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-rwxr-xr-x599/CH6/EX6.1/example6_1.sce33
-rwxr-xr-x599/CH6/EX6.10/example6_10.sce25
-rwxr-xr-x599/CH6/EX6.11/example6_11.sce22
-rwxr-xr-x599/CH6/EX6.12/example6_12.sce23
-rwxr-xr-x599/CH6/EX6.13/example6_13.sce28
-rwxr-xr-x599/CH6/EX6.14/example6_14.sce26
-rwxr-xr-x599/CH6/EX6.15/example6_15.sce35
-rwxr-xr-x599/CH6/EX6.16/example6_16.sce21
-rwxr-xr-x599/CH6/EX6.17/example6_17.sce38
-rwxr-xr-x599/CH6/EX6.2/example6_2.sce46
-rwxr-xr-x599/CH6/EX6.3.a/example6_3_a.sce32
-rwxr-xr-x599/CH6/EX6.3.b/example6_3_b.sce30
-rwxr-xr-x599/CH6/EX6.4/example6_4.sce37
-rwxr-xr-x599/CH6/EX6.5.a/example6_5_a.sce50
-rwxr-xr-x599/CH6/EX6.5.b/example6_5_b.sce16
-rwxr-xr-x599/CH6/EX6.5.c/example6_5_c.sce18
-rwxr-xr-x599/CH6/EX6.6/example6_6.sce24
-rwxr-xr-x599/CH6/EX6.7/example6_7.sce26
-rwxr-xr-x599/CH6/EX6.8.a/example6_8_a.sce20
-rwxr-xr-x599/CH6/EX6.8.b/example6_8_b.sce31
-rwxr-xr-x599/CH6/EX6.9/example6_9.sce22
21 files changed, 603 insertions, 0 deletions
diff --git a/599/CH6/EX6.1/example6_1.sce b/599/CH6/EX6.1/example6_1.sce
new file mode 100755
index 000000000..422ea7a86
--- /dev/null
+++ b/599/CH6/EX6.1/example6_1.sce
@@ -0,0 +1,33 @@
+
+clear;
+clc;
+printf("\t Example 6.1\n");
+
+ //air leaves the pre-heater of the dryer at 325K
+
+H1=.005; //humidity of incoming air per kg of dry air
+T1=25; //wet bulb temperature
+ //moisture is removed along constant wet bulb temp. till 60per R.H is reached
+// from the chart ,humidity of ai rleaving first shelf =.016 kg water /kg dry air.
+
+//dry bulb temp. of exit air is at 27 degree aand is at humidity of .016 kg water/kg dry air.the air is again heated to 52 degree dry bulb temp. in 2nd heater .
+
+ //so air leaves heater at 52 degree and humidity of .016 kg water/kg dry air. when it leaves the 2nd shelf the correspondin dry bulb temp. is 34 degree and humidity is .023 kg water/kg dry air. the air enters the 3rd shelf after preheating to 52 degree .
