiting reagent
# 0.030 lbmol of Fe2S3 is required to react with 0.010 lbmol of H2SO4\
v_r=1600#volume of room,ft**3
n_H2SO4=0.010# lbmol of H2SO4
Stoi_c_H2SO4=3#stoichiometric coeff. of H2SO4
Stoi_c_H2S=3#stoichiometric coeff. of H2S
n_H2S=n_H2SO4*(Stoi_c_H2S/Stoi_c_H2SO4)#lbmol of H2S
print "\n lbmol of H2S n_H2S=%.2f lbmol"%(n_H2S)#
m_H2S=n_H2S*Mw_H2S#conversion of moles into mass of H2S
print "\n mass of H2S m_H2S=%.2f lb"%(m_H2S)#
#at 32 degF and i atm pressure an ideal gas occupies 359 ft**3 volume then,at 51 deg F occupies
T_r=51#temperature of air in the room
T_st=32#standard temperature
v_st=359#standard volume
print "\n stand. temperature T_st=%.2f F\n temperature of air in room T_r=%.2f F\n stand. volume v_st=%.2f ft**3"%(T_st,T_r,v_st)#
V_a=v_st*(460+T_r)/(460+T_st)#volume of air
print "\n volume of air at 51deg F V_a=%.2f ft**3"%(V_a)#
#the final concentration of H2S in the room in ppm C_H2S
C_H2S=m_H2S*(V_a/Mw_air)*1e+6/(v_r)
print "\n conc. of H2S in ppm C_H2S=%.2f ppm"%(C_H2S)#
Example 20.3 page no 270

lbmol of H2S n_H2S=0.01 lbmol

mass of H2S m_H2S=0.34 lb

stand. temperature T_st=32.00 F
temperature of air in room T_r=51.00 F
stand. volume v_st=359.00 ft**3

volume of air at 51deg F V_a=372.86 ft**3

conc. of H2S in ppm C_H2S=2732.19 ppm

Example 20.4 Page no 271¶

In [3]:
from __future__ import division
print "Example 20.4 page no 271\n\n"
#vinyl chloride application
#calculation of density by using ideal gas law
Mw=78#molecular weight of vinyl chloride
R=82.06#gas constant,cm**3.atm/mol.K
T=298#temperature,K
P=1#pressure ,atm
rho=P*Mw/(R*T)#density of vinyl chloride
print "\n rho density of vinyl chloride=%.2f g/cm**3"%(rho)#
#given
m_dot=10#mass flow rate,g/min
q=m_dot/rho#volumatric flow rate
print "\n vol. flow rate q=%.2f cm**3/min"%(q)#
q_acfm=0.1107#vol flow rate in acfm
#cal. the air flow rate in acfm q_air required to meet the 1.0 ppm constraint with the equation
q_air=q_acfm/1e-6
print "\n vol.flow rate q_air=%.2f acfm"%(q_air)#
S_factor=10#correct for mixing by employing a saftey factor
#apply saftey factor to calculate the actual air flow rate for dilution ventilation
q_dil=S_factor*q_air
print "\n air flow rate for dilution q_dil=%.2f acfm"%(q_dil)#
#now consider the local exhaust ventilation by first calculating the face area
H=30#height of hood,in
W=25#width of hood,in
S=H*W/144#surface area of hood ,ft**2
#the air flow rate in acfm q_air,exh required for a face velocity of 100 ft/min is then
v=100#face velocity,ft/min
q_exh=round(S*v)
print "\n air flow rate q_exh=%.2f acfm"%(q_exh)#
Example 20.4 page no 271

rho density of vinyl chloride=0.00 g/cm**3

vol. flow rate q=3135.11 cm**3/min

vol.flow rate q_air=110700.00 acfm

air flow rate for dilution q_dil=1107000.00 acfm

air flow rate q_exh=521.00 acfm

Example 20.7 Page no 276¶

In [4]:
from __future__ import division
print "\n Example 20.7 page no 276\n\n"
#refer to illustrative Example 20.5
#(1)
#we have to calculate minimum air ventilation flow rate into the room containing 10 ng/m**3 of a toxic chemical
#ng means nanograms
rV=250#chemical generated in the laboratory,ng/min
c_o=10#room containg toxic chemical of 10ng/m**3
c=35#limit of chemical concentration,ng/m**3
#applicable modal in this case
#q_o(c_o-c) + rV  =V*dc/dt
#substituting gives
q_o=(-rV)/(c_o-c)#minimum air ventilation flow rate
print "\n q_o min. air ventilation flow rate=%.2f m**3/min"%(q_o)#
Example 20.7 page no 276

q_o min. air ventilation flow rate=10.00 m**3/min

Example 20.8 Page no 277¶

In [5]:
from __future__ import division
from math import exp
from scipy.optimize import fsolve
print " Example 20.8 page no 277\n\n"
#refer to example no 20.5 and 20.7
V=142#volume of room,m**3
q=12.1# flow rate of air,m**3/min
tou=V/q#time ,min
r=30#rate of generation of chemical,ng/min
k=r/V#ng/(m**3.min)
c_i=85#intial concentration in laboratory,ng/m**3
c_o=10#given concentration in room
c=20.7#final concentration in room
#by using  trial and error mthod we get
def f(t):
y=c_i*(exp(-t/tou))+ (c_o+k*tou)*(1-exp(-t/tou)) - c
return y
t=fsolve(f,30)#
#by using trail and error method we get
t=29
print "\n t=%.2f min "%(t)#
Example 20.8 page no 277

t=29.00 min