initial commit / add all books

author: priyanka 2015-06-24 15:03:17 +0530
committer: priyanka 2015-06-24 15:03:17 +0530
commit: b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch)
tree: ab291cffc65280e58ac82470ba63fbcca7805165 /1073
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164 files changed, 4652 insertions, 0 deletions
diff --git a/1073/CH2/EX2.1/2_1.sce b/1073/CH2/EX2.1/2_1.sce
new file mode 100755
index 000000000..e10643023
--- /dev/null
+++ b/1073/CH2/EX2.1/2_1.sce
@@ -0,0 +1,23 @@
+clc;
+clear;
+printf("Example 2.1 \n Page no. 2.18\n Part-(a)")
+A=1; //sq metre
+printf("Area of heat transfer,A=%f m^2\n",A)
+Q=450; // W/ sq mtre
+printf("Rate of heat loss/unit area=%f W/m^2\n",Q)
+dT=400; // K
+printf("Temperature difference across insulation layer\t,dT=%f K\n",dT)
+k=0.11 //W/(m.K)
+printf("For asbestos,k=%f\n",k)
+//Q=(k* A*dT)/x
+x=(k*A*dT)/Q
+X=x*1000;
+
+//for fire clay insulation
+k=0.84; // W/(m.K)
+printf("For fire clay insulation,k=%f W/(m.K)\n",k);
+x=(k*A*dT)/Q;
+X=x*1000;
+printf("Ans.(A).Thickness of asbestos is: %f m=%f mm\n",x,X)
+printf("Ans.(B)Thickness of fire clay insulation is: %f m =%f mm\n",x,X)
+
diff --git a/1073/CH2/EX2.10/2_10.sce b/1073/CH2/EX2.10/2_10.sce
new file mode 100755
index 000000000..2e0ca6d0c
--- /dev/null
+++ b/1073/CH2/EX2.10/2_10.sce
@@ -0,0 +1,27 @@
+clear;
+clc;
+//Example 2.10
+printf("Example 2.10")
+A= 1 //sq m
+x1=0.15
+x2=0.01
+x4=0.15
+T1=973  //[K]
+T2=288  //[K]
+dT=T1-T2  //[K]
+//Thermal conductivities
+k1=1.75  
+k2=16.86
+k3=0.033
+k4=5.23
+//in absence of air gap,sum of thermal resistances 
+sR=(x1/k1*A)+(x2/k2*A)+(x4/k4*A)
+Q= dT/sR
+printf("Heat lost per sq meter is %d W/sq m",Q);
+//When heat loss,Q=1163,then new resistance =sR1
+Q1=1163  //[W/sq m]
+sR1=dT/Q1
+//width of air gap be w then
+w=(sR1-sR)*k3*A  // [m]
+w=w*1000  //in [mm]
+printf("Width of air gap is %f mm",w);
diff --git a/1073/CH2/EX2.11/2_11.sce b/1073/CH2/EX2.11/2_11.sce
new file mode 100755
index 000000000..c467999d5
--- /dev/null
+++ b/1073/CH2/EX2.11/2_11.sce
@@ -0,0 +1,37 @@
+clear;
+clc;
+//Example 2.11
+printf("Example 2.11");
+d1=300  //[mm]
+r1=d1/2  // [mm]
+r1=r1/1000  //[m]
+r2=r1+0.05  //[m]
+r3=  r2+0.04  //[m]
+x1=0.05  //[m]
+x2=0.04  //[m]
+k1=0.105  //W/(m.K)
+k2=0.07   //W/(m.K)
+rm1=  (r2-r1)/log(r2/r1);  // [m]
+rm2=(r3-r2)/log(r3/r2);     //[m]
+L=1  //let
+A1=%pi*rm1*L   //let L=1
+R1=x1/(k1*A1);
+A2=%pi*rm2*L
+R2=x2/(k2*A2)
+T1=623  //[K]
+T2=323  //[K]
+dT=T1-T2  //[K]
+//Part a
+Q_by_L=  dT/(R1+R2)   //Heat loss
+printf("Heat loss is %f W/m",Q_by_L);
+//Part b:
+P=2*%pi*(r1+x1+x2)  //[m]
+Q_by_L_peri=Q_by_L/P  // [W/sq m]
+
+printf("Heat lost per sq meter of outer insulation is %f W/sq m",Q_by_L_peri);
+R1=x1/(k1*A1) 
+sR=0.871+0.827
+dT1=dT*R1/sR
+printf("Temperature between two layers of insulation=%f K",(T1-dT1) );
+
+  
diff --git a/1073/CH2/EX2.12/2_12.sce b/1073/CH2/EX2.12/2_12.sce
new file mode 100755
index 000000000..c163be406
--- /dev/null
+++ b/1073/CH2/EX2.12/2_12.sce
@@ -0,0 +1,29 @@
+//Example 2.12
+clear;
+clc;
+printf("Example 2.12\n")
+//Given
+x1=0.01  //[m]
+x2=0.15  //[m]
+x3=0.15  //[m]
+T1=973  //[K]
+T2=423  //[K]
+dT=T1-T2;
+//Thermal conductivities
+k1=16.86  //[W/m.K]
+k2=1.75  //[W/m.K]
+k3=5.23  //[W/m.K]
+k_air=0.0337  // [W/m.K]
+A=1  //[sq m]
+sigma_R=(x1/(k1*A)+x2/(k2*A)+x3/(k3*A))
+Q=dT/sigma_R  //Heat flow in  [W
+Tm= Q*x3/k3  //Temperature drop in magnesite brick
+//Interface temperature=iT
+iT=T2+Tm  //[K]
+sigma_xbyk= A*dT/1163   //with air gap for reducing heat loss to 1163 per sq m
+x_by_k=sigma_xbyk-sigma_R   //x/k for air
+t=x_by_k*k_air
+t=t*1000;
+printf("Width of the air gap is %f mm",t);
+
+ 
diff --git a/1073/CH2/EX2.13/2_13.sce b/1073/CH2/EX2.13/2_13.sce
new file mode 100755
index 000000000..8e980f6f4
--- /dev/null
+++ b/1073/CH2/EX2.13/2_13.sce
@@ -0,0 +1,43 @@
+//Example 2.13
+printf("Example 2.13 \n");
+
+L=1   //assume  [m]
+
+k1=43.03   //[W/(m.K)
+
+k2=0.07    //(W/m.K)
+
+T1=423    //inside temperature [K]
+
+T2=305     // [K]
+
+r1=0.0525    //[mm]
+
+r2=0.0575;   //[m]
+
+r3=0.1075 //[m]
+//r3=r3/1000;  //[m]
+Q=(2*%pi*L*(T1-T2))/(((log(r2/r1))/k1)+((log(r3/r2))/k2));     //Heat loss per metre 
+
+printf("Heat flow per metre of pipe is %f W/m",Q);
+
+printf("Part 2\n");
+//T=Temperature of outer surface
+T=T1-(Q*log(r2/r1))/(k1*2*%pi*L);
+
+printf("Temperature at outer surface of steel pipe: %f K",T);
+
+printf("\nPart iii\n");
+id=0.105    //inside diametre in [m]
+
+A=%pi*id*1  //inside area in [sq m]
+
+C=Q/(A*(T1-T2));   //conductance per length
+
+printf("Conductance per m length based on inside area is %f W/K",C)
+  
+  
+
+
+  
+  
+\ No newline at end of file
diff --git a/1073/CH2/EX2.15/2_15.sce b/1073/CH2/EX2.15/2_15.sce
new file mode 100755
index 000000000..5d3837479
--- /dev/null
+++ b/1073/CH2/EX2.15/2_15.sce
@@ -0,0 +1,17 @@
+//Example 2.15
+printf("Example 2.15 \n")
+A=1  // [sq m]
+  x1=0.1    //m
+  x2=0.04
+  k1=0.7
+  k2=0.48
+  sigma=x1/(k1*A)+x2/(k2*A)   //K/W
+  //Q=4.42*dT
+  //Q=dT/sigma
+  //with rockwool insulation added,Q_dash=0.75*Q
+ k3=0.065     // W/(m.K)
+  //Q_dash=dT/sigma+x3/k3*A
+  //On solving Q and Q_dash we get
+  x3=((1/(0.75*4.42))-sigma)*k3    //[m]
+  x3=x3*1000     // [mm]
+  printf("Thickness of rockwool insulation required=%f mm",x3)
+\ No newline at end of file
diff --git a/1073/CH2/EX2.16/2_16.sce b/1073/CH2/EX2.16/2_16.sce
new file mode 100755
index 000000000..c41b06d19
--- /dev/null
+++ b/1073/CH2/EX2.16/2_16.sce
@@ -0,0 +1,30 @@
+clc;
+clear;
+//Example 2.16,Page no 2.36
+d1=40;   // Diameter of pipe[mm]
+r1=(d1/2)/1000  //Outside radius in [m]
+t1=20;    //Insulation 1 thickness in [mm]
+t1=t1/1000    //[m]
+t2=t1;         //Insulation 2 thickness in[m]
+r2=r1+t1;     //radius after 1st insulation  in [m]
+r3=r2+t2;     //Radius after second insulation in [m]
+
+//Since Scilab does not handles symbolic constants,we will assume some values:
+//(1)
+printf("Let the layer M-1 be nearer to the surface")
+L=1;         //[m]
+T1=10;       //Temperature of inner surface of pipe [K]
+T2=5;        //Temperature of outer surface of insulation [K]
+k=1;         //Thermal conductivity
+k1=k;        //For M-1 material
+k2=3*k;      //For material M-2
+Q1=(T1-T2)/(log(r2/r1)/(2*%pi*L*k1)+log(r3/r2)/(2*%pi*L*k2))
+
+//(2)
+printf("Let the layer of material M-2 be nearer to the surface");
+Q2=(T1-T2)/(log(r2/r1)/(2*%pi*L*k2)+log(r3/r2)/(2*%pi*L*k1))
+printf("Q1=%f and Q2= %f \n For dummy variables unity...\nFor any value of k,T1 and T2,Q1 is always less than Q2",Q1,Q2);
+printf("\n So,M-1 near the surface is advisable(i.e Arrangement one will result i ,ess heat loss\n)");
+per_red=(Q2-Q1)*100/Q2
+printf("Percent reduction in heat loss is %f percent",per_red)
+printf("\nNOTE:Slight variation in answers due to less precise calculation in book.If performed manually,this answer stands to be correct")
diff --git a/1073/CH2/EX2.17/2_17.sce b/1073/CH2/EX2.17/2_17.sce
new file mode 100755
index 000000000..cb6cd0d3f
--- /dev/null
+++ b/1073/CH2/EX2.17/2_17.sce
@@ -0,0 +1,18 @@
+//Example2.17
+T1=523   //[K]
+  T2=323   //[K]
+  r1=0.05   //[m]
+  r2=0.055   //[m]
+  r3=0.105   //[m]
+  r4=0.155   //[m]
+  k1=50      //[W/(m.K)]
+  k2=0.06    //[W/(m.K)]
+  k3=0.12    //W/(m.K)
+  //CASE 1
+  Q_by_L1=2*%pi*(T1-T2)/((log(r2/r1))/k1+(log(r3/r2))/k2+(log(r4/r3))/k3)   //[W/m]
+  printf("Heat loss=%f W/m",Q_by_L1)
+  //Case 2
+  Q_by_L2=2*%pi*(T1-T2)/((log(r2/r1))/k1+(log(r3/r2))/k3+(log(r4/r3))/k2)
+ perct=(Q_by_L2-Q_by_L1)*100/Q_by_L1
+  printf("If order is changed then heat loss=%f W/m",Q_by_L2) 
+  printf("\n loss of heat is increased by %f percent by putting material with higher thermal conductivity near the pipe surface",perct)
+\ No newline at end of file
diff --git a/1073/CH2/EX2.18/2_18.sce b/1073/CH2/EX2.18/2_18.sce
new file mode 100755
index 000000000..ed67b96d1
--- /dev/null
+++ b/1073/CH2/EX2.18/2_18.sce
@@ -0,0 +1,29 @@
+
+clc;
+clear;
+//Example 2.18,Page no 2.38
+//Given
+//Assume:
+L=1    //[m]
+r1=0.10    //[m]    Outside radius od pipe
+ia=0.025    //inner insulaiton [m]
+
+r2=r1+ia    //Outer radius of inner insulation
+r3=r2+ia    //Outer radius of outer insulation
+//CASE 1:'a' near the pipe surface
+//let k1=1
+k1=1;    //Thermal conductivity of A[W/m.K]
+//and k2=3k1=3
+k2=3;    //Thermal conductivity of B[W/m.K]
+//Let dT=1
+dT=1
+Q1=dT/(log(r2/r1)/(2*%pi*k1*L)+log(r3/r2)/(2*%pi*k2*L))
+Q1=22.12    //Approximate
+//CASE 2:'b' near the pipe surface 
+Q2=dT/(log(r2/r1)/(2*%pi*k2*L)+log(r3/r2)/(2*%pi*k1*L))
+Q2=24.39       //Approximation
+printf("ANSWER-(i)\nQ1=%f W \nQ2= %f W \nQ1 is less than Q2.i.e arrangement A near the pipe surface and B as outer layer gives less heat loss\n",Q1,Q2);
+percent=(Q2-Q1)*100/Q1;    //percent reduction in heat loss
+printf("\nANSWER-(ii) \nPercent reduction in heat loss (with near the pipe surface)=%f percent",percent);
+
+
diff --git a/1073/CH2/EX2.19/eg2_19.sce b/1073/CH2/EX2.19/eg2_19.sce
new file mode 100755
index 000000000..02b6839ac
--- /dev/null
+++ b/1073/CH2/EX2.19/eg2_19.sce
