Chapter 6 : Drying¶
Example 6.1¶
In [22]:
H1=.005; #humidity of incoming air per kg of dry air
T1=25.; #wet bulb temperature
print "\n the solid temp. correspond to wbt and they are 23, 27,32 and 34 degree respectively"
#part(ii)
Ybar=.032; #kg water/kg dry air#final moist air condotions
T2=42.; #dry bulb temperature
Mair=28.84; #molecular weight of air
Mwater=18; #molecular weight of water
pt=1.013*10**5; #total pressure in pascal
Vh=8315*((1/Mair)+(Ybar/Mwater))*((T2+273)/pt);
r=300; #flow rate of moist air leaving the dryer
a=r*60/Vh; #amount of dry air leaving /hr
w=a*(Ybar-0.005); # water removed /hr
# Result
print "\n the water removed /hr is :%fkg /hr"%w
#end
Example 6.2¶
In [1]:
%matplotlib inline
Ls=262.5; #mass of bone dry solid ais the drying surface
A=262.5/8; #both upper surafce and lower surface are exposed
Nc=0.3; #in kg/m**2*hr
x2=.06; #moisture content on wet basis finally after drying
x1=.25; #moisture content on wet basis finally after drying
Xcr=0.20; #crtical moisture content
X1=x1/(1-x1); #moisture content on dry basis intially
X2=x2/(1-x2); #moisture content on dry basis finally after drying
Xbar=0.025; #equillibrium moisture
# Calculation
t1=Ls/(A*Nc) *(X1-Xcr); #so for constant rate period
#for falling rate period we find time graphically
p = [.20 ,.18, .16, .14, .12, .10, .09, .08, .07, .064];
a = [3.3, 5.56, 6.25, 7.14, 8.32, 10.00, 11.11, 12.5, 14.29, 15.625];
from matplotlib.pyplot import *
# Result
plot(p,a);
title("Fig.6.18 Example2 1/N vs X for fallling rate period");
xlabel("X-- Moisture content, X(kg/kg)");
ylabel("Y-- 1/N, hr,m**2/kg");
Area=1.116; #area under the curve
t2=Area *Ls/A; #falling rate period we find time graphically
ttotal=t1+t2; #total time for drying
print "\n the total time for drying the wet slab on wet basis is :%f min"%ttotal
#end
Example 6.3.a¶
In [2]:
#part(i)
# Variable Declaration
p = [1.0,0.6,.44,0.4,.36,.224, 0.14];
a = [5.0,5.0, 4.5, 4.0, 3.5, 2.00, 1.00];
i=0; #looping for calc. of 1/N
t = [0,0,0,0,0,0,0]
# Calculation
while(i<7): #looping begins
t[i]=1./(a[i]);
i=i+1;
from matplotlib.pyplot import *
# Result
plot(p,a);
title("Fig.6.19(a) Example3 Drying Rate curve");
xlabel("X-- Moisture content, X(kg/kg) ---->");
ylabel("Y-- Drying Rate, N(kg/hr.m**2 ---->");
#xset('window',1);
plot(p,t,"o-");
title("Fig.6.19(b) Example3 1/N vs X");
xlabel("X-- Moisture content, X(kg/kg) --->");
ylabel("Y-- 1/N, hr,m**2/kg --->");
show()
#from X=0.6 to 0.44 ,falling rate is non linear and from X=.44 to .14 falling rate is linear
print "\n from the graph we get critical moisture content as 0.6 kg moisture/kg dry solid"
#end
Example 6.3.b¶
In [24]:
# Variable Declaration
#part(ii)
w1=5.; #wet of wet solid
c1=.5/(1-.5); #moisture content per kg wet solid
w2=5.*0.5; #moisture for 5kg wet solid
w3=w1-w2; #weight of dry solid
xbar=0.05; #equillibrium moisture content
Xbar=xbar/(1-xbar); #equillibrium moisture content
Ls=2.5; #mass of bone dry solid ais the drying surface
A=5.; #both upper surafce and lower surface are exposed
Nc=0.6; #in kg/m**2*hr