+
+//similarly fro 3rd shelf , exit air has a humidity of .028 kg water/kg dry air and adry bulb temp. is 39 degree. the air is leaving the 4rth shelf has a humidity of .016 kg water/kg dry air and adry bulb temp. of 42 degree(the figure is only indicative and doed not correspond toactual one)
+
+printf("\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
+printf("\n the water removed /hr is :%fkg /hr",w);
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.10/example6_10.sce b/599/CH6/EX6.10/example6_10.sce
new file mode 100755
index 000000000..e5a1a3c91
--- /dev/null
+++ b/599/CH6/EX6.10/example6_10.sce
@@ -0,0 +1,25 @@
+
+clear;
+clc;
+printf("\t Example 6.10\n");
+ //determine the drying condition of sample 0.3*0.3 size.sheet lost weight at rate of 10^-4 kg/s until the moisture fell to 60 percent
+
+//Ls/ A*Nc is unknown;
+
+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
+
+ //let Ls / A*Nc be = p
+p=poly([0],'p'); //calc. of Ls / A*Nc be = p value
+x=roots(tbar-p * ((X1-Xcr)+(Xcr-Xbar)*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)*log((Xcr-Xbar)/(X2-Xbar)));
+printf("\n the time for drying the sheets from 33 to 9 percent moisture under same drying conditions is :%f hr",tbar);
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.11/example6_11.sce b/599/CH6/EX6.11/example6_11.sce
new file mode 100755
index 000000000..697f30b05
--- /dev/null
+++ b/599/CH6/EX6.11/example6_11.sce
@@ -0,0 +1,22 @@
+
+clear;
+clc;
+printf("\t Example 6.11\n");
+
+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
+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
+
+ //initial moisture is more than Xcr, so there is constant rate drying period and only falling rate peroid is observed
+tbar=(Ls/(A*Nc))*((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
+printf("\n the time for drying the sheets from 65 to 30 percent moisture under same drying conditions is :%f hour",tbar);
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.12/example6_12.sce b/599/CH6/EX6.12/example6_12.sce
new file mode 100755
index 000000000..4baa3447e
--- /dev/null
+++ b/599/CH6/EX6.12/example6_12.sce
@@ -0,0 +1,23 @@
+
+clear;
+clc;
+printf("\t Example 6.12\n");
+
+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
+
+//tbar=(Ls/(A*Nc))*((Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
+ // but initial moisture is more than Xcr, so there is constant rate drying period and only falling rate peroid is observed
+tbar=(Ls/(A*Nc))*((Xcr-Xbar)*log((X1-Xbar)/(X2-Xbar)));
+printf("\n the time for drying the sheets from 15 to 8.5 percent moisture under same drying conditions is :%f min",tbar*60);
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.13/example6_13.sce b/599/CH6/EX6.13/example6_13.sce
new file mode 100755
index 000000000..a6a057540
--- /dev/null
+++ b/599/CH6/EX6.13/example6_13.sce
@@ -0,0 +1,28 @@
+
+clear;
+clc;
+printf("\t Example 6.13\n");
+
+//Ls/ A*Nc is unknown;
+
+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=roots(tbar-p * ((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar))));
+
+//new X1 AND X2 are now given as follows
+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)*log((Xcr-Xbar)/(X2-Xbar)));
+printf("\n the time for drying the sheets from 30 to 10 percent moisture under same drying conditions is :%f hr",tbar);
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.14/example6_14.sce b/599/CH6/EX6.14/example6_14.sce
new file mode 100755
index 000000000..1ca010477
--- /dev/null
+++ b/599/CH6/EX6.14/example6_14.sce
@@ -0,0 +1,26 @@
+
+clear;
+clc;
+printf("\t Example 6.14\n");
+
+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
+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
+
+//tbar=(Ls/(A*Nc))*((Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