@@ -0,0 +1,15 @@
+clc
+clear
+printf("Example 2.19.Page no.2.39")
+//Given
+x1=0.224 // m
+k1=1.3 // W/(m.K)
+k2=0.346 // W/(m.K)
+T1=1588 // K
+T2= 299 // K
+QA=1830 // W/ sq metre //heat loss
+//solution
+printf("Q/A=(T1-T2)/x1/k1+x2/k2");
+x2=k2*((T1-T2)*1/(QA)-(x1/k1))
+x2=x2*1000;
+printf("Thickness of insulation=%f mm",x2)
diff --git a/1073/CH2/EX2.2/2_2.sce b/1073/CH2/EX2.2/2_2.sce
new file mode 100755
index 000000000..c9b67c926
--- /dev/null
+++ b/1073/CH2/EX2.2/2_2.sce
@@ -0,0 +1,35 @@
+
+clc;
+printf("Example 2.2,\nPage no.2.18\n");
+L=1 // m
+printf("Length of ppipe,L = %f m\n",L);
+r1=(50/2) // in mm
+r1=r1/1000 // in m
+printf("r1=%f m\n",r1);
+r2=(25+3)/1000 // m
+printf("r2=%f m\n",r2)
+rm1=(r2-r1)/log(r2/r1);
+printf("rm1=%f m\n",rm1)
+k1=45 //W/(m.K)
+R1=(r2-r1)/(k1*(2*%pi*rm1*L)) // K/W
+printf("Thermal resistance of wall pipe=R1=%f K/W\n",R1);
+printf("For inner lagging:\n") ;
+k2=0.08 //W/(m.K)
+ri1=0.028 //m
+ri2=(ri1+r1) // m
+rmi1=(ri2-ri1)/log(ri2/ri1)
+R2=(ri2-ri1)/(k2*2*%pi*rmi1*L)
+printf("Thermal resistance of inner lagging=R2=%f K/W",R2);
+printf("For  outer lagging:\n") ;
+k3=0.04 //W/(m.K)
+ro1=0.053 //m
+ro2=(ro1+0.04) // m
+rmo1=(ro2-ro1)/log(ro2/ro1)
+R3=(ro2-ro1)/(k3*2*%pi*rmo1*L)
+printf("Thermal resistance of inner lagging=R2=%f K/W\n",R3);
+R=R1+R2+R3
+Ti=550 //K //inside
+To=330 //K  // outside
+dT=Ti-To; //Temperature difference
+Q=dT/R
+printf("Rate of heat loss per metre of pipe,Q=%f W/m",Q)
diff --git a/1073/CH2/EX2.20/2_20.sce b/1073/CH2/EX2.20/2_20.sce
new file mode 100755
index 000000000..17898e54e
--- /dev/null
+++ b/1073/CH2/EX2.20/2_20.sce
@@ -0,0 +1,31 @@
+//Example 2.20
+//Given
+//for clay
+k1=0.533    //[W/(m.K)]
+//for red brick
+k2=0.7      //[W/m.K]
+//Case 1
+  A=1     //Area
+  x1=0.125    //[m]
+  x2=0.5      //[m]
+  //Resistances
+  r1=x1/(k1*A) //Res of fire clay [K/W]
+  r2=x2/(k2*A)  //Res of red brick[K/W]
+  r=r1+r2
+  //Temperatures
+  T1=1373    //[K]
+  T2=323     //[K]
+  Q=(T1-T2)/r //[W/sq m]
+  Tdash=T1-Q*r1  //[K]
+//Case2
+ // Heat loss must remain  unchanged,Thickness of red brick also reduces to its half
+  x3=x2/2   //[m]
+  r3=x3/(k2*A)  //[K/W]
+  Tdd= T2+(Q*r3)    //[K]
+  //Thickness of diatomite be x2,km be mean conductivity
+  Tm=(Tdash+Tdd)/2   //[K]
+  km=0.113+(0.00016*Tm)  //[W/(m.K]
+  x2=km*A*(Tdash-Tdd)/Q  //[m]
+  x2=x2*1000      //[mm]
+  printf("Thickness of diatomite layer=%f mm",x2)
+  
+\ No newline at end of file
diff --git a/1073/CH2/EX2.21/2_21.sce b/1073/CH2/EX2.21/2_21.sce
new file mode 100755
index 000000000..da1f59cfa
--- /dev/null
+++ b/1073/CH2/EX2.21/2_21.sce
@@ -0,0 +1,18 @@
+//Exaample2.21
+//Given
+k1=0.7   //common brick   W/((m.K)
+k2=0.48  //gypsum layer [W/(m.K)
+k3=0.065 //Rockwool    [W/m.K]
+//Heat loss with insulatiob will be 20% of without insulation
+A=1      //sq m
+x1=0.1   //[m]
+x2=0.04 //[m]
+R1=x1/(k1*A)   //K/W
+R2=x2/(k2*A)   //K/W
+R=R1+R2        //K/W
+//R3=x3/(k3*A)
+QbyQd=0.2
+sigRbyRd=QbyQd
+x3=(R/QbyQd-R)/15.4 //m
+x3=x3*1000          //[mm]
+printf("Thickness of rockwool insulation =%f mm",x3)
+\ No newline at end of file
diff --git a/1073/CH2/EX2.22/2_22.sce b/1073/CH2/EX2.22/2_22.sce
new file mode 100755
index 000000000..4523772f5
--- /dev/null
+++ b/1073/CH2/EX2.22/2_22.sce
@@ -0,0 +1,35 @@
+clc;
+clear;
+//Example 2.22
+Ts=451;      //Steam temperature in [K]
+Ta=294;      //Air temperature in [K]
+Di=25;   //Internal diameter of pipe  [mm]
+Di=Di/1000;              //[m]
+od=33;   //Outer diameter of pipe   [mm] 
+od=od/1000;              //[m]
+hi=5678; //Inside heat transfer coefficient [W/(m^2.K)]
+ho=11.36; //Outsideheat transfer coefficient  [W/(sq m.K)]
+xw=(od-Di)/2;    //Thickness of steel pipe  [m]
+k2=44.97;        //k for steel in W/(m.K)
+k3=0.175;        //k for rockwool in W/(m.K)
+ti=38/1000;           //thickness of insulation in [m]
+r1=Di/2;         //[m]
+r2=od/2;         //[m]
+rm1=(r2-r1)/log(r2/r1);      //[m]
+r3=r2+ti;            //[m]
+rm2=(r3-r2)/log(r3/r2);  //[m]
+Dm1=2*rm1;               //[m]
+Dm2=2*rm2;               //[m]
+//Rate of heat loss = dT/(sigma_R)
+L=1;             //[m]
+R1=1/(hi*%pi*Di*L);  //[K/W]
+R2=xw/(k2*%pi*Dm1*L);
+R3=(r3-r2)/(k3*%pi*Dm2*L);
+Do=(od+2*ti) ;          //[mm]
+R4=1/(ho*%pi*Do*L);              //[m]
+sigma_R=R1+R2+R3+R4;
+//Heat loss
+dT=Ts-Ta;                //[K]
+Q=dT/sigma_R;            //Heat loss [W/m]
+printf("\nAns:Rate of heat loss is %f W/m",Q);
+printf("\n NOTE:Slight variation in final answer due to lack of precision in calculation of R1,R2,R3 and R4.In book an approximate values of these is taken\n ")
diff --git a/1073/CH2/EX2.23/2_23.sce b/1073/CH2/EX2.23/2_23.sce
new file mode 100755
index 000000000..c8dd64665
--- /dev/null
+++ b/1073/CH2/EX2.23/2_23.sce
@@ -0,0 +1,21 @@
+clc;
+//Example 2.23
+T1=913     //[K]
+T=513     //[K]
+T2=313      //[K]
+//Q=(T1-T)/(x/(k*A))
+//Q=(T-T2)/(1/(h*A))
+//x=2k/h
+//Q=(T1-T2)/(x/(kA)+1/(h*A))
+//Therefore,Q=hA/3*(T1-T2)
+//With increase in thickness(100%)
+//x1=4*k/h
+//Q2=(T1-T2)/(x1/k*A+1/(h*A))
+//Q2=(h*A)/5)*(T1-T2)
+//Now
+h=1;    //Assume
+A=1;    //Assume for calculation
+Q1=(h*A/3)*(T1-T2)
+Q2=((h*A)/5)*(T1-T2)
+percent=(Q1-Q2)*100/Q1      //Percent reduction in heat loss
+printf("\nTherefore,Percentage reduction in heat loss is %d percent",percent);
diff --git a/1073/CH2/EX2.24/2_24.sce b/1073/CH2/EX2.24/2_24.sce
new file mode 100755
index 000000000..e95a70e47
--- /dev/null
+++ b/1073/CH2/EX2.24/2_24.sce
@@ -0,0 +1,24 @@
+clc;
+clear;
+printf("Example 2.24\n Page no. 2.47");
+//given
+L=1//m
+thp=2//Thickness of pipe; in mm
+thi=10//Thickness of insulation; in mm
+T1=373//K
+T2=298//K
+id=30//mm
+r1=id/2//mm
+r2=r1+thp//mm
+r3=r2+thi//mm
+//In S.I units
+r1=r1/1000 //m
+r2=r2/1000//m
+r3=r3/1000//m
+k1=17.44//W/(m.K)
+k2=0.58//W/(m.K)
+hi=11.63//W/(sq m.K)
+ho=11.63//W/(sq m.K)
+//Solution
+Q=(2*%pi*L*(T1-T2))/(1/(r1*hi)+(log(r2/r1))/k1+((log(r3/r2))/k2)+(1/(0.02*ho)))
+printf("ANSWER: \n Rate of heat loss,Q=%f W",Q);
diff --git a/1073/CH2/EX2.25/2_25.sce b/1073/CH2/EX2.25/2_25.sce
new file mode 100755
index 000000000..a9c1b7d59
--- /dev/null
+++ b/1073/CH2/EX2.25/2_25.sce
@@ -0,0 +1,16 @@
+clc;
+clear;
+//Examplr 2.25
+h=8.5 ;      //[W/sq m.K]
+dT=175 ;     //[K]
+r2=0.0167;           //[m]
+Q_by_l=h*2*%pi*r2*dT        //[W/m]
+k=0.07 ;         //For insulating material in  [W/m.K]
+//for insulated pipe--50% reduction in heat loss
+Q_by_l1=0.5*Q_by_l      //[w/m]
+deff('[x]=f(r3)','x=Q_by_l1-dT/((log(r3/r2))/(2*%pi*k)+1/(2*%pi*r3*h))')
+
+//by trial and error method we get:
+r3=fsolve(0.05,f)
+t=r3-r2         //thickness of insulation in [m]
+printf('\n Hence,required thickness of insulation is %f m=%f mm or %d m",t,t*1000,round(t*1000));
diff --git a/1073/CH2/EX2.26/2_26.sce b/1073/CH2/EX2.26/2_26.sce
new file mode 100755
index 000000000..6e7178b69
--- /dev/null
+++ b/1073/CH2/EX2.26/2_26.sce
@@ -0,0 +1,19 @@
+
+//Example 2.26
+//Calculate heat loss per metre length
+//Given
+id=0.1    //internal diameter in[m]
+od=0.12    //outer diameter in [m]
+T1=358    //Temperature of fluid   [K]
+T2=298    //Temperature of surrounding   [K]
+t=0.03    //thickness of insulation     [m]
+k1=58    //[W/m.K]
+k2=0.2    //W/(m.K) insulating material
+h1=720    //inside heat transfer coeff [W/sq m .K]
+h2=9      //W/sq m.K
+r1=id/2    //[m]
+r2=od/2    //[m]
+r3=r2+t    //[m]
+//Heat loss per meter=Q_by_L
+Q_by_L=(T1-T2)/(1/(2*%pi*r1*h1)+log(r2/r1)/(2*%pi*k1)+log(r3/r2)/(2*%pi*k2)+1/(2*%pi*r3*h2)); //W/m
+printf("Heat loss per metre length of pipe=%f W",Q_by_L)
diff --git a/1073/CH2/EX2.27/2_27.sce b/1073/CH2/EX2.27/2_27.sce
new file mode 100755
index 000000000..345471d77
--- /dev/null
+++ b/1073/CH2/EX2.27/2_27.sce
@@ -0,0 +1,22 @@
+
+clc;
+clear;
+//Example 2.26
+//Given:
+T1=573;          //[K]
+T2=323;         //[K]
+T3=298;         //[K]
+h1=29;           // Outside heat transfer coefficients [W/sq m.K]
+h2=12;          //[W/sq m.K]
+r1=0.047;            //Internal radius [m]
+r2=0.05;             //Outer radius[m]
+k1=58 ;              //[W/m.K]
+k2=0.052;            //[W/m.K]
+//Q=(T1-T2)/(1/(r1*h1)+log(r2/r1)/k1+log(r3/r2)/k2)=(T2-T3)/(1/(r3*h2))
+deff('[x]=f(r3)','x=(T1-T2)/(1/(r1*h1)+log(r2/r1)/k1+log(r3/r2)/k2)-(T2-T3)/(1/(r3*h2))')
+//by trial and error method :
+r3=fsolve(0.05,f)
+t=r3-r2             //Thickness of insulation in [m]
+//Q=h2*2*%pi*r3*L*(T2-T3)
+Q_by_l=h2*2*%pi*r3*(T2-T3)      //[W/m]
+printf("\n Thicknesss of insulation is %d mm \n Rate of heat loss per unit length is %f W/m",round(t*1000),Q_by_l);
diff --git a/1073/CH2/EX2.28/2_28.sce b/1073/CH2/EX2.28/2_28.sce
new file mode 100755
index 000000000..ada36900a
--- /dev/null
+++ b/1073/CH2/EX2.28/2_28.sce
@@ -0,0 +1,24 @@
+
+clc;
+clear;
+//Example 2.28
+//Calculate heat loss per sq m and temperature of outside surface
+//Given
+A=1 //assume [sq m]
+x1=0.006    //[m]
+x2=0.075    //[m]
+x3=0.2    //[m]
+k1=39    //[W/m.K]
+k2=1.1    //[W/m.K]
+k3=0.66    //[W/m.K]
+h0=65    //W/sq m .K
+T1=900    //K
+T2=300    //K
+sigma_R=(x1/(k1*A)+x2/(k2*A)+x3/(k3*A)+1/(h0*A));