#from X=0.6 to 0.44 ,falling rate is non linear and from X=.44 to .14 falling rate is linear
X2=.15/(1-.15);
Xcr=.6; #kg moisture per kg dry solid
#so we can find time fro drying from 0.6 to .44 graphically and then for X=.44 to .1765
X1=1.; #moisture content on dry basis intially
# Calculation
t1=Ls/(A*Nc) *(X1-Xcr); #time taken for constant drying rate(fromX=1 to .6)
X1=.44; #moisture content on dry basis
import math
t2=(Ls/(A*Nc))*((Xcr-Xbar)*math.log((X1-Xbar)/(X2-Xbar)));
t3=0.0336*Ls/Nc; #fro graph we get from X=.6 to .44
ttotal=t1+t2+t3; #total time for drying the wet slab
# Result
print "\n the total time for drying the wet slab to 15 percent moisture on wet basis is :%f min"%(ttotal*60)
#end
Example 6.4¶
In [3]:
#let Ls/A=p
# Variable Declaration
p=48.; #mass of bone dry solid ais the drying surface
v=1.5*1.5*.5; #volume of material
Nc=1.22; #in kg/m**2*hr
Xcr=0.2; #crtical moisture content
X1=0.25; #moisture content on dry basis intially
X2=0.08; #moisture content on dry basis finally after drying
Xbar=0.025; #equillibrium moisture
#tbar=(Ls/(A*Nc))*((Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
# Calculation
t1=p/(Nc) * (X1-Xcr); #time taken for constant drying rate period
# equillibrium relation is given under
p = [.18,.15,.14,.11,.07,.05];
a = [.8772,1.11,1.25,1.7857,4.545,20];
from matplotlib.pyplot import *
# Result
plot(p,a);
title("Fig.6.20 Example4 1/N vs X for fallling rate period");
xlabel("X-- Moisture content, X(kg/kg) ---->");
ylabel("Y-- 1/N, hr,m**2/kg ---->");
a=14*.025*1; #area under the curve
t2=a*48; #time taken for varying drying period
ttotal=t1+t2; #total time taken
print "\n total time for drying the material from 25 to 8 percent moisture under same drying conditions is :%f hr"%(ttotal)
#end
Example 6.5.a¶
In [4]:
# Variable Declaration
w=[5.314,5.238,5.162,5.124,5.048,4.972,4.895,4.819,4.743,4.667,4.524,4.468,4.426,4.340,4.120]
t=[0.0,0.4,0.8,1.0,1.4,1.8,2.2,2.6,3.0,3.4,4.2,4.6,5.0,6.0]
#part(i)
x=4.120; #weight of the dried material
print "\n moisture content (dry basis) "
i=0;
p = [] #looping starts
# Calculation
while(i<15): #calculation of moisture content
p.append((w[i]-x)/x);
print "\n :%f"%p[i]
i=i+1;
print "Drying rate kg/hr*m**2"
i=1;
a = []
for i in range(15):
a.append(0)
while(i<14):
a[i] =((p[i-1]-p[i])*4.12/(t[i]-t[i-1]))
print " %f "%a[i]
i=i+1;
from matplotlib.pyplot import *
a[0]=.19;
a[14]=0;
# Result
print "\n\n from the above data it is clear that critical moisture content Xcr=0.11"
plot(p,a);
title("Fig.6.19(a) Example3 Drying Rate curve");
xlabel("X-- Moisture content, X(kg/kg) ---->");
ylabel("Y-- Drying Rate, N(kg/hr.m**2 ---->");
show()
#end
Example 6.5.b¶
In [27]:
# Variable Declaration
#part(ii)
w1=4.934; #weight after two hours
w0=5.314; #initial weight
w2=w0-w1; # water evaporated in 2 hrs
H1=.01; #humidty of incoming air
H2=.03; #humidity of leaving air
yout=.03;
yin=.01;
# Calculation
Gs=w2/(yout- yin); #water carried away
# Result
print "\n the amount of air required in 2hours is :%f kg"%Gs
#end
Example 6.5.c¶
In [28]:
# Variable Declaration
#part(iii)