+ // but initial moisture is more than Xcr, so there is constant rate drying period and only falling rate peroid is observed
+tbar=Ls/(A*Nc) * ((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
+printf("\n the time for drying the sheets from 45 to 5 percent moisture under same drying conditions is :%f min",tbar);
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.15/example6_15.sce b/599/CH6/EX6.15/example6_15.sce
new file mode 100755
index 000000000..b0f313354
--- /dev/null
+++ b/599/CH6/EX6.15/example6_15.sce
@@ -0,0 +1,35 @@
+
+clear;
+clc;
+printf("\t Example 6.15\n");
+
+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
+
+ //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
+printf("\n :%f kW of heat required for 2000kg/hr of dry air",total);
+ans1=a2+a1; //heat needed for evaporation
+printf("\n heat needed fro evaporation is :%f",ans1/3600);
+ans2=(ans1/3600)/total; //fraction of this heat needed for evaporation
+printf("\n fraction of this heat needed for evaporation:%f",ans2);
+
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.16/example6_16.sce b/599/CH6/EX6.16/example6_16.sce
new file mode 100755
index 000000000..f235578da
--- /dev/null
+++ b/599/CH6/EX6.16/example6_16.sce
@@ -0,0 +1,21 @@
+
+clear;
+clc;
+printf("\t Example 6.16\n");
+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
+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
+printf("\n surface area of the roller required to produce aproduction rate of 20 kg product per hour:%f m^2",A/3.600);
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.17/example6_17.sce b/599/CH6/EX6.17/example6_17.sce
new file mode 100755
index 000000000..0667836dd
--- /dev/null
+++ b/599/CH6/EX6.17/example6_17.sce
@@ -0,0 +1,38 @@
+
+clear;
+clc;
+printf("\t Example 6.17\n");
+ //moisture content reduces from 25 to 2
+r=7.5*10^-5; //constant drying rate in kg/s
+A1=.3*.3**2; // area of the sppecimen
+Nc=r/A1; //drying rate
+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
+
+ //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)*log((Xcr-Xbar)/(Xfinal-Xbar)));
+printf("\n the time for drying the sheets from 25 to 2 percent moisture under same drying conditions is :%f hours",DT/3600);
+
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.2/example6_2.sce b/599/CH6/EX6.2/example6_2.sce
new file mode 100755
index 000000000..1729d9aad
--- /dev/null
+++ b/599/CH6/EX6.2/example6_2.sce
@@ -0,0 +1,46 @@
+
+clear;
+clc;
+printf("\t Example 6.2\n");
+//table X*100,(kgmoisture/kg dry solid) N*100 (kg moisture evaporated /hr*m^2)
+
+// 35 30
+// 25 30
+// 20 30
+// 18 26.6
+// 16 23.9
+// 14 20.8
+// 12 18
+// 10 15
+// 9 9.7
+// 8 7
+// 7 4.3
+// 6.4 2.511111
+
+
+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
+
+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];
+
+plot(p,a,"o-");
+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
+printf("\n the total time for drying the wet slab on wet basis is :%f min",ttotal);
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.3.a/example6_3_a.sce b/599/CH6/EX6.3.a/example6_3_a.sce
new file mode 100755
index 000000000..213bd1e47
--- /dev/null
+++ b/599/CH6/EX6.3.a/example6_3_a.sce
@@ -0,0 +1,32 @@
+
+clear;
+clc;
+printf("\t Example 6_3_a\n");
+ //part(i)
+//table wt of wet slab,kg -- 5.0 4.0 3.6 3.5 3.4 3.06 2.85
+// drying rate,kg/m^2s-- 5.0 5.0 4.5 4.0 3.5 2.00 1.00
+// X,Dry basis -- 1.0 0.6 .44 0.4 .36 .224 0.14
+// equillibrium relation is given under
+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=1; //looping for calc. of 1/N
+while(i<8) //looping begins
+t(i)=1/(a(i));
+i=i+1;
+end //as 1/N plot is needed
+
+plot(p,a,"o-");
+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 --->");