+//To calculate heat loss/sq m area
+Q=(T1-T2)/sigma_R    //[W/sq m]
+printf("Heat loss per sq metre area is: %f W/sq m",Q);
+//Q/A=T-T2/(1/h0), where T=Temp of outside surface
+//So, T=T2+Q/(A*h0)
+T=Q/(A*h0)+T2    //[K]
+printf("Temperature of utside surface of furnace is: %f K (%f degree C)",T,T-273)
diff --git a/1073/CH2/EX2.29/2_29.sce b/1073/CH2/EX2.29/2_29.sce
new file mode 100755
index 000000000..9aaf3d213
--- /dev/null
+++ b/1073/CH2/EX2.29/2_29.sce
@@ -0,0 +1,22 @@
+
+clear;
+clc;
+//Example 2.29
+//Determine necessary thickness of insulation brick
+//Given
+A=1    //Assume [sq m]
+x1=0.003    //[m]
+x3=0.008    //[m]
+k1=30    //[W/m.K]
+k2=0.7    //[W/m.K]
+k3=40    //[W/m.K]
+T1=363    //[K]
+T=333    //[K]
+T2=300    //[K]
+h0=10    //W/sq m.K
+//Q=(T1-T2)/(x1/(k1*A)+x2/(k2*A)+x3/(k3*A)+1/(h0*A))
+//Also,Q=(T-T2)/(1/(h0*A))
+//So, (T1-T2)/((x1/(k1*A)+x2/(k2*A)+x3/(k3*A)+1/(h0*A))=(T-T2)/(1/(h0*A))
+//or,x2=k2*A((T1-T2)/((T-T2)*h0*A)-1/(h0*A)-x1/(k1*A)-x3/(k3*A))
+x2=k2*A*((T1-T2)/((T-T2)*h0*A)-1/(h0*A)-x1/(k1*A)-x3/(k3*A)); //[m]
+printf("Thicknessof insulating brick required is %f mm",x2*1000);
diff --git a/1073/CH2/EX2.3/2_3.sce b/1073/CH2/EX2.3/2_3.sce
new file mode 100755
index 000000000..cb47c9e56
--- /dev/null
+++ b/1073/CH2/EX2.3/2_3.sce
@@ -0,0 +1,18 @@
+clear;
+clc;
+printf("Example 2.3")
+//Given
+r1=44 // [mm]
+r1=r1/1000  //[m]
+r2=0.094   // [m]
+r3=0.124  // [m]
+T1=623  //Temperature at outer surface of wall in[K]
+T3=313  //Temperature at outer surface of outer insulation  [K]
+k1=0.087  //Thermal conductivity of insulation layer 1..in [W/m.K]
+k2=0.064  //Thermal conductivity of insulation layer 2  [W/m.K]
+l=1    // Length of pipe   [m]
+rm1=(r2-r1)/log(r2/r1)  //log mean radius of insulation layer 1 [m]
+rm2=(r3-r2)/log(r3/r2)  //log mean radius of insulation layer 2[m]
+//Putting values in following eqn:
+Q= (T1-T3)/((r2-r1)/(k1*2*%pi*rm1*l)+(r3-r2)/(k2*2*%pi*rm2*l));
+printf("Heat loss per meter pipe is %f W/m",Q)
diff --git a/1073/CH2/EX2.30/2_30.sce b/1073/CH2/EX2.30/2_30.sce
new file mode 100755
index 000000000..1d2897831
--- /dev/null
+++ b/1073/CH2/EX2.30/2_30.sce
@@ -0,0 +1,30 @@
+
+clear;
+clc;
+//Example 2.30
+//Given
+hi=75        //[W/sq m.K)
+x1=0.2    //m
+x2=0.1    //[m]
+x3=0.1    //[m]
+T1=1943    //[K]
+k1=1.25    //W/m.K
+k2=0.074    ///W/m.K
+k3=0.555    //W/m.K
+T2=343    //K
+A=1    //assume [sq m]
+sigma_R=1/(hi*A)+x1/(k1*A)+x2/(k2*A)+x3/(k3*A);
+//Heat loss per i sq m
+Q=(T1-T2)/sigma_R    //[W]
+//if T=temperature between chrome brick and koalin brick then 
+//Q=(T1-T)/(1/(hi*A)+x1/(k1*A))
+//or T=T1-(Q*(1/(hi*A)+x1/(k1*A)))
+T=T1-(Q*(1/(hi*A)+x1/(k1*A)));    //[K]
+printf("Temperature at inner surface of middle layer=%f K(%f degree C)",T,T-273);
+//if Tdash=temperature at the outer surface of middel layer,then
+//Q=(Tdash-T2)/(x3/(k1*A))
+//or Tdash=T2+(Q*x3/(k3*A))
+Tdash=T2+(Q*x3/(k3*A))    //[K]
+printf("Temperature at outer surface of middle layer=%f K (%f degree C)",Tdash,Tdash-273);
+
+
diff --git a/1073/CH2/EX2.31/2_31.sce b/1073/CH2/EX2.31/2_31.sce
new file mode 100755
index 000000000..0545e0ac5
--- /dev/null
+++ b/1073/CH2/EX2.31/2_31.sce
@@ -0,0 +1,24 @@
+
+clear;
+clc;
+//Example 2.31
+//Calculate:(a) Heat loss per unit length  
+//(b)Reduction in heat loss
+//Given
+hi=10    //W/sq m.K
+h0=hi    //W/sq.m.K
+r1=0.09    //m
+r2=0.12    //m
+t=0.05    //thickness of insulation [m]
+k1=40    //W/m.K
+k2=0.05    //W/m.K
+T1=473    //K
+T2=373    //K
+Q_by_L=2*%pi*(T1-T2)/(1/(r1*hi)+log(r2/r1)/k1+1/(r2*h0));    //W/m
+printf("Ans (a) Heat loss=%f W/m ",Q_by_L)
+//After addition of insulation:
+r3=r2+t;    //radius of outer surface of insulaiton
+Q_by_L1=2*%pi*(T1-T2)/(1/(r1*hi)+log(r2/r2)/k1+log(r3/r2)/k2+1/(r3*h0));    // W
+Red=Q_by_L-Q_by_L1    //Reduciton in heat loss in [W/m]
+percent_red=(Red/Q_by_L)*100    //% Reduction in heat loss
+printf("Ans (b) Percent reduction in heat loss is %f percent",percent_red)
diff --git a/1073/CH2/EX2.32/2_32.sce b/1073/CH2/EX2.32/2_32.sce
new file mode 100755
index 000000000..85617e2c1
--- /dev/null
+++ b/1073/CH2/EX2.32/2_32.sce
@@ -0,0 +1,22 @@
+
+clear;
+clc;
+//Example 2.32
+//Determine: i-Heat flux across the layers and
+//ii-Interfacial temperature between the layers
+
+//Given
+T1=798    //K
+T2=298    //K
+x1=0.02    //m
+x2=x1    //m
+k1=60    //W/m.K
+k2=0.1    //W/m.K
+hi=100    //W/sq m.K
+h0=25    //W/sq m.K
+Q_by_A=(T1-T2)/(1/hi+x1/k1+x2/k2+1/h0);    //W/sq m
+printf("Ans (i)- Heat flux across the layers is %f W/sq m",Q_by_A);
+//If Tis the interfacial temperature between steel plate and insulating material
+//Q_by_A=(T-T2)/(x2/k2+1/h0)
+T=Q_by_A*(x2/k2+1/h0)+T2
+printf("Ans-(ii)-Interfacial temperature between layers is %f K (%f degree C)",T,T-273);
diff --git a/1073/CH2/EX2.33/2_33.sce b/1073/CH2/EX2.33/2_33.sce
new file mode 100755
index 000000000..05223a212
--- /dev/null
+++ b/1073/CH2/EX2.33/2_33.sce
@@ -0,0 +1,41 @@
+
+
+clc;
+clear;
+//Example 2.33
+//Determine Temperature at the outer surface of wall and convective conductance on the outer wall
+    //Temperature of hot gas:
+T1=2273    //K
+    //Ambient aur temperature:
+T4=318    //K
+    //Heat flow by radiation from gases to inside surface of wall:
+Qr1_by_A=23260    //[W/sq m]
+    //Heat transfer coefficient on inside wall:
+hi=11.63    //W/sq m.K
+    //Thermal conductivity of wall:
+K=58    //W.sq m/K
+    //Heat flow by radiation from external surface to ambient:
+Qr4_by_A=9300   //W/sq m.
+    //Inside Wall temperature:
+T2=1273    //K
+
+Qr1=Qr1_by_A    //W  for
+A=1    //sq m
+
+Qc1_by_A=hi*(T1-T2)    //W/sq m
+Qc1=Qc1_by_A    //for A=1 sq m
+    //Thermal resistance:
+R=1/K    //K/W per sq m
+//Now Q=(T2-T3)/R,i.e 
+//External wall temp T3=T2-Q*R
+//Q entering wall=
+Q_enter=Qr1+Qc1    //W
+T3=T2-Q_enter*R    //K
+T3=673      //Approximate
+//Heat loss due to convection:
+Qc4_by_A=Q_enter-Qr4_by_A    //W/sq m
+//Qc4_by_A=h0*(T3-T4)
+//or  h0=Qc4_by_A/(T3-T4)
+h0=Qc4_by_A/(T3-T4)    //W/sq m.K
+//Result
+printf("Convective conductance is: %f W/sq m.K",h0)
diff --git a/1073/CH2/EX2.34/2_34.sce b/1073/CH2/EX2.34/2_34.sce
new file mode 100755
index 000000000..0db6c8ebf
--- /dev/null
+++ b/1073/CH2/EX2.34/2_34.sce
@@ -0,0 +1,19 @@
+clc;
+clear;
+//Example 2.34
+//Given
+T1=473    //[K]
+T2=293    //[K]
+k=0.17    //W/(m.K)
+h=3    //W/(sq m.K)
+h0=h    //W/sq m.K
+rc=k/h    //m
+r1=0.025    //Inside radius of insulaiton [mm]    
+q_by_l1=2*%pi*(T1-T2)/(log(rc/r1)/k+1/(rc*h0))    //Heat transfer with insulation in W/m
+//Without insulation:
+q_by_l2=h*2*%pi*r1*(T1-T2)    //W/m
+inc=(q_by_l1-q_by_l2)*100/q_by_l2    //Increase of heat transfer
+printf("When covered with insulation,\n heat loss=%f W \n When without insulation,heat loss= %f W \n percent increase =%f percent",q_by_l1,q_by_l2,inc);
+k=0.04    //Fibre glass insulaiton W/(sq m.K)
+rc=k/h    //Critical radius of insulaiton
+printf("In this case the avlue of rc=%f m is less than the outside radius of pipe (%f),\n So additon of any fibre glass would cause a decrease in the heat transfer \n",rc,r1)    
diff --git a/1073/CH2/EX2.36/2_36.sce b/1073/CH2/EX2.36/2_36.sce
new file mode 100755
index 000000000..a9c0ecb1d
--- /dev/null
+++ b/1073/CH2/EX2.36/2_36.sce
@@ -0,0 +1,19 @@
+
+clear;
+clc;
+//Example 2.36
+//Calculate the heat loss per metre of pipe and outer surface temperature
+//Given
+k=1    //Thermal conductivity in [W/sq m.K]
+h=8    //Het transfer coeff in W/sq m.K
+rc=k/h  //Critical radius in m
+T1=473    //K
+T2=293    //K
+r1=0.055    //Outer radius =inner radius in [m]
+Q_by_L=2*%pi*(T1-T2)/(log(rc/r1)/k+1/(rc*h))
+printf("Heat loss per meter of pipe is %f W/m",Q_by_L)
+//For outer surface
+//Q_by_L=2*%pi*(T-T2)/(1/rc*h)
+// implies that, T=T2+Q_by_L/(rc*2*%pi)
+T=T2+Q_by_L/(rc*2*%pi*h)    //K
+printf("Outer surface temperature is: %f K(%f degree C)",T,T-273)
diff --git a/1073/CH2/EX2.37/2_37.sce b/1073/CH2/EX2.37/2_37.sce
new file mode 100755
index 000000000..159947812
--- /dev/null
+++ b/1073/CH2/EX2.37/2_37.sce
@@ -0,0 +1,25 @@
+
+clc;
+clear;
+//Example 2.37
+//Calculate the time required for a ball to attain a temperature of 423 K
+//Given
+k_steel=35    //W/m.K
+Cp_steel=0.46    //kJ/(kg*K)
+Cp_steel=Cp_steel*1000    //J/(kg*K)
+h=10    //W/sq m.K
+rho_steel=7800    //kg/cubic m
+dia=50    //mm
+dia=dia/1000    //m
+R=dia/2      //radius in m
+A=4*%pi*R^2  //Area  in sq m
+V=A*R/3      //Volume in cubic meter
+Nbi=h*(V/A)/k_steel
+//As Nbi<0.10,internal temp gradient is negligible
+T=423    //K
+T0=723    //K
+T_inf=373    //K
+//(T-T_inf)/(T0-T_inf)=e^(-h*At/rho*Cp*V)
+t=-rho_steel*Cp_steel*R*log((T-T_inf)/(T0-T_inf))/(3*h);    //s
+printf("Time required for a ball to attain a temperature of 423 K is %f s= %f h",t,t/(3600))
+
diff --git a/1073/CH2/EX2.38/2_38.sce b/1073/CH2/EX2.38/2_38.sce
new file mode 100755
index 000000000..513f4198f
--- /dev/null
+++ b/1073/CH2/EX2.38/2_38.sce
@@ -0,0 +1,20 @@
+
+clc;
+clear;
+//Example 2.38
+//Given
+dia=50    //mm
+dia=dia/1000    //m
+r=dia/2    //radius in m
+h=115    //W/sq m.K
+rho=8000 //kg/cubic m