#let us choose the consistency of 11 and 13 readings
Xbar=0; #equillibrium moisture content
Ls=4.12; #mass of bone dry solid ais the drying surface
A=1.; #both upper surafce and lower surface are exposed
Nc=0.19; #in kg/m**2*hr
X1=.098; #moisture content on dry basis intially
Xcr=.11; #kg moisture per kg dry solid
X2=0.074; #moisture content on dry basis finally
# Calculation
import math
tfall=(Ls/(A*Nc))*((Xcr-Xbar)*math.log((X1-Xbar)/(X2-Xbar)));
# Result
print "\n from this data we get time as :%f hour"%tfall
print "\n the actual time is 0.8 hours"
#end
Example 6.6¶
In [29]:
# Variable Declaration
Xcr=0.55; #crtical moisture content
X1=1; #moisture content on dry basis intially
X2=.1; #moisture content on dry basis finally after drying
Xbar=.06; #equillibrium moisture
#since eqn 1 is tobe divided by eqn 2 so let the value of Ls/A*Nc be = 1 as it will be cancelled
p=1; #let Ls/A*Nc be =p
#p=poly([0],'p'); #calc. of time 1
tbar=1; #since the eqns are independent of tbar
# Calculation
import math
t1=0.5960514 #roots(tbar- p*((X1-Xcr)+(Xcr-Xbar)*math.log((Xcr-Xbar)/(X2-Xbar)))); #------eqn1
X2bar=.16;
#p=poly([0],'p'); #calc. of time 2
t2=0.8138516 #roots(tbar- p*((X1-Xcr)+(Xcr-Xbar)*math.log((Xcr-Xbar)/(X2bar-Xbar))));#------eqn2
#let t1/t2 be = k
k=t1/t2;
ans=1./k-1; #reduction in time for drying
# Result
print "\n the reduction in time for drying is :%f percent"%(ans*100)
#end
Example 6.7¶
In [30]:
#Ls/ A*Nc is unknown;
# Variable Declaration
Xcr=0.14; #crtical moisture content
X1=.3/(1-.3); #moisture content on dry basis intially
X2=0.1/(1-0.1); #moisture content on dry basis finally after drying
Xbar=.04; #equillibrium moisture
tbar=5; #time needed to dry from 30 to 6 percent on bone dry basis
#let Ls / A*Nc be = p
#p=poly([0],'p'); #calc. of Ls / A*Nc be = p value
# Calculation and Result
x= 15.495992 #roots(tbar-p * ((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar))));
print "\n the value of Ls/ A*Nc is :%f"%x
import math
#new X1 AND X2 are now given as follows
X1=0.3/(1-.3); #new moisture content on dry basis intially
X2=0.064; #new moisture content on dry basis finally after drying
tbar=x * ((X1-Xcr)+(Xcr-Xbar)*math.log((Xcr-Xbar)/(X2-Xbar)));
print "\n the time for drying the sheets from 30 to 6 percent moisture under same drying conditions is :%f hr"%tbar
#end
Example 6.8.a¶
In [31]:
# Variable Declaration
Ls=1000.; #mass of bone dry solid ais the drying surface
A=55.; #both upper surafce and lower surface are exposed
v=.75; #velocity of air
Nc=.3*10**-3; #in kg/m**2*s
x2=.2; #moisture content on wet basis finally after drying
Xcr=0.125; #crtical moisture content
X1=0.15; #moisture content on dry basis intially
X2=0.025; #moisture content on dry basis finally after drying
Xbar=0.0; #equillibrium moisture
import math
# Calculation
tbar=(Ls/(A*Nc))*((X1-Xcr)+(Xcr-Xbar)*math.log((Xcr-Xbar)/(X2-Xbar)));
# Result
print "the time for drying the sheets from .15 to .025 kg water /kg of dyr solid moisture under same drying conditions is :%f hour"%(tbar/3600);
#end
Example 6.8.b¶
In [32]:
# Variable Declaration
Ls=1000.; #mass of bone dry solid ais the drying surface
A=55.; #both upper surafce and lower surface are exposed
v=.75; #velocity of air
Nc=.3*10**-3; #in kg/m**2*s
x2=.2; #moisture content on wet basis finally after drying
Xcr=0.125; #crtical moisture content
X1=0.15; #moisture content on dry basis intially
X2=0.025; #moisture content on dry basis finally after drying
Xbar=0.0; #equillibrium moisture
tbar=3.8077; #time to dry material ,calculated from previous part
V1=.75; #old velocity
V2=4.; #new velocity
import math
# Calculation
Nc2=Nc*(V2/V1)**.71; #in kg/m**2*s
t2=(Ls/(A*Nc2))*((X1-Xcr)+(Xcr-Xbar)*math.log((Xcr-Xbar)/(X2-Xbar))); #if air velocity is increased to 4
t=tbar-t2/3600; #time saved
# Result
print "\n the time saved , if air velocity is increased to 4 m/s: %f"%t
#end
Example 6.9¶
In [34]:
x1=.75; #moisture content on wet basis
xbar=0.1; #equilllibrium moisture on dry basis
xcr=0.6; #critical moisture content
Ls=0.90; #mass of bone dry solid ais the drying surface
A=0.3*0.3*2; #both upper surafce and lower surface are exposed
#A*Nc=10**-4;
x2=.2; #moisture content on wet basis finally after drying
# Calculation
Xcr=0.6/0.4; #crtical moisture content
X1=3; #moisture content on dry basis intially
X2=0.25; #moisture content on dry basis finally after drying
Xbar=0.1/0.9; #equillibrium moisture
import math
tbar=Ls/(10**-4) * ((X1-Xcr)+(Xcr-Xbar)*math.log((Xcr-Xbar)/(X2-Xbar)));
# Result
print "\n the time for drying the sheets from 75 to 25 percent moisture under same drying conditions is :%f hr"%(tbar/3600);
#end
Example 6.10¶
In [35]:
# Variable Declaration
Xcr=0.16; #crtical moisture content
X1=.33; #moisture content on dry basis intially
X2=0.09; #moisture content on dry basis finally after drying
Xbar=.05; #equillibrium moisture
tbar=7; #time needed to dry from 33 to 9 percent on bone dry basis
import math
#let Ls / A*Nc be = p
#p=poly([0],'p'); #calc. of Ls / A*Nc be = p value
# Calculation
x= 24.886579 #roots(tbar-p * ((X1-Xcr)+(Xcr-Xbar)*math.log((Xcr-Xbar)/(X2-Xbar))));
#new X1 AND X2 are now given as follows
X1=0.37; #new moisture content on dry basis intially
X2=0.07; #new moisture content on dry basis finally after drying
tbar=x * ((X1-Xcr)+(Xcr-Xbar)*math.log((Xcr-Xbar)/(X2-Xbar)));
# Result
print "\n the time for drying the sheets from 33 to 9 percent moisture under same drying conditions is :%f hr"%tbar
#end
Example 6.11¶
In [36]:
# Variable Declaration
d=.22; #density of dry pulp in g/cc;
x1=.65; #moisture content on wet basis
x2=.3; #moisture content on wet basis
Ls=2.5; #mass of bone dry solid ais the drying surface in kg
A=1.5*1.5*2; #both upper surafce and lower surface are exposed
v=1.5*1.5*.5; #volume of material
Nc=1.4; #in kg/m**2*hr
Xcr=1.67; #crtical moisture content
# Calculation
X1=x1/(1-x1); #moisture content on dry basis intially
X2=x2/(1-x2); #moisture content on dry basis finally after drying
Xbar=0.0; #equillibrium moisture
import math
tbar=(Ls/(A*Nc))*((X1-Xcr)+(Xcr-Xbar)*math.log((Xcr-Xbar)/(X2-Xbar)));
# Result
print "\n the time for drying the sheets from 65 to 30 percent moisture under same drying conditions is :%f hour"%tbar