+//from X=0.6 to 0.44 ,falling rate is non linear and from X=.44 to .14 falling rate is linear
+
+printf("\n from the graph we get critical moisture content as 0.6 kg moisture/kg dry solid");
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.3.b/example6_3_b.sce b/599/CH6/EX6.3.b/example6_3_b.sce
new file mode 100755
index 000000000..2cb4e9f1c
--- /dev/null
+++ b/599/CH6/EX6.3.b/example6_3_b.sce
@@ -0,0 +1,30 @@
+
+
+clear;
+clc;
+printf("\t Example 6_3_b\n");
+
+
+ //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
+t1=Ls/(A*Nc) *(X1-Xcr); //time taken for constant drying rate(fromX=1 to .6)
+X1=.44; //moisture content on dry basis
+t2=(Ls/(A*Nc))*((Xcr-Xbar)*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
+printf("\n the total time for drying the wet slab to 15 percent moisture on wet basis is :%f min",ttotal*60);
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.4/example6_4.sce b/599/CH6/EX6.4/example6_4.sce
new file mode 100755
index 000000000..82153b271
--- /dev/null
+++ b/599/CH6/EX6.4/example6_4.sce
@@ -0,0 +1,37 @@
+
+clear;
+clc;
+printf("\t Example 6.4\n");
+
+//table X-- .30 .20 .18 .15 .14 .11 .07 .05
+// N-- 1.22 1.22 1.14 .90 .80 .56 .22 .05
+//let Ls/A=p
+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)));
+
+t1=p/(Nc) * (X1-Xcr); //time taken for constant drying rate period
+//table X-- .18 .15 .14 .11 .07 .05
+// 1/N-- .8772 1.11 1.25 1.7857 4.545 20
+
+// equillibrium relation is given under
+p = [.18 .15 .14 .11 .07 .05];
+a = [.8772 1.11 1.25 1.7857 4.545 20];
+
+plot(p,a,"o-");
+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
+printf("\n total time for drying the material from 25 to 8 percent moisture under same drying conditions is :%f hr",ttotal);
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.5.a/example6_5_a.sce b/599/CH6/EX6.5.a/example6_5_a.sce
new file mode 100755
index 000000000..fb8dca9e3
--- /dev/null
+++ b/599/CH6/EX6.5.a/example6_5_a.sce
@@ -0,0 +1,50 @@
+
+clear;
+clc;
+printf("\t Example 6_5_a\n");
+
+//table 6.5.1
+//S.NO. Time (Hr) weight of wet material(kg)
+// 0.0 5.314
+// 0.4 5.238
+// 0.8 5.162
+// 1.0 5.124
+// 1.4 5.048
+// 1.8 4.972
+// 2.2 4.895
+// 2.6 4.819
+// 3.0 4.743
+// 3.4 4.667
+// 4.2 4.524
+// 4.6 4.468
+// 5.0 4.426
+// 6.0 4.340
+// infinite 4.120
+
+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
+ printf("\n moisture content (dry basis) ");
+i=1; //looping starts
+while(i<16) //calculation of moisture content
+ p(i)=(w(i)-x)/x;
+ printf("\n :%f",p(i));
+i=i+1;
+end
+ printf("\n \n Drying rate kg/hr*m^2");
+i=2;
+while(i<15)
+ a(i)=(p(i-1)-p(i))*4.12/(t(i)-t(i-1));
+ printf("\n :%f ",a(i));
+ i=i+1;
+end
+a(1)=.19;
+a(15)=0;
+printf("\n\n from the above data it is clear that critical moisture content Xcr=0.11");
+plot(p,a,"o-");
+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 ---->");
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.5.b/example6_5_b.sce b/599/CH6/EX6.5.b/example6_5_b.sce
new file mode 100755
index 000000000..7f56a2504
--- /dev/null
+++ b/599/CH6/EX6.5.b/example6_5_b.sce
@@ -0,0 +1,16 @@
+
+clear;
+clc;
+printf("\t Example 6_5_b\n");
+
+//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;
+Gs=w2/(yout- yin); //water carried away
+printf("\n the amount of air required in 2hours is :%f kg",Gs);
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.5.c/example6_5_c.sce b/599/CH6/EX6.5.c/example6_5_c.sce
new file mode 100755
index 000000000..e67b527e6
--- /dev/null
+++ b/599/CH6/EX6.5.c/example6_5_c.sce
@@ -0,0 +1,18 @@
+
+clear;
+clc;
+printf("\t Example 6_5_c\n");
+
+//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
+tfall=(Ls/(A*Nc))*((Xcr-Xbar)*log((X1-Xbar)/(X2-Xbar)));
+printf("\n from this data we get time as :%f hour",tfall);