+Cp=0.42    //kJ/kg.K
+Cp=Cp*1000    //J/(kg*K)
+A=4*%pi*r^2    //Area in sq m
+V=A*r/3    //Volume in cubic m
+T=423    //K
+T_inf=363    //K
+T0=723    //K
+//(T-T_inf)/(T0-T_inf)=e^(-3ht/(rho*Cp*r))
+t=-rho*Cp*r*log((T-T_inf)/(T0-T_inf))/(3*h);    //Time in seconds
+printf("Time taken by centre of ball to reach a temperature of 423 K is %f s (=%f minutes",t,t/60);
diff --git a/1073/CH2/EX2.39/2_39.sce b/1073/CH2/EX2.39/2_39.sce
new file mode 100755
index 000000000..575b14bb7
--- /dev/null
+++ b/1073/CH2/EX2.39/2_39.sce
@@ -0,0 +1,20 @@
+
+clc;
+clear;
+//Example 2.39
+//Given
+h=11.36    //W/sq m.K
+k=43.3    //w/(m.K)
+r=25.4    //radius in mm
+r=r/1000    // radius in m
+A=4*%pi*r^2    //Area of sphere [sq m]
+V=A*r/3    //Volume in [cubic m]
+rho=7849    //kg/cubic m
+Cp=0.4606*10^3    //J/kg.K
+t=1    //hour
+t=t*3600    //seconds
+T_inf=394.3    //[K]
+T0=700    //[K]
+// (T-T_inf)/(T0-T_inf)=e^(-3*h*t/rho*Cp*V)
+T=T_inf+(T0-T_inf)*(%e^((-h*A*t)/(rho*Cp*V)));
+printf("Temperature of ball after 1 h= %f K (%f degree C)",T,T-273)
diff --git a/1073/CH2/EX2.4/2_4.sce b/1073/CH2/EX2.4/2_4.sce
new file mode 100755
index 000000000..e9b5555c4
--- /dev/null
+++ b/1073/CH2/EX2.4/2_4.sce
@@ -0,0 +1,24 @@
+clc;
+clear;
+//Example 2.4
+printf("Example 2.4")
+//Given
+A=1 //Heat transfer area  [sq m]
+x1=0.229  // thickness of fire brick in [m]
+x2=0.115  // thickness of insulating brick in [m]
+x3=0.229  // thickness of building brick in [m]
+k1=6.05  //thermal conductivity of fir brick  [W/(m.K)]
+k2=0.581  //thermal conductivity of insulating brick   [W/m.K]
+k3=2.33  //thermal conductivity of building brick   [W/m.K]
+T1=1223  // inside temperature [K]
+T2=323 // Outside temperature[K]
+dT=T1-T2  //Overall temp drop [K]
+R1=(x1/k1*A) //thermal resistance 1
+R2=(x2/k2*A) // Thermal resistance 2
+R3=(x3/k3*A) //Thermal resistance 3
+Q=dT/(R1+R2+R3) //w/SQ m
+Ta=-((Q*R1)-T1)  //from  Q1=Q=(T1-Ta)/(x1/k1*A)
+//Similarly
+Tb=(Q*R3)+T2;
+printf("Interface temperature:\n i-Between  FB-IB=%f K  \nii-Between IB-PB=%fK",Ta,Tb);
+
diff --git a/1073/CH2/EX2.40/2_40.sce b/1073/CH2/EX2.40/2_40.sce
new file mode 100755
index 000000000..d0e637814
--- /dev/null
+++ b/1073/CH2/EX2.40/2_40.sce
@@ -0,0 +1,27 @@
+clc;
+clear;
+//Example 2.40
+//Given
+rho=9000; //kg/cubic m
+Cp=0.38;  //kJ/(kg.K)
+Cp=Cp*1000    //J/(kg.K)
+k=370;    //W/m.K
+h=90;     //W/sq m.K
+l=400;    //mm
+l=l/1000 ;   //length of copper slab
+t=5/1000 ;   //thickness in [m]
+A=2*l^2     //Area of slab
+V=t*l^2       //Volume in [cubic m]
+L_dash=V/A    //[m]
+//for slab of thickness 2x
+//L_dash=x
+L_dash=0.025 ;   //[m]
+Nbi=h*L_dash/k    //< 0.10
+var=h*A/(rho*Cp*V)
+//As Nbi<0.10,we can apply lumped capacity analysis
+T=363    //[K]
+T_inf=303    //[K]
+T0=523    //[K]
+t=-(log((T-T_inf)/(T0-T_inf)))/var
+printf("Time at which slab temperature becomes 363 K is %f s",t)
+printf("CALCULATION MISTAKE IN BOOK IN LAST LINE")
diff --git a/1073/CH2/EX2.41/2_41.sce b/1073/CH2/EX2.41/2_41.sce
new file mode 100755
index 000000000..73bdd967e
--- /dev/null
+++ b/1073/CH2/EX2.41/2_41.sce
@@ -0,0 +1,22 @@
+
+clc;
+clear;
+//Example 2.41
+//Given
+rho=9000    //kg/cubic meter
+Cp=0.38    //kJ/(kg.K)
+Cp=Cp*1000    //J/kg.K
+k=370    //W/(m.K)
+T0=483    //K
+T_inf=373    //K
+delta_T=40    //K
+T=T0-delta_T    //K
+t=5    //time in [minutes]
+t=t*60    //[seconds]
+//A=2A.....Two faces
+//V=A.2x
+//2x=thickness of slab=30    mm=0.03    m
+x=0.015    //[m]
+th=2*x    //thickness of slab
+h=-rho*Cp*x*log((T-T_inf)/(T0-T_inf))/t
+printf("Heat transfer coefficient is: %f W/(sq m.K)",h)
diff --git a/1073/CH2/EX2.42/2_42.sce b/1073/CH2/EX2.42/2_42.sce
new file mode 100755
index 000000000..3d93507ad
--- /dev/null
+++ b/1073/CH2/EX2.42/2_42.sce
@@ -0,0 +1,24 @@
+
+clear;
+clc;
+//Example 2.42
+//Given
+rho=7800    //[kg per cubic m]
+h=100    //W/(sq m.K)  Convective heat transfer coeff
+Cp=460    //J/(kg.K)
+k=40    //W/(m.K)
+L=1    //[m] length ofrod
+D=10    //mm    
+D=D/1000    //diameter in[m]
+R=D/2    //raidus in [m]
+//For cylindrical rod:
+A=2*%pi*R*L    //Area in [sq m]
+V=%pi*R^2*L    //Volume in [cubic m]
+L_dash=V/A    //[m]
+Nbi=h*L_dash/k    //Biot number
+//N_bi<0.10,Hence lumped heat capavity is possible
+T=473    //[K]    
+T_inf=393    //[K]
+T0=593    //[K]
+t=-rho*Cp*V*log((T-T_inf)/(T0-T_inf))/(h*A)
+printf("Time required to reach temperature %f is %f s",T,t);
diff --git a/1073/CH2/EX2.43/2_43.sce b/1073/CH2/EX2.43/2_43.sce
new file mode 100755
index 000000000..81d2d4855
--- /dev/null
+++ b/1073/CH2/EX2.43/2_43.sce
@@ -0,0 +1,25 @@
+
+clear;
+clc;
+//Example 2.43   
+//Given
+rho=8600    //[kg/cubic m]
+Cp=0.42    //kJ/(kg.K)
+Cp=Cp*1000    //J/(kg.K)
+dia=0.71    //[mm]
+dia=dia/1000    //[dia in m]
+R=dia/2    //radius [m]
+h=600    //convective coeff W/(sq m.K)
+//Let length =L=1
+L=1         //[m]
+A=2*%pi*R*L;
+V=%pi*(R^2)*L;
+tao=(rho*Cp*V)/(h*A);
+printf("Time constant of the thermocouple is %f s",tao);
+//at
+t=tao
+//From  (T-T_inf)/(T0-T_inf)=e^(-t/tao)
+ratio=%e^(-t/tao)    //Ratio of thermocouple difference to initial temperature difference
+printf("At the end of the time period t=tao=%f s ,Temperature difference b/n the thermocouple and the gas stream would be %f of the initial temperature difference",tao,ratio);
+printf("\n It should be reordered after %f s",4*tao);
+
diff --git a/1073/CH2/EX2.44/2_44.sce b/1073/CH2/EX2.44/2_44.sce
new file mode 100755
index 000000000..6e006dc32
--- /dev/null
+++ b/1073/CH2/EX2.44/2_44.sce
@@ -0,0 +1,31 @@
+
+clc;
+clear;
+//Example 2.44
+rho=8000    //kg/cubic m
+Cp=420    //J/(kg.K)
+h_hot=60    // for hot stream W/(sq m.K)    
+dia=4    //[mm]
+t=10;
+r=dia/(2*1000)    //radius in [m]
+//For sphere
+V=(4/3)*%pi*r^3    //Volume in [cubic m]
+A=4*%pi*r^2        //Volume in [sq m]
+tao=rho*Cp*V/(h_hot*A)    // Time constant in [s]
+ratio=%e^(-t/tao)    // %e^(-t/tao)=(T-T-inf)/(T0-T_inf)
+T_inf=573    //[K]
+T0=313    //[K]
+T=T_inf+ratio*(T0-T_inf)
+//ANS-[i]
+printf("\n Answer: Time constant of thermocouple is %f s",tao);
+
+//IN STILL AIR:
+h_air=10    //W/(sq m .K)
+tao_air=rho*Cp*V/(h_air*A)    //[s]
+t_air=20    //[s]
+ratio_air=%e^(-t_air/tao_air)
+T_inf_air=303    //[K]
+T0_air=T;
+T_air=T_inf_air+ratio_air*(T0_air-T_inf_air)
+//ANS-[ii]
+printf("Temperature attained by junction 20 s after removing from the hot air stream is:%d K",round(T_air))
diff --git a/1073/CH2/EX2.45/2_45.sce b/1073/CH2/EX2.45/2_45.sce
new file mode 100755
index 000000000..f762e58a0
--- /dev/null
+++ b/1073/CH2/EX2.45/2_45.sce
@@ -0,0 +1,27 @@
+clc;
+clear;
+//Example 2.45
+T_inf=390;               //[K]
+U=600;                   //[W/sq m.K]
+Ac=1;                     //[sq m]
+Av=10                       //Vessel area in [sq m]
+m=1000;                  //[kg]
+Cp=3.8*10^3;            //[J/kg.K]
+To=290;                  //[K]
+T=360;                   //[K]
+h=8.5                   //[W/sq m.K]
+//Heat gained from the steam=Rate of increase of internal energy
+//U*A*(T_inf-T)=m*Cp*dT
+deff('[x]=f(t)','x=log((T_inf-To)/(T_inf-T))-U*Ac*t/(m*Cp)');
+t=fsolve(1,f);          //[in s]
+t=round(t)    //[in s]
+Ts=290;
+printf("\nTime taken to heat the reactants over the same temperature range is %f h",t);
+function t1=g(T),t1=m*Cp/(U*Ac*(T_inf-T)-h*Av*(T-Ts)),endfunction
+t1=intg(To,T,g);
+deff('[m]=fx(Tmax)','m=U*Ac*(T_inf-Tmax)-h*Av*(Tmax-Ts)')
+T_max=fsolve(1,fx)
+printf("\nANS: In CASE 1\nTime  taken to heat the reactants = %f s .ie %f h \n",t,t/3600);
+printf("\nANS: In CASE 2 \n Time  taken to heat the reactants = %f s\n",t1);
+printf("\nANS.:  Maximum temperature at which temperature can be raised is %f K\n",T_max);
+
diff --git a/1073/CH2/EX2.46/2_46.sce b/1073/CH2/EX2.46/2_46.sce
new file mode 100755
index 000000000..290ccb426
--- /dev/null
+++ b/1073/CH2/EX2.46/2_46.sce
@@ -0,0 +1,15 @@
+
+clc;
+clear;
+//Example 2.46    
+dia=3    //[mm]
+dia=dia/1000    //[m]
+r=dia/2    //radius in[m]
+k=150    //W/(m.K)
+h=300    //W/(sq m.K)
+T0=413    //[K]
+T_inf=288    //[K]
+A=%pi*(r^2)    //Area in [sq m]
+P=%pi*dia      //[W/sq m.K]
+Q=(T0-T_inf)*sqrt(h*P*k*A)   //Heat dissipated in [W]
+printf("Heat dissipated by the rod is %f W",Q)
diff --git a/1073/CH2/EX2.47/2_47.sce b/1073/CH2/EX2.47/2_47.sce
new file mode 100755
index 000000000..5ed83f5e8
--- /dev/null
+++ b/1073/CH2/EX2.47/2_47.sce
@@ -0,0 +1,30 @@
+
+clc;
+clear;
+//Example 2.47   
+//Given
+k=200    //W/(m.K)
+h=15    //W/(sq m.K)
+T0=523    //[K]
+T_inf=288    //[K]
+theta_0=T0-T_inf    
+dia=25    //diameter[mm]
+dia=dia/1000    //diameter[m]
+r=dia/2    //radius in [m]
+P=%pi*dia    //[m]
+A=%pi*r^2    //[sq m]
+//For insulated fin:
+m=sqrt(h*P/(k*A))
+L=100    //length of rod in [mm]
+L=L/1000    //length of rod in [m]
+Q=theta_0*tanh(m*L)*sqrt(h*P*k*A)    //Heat loss 
+//ANSWER-1
+printf("Heat loss by the insulated rod is %f W \n",Q)
+nf=tanh(m*L)/(m*L)    //Fin efficiency for  insulated fin
+//ANSWER-2
+printf("Fin efficiency is %f percent \n",nf*100)
+//At the end of the fin: theta/theta_0=(cosh[m(L-x)]/cosh(mL))
+//at x=L, theta/theta_0=1/(cosh(mL)