#end
Example 6.12¶
In [37]:
# Variable Declaration
d=.22; #density of dry pulp in g/cc;
Ls=1.125*10**-2*.22*10**3; #mass of bone dry solid ais the drying surface
A=1.5*1.5*2; #both upper surafce and lower surface are exposed
v=1.5*1.5*.5; #volume of material
Nc=1.4; #in kg/m**2*hr
x2=.2; #moisture content on wet basis finally after drying
Xcr=0.46; #crtical moisture content
X1=0.15; #moisture content on dry basis intially
X2=0.085; #moisture content on dry basis finally after drying
Xbar=0.025; #equillibrium moisture
import math
#tbar=(Ls/(A*Nc))*((Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
# Calculation
tbar=(Ls/(A*Nc))*((Xcr-Xbar)*math.log((X1-Xbar)/(X2-Xbar)));
# Result
print "\n the time for drying the sheets from 15 to 8.5 percent moisture under same drying conditions is :%f min"%(tbar*60)
#end
Example 6.13¶
In [38]:
# Variable Declaration
import math
Xcr=0.14; #crtical moisture content
x1=0.3; #moisture content on wet basis
x2=0.1; #moisture content on wet basis
X1=x1/(1-x1); #moisture content on dry basis intially
X2=x2/(1-x2); #moisture content on dry basis finally after drying
Xbar=0.04; #equillibrium moisture
tbar=5; #time needed to dry from 30 to 10 percent on bone dry basis
#let Ls / A*Nc be = p
#p=poly([0],'p'); #calc. of Ls / A*Nc be = p value
x= 15.495992 #roots(tbar-p * ((X1-Xcr)+(Xcr-Xbar)*math.log((Xcr-Xbar)/(X2-Xbar))));
#new X1 AND X2 are now given as follows
# Calculation
x1=.3; #new moisture content on wet basis
x2=0.06; #new moisture content on wet basis
X1=x1/(1-x1); #new moisture content on dry basis intially
X2=x2/(1-x2); #new moisture content on dry basis finally after drying
tbar=x * ((X1-Xcr)+(Xcr-Xbar)*math.log((Xcr-Xbar)/(X2-Xbar)));
# Result
print "\n the time for drying the sheets from 30 to 10 percent moisture under same drying conditions is :%f hr"%tbar
#end
Example 6.14¶
In [39]:
d=450.; #density of dry pulp in kg/m**3;
thickness=0.05; #thickness in m**2
Ls=d*thickness; #mass of bone dry solid ais the drying surface
A=1.; #area in m**2
v=1*5*10**-2; #volume of material
Nc=4.8; #in kg/m**2*hr
xcr=.2;
xbar=0.02;
x1=.45; #new moisture content on wet basis
x2=0.05; #new moisture content on wet basis
# Calculation
X1=x1/(1-x1); #new moisture content on dry basis intially
X2=x2/(1-x2); #new moisture content on dry basis finally after drying
Xbar=xbar/(1-xbar); #crtical moisture content
Xcr=xcr/(1-xcr); #equillibrium moisture
import math
#tbar=(Ls/(A*Nc))*((Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
tbar=Ls/(A*Nc) * ((X1-Xcr)+(Xcr-Xbar)*math.log((Xcr-Xbar)/(X2-Xbar)));
# Result
print "\n the time for drying the sheets from 45 to 5 percent moisture under same drying conditions is :%f min"%tbar
#end
Example 6.15¶
In [40]:
# Variable Declaration
t1=20.; #ambient air temperature
t2=70.; #exhaust air temperature
r1=150.; #evaporation of water
r2=.25; #outlet solid moisture content
t3=15.; #inlet solid temperature
t4=65.; #outlet solid temperature
p=5.; #power demand in KW
h=18.; #heat loss in kj