+printf("\n the actual time is 0.8 hours");
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.6/example6_6.sce b/599/CH6/EX6.6/example6_6.sce
new file mode 100755
index 000000000..729b6d22a
--- /dev/null
+++ b/599/CH6/EX6.6/example6_6.sce
@@ -0,0 +1,24 @@
+
+clear;
+clc;
+printf("\t Example 6.6\n");
+ //wooden cloth is dried from 100 to 10 and then th efinal moisture content is changed to 16 percent from 10;
+
+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
+t1=roots(tbar- p*((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)))); //------eqn1
+X2bar=.16;
+p=poly([0],'p'); //calc. of time 2
+t2=roots(tbar- p*((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2bar-Xbar))));//------eqn2
+
+ //let t1/t2 be = k
+k=t1/t2;
+ans=1/k-1; //reduction in time for drying
+printf("\n the reduction in time for drying is :%f percent",ans*100);
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.7/example6_7.sce b/599/CH6/EX6.7/example6_7.sce
new file mode 100755
index 000000000..f6aea3ed1
--- /dev/null
+++ b/599/CH6/EX6.7/example6_7.sce
@@ -0,0 +1,26 @@
+
+clear;
+clc;
+printf("\t Example 6.7\n");
+ //assume rate of drying in the falling rate period is directly prporional to the free moisture content
+
+//Ls/ A*Nc is unknown;
+
+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
+x=roots(tbar-p * ((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar))));
+printf("\n the value of Ls/ A*Nc is :%f",x);
+
+//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)*log((Xcr-Xbar)/(X2-Xbar)));
+printf("\n the time for drying the sheets from 30 to 6 percent moisture under same drying conditions is :%f hr",tbar);
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.8.a/example6_8_a.sce b/599/CH6/EX6.8.a/example6_8_a.sce
new file mode 100755
index 000000000..4e5612b19
--- /dev/null
+++ b/599/CH6/EX6.8.a/example6_8_a.sce
@@ -0,0 +1,20 @@
+
+clear;
+clc;
+printf("\t Example 6_8_a\n");
+
+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=(Ls/(A*Nc))*((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
+
+printf("\n 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 \ No newline at end of file
diff --git a/599/CH6/EX6.8.b/example6_8_b.sce b/599/CH6/EX6.8.b/example6_8_b.sce
new file mode 100755
index 000000000..1e36c1caa
--- /dev/null
+++ b/599/CH6/EX6.8.b/example6_8_b.sce
@@ -0,0 +1,31 @@
+
+clear;
+clc;
+printf("\t Example 6_8_b\n");
+
+//tbar=(Ls/(A*Nc))*((Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
+
+//part(i)
+ // assuming only surface evaporation and assuming air moves parellel to surface
+
+//Nc=G^0.71; G=V*d
+//so NC = k* V^.71
+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
+Nc2=Nc*(V2/V1)^.71; //in kg/m^2*s
+t2=(Ls/(A*Nc2))*((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar))); //if air velocity is increased to 4
+t=tbar-t2/3600; //time saved
+
+printf("\n the time saved , if air velocity is increased to 4 m/s: %f",t);
+
+//end \ No newline at end of file
diff --git a/599/CH6/EX6.9/example6_9.sce b/599/CH6/EX6.9/example6_9.sce
new file mode 100755
index 000000000..393ec1fda
--- /dev/null
+++ b/599/CH6/EX6.9/example6_9.sce
@@ -0,0 +1,22 @@
+
+clear;
+clc;
+printf("\t Example 6.9\n");
+
+ //determine the drying condition of sample 0.3*0.3 size.sheet lost weight at rate of 10^-4 kg/s until the moisture fell to 60 percent
+
+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
+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
+tbar=Ls/(10^-4) * ((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
+printf("\n the time for drying the sheets from 75 to 25 percent moisture under same drying conditions is :%f hr",tbar/3600);
+
+//end \ No newline at end of file
generated by cgit (git 2.34.1) at 2025-03-13 09:15:59 +0000