+T=T_inf+(T0-T_inf)*(1/cosh(m*L))    //[K]
+//ANSWER-3
+printf("Temperature at the end of the fin is %f K \n",T)
diff --git a/1073/CH2/EX2.49/2_49.sce b/1073/CH2/EX2.49/2_49.sce
new file mode 100755
index 000000000..2958a99b4
--- /dev/null
+++ b/1073/CH2/EX2.49/2_49.sce
@@ -0,0 +1,25 @@
+
+clc;
+clear;
+//Example 2.49
+//Given
+k=300    //W/(m.K)
+h=20    //W.(sq m.K)
+P=0.05    //[m]
+A=2    //[sq cm]
+A=A/10000    //[sq m]
+T0=503    //[K]
+T_inf=303    //[K]
+theta_0=T0-T_inf    //[K]
+m=sqrt(h*P/(k*A))
+//CASE 1:  6 Fins of 100 mm length
+L1=0.1    //Length of fin in [m]
+Q=sqrt(h*P*k*A)*theta_0*tanh(m*L1)    //[W]
+//For 6 fins
+Q=Q*6    //for 6 fins [W]
+//CASE 2:  10 fins of 60 mm length
+L2=60    //[mm]
+L2=L2/1000    //[m]
+Q2=sqrt(h*P*k*A)*theta_0*tanh(m*L2);    //[W]
+Q2=Q2*10    //For 10 fins
+printf("As,Q for 10 fins of 60 mm length( %f W) is more than Q for 6 fins of 100 mm length (%f W).\n The agreement-->10 fins of 60 mm length is more effective",Q2,Q); 
diff --git a/1073/CH2/EX2.5/2_5.sce b/1073/CH2/EX2.5/2_5.sce
new file mode 100755
index 000000000..206d15307
--- /dev/null
+++ b/1073/CH2/EX2.5/2_5.sce
@@ -0,0 +1,29 @@
+clc;
+clear;
+//Example 2.5
+printf("Example 2.5\nPage 2.23")
+//Given
+A=1;  //let [sq m]
+x1=0.23;   //thickness of fir brick layer[m]
+x2=0.115;   // [m]
+x3=0.23;   //[m]
+T1=1213;  //Temperature of furnace [K]
+T2=318;   //Temperature of furnace  [K]
+dT=T1-T2;  //[K]
+k1=6.047;   //W/(m.K) (fire brick)
+k2=0.581;   //W/(m.K) (insulating brick)
+k3=2.33;   //W/(m.K) (building brick)
+Q_by_A=dT/((x1/k1)+(x2/k2)+(x3/k3))  //Heat lost per unit Area in Watt
+
+R1=(x1/k1)  //Thermal resistance
+R1=0.04     //Approximate
+R2=(x2/k2)
+R2=0.2025      //Approximate
+R3=(x3/k3)
+R3=0.1      //Approximate
+Ta=T1-((dT*R1)/(R1+R2+R3))
+Tb=((dT*R3)/(R1+R2+R3))+T2
+Tb=565          //Approximation
+printf("\nAnswer:Heat loss per unit area is %f W=%f J/s\n",Q_by_A,Q_by_A);
+printf("\nAnswer:\n Ta=%f K =Temperature at the interface between fire brick and insulating brick\n Tb=%d K Temperature at the interface between insulating and building brick\n",Ta,Tb)
+
diff --git a/1073/CH2/EX2.50/2_50.sce b/1073/CH2/EX2.50/2_50.sce
new file mode 100755
index 000000000..ee60fd341
--- /dev/null
+++ b/1073/CH2/EX2.50/2_50.sce
@@ -0,0 +1,35 @@
+clc;
+clear;
+//Example 2.50   
+//Given
+h_oil=180    //W/(sq m.K)
+h_air=15    //W/(sq m.K)
+T_oil=353    //[K]
+T_air=293    //[K]
+delta_T=T_oil-T_air;    //[K]
+k=80    //Conductivity in  [W/(m.K)]
+for_section=11*10^-3    //[m]
+L=25    //[mm]
+L=L/1000    //[m]
+W=1    //[m] Width,..let
+t=1    //[mm] 
+t=t/1000    //[m]
+A=W*t    //[m]
+P=2*t
+Af=2*L*W    //sq m
+N=1
+Ab=for_section-A    //[sq m]
+//CASE 1: Fin on oil side only
+m=sqrt(h_oil*P/(k*A)) 
+nf_oil=tanh(m*L)/(m*L)
+Ae_oil=Ab+nf_oil*Af*N    //[sq m]
+Q=delta_T/(1/(h_oil*Ae_oil)+1/(h_air*for_section))    //[W]
+printf("In oil side,Q=%f W\n",Q);
+//CASE 2: Fin on  air side only
+m=sqrt(h_air*P/(k*A))
+nf_air=tanh(m*L)/(m*L)
+nf_air=0.928        //Approximation
+Ae_air=Ab+nf_air*Af*N    //[sq m]
+Q=delta_T/(1/(h_oil*for_section)+1/(h_air*Ae_air))    //[W]
+printf("In air side,Q=%f W",Q);
+printf("\n From above results we see that more heat transfer takes place if fins are provided on the air side");
diff --git a/1073/CH2/EX2.51/2_51.sce b/1073/CH2/EX2.51/2_51.sce
new file mode 100755
index 000000000..229dfb937
--- /dev/null
+++ b/1073/CH2/EX2.51/2_51.sce
@@ -0,0 +1,46 @@
+
+clc;
+clear;
+//Example 2.51
+//Given
+k=75    //Thermal conductivity [W/(m.K)]
+T_water=363    //[K]
+T_air=303    //[K]    
+dT=T_water-T_air    //delta T
+h1=150    // for water[W/(sq m.K)]
+h2=15       //for air [W/(sq m.K)]
+W=0.5    //Width of wall[m]
+L=0.025    //[m]
+Area=W^2    //Base Area [sq m]
+t=1    //[mm]
+t=t/1000    //[m]
+pitch=10    //[mm]
+pitch=pitch/1000    //[m]
+N=W/pitch    //[No of fins]
+//Calculations
+A=N*W*t    //Total cross-sectional area of fins   in [sq m]
+Ab=Area-A    //[sq m]
+Af=2*W*L    //Surface area of fins    [sq m]
+
+//CASE 1: HEAT TRANSFER WITHOUT FINS
+A1=Area    //[sq m]
+A2=A1    //[sq m]
+Q=dT/(1/(h1*A1)+1/(h2*A2));        //[W]
+printf("\nWithout fins,Q=%f W\n",Q);
+//CASE 2: Fins on the water side
+P=2*(t+W);
+A=0.5*10^-3;
+m=sqrt(h1*P/(k*A))
+nfw=tanh(m*L)/(m*L)   //Effeciency on water side
+Aew=Ab+nfw*Af*N    //Effective area on the water side    [sq m]
+Q=dT/(1/(h1*Aew)+1/(h2*A2));        //[W]
+printf("\n With fins on water side,Q=%f W \n",Q);
+//CASE 3: FINS ON THE AIR SIDE
+m=sqrt(h2*P/(k*A))
+nf_air=tanh(m*L)/(m*L)    //Effeciency
+Aea=Ab+nf_air*Af*N    //Effective area on air side
+Q=dT/(1/(h1*A1)+1/(h2*Aea));        //[W]
+printf("\n With Fins on Air side,Q=%f W \n",Q)
+//BOTH SIDE:
+Q=dT/(1/(h1*Aew)+1/(h2*Aea));        //[W]
+printf("\n With Fins on both side,Q=%f W \n",Q);
diff --git a/1073/CH2/EX2.7/eg2_7.sce b/1073/CH2/EX2.7/eg2_7.sce
new file mode 100755
index 000000000..858d608ca
--- /dev/null
+++ b/1073/CH2/EX2.7/eg2_7.sce
@@ -0,0 +1,22 @@
+clc
+printf("Example 2.7,Page no 2/26 \n");
+printf("Part-(a)\n");
+A=1; // sq metre
+x1=114 // mm
+x1=114/1000 // metre
+k1=0.138 // W/(m.K)
+R1= x1/(k1*A)
+x2=229 //mm
+x2= x2/1000  // metre
+k2=1.38  // W/m.K
+R2=x2/(k2*A)
+dT=1033-349
+//Heat loss
+Q=dT/(R1+R2)
+printf("ANSWER:Heat loss  from 1 sq metre wall=%f W",Q);
+printf("Part(b)\n");
+//contact resistance=cr
+cr=0.09 //K/W
+R=R1+R2+cr
+Q=dT/R
+printf("ANSWER:Heat loss from 1 sq metre when resistance present=%f W",Q);
diff --git a/1073/CH2/EX2.8/2_8.sce b/1073/CH2/EX2.8/2_8.sce
new file mode 100755
index 000000000..ad4ae7e54
--- /dev/null
+++ b/1073/CH2/EX2.8/2_8.sce
@@ -0,0 +1,21 @@
+clear;
+clc;
+//Example 2.8
+printf("Example 2.8 \n")
+//Given:
+x1=0.02  //[m]
+x2=0.01  //[m]
+x3=0.02  //[m]
+k1=0.105  //W/(m.k)
+k3=k1  //W/(m.K)
+k2=0.041  //W/(m.K)
+T1=303
+T2=263
+dT=T1-T2   //[K]
+Q_by_A=dT/((x1/k1)+(x2/k2)+(x3/k3))
+R=0.625  //K/W
+Tx=293   //K
+Rx=0.9524   //K/W
+x=R*(T1-Tx)/(dT*Rx)
+x=x*100  //mm
+printf("The temperature of 293 K will be reached at point %f  mm from the outermost wall surface of the ice-box",x)
diff --git a/1073/CH2/EX2.9/2_9.sce b/1073/CH2/EX2.9/2_9.sce
new file mode 100755
index 000000000..507c9e8f0
--- /dev/null
+++ b/1073/CH2/EX2.9/2_9.sce
@@ -0,0 +1,21 @@
+clc
+printf("Example 2.9,Page 2.28\n");
+//Given
+ID=50 //mm;
+dT=(573-303);
+printf("Internal diameter,ID=%f mm",ID);
+r1=ID/2 //mm
+r1=r1/1000 // metres
+OD=150 // mm
+printf("Outer diameter,OD=%f mm",OD);
+r2=OD/2 // mm
+r2=75/1000 // m
+//Thermal conductivity
+k=17.45 // W/(m.K) 
+//Solution
+printf("Q/A=dT/(r2-r1)/k\n");
+A1=4*%pi*(r1^2);
+A2=4*%pi*(r2^2);
+A=sqrt(A1*A2)
+Q=(A*k*dT)/(r2-r1)
+printf("ANSWER:\nHeat loss=Q=%f W",Q);
diff --git a/1073/CH3/EX3.1/3_1.sce b/1073/CH3/EX3.1/3_1.sce
new file mode 100755
index 000000000..12e938423
--- /dev/null
+++ b/1073/CH3/EX3.1/3_1.sce
@@ -0,0 +1,31 @@
+clc;
+clear;
+//Example 3.1
+mu=10^-3        //N.s/m^2
+//At distance y from surface
+//ux=a+by+cy^2+dy^3
+//At y=0,ux=0 therefore a=0
+//i.e tao=0
+//At edge of boundary layer,ie y=del
+//ux=u_inf
+//At y=o,c=0
+//At y=del,ux=b*del+d*del^3
+
+//Therefore, b=-3*d*del^3
+//d=-u_inf/(2*del^2)
+//b=3*u_inf/(2*del)
+
+//For velocity profile,we have:
+//del/x=4.64*(Nre_x)^(-1/2)
+
+//Evaluate N re_x
+
+x=75;        //[mm]
+x=x/1000;        //[m]
+u_inf=3;            //[m/s]
+rho=1000            //[kg/m^3] for air
+Nre_x=u_inf*rho*x/mu        //Reynold number
+//Substituting the value,we get
+del=x*4.64*(Nre_x^(-1/2))       //[m]
+printf("\nBoundary layer thickness is del=%f m or %f mm",del,del*1000);
+printf("\nWrong units in answer of book,m and mm are wrongly interchanged");
diff --git a/1073/CH3/EX3.10/3_10.sce b/1073/CH3/EX3.10/3_10.sce
new file mode 100755
index 000000000..33f13eb35
--- /dev/null
+++ b/1073/CH3/EX3.10/3_10.sce
@@ -0,0 +1,26 @@
+clc;
+clear;
+//Example 3.10
+v=16.96*10^-6    //[sq m./s]
+rho=1.128    //[kg/cubic m]
+Npr=0.699    //Prandtl number
+k=0.0276    //[W/m.K]
+u_inf=15    //[m/s]
+L=0.2    //[m]
+Nre_l=L*u_inf/v    //Reynold's number
+//Since this is less than 3*10^5,the boundary layer is laminar over entire length
+Nnu=0.664*sqrt(Nre_l)*(Npr^(1.0/3.0))
+h=Nnu*k/L    //[W/sq m.K]
+A=L^2    //Area in [sq m]
+Tw=293    //[K]
+T_inf=333    //[K]
+//Rate of heat transfer from BOTH sides is:
+Q=2*h*A*(T_inf-Tw)    //[W]
+printf("Rate of heat transfer from both sides of plate is %f W\n",Q);
+//ii-With turbulent boundary layer from the leading edge:
+h=k*0.0366*(Nre_l^(0.8))*(Npr^(1.0/3.0))/L       //[W/(sq m.K)]
+//Heat transfer from both sides is :
+Q=2*h*A*(T_inf-Tw)            //[W]
+printf("\nThese calculations sho that the that transfer rate is approximately doubled if boundary layer is turbulent from the leading edge \n");  
+
+
diff --git a/1073/CH3/EX3.11/3_11.sce b/1073/CH3/EX3.11/3_11.sce
new file mode 100755
index 000000000..adf37571b
--- /dev/null
+++ b/1073/CH3/EX3.11/3_11.sce
@@ -0,0 +1,21 @@
+clc;
+clear;
+//Example 3.11
+mu=1.906*10^-5    //[kg/(m.s)]
+k=0.02723    //W/m.K
+Cp=1.007    //[kJ/(kg.K)]
+rho=1.129    //[kg/cubic m]
+Npr=0.70