h1=1.; #mean specific heat of dry air in kj/kg*K
h2=1.25; #mean specific heat of dry material in kj/kg*K
h3=4.18; #mean specific heat of moisture in kj/kg*K
e=2626.; #enthalpy of saturated water vapour in kj/kg
# Calculation
#basis is 1hr
a1=r1*h3*(t4-t3); #heat required for heating 150 kg of water from 15 to 65
a2=r1*e; #heat required for 150 kg water evaporation
a3=2000*h1*(t2-t1); #heat required for heating air from 20 to 70
a4=r2*h3*(t4-t3); #heat required for heating moisture from 15 to 65
a5=120*h2*(t4-t3); #heat required for heating dry solid from 15 to 65
hlost=h*3600; #heat lost in kj
total=(a2+a3+a4+a5+hlost)/3600.; #total heat lost
# Result
print "\n :%f kW of heat required for 2000kg/hr of dry air"%total
ans1=a2+a1; #heat needed for evaporation
print "\n heat needed fro evaporation is :%f"%(ans1/3600)
ans2=(ans1/3600.)/total; #fraction of this heat needed for evaporation
print "\n fraction of this heat needed for evaporation:%f"%(ans2)
#end
Example 6.16¶
In [41]:
m1=.12; #initial moisture content
dT=85.; #product of 85 degree is used in design purpose
U=1700.; #overall heat transfer coefficient
m2=.4; #final moisture content
r=20.; #production rate
#4 kg of moisture is present in 100 kg product
# Calculation
t=4*20./100; # moisture content in 20 kg moisture
w=20-t; #dry solid weight
i=w*m1/(1-m1); #initial moisture content
j=i-t #water evaporated
ds=2296.1; #latent heat for vaporisation at 85 degree in kj/kg
h=j*ds; #heat required (assuming th esolid mix. enters at 85)
#U*A*dT = j*ds
A=h/(U*dT); #surface area of the roller required to produce aproduction rate of 20 kg product per hour
# Result
print "\n surface area of the roller required to produce aproduction rate of 20 kg product per hour:%f m**2"%(A/3.600);
#end
Example 6.17¶
In [42]:
# Variable Declaration
r=7.5*10**-5; #constant drying rate in kg/s
A1=.3*.3**2; # area of the sppecimen
Nc=r/A1; #drying rate
# Calculation
Xcr=.15/0.85; #.15 is the critical moisture content
Xo=.25/.75; #.25 is the initial moisture content
Xfinal=.02/0.98; #.02 is the final moisture content
Xbar=0; #equillibrium moisture content
A=1.2*.6*2; #area of the new solid
Ls=28.8; #bone dry weight of new solid
v1=.3*.3*.006; #volume of the old solid;
v2=.6*1.2*.012; #volume of the new solid
w2=1.8; #weight of the old solid
w3=864*10**-5*1.8*10**-5/54; #weight of the bone dry solid
#Nc is prporional to =(t-ts) = (G)**0.71---- whrere G is the mass flow rate
v1=3.; #old velocity
Tg=52.; #old dry bulb temperature
Tw=21.; #wet bulb temperature
H=.002; #humidity
SH=0.015; #saturated humidity
vnew=5. #new velocity
Tgnew=66.; #new DBT
Twnew=24.; #new WBT
Hnew=.004; #new humidity
SH=.020; #new satuurated humidity
import math
#hence drying rate of air under new condition
Nc=4.167*10**-4*((vnew/v1)*(273+Tg)/(273+Tgnew))**0.71 * ((.019-H)/(.015-H));#drying rate of air under new condition in kg/m**2*s
DT=Ls/(A*Nc) * ((Xo-Xcr)+(Xcr-Xbar)*math.log((Xcr-Xbar)/(Xfinal-Xbar)));
# Result
print "\n the time for drying the sheets from 25 to 2 percent moisture under same drying conditions is :%f hours"%(DT/3600)
#end