+Mavg=29
+u_inf=35    //[m/s]
+L=0.75    //[m]
+Tm=313    //[K]
+P=101.325    //[kPa]
+Nre_l=rho*u_inf*L/mu    //Reynold's number >5*10^5
+Nnu=0.0366*Nre_l^(0.8)*Npr^(1.0/3.0);
+h=Nnu*k/L    //[W/s m.K]
+A=1*L    //[sq m]
+Tw=333    //[K]
+T_inf=293    //[K]
+Q=h*A*(Tw-T_inf);    //[W]
+printf("Heat transfer from the plate is %f W",Q);
+\ No newline at end of file
diff --git a/1073/CH3/EX3.12/3_12.sce b/1073/CH3/EX3.12/3_12.sce
new file mode 100755
index 000000000..9044b6c09
--- /dev/null
+++ b/1073/CH3/EX3.12/3_12.sce
@@ -0,0 +1,16 @@
+clc;
+clear;
+//Example 3.12
+v=18.23*10^-6    //sq m/s
+k=0.02814    //[W/m.K]
+D=0.012    //[m]
+r=0.006    //[m]
+u_inf=4    //[m/s]
+Nre=D*u_inf/v    //Reynold's number
+Nnu=0.37*Nre^(0.6);
+h=Nnu*(k/D)    
+A=4*%pi*r^2    //Area of sphere in [sq m]
+Tw=350    //[K]
+T_inf=300    //[K]
+Q=h*A*(Tw-T_inf)    //Heat lost by sphere in [W]
+printf("\n Heat lost by sphere is %f W",Q);
+\ No newline at end of file
diff --git a/1073/CH3/EX3.13/3_13.sce b/1073/CH3/EX3.13/3_13.sce
new file mode 100755
index 000000000..2d9a907ba
--- /dev/null
+++ b/1073/CH3/EX3.13/3_13.sce
@@ -0,0 +1,20 @@
+clc;
+clear;
+//Exmaple 3.13
+v=15.69*10^-6    //[sq m./s]
+k=0.02624    //[W/m.K]
+Npr=0.708    //Prandtl number
+mu=2.075*10^-5    //kg/m.s
+u_inf=4    //[m/s]
+mu_inf=1.8462*10^-5    //[m/s] velocity
+Tw=350    //[K]
+T_inf=300    //[K]
+D=0.012    //[m]
+r=D/2    //Radius in [m]
+Nre=u_inf*D/v    //Reynold's numbe
+Nnu=2+(0.4*Nre^(1.0/2.0)+0.06*Nre^(2.0/3.0))*Npr^(0.4)*(mu_inf/mu)^(1.0/4.0)
+h=Nnu*k/D    //[W/sq m.K]
+
+A=4*%pi*r^2    //Area in [sq m]
+Q=h*A*(Tw-T_inf);
+printf("\n Heat lost by the sphere is %f W",Q);
+\ No newline at end of file
diff --git a/1073/CH3/EX3.14/3_14.sce b/1073/CH3/EX3.14/3_14.sce
new file mode 100755
index 000000000..2876908b6
--- /dev/null
+++ b/1073/CH3/EX3.14/3_14.sce
@@ -0,0 +1,21 @@
+clc;
+clear;
+//Example 3.14
+v=2.08*10^-5    //[sq m/s]
+k=0.03    //W/(m.K)
+Npr=0.697    //Prandtl number
+D=0.06    //[m]
+u_inf=0.3    //[m/s]
+Nre=D*u_inf/v    //Reynolds number
+//Average nusselt number is given by:
+Nnu=0.37*(Nre^0.6);
+h=Nnu*k/D    //W/sq m.K
+Tw=400    //[K]
+T_inf=300    //[K]
+D=0.06    //[m]
+r=0.03    //[m]
+A=4*%pi*r^2    //Area in [sq m]
+Q=h*A*(Tw-T_inf)    //[W]
+per=Q*100/100    //Percent of heat lost by forced convection
+printf("Heat transfer rate is %f W,And percentage of power lost by convectio is: %f percent ",Q,per);
+ 
+\ No newline at end of file
diff --git a/1073/CH3/EX3.15/3_15.sce b/1073/CH3/EX3.15/3_15.sce
new file mode 100755
index 000000000..04ad2bf9a
--- /dev/null
+++ b/1073/CH3/EX3.15/3_15.sce
@@ -0,0 +1,21 @@
+
+clc;
+clear;
+//Example 3.15    
+u_inf=50    //velocity in [m/s]
+mu=2.14*10^-5    //[kg/(m.s)]
+rho=0.966    //[kg/cubic m]
+k=0.0312    //[W/(m.K)]
+Npr=0.695    //Prandtl number
+D=0.05    //Diameter in [m]
+Nre=D*u_inf*rho/mu ;  //Reynold's number
+printf("%f",Nre)
+Nnu=0.0266*Nre^0.805*Npr^(1/3);
+h=Nnu*k/D   ; //W/sq m.K
+h=171.7     //Approximation
+printf("\n%f",h)
+Tw=423    //[K]
+T_inf=308    //[K]
+//Heat loss per unit length is :
+Q_by_l=h*%pi*D*(Tw-T_inf);  //[W]
+printf("Heat lost per unit length of cylinder is %f W(approx)",round(Q_by_l));
diff --git a/1073/CH3/EX3.16/3_16.sce b/1073/CH3/EX3.16/3_16.sce
new file mode 100755
index 000000000..85389ff5b
--- /dev/null
+++ b/1073/CH3/EX3.16/3_16.sce
@@ -0,0 +1,34 @@
+
+clc;
+clear;
+//Example 3.16
+v=20.92*10^-6    //sq m/s
+k=3*10^-2    //W/(m.K)
+Npr=0.7
+u_inf=25    //[m/s]
+d=50    //[mm]
+d=d/1000    //[m]
+Nre=u_inf*d/v    //Reynold's number
+Tw=397    //[K]
+T_inf=303    //[K]
+
+//Case 1: Circular tube
+
+Nnu=0.0266*Nre^(0.805)*Npr^(1.0/3.0);
+h=Nnu*k/d    //[W/sq m.K]
+A=%pi*d    //Area in [sq m]
+Q=h*A*(Tw-T_inf)    //[W]
+Q_by_l1=h*%pi*d*(Tw-T_inf)    //[W/m]
+
+//Case 2:Square tube
+A=50*50    //Area in [sq mm]
+P=2*(50+50)    //Perimeter [mm]
+l=4*A/P       //[mm]
+l=l/1000        //[m]
+Nnu=0.102*(Nre^0.675)*(Npr^(1.0/3.0))
+h=Nnu*k/d    //W/(sq m.K)
+A=4*l*l        //[sq m]
+
+Q=h*A*(Tw-T_inf)
+Q_by_l2=Q/l    //[W/m]
+printf("\nRate of heat flow from the square pipe=%f W/m  \n which is more than that from the circular pipe which is equal to %f W/m",Q_by_l2,Q_by_l1);
diff --git a/1073/CH3/EX3.17/3_17.sce b/1073/CH3/EX3.17/3_17.sce
new file mode 100755
index 000000000..47d4e87bc
--- /dev/null
+++ b/1073/CH3/EX3.17/3_17.sce
@@ -0,0 +1,27 @@
+
+clc;
+clear;
+//Example 3.17
+mu=0.8    //Viscosity of flowing fluid [N.s/sq m]
+rho=1.1    //Density of flowinf fluid [g/cubic cm]
+rho=rho*1000    //Density in [kg/cubic m]
+Cp=1.26    //Specific heat [kJ/kg.K]
+Cp=Cp*10^3    // in[J/(kg.K)]
+k=0.384    //[W/(m.K)]
+mu_w=1    //Viscosity at wall temperature [N.s/sq m]
+L=5    //[m]
+vfr=300    //Volumetric flow rate in [cubic cm/s]
+vfr=vfr*10^-6    //[cubic m/s]
+mfr=vfr*rho    //Mass flow rate of flowinf fluid [kg/s]
+Di=20    //Inside diameter in[mm]
+Di=Di/1000    //[m]
+Area=(%pi/4)*Di^2    //Area of cross-section [sq m]
+u=vfr/Area    //Veloctiy in [m/s]
+Nre=Di*u*rho/mu    //Reynold's number
+//As reynold's number is less than 2100,he flow is laminar
+Npr=Cp*mu/k    //Prandtl number
+Nnu=1.86*(Nre*Npr*Di/L)^(1.0/3.0)*(mu/mu_w)^(0.14)
+hi=Nnu*k/Di    //inside heat transfer coefficient [W/sq m.K]
+printf("Inside heat transfer coefficient is %f W/(sq m.K)",hi);
+//Note:
+printf("\n The answer given in book..ie 1225 is wrong.please redo the calculation of last line manually to check\n"); 
diff --git a/1073/CH3/EX3.18/3_18.sce b/1073/CH3/EX3.18/3_18.sce
new file mode 100755
index 000000000..9a94ce93f
--- /dev/null
+++ b/1073/CH3/EX3.18/3_18.sce
@@ -0,0 +1,24 @@
+clc;
+clear;
+//Example 3.18
+m=5500    //Mass flow rate in [kg/h]
+m=m/3600    //[kg/s]
+rho=1.07    //Density of fluid in [g/cm^3]
+rho=rho*1000    //[kg/m^3]
+vfr=m/rho    //Volumetric flow rate in [m^3/s]
+Di=40    //Diameter of tube [mm]
+Di=Di/1000    //[m]
+A=(%pi/4)*Di^2    //Area of cross-section in [sq m]
+u=vfr/A    //Velocity of flowing fluid    [m/s]
+rho=1070    //Density in [kg/m^3]
+mu=0.004    //Viscosity in [kg/m.s]
+Nre=Di*u*rho/mu
+Nre=12198       //Approx
+//Since this reynold's number is less than 10000,the flow is turbulent
+Cp=2.72    //Specific heat in [kJ/kg.K]
+Cp=Cp*10^3    //Specific heat in [J/kg.K]
+k=0.256    //thermal conductivity in [W/m.K]
+Npr=Cp*mu/k    //Prandtl number
+Nnu=0.023*(Nre^0.8)*(Npr^0.4)    //Nusselt number
+hi=k*Nnu/Di    //Inside heat transfer coefficient in [W/m^2.K]
+printf("Inside heat transfer coefficient is %f W/sq m.K",hi);
diff --git a/1073/CH3/EX3.19/3_19.sce b/1073/CH3/EX3.19/3_19.sce
new file mode 100755
index 000000000..0689b2981
--- /dev/null
+++ b/1073/CH3/EX3.19/3_19.sce
@@ -0,0 +1,30 @@
+
+clc;    
+clear;
+//Example 3.19
+
+//DATA:
+rho=984.1    //Density of water [kg/m^3]
+Cp=4187    //Specific heat in [J/kg.K]
+mu=485*10^-6    //Viscosity at 331 K[Pa.s]
+k=0.657    //[W/(m.K)]
+mu_w=920*10^-6    //Viscosity at 297 K [Pa.s]
+//Solution
+D=16    //Diameter in [mm]
+D=D/1000    //Diameter in [m]
+u=3    //Velocity in [m/s]
+rho=984.1    //[kg/m^3]
+Nre=D*u*rho/mu    //Reynolds number
+Nre=round(Nre)
+Npr=Cp*mu/k    //Prandtl number
+
+//Dittus-Boelter equation (i)
+Nnu=0.023*(Nre^0.8)*(Npr^0.3)    //nusselt number
+h=k*Nnu/D    //Heat transfer coefficient [W/m^2.K]
+printf("\nANSWER-(i) \nBy Dittus-Boelter equation we get h=%f W/sq m.K\n\n\n",h);
+
+//sieder-tate equation (ii)
+Nnu=0.023*(Nre^0.8)*(Npr^(1.0/3.0))*((mu/mu_w)^0.14)    //Nusselt number
+h=k*Nnu/D    //Heat transfer coefficient in [W/sq m.K]
+printf("\nAnswer-(ii)\n-By Sieder-Tate equation we get h=%f W/sq m.K\n",h);
+printf("\nNOTE:Calculation mistake in book in part 2 ie sieder tate eqn\n")
diff --git a/1073/CH3/EX3.2/3_2.sce b/1073/CH3/EX3.2/3_2.sce
new file mode 100755
index 000000000..d84452ac5
--- /dev/null
+++ b/1073/CH3/EX3.2/3_2.sce
@@ -0,0 +1,17 @@
+clc;
+clear;
+//Example3.2
+//Given
+mu=15*10^-6    //sq m /s
+v=2    //m/s
+L=2    //[m] length of plate
+Nre_x=3*10^5
+xc=Nre_x*mu/v    //critical length at whihc the transition takes place
+//Since xc is less than 2 m.Therefore the flow is laminar
+//at any distance x,.it is calculated from
+//del/x=4.64/(sqrt(NRe,x))
+//At x=L=2 m
+Nre_l=v*L/mu
+del_l=4.64*L/sqrt(Nre_l)
+del_l=del_l*1000    //[mm]
+printf("Boundary layerthickness at the trailing edge is %f mm",del_l);
+\ No newline at end of file
diff --git a/1073/CH3/EX3.20/3_20.sce b/1073/CH3/EX3.20/3_20.sce
new file mode 100755
index 000000000..9cdd0dbd3
--- /dev/null
+++ b/1073/CH3/EX3.20/3_20.sce
@@ -0,0 +1,22 @@
+clc;
+clear;
+//Example 3.20
+m_dot=2250    //Mass flow arte in [kg/h]
+Cp=3.35    //Specific heat in [kJ/(kg.K)]
+dT=316-288.5    //Temperature drop for oil    [K]
+Q=Cp*m_dot*dT    //Rate of heat transfer in [kJ/h]
+Q=round(Q*1000/3600)    //[J/s] or[W]
+Di=0.04    //Inside diameter [m]
+Do=0.048    //Outside diamter in [m]
+hi=4070    //for steam [W/sq m.K]
+ho=18.26    //For oil [W/sq m.K]
+Rdo=0.123    //[sq m.K/W]
+Rdi=0.215    //[sq m.K/W]
+Uo=1/(1/ho+Do/(hi*Di)+Rdo+Rdi*(Do/Di))    //[W/m^2.K]
+Uo=2.3
+dT1=373-288.5    //[K]
+dT2=373-316    //[K]
+dTm=(dT1-dT2)/log(dT1/dT2)    //[K]
+Ao=Q/(Uo*dTm)    //Heat transfer area in [m^2]
+printf("Heatr transfer area is:%f m^2",Ao);
+
diff --git a/1073/CH3/EX3.21/3_21.sce b/1073/CH3/EX3.21/3_21.sce
new file mode 100755
index 000000000..61570ca63
--- /dev/null
+++ b/1073/CH3/EX3.21/3_21.sce
@@ -0,0 +1,42 @@
+
+clc;
+clear;
+//Example 3.21
+k_tube=111.65    //[W/m.K]
+W=4500    //[kg/h]
+rho=995.7    //[kg/sq m]
+Cp=4.174    //[kJ/(kg.K)]
+k=0.617    //[W/(m.K)]
+v=0.659*10^-6    //Kinematic viscosity [sq m/s]
+m_dot=1720    //kg/h
+T1=293    //Initial temperature in [K]
+T2=318    //Final temperature in [K]
+dT=T2-T1    //[K]
+Q=m_dot*Cp*dT    //Heat transfer rate in [kJ/h]
+Q=Q*1000/3600    //[J/s] or [W]
+Di=0.0225    //[m]
+u=1.2    //[m/s]
+//Nre=Di*u*rho/mu or
+Nre=Di*u/v    //Reynolds number
+//As Nre is greater than 10000,Dittus Boelter equation is applicable
+Cp=Cp*10^3    //J/(kg.K)
+mu=v*rho    //[kg/(m.s)]
+Npr=Cp*mu/k    //Prandtl number
+//Dittus-Boelter equation for heating is 
+Nnu=0.023*(Nre^0.8)*(Npr^0.4)
+hi=k*Nnu/Di    //Heat transfer coefficient  [W/(sq m.K)]
+Do=0.025    //[m]
+Dw=(Do-Di)/log(Do/Di)    //Log mean diameter in [m]
+ho=4650    //[W/sq m.K]
+k=111.65    //[W/m.K]
+xw=(Do-Di)/2    //[m]
+Uo=1/(1/ho+Do/(hi*Di)+xw*Do/(k*Dw))    //Overall heat transfer coefficient in W/(m^2.K)
+T_steam=373    //Temperature of condensing steam in [K]
+dT1=T_steam-T1+10    //[K]
+dT2=T_steam-T2+10    //[K]
+dTm=(dT1-dT2)/log(dT1/dT2)    //[K]
+Ao=Q/(Uo*dTm)//Area in [m^2]
+L=4    //length of tube [m]
+n=Ao/(%pi*Do*L)    //number of tubes
+printf("No. of tubes required=%d\n",round(n));
+printf("\n NOTE: there is an error in book in calculation of dT1 and dT2,\n 373-293 is written as 90,instead of 80...similarly in dT2,\nSo,in compliance with the book,10 is added to both of them") 
diff --git a/1073/CH3/EX3.22/3_22.sce b/1073/CH3/EX3.22/3_22.sce
new file mode 100755
index 000000000..034f1f602
--- /dev/null
+++ b/1073/CH3/EX3.22/3_22.sce
@@ -0,0 +1,23 @@
+clc;
+clear;
+//Example 3.22
+m_dot=25000    //massflow rate of water [kg/h]
+rho=992.2    //[kg/m^3]
+k=0.634    //[W/m.K]
+vfr=m_dot/rho    //[m^3/h]
+Npr=4.31    //Prandtl numberl
+Di=50    //[mm]
+Di=0.05    //[m]
+dT=10    //[K] as the wall is at a temperature of 10 K above the bulk temperature
+u=(vfr/3600)/(%pi*(Di/2)^2)    //Velocity of water in [m/s]
+u=3.56          //Approximation
+//Nre=Di*u*rho/mu=Di*u/v    as v=mu/rho
+v=0.659*10^-6;    //[m^2/s]
+Nre=Di*u/v    //Reynolds number
+//As it is less than 10000,the flow is in the turbulent region for heat transfer and Dittus Boelter eqn is used
+Nnu=0.023*(Nre^0.8)*(Npr^0.4);    //Nusselt number
+hi=Nnu*k/Di    //Heat transfer coefficiet in [W/sq m.K]
+q_by_l=hi*%pi*Di*dT   //Heat transfer per unit length[kW/m]
+printf("Average value of convective film coefficient is hi= %d W/sq m.K \nHeat transferred per unit length is Q/L=%f kW/m",round(hi),q_by_l/1000);
+ 
+
diff --git a/1073/CH3/EX3.23/3_23.sce b/1073/CH3/EX3.23/3_23.sce
new file mode 100755
index 000000000..8298dd6f2
--- /dev/null
+++ b/1073/CH3/EX3.23/3_23.sce
@@ -0,0 +1,36 @@
+clc;
+clear;
+//Example 3.23
+vfr=1200  ;  //Water flow rate in [l/h]
+rho=0.98 ;   //Density of water in g/[cubic cm]
+m_dot=vfr*rho    //Mass flow rate of water [kg/h]
+m_dot2=m_dot/3600    //[kg/s]
+Cp=4.187*10^3 ;   //[J/kg.K]
+Di=0.025  ;  //Diameter in [m]
+mu=0.0006 ;   //[kg/(m.s)]
+Ai=%pi*((Di/2)^2)    //Area of cross-section in [m^2]
+Nre=(Di/mu)*(m_dot2/Ai)    //Reynolds number
+k=0.63 ; //for metal wall in [W/(m.K)]
+Npr=Cp*mu/k;    //Prandtl number
+//Since Nre>10000
+//therefore ,Dittus boelter eqn for heating is 
+Nnu=0.023*(Nre^(0.8))*(Npr^(0.4))
+ho=5800   ; //Film heat coefficientW/(m^2.K)
+hi=Nnu*k/Di    //Heat transfer coeffcient in [W/(sq m.K)]
+Do=0.028 ;   //[m]
+Di=0.025 ;   //[m]
+xw=(Do-Di)/2;    //[m]
+Dw=(Do-Di)/log(Do/Di);    //[m]
+k=50  ;  //for metal wall in [W/(m.K)]
+Uo=1/(1/ho+Do/(hi*Di)+xw*Do/(k*Dw));    //in [W/sq m.K]
+dT=343-303 ; //[K]
+dT1=393-303 ;   //[K]
+dT2=393-343 ;   //[K]
+dTm=(dT1-dT2)/log(dT1/dT2);    //[K]
+Cp=Cp/1000;  //[in [kJ/kg.K]]
+Q=m_dot*Cp*dT;    //Rate of heat transfer in [kJ/h]
+Q=Q*1000/3600;    //[J/s] or [W]
+Ao=Q/(Uo*dTm);    //Heat transfer area in [sq m]
+//Also,..Ao=%pi*Do*L    ..implies that
+L=Ao/(%pi*Do)    //[m]
+printf("Length of tube required is %f m",round(L));
diff --git a/1073/CH3/EX3.24/3_24.sce b/1073/CH3/EX3.24/3_24.sce
new file mode 100755
index 000000000..9c27cf1ba
--- /dev/null
+++ b/1073/CH3/EX3.24/3_24.sce
@@ -0,0 +1,20 @@
+clc;
+clear;
+//Example 3.24
+//1.For initial conditions:
+T=360;           //[K]
+T1=280;         //[K]
+T2=320;         //[K]
+dT1=T-T1;       //[K]
+dT2=T-T2;       //[K]
+//Q1=m1_dot*Cp1*(T2-T1)
+Cp1=4.187           //Heat capacity 
+dTlm=(dT1-dT2)/log(dT1/dT2) //[K]
+m1_by_UA=dTlm/(Cp1*(T2-T1)) 
+//For final conditions :
+//m2_dot=m1_dot
+//U2=U1
+//A2=5*A1
+deff('x=f(t)','x=m1_by_UA*Cp1*(t-T1)-5*((dT1-(T-t))/log(dT1/(T-t)))')
+T=fsolve(350.5,f)
+printf("\nOutlet temperature of water is %f K\n",T);
diff --git a/1073/CH3/EX3.25/3_25.sce b/1073/CH3/EX3.25/3_25.sce
new file mode 100755
index 000000000..99160a199
--- /dev/null
+++ b/1073/CH3/EX3.25/3_25.sce
@@ -0,0 +1,31 @@
+clc;
+clear;
+//Example 3.25
+mo_dot=60   //Mass flow rate of oilin [g/s]
+mo_dot=6*10^-2  //[kg/s]
+Cpo=2.0     //Specific heat of oil in [kJ/(kg.K)]
+T1=420  //[K]
+T2=320  //[K]
+Q=mo_dot*Cpo*(T1-T2)    //Rate of heat flow in [kJ/s]
+mw_dot=mo_dot   //Mass flow rate of water   //kg/s
+t1=290  //[K]
+Cpw=4.18    //[kJ/(kg.K)]
+//For finding outlet temperature of water
+t2=t1+Q/(mw_dot*Cpw)    //[K]
+dT1=T1-t2   //[K]
+dT2=T2-t1   //[K]
+dTm=(dT1-dT2)/log(dT1/dT2)  //[K]
+ho=1.6  //Oil side heat transfer coefficient in [kW/(sq m.K)]
+hi=3.6  //Water side heat transfer coeff in [kW/(sq m.K)]
+//Overall heat transfer coefficient is:
+U=1/(1/ho+1/hi) //[kW/(m^2.K)]
+
+A=Q/(U*dTm) //[sq m]
+Do=25   //[mm]
+Do=Do/1000  //[m]
+L=A/(%pi*Do)    //Length of tube in [m]
+printf("\nOutlet temperature of water is %f K \n",round(t2));
+printf("Area of heat transfer required is %f sq m\n",A);
+printf("Length of tube required is %f m",L)
+
+
diff --git a/1073/CH3/EX3.26/3_26.sce b/1073/CH3/EX3.26/3_26.sce
new file mode 100755
index 000000000..195f602d0
--- /dev/null
+++ b/1073/CH3/EX3.26/3_26.sce
@@ -0,0 +1,24 @@
+
+clc;
+clear;
+//Example 3.26
+k=0.14  // for oil[W/m.K]
+Cp=2.1 // for oil [kJ/kg.K]
+Cp=Cp*10^3  //J/kg.K
+mu=154  //[mN.s/sq m]
+mu_w=87 //(mn.s/sq m)
+L=1.5   //[m]
+m_dot=0.5   //Mass flow rate of oil[kg/s]
+Di=0.019 //Diameter of tube [m]
+mean_T=319  //Mean temperature of oil [K]
+mu=mu*10^-3 //[N.s/sq m] or [kg/(m.s)]
+A=%pi*(Di/2)^2   //[sq m]
+G=m_dot/A   //Mass velocity in [kg/sq m.s]
+Nre=Di*G/mu  //Reynolds number
+//As Nre<2100,the flow is laminar
+mu_w=mu_w*10^-3 //[N.s/sq m] or kg/(m.s)
+//The sieder tate equation is 
+hi=(k*(2.0*((m_dot*Cp)/(k*L))^(1.0/3.0)*(mu/mu_w)^(0.14)))/Di   //Heat transfer coeff in [W/sq m.K]
+printf("\n The inside heat transfer coefficient is %f W/(m^2.K) ",hi);
+
+printf('\nNOTE:Calculation mistake in last line.ie in the calculation of hi in book,please perform the calculation manually to check the answer\n")
diff --git a/1073/CH3/EX3.27/3_27.sce b/1073/CH3/EX3.27/3_27.sce
new file mode 100755
index 000000000..87111e48a
--- /dev/null
+++ b/1073/CH3/EX3.27/3_27.sce
@@ -0,0 +1,25 @@
+clc;
+clear;
+//Example 3.27
+
+m_dot=0.217 //Water flow rate in [kg/s]
+Do=19   //Outside diameter in [mm]
+rho=1000    //Density
+t=1.6   //Wall thickness in [mm]
+Di=Do-2*t   //i.d of tube in [mm]
+Di=Di/1000  //[m]
+Do=Do/1000  //[m]
+Ai=%pi*(Di/2)^2 //Cross-sectional area in sq m
+u=m_dot/(rho*Ai)  //Water velocity through tube  [m/s]
+u=1.12  //approx in book
+Di=0.0157   //apprx in book
+T1=301  //Inlet temperature of water in [K]
+T2=315  //Outlet temperature of water in [K]
+T=(T1+T2)/2 //[K]
+hi=(1063*(1+0.00293*T)*(u^0.8))/(Di^0.20) //Inside heat transfer coefficient W/(sq m.K)
+hi=5084     //Approximation
+printf("%f",hi);
+hio=hi*(Di/Do)  //Inside heat transfer coeff based on outside diameter  in W/(sq m.K)
+printf("%f",hio);
+printf("Based on outside temperature,Inside heat transfer coefficient is  %d W/(m^2.K) or %f kW/(m^2.K)",round(hio),round(hio)/1000);
+
diff --git a/1073/CH3/EX3.28/3_28.sce b/1073/CH3/EX3.28/3_28.sce
new file mode 100755
index 000000000..2e0ad63ea
--- /dev/null
+++ b/1073/CH3/EX3.28/3_28.sce
@@ -0,0 +1,31 @@
+clc;
+clear;
+//Example 3.28
+mair_dot=0.90   //[kg/s]
+T1=283  //[K]
+T2=366  //[K]
+dT=(T1+T2)/2    //[K]
+Di=12   //[mm]
+Di=Di/1000  //[m]
+G=19.9  //[kg/(sq m.s)]
+mu=0.0198   //[mN.s/(sq m)]
+mu=mu*10^-3 //[N.s/sq m] or [kg/(m.s)]
+Nre=Di*G/mu //Reynolds number
+//It is greater than 10^4
+k=0.029 //W/(m.K)
+Cp=1    //[kJ/kg.K]
+Cp1=Cp*10^3  //[J/kg.K]
+Npr=Cp1*mu/k //Parndtl number
+//Dittus-Boelter equation is
+hi=0.023*(Nre^0.8)*(Npr^0.4)*k/Di   //[W/sq m.K]
+ho=232  //W/sq m.K
+U=1/(1/hi+1/ho) //Overall heat transfer coefficient  [W/m^2.K]
+Q=mair_dot*Cp*(T2-T1)   //kJ/s
+Q=Q*10^3    //[J/s] or [W]
+T=700   //[K]
+dT1=T-T2    //[K]
+dT2=T2-T1   //[K]
+dTm=(dT1-dT2)/log(dT1/dT2)  //[K]
+//Q=U*A*dTm
+A=Q/(U*dTm) //Area in sq m
+printf("Heat transfer area of equipment is %f sq m",A);
diff --git a/1073/CH3/EX3.29/3_29.sce b/1073/CH3/EX3.29/3_29.sce
new file mode 100755
index 000000000..9c0b6ebfd
--- /dev/null
+++ b/1073/CH3/EX3.29/3_29.sce
@@ -0,0 +1,27 @@
+clc;
+clear;
+//Example 3.29
+v=18.41*10^-6   //[sq m./s]
+k=28.15*10^-3   //[W/m.K]
+Npr=0.7 //Prandtl number
+Beta=3.077*10^-3    //K^-1
+g=9.81  //m/s^2
+Tw=350  //[K]
+T_inf=300   //[K]
+dT=Tw-T_inf //[K]
+L=0.3   //[m]
+//1.Free Convection
+Ngr=(g*Beta*dT*L^3)/(v^2) //Grashof number
+Npr=0.7 //Prandtl number
+
+Nnu=0.59*(Ngr*Npr)^(1.0/4.0)    //Nusselt number
+h=Nnu*k/L   //Average heat transfer coefficient [W/sq m K]
+printf("\n In free convection,heat transfer coeff,h=%f W/(sq m.K)\n",h)
+//2.Forced Convestion
+u_inf=4 //[m/s]
+Nre_l=u_inf*L/v
+Nnu=0.664*(Nre_l^(1/2))*(Npr^(1.0/3.0))     //Nusselt number
+h=Nnu*k/L   //[W/sq m.K]
+printf("\n In forced convection,heat transfer coeff,h=%f W/(sq m.K)\n",h)
+printf("\n From above it is clear that heat transfer coefficient in forced convection is much larger than that in free convection \n  ");
+
diff --git a/1073/CH3/EX3.3/3_3.sce b/1073/CH3/EX3.3/3_3.sce
new file mode 100755
index 000000000..2affc33cf
--- /dev/null
+++ b/1073/CH3/EX3.3/3_3.sce
@@ -0,0 +1,16 @@
+clc;
+clear;
+//Example 3.3
+//Given
+mu=15*10^-6    //Kinematic viscosity in [sq m /s]
+x=0.4    //[m]
+u_inf=3    //[m/s]
+//At x=0.4 m,
+Nre_x=u_inf*x/mu    ;
+printf("Since Nre,x (%f)is Less than 3*10^5,..the boundary layer is laminar",Nre_x);
+del=4.64*x/sqrt(Nre_x)    //[m]
+del=del*1000    //[mm]
+printf("\nThickness of boundary layer at x=%f m =%f mm\n",x,del);
+Cf_x=0.664/sqrt(Nre_x);
+printf("Local skin friction coefficient is :%f",Cf_x);
+
diff --git a/1073/CH3/EX3.30/3_30.sce b/1073/CH3/EX3.30/3_30.sce
new file mode 100755
index 000000000..996eeead8
--- /dev/null
+++ b/1073/CH3/EX3.30/3_30.sce
@@ -0,0 +1,24 @@
+
+clc;
+clear;
+//Example 3.30
+k=0.02685   //W/(m.K)
+v=16.5*10^-6  //kg/(m.s)
+Npr=0.7 //Prandtl number
+Beta=3.25*10^-3 //K^-1
+g=9.81  //m/(s^2)
+Tw=333; //[k]
+T_inf=283   //[K]
+dT=Tw-T_inf  //[K]
+L=4 //Length/height  of plate [m]
+Ngr=(g*Beta*dT*(L^3))/(v^2)   //Grashoff number
+//Let const=Ngr*Npr
+const=Ngr*Npr
+//Sice it is >10^9
+Nnu=0.10*(const^(1.0/3.0))   //Nusselt number
+h=Nnu*k/L   //W/(sq m.K)
+h=4.3       //Approx in book
+W=7 //width in [m]
+A=L*W   //Area of heat transfer in [sq m]
+Q=h*A*dT    //[W]
+printf("\nHeat transferred is %d W\n",Q)
diff --git a/1073/CH3/EX3.31/3_31.sce b/1073/CH3/EX3.31/3_31.sce
new file mode 100755
index 000000000..ab18e5eef
--- /dev/null
+++ b/1073/CH3/EX3.31/3_31.sce
@@ -0,0 +1,21 @@
+clc;
+clear;
+//Example 3.31
+v=18.97*10^-6   //m^2/s
+k=28.96*10^-3   //W/(m.K)
+Npr=0.696
+D=100   //Outer diameter [mm]
+D=D/1000    //[m]
+Tf=333  //Film temperature in [K]
+Tw=373  //[K]
+T_inf=293   //[K]
+dT=Tw-T_inf //[K]
+Beta=1/Tf   //[K^-1]
+g=9.81  //[m/s^2]
+L=3 //Length of pipe [m]
+Ngr=(g*Beta*dT*(L^3))/(v^2) //Grashof number
+Nra=Ngr*Npr
+Nnu=0.10*(Ngr*Npr)^(1.0/3.0)    //nusselt number for vertical cylinder
+h=Nnu*k/L   //W/(sq m.K)
+Q_by_l=h*%pi*D*dT   //Heat loss per metre length [W/m]
+printf("\n Hence,Heat loss per metre length is %f W/m \n",Q_by_l);
diff --git a/1073/CH3/EX3.32/3_32.sce b/1073/CH3/EX3.32/3_32.sce
new file mode 100755
index 000000000..6b2bc2fad
--- /dev/null
+++ b/1073/CH3/EX3.32/3_32.sce
@@ -0,0 +1,23 @@
+clc;
+clear;
+//Example 3.32
+k=0.630 //W/(m.K
+Beta=3.04*10^-4 //K^-1
+rho=1000    //kg/m^3
+mu=8.0*10^-4    //[kg/(m.s)]
+Cp=4.187    //kJ/(kg.K)
+g=9.81  //[m/(s^2)]
+Tw=313  //[K]
+T_inf=298   //[K]
+dT=Tw-T_inf //[K]
+D=20    //[mm]
+D=D/1000    //[m]
+Ngr=9.81*(rho^2)*Beta*dT*(D^3)/(mu^2)   //Grashoff number
+Cp1=Cp*1000 //[J/kg.K]
+Npr=Cp1*mu/k //Prandtl number
+//Average nusselt number is
+Nnu=0.53*(Ngr*Npr)^(1.0/4.0)
+h=Nnu*k/D  //[W/ sqm.K]
+Q_by_l=h*%pi*D*dT   //Heat loss per unit length [W/m]
+printf("\nHeat loss per unit length of the heater is %f W/m",Q_by_l);
+
diff --git a/1073/CH3/EX3.33/3_33.sce b/1073/CH3/EX3.33/3_33.sce
new file mode 100755
index 000000000..9735551d9
--- /dev/null
+++ b/1073/CH3/EX3.33/3_33.sce
@@ -0,0 +1,19 @@
+clc;
+clear;
+//Example 3.33
+k=0.03406   //[W/(m/K)]
+Beta=2.47*10^-3 //K^-1
+Npr=0.687   //Prandtl number
+v=26.54*10^-6   //m^2/s
+g=9.81  //[m/s^2]
+Tw=523  //[K]
+T_inf=288   //[K]
+dT=Tw-T_inf //[K]
+D=0.3048    //[m]
+Ngr=(g*Beta*dT*(D^3))/(v^2)     //Grashof number
+Nra=Ngr*Npr 
+//For Nra less than 10^9,we have for horizontal cylinder
+Nnu=0.53*(Nra^(1.0/4.0))    //Nusselt number
+h=Nnu*k/D   //[W/sq m.K]
+Q_by_l=h*%pi*D*dT;  //W/m
+printf("Heat loss of heat transfer per meter lengh is %f W/m",Q_by_l);
diff --git a/1073/CH3/EX3.34/3_34.sce b/1073/CH3/EX3.34/3_34.sce
new file mode 100755
index 000000000..0ad13de11
--- /dev/null
+++ b/1073/CH3/EX3.34/3_34.sce
@@ -0,0 +1,31 @@
+
+clc;
+clear;
+//Example 3.34
+rho=960.63  //Density in [kg/m^3]
+Cp=4.216*10^3   //Specific heat in [J/(kg.K)]
+D=16    //Diameter in [cm]
+D=D/100 //[m]
+k=0.68  //Thermal conductivity in [W/m.K]
+A=(%pi*(D/2)^2)
+L=A/(%pi*D)   //Length=A/P  in [m]
+Beta=0.75*10^-3  //[K^-1]
+alpha=1.68*10^-7    //[m^2/s]
+g=9.81  //[m/s^2]
+Tw=403  //[K]
+T_inf=343   //[K]
+dT=Tw-T_inf //[K]
+v=0.294*10^-6   //[m^2/s]
+Nra=(g*Beta*(L^3)*dT)/(v*alpha) 
+
+//1.For Top surface
+Nnu=0.15*(Nra)^(1.0/3.0)   //Nusselt number
+ht=Nnu*k/L  //Heat transfer coeff for top surface in W/(m^2.K)
+ht=round(ht)
+//2.For bottom surface
+Nnu=0.27*Nra^(1.0/4.0)  //Nusselt number
+hb=Nnu*k/L  //[W/sq m.K]
+hb=round(hb)
+Q=(ht+hb)*A*dT; //[W]
+printf("The rate of heat input is %f W",Q)
+
diff --git a/1073/CH3/EX3.35/3_35.sce b/1073/CH3/EX3.35/3_35.sce
new file mode 100755
index 000000000..a6bb9444b
--- /dev/null
+++ b/1073/CH3/EX3.35/3_35.sce
@@ -0,0 +1,29 @@
+clc;
+clear;
+//Example 3.35
+v=2*10^-5   //[m^2/s]
+Npr=0.7 //Prandtl number
+k=0.03  //[W/m.K]
+D=0.25   //Diameter in [m]
+L=0.90*D    //Characteristic length,let  [m]
+T1=298  //[K]
+T2=403  //[K]
+dT=T2-T1    //[K]
+Tf=(T1+T2)/2  //[K]
+Beta=1/Tf   //[K^-1]
+A=%pi*(D/2)^2   //Area in[sq m]
+g=9.81  //[m/s^2]
+
+//Case 1: Hot surface facing up
+Ngr=g*Beta*dT*(L^3)/(v^2)   //Grashoff number
+Nnu=0.15*((Ngr*Npr)^(1.0/3.0))    //Nusselt number
+h=Nnu*k/L   //[W/sq m.K]
+Q=h*A*dT    //[W]
+printf("\n Heat transferred when hot surface is facing up is %f W\n",Q);
+
+
+//Case 2:For hot surface facing down
+Nnu=0.27*(Ngr*Npr)^(1.0/4.0);   //Grashof Number
+h=Nnu*k/L   //[W/sqm.K]
+Q=h*A*dT    //[W]
+printf("\n Heat transferred when hot surface is facing down is %f W\n",Q);
diff --git a/1073/CH3/EX3.36/3_36.sce b/1073/CH3/EX3.36/3_36.sce
new file mode 100755
index 000000000..9036ece1c
--- /dev/null
+++ b/1073/CH3/EX3.36/3_36.sce
@@ -0,0 +1,33 @@
+
+
+clc;
+clear;
+//Example 3.36
+rho=960 //[kg/m^3]
+Beta=0.75*10^-3 //[K^-1]
+k=0.68  //[W/m.K]
+alpha=1.68*10^-7    //[m^2/s]
+v=2.94*10^-7    //[m^2/s]
+Cp=4.216    //[kJ/kg.K]
+Tw=403  //[K]
+T_inf=343   //[K]
+dT=Tw-T_inf     //[K]
+g=9.81  //[m/s^2]
+l=0.8   //[m]
+W=0.08  //[m]
+A=l*W   //Area in [m^2]
+P=2*(0.8+0.08)   //Perimeter in [m]
+L=A/P   //Characteristic dimension/length,L in   [m]
+Nra=g*Beta*L^3*dT/(v*alpha) 
+
+//(i) for natural convection,heat transfer from top/upper surface heated 
+Nnu=0.15*(Nra^(1.0/3.0))    //Nusselt number
+ht=Nnu*k/L  //[W/m^2.K]
+ht=2115.3   //Approximation in book,If done manually then answer diff
+//(ii)For the bottom/lower surface of the heated plate
+Nnu=0.27*(Nra^(1.0/4.0))    //Nusselt number
+hb=Nnu*k/L  //[W/(m^2.K)]
+hb=round(hb)
+//Rate of heat input is equal to rate of heat dissipation from the upper and lower surfaces of the plate
+Q=(ht+hb)*A*(Tw-T_inf)   //[W]
+printf("\n Rate of heat input is equal to heat dissipation =%f W",Q);
diff --git a/1073/CH3/EX3.37/3_37.sce b/1073/CH3/EX3.37/3_37.sce
new file mode 100755
index 000000000..23d855086
--- /dev/null
+++ b/1073/CH3/EX3.37/3_37.sce
@@ -0,0 +1,45 @@
+clc;
+clear;
+//Example 3.37
+k=0.03  //W/(m.K)
+Npr=0.697   //Prandtl number
+v=2.076*10^-6   //m^2/s
+Beta=0.002915   //K^-1
+D=25 ;    //[Diameter in cm]
+D=D/100 //[m]
+Tf=343  //Film temperature in [K]
+A=%pi*(D/2)^2   //Area in [m^2]
+P=%pi*D //Perimeter [m]
+T1=293  //[K]
+T2=393  //[K]
+g=9.81  //[m/s^2]
+
+//Case (i) HOT SURFACE FACING UPWARD
+L=A/P   //Characteristic length in [m]
+Beta=1/Tf;  //[K^-1]
+dT=T2-T1    //[K]
+Ngr=(g*Beta*dT*(L^3))/(v^2) //Grashoff number
+Nra=Ngr*Npr 
+Nnu=0.15*(Nra^(1.0/3.0))    //Nusselt number
+h=Nnu*k/L   //[W/m^2.K]
+Q=h*A*dT    //[W]
+printf("\nHeat transferred when disc is horizontal with hot surface facing upward is %f W\n",Q);
+
+//Case-(ii) HOT FACE FACING DOWNWARD
+Nnu=0.27*(Nra^(1/4))    //Nusselt number
+h=Nnu*k/L   //W/(m^2.K)
+Q=h*A*dT    //[W]
+printf("\nHeat transferred when disc is horizontal with hot surface facing downward is %f W\n",Q);
+
+
+//Case-(iii)-For disc vertical 
+L=0.25  //Characteristic length[m]  
+D=L //dia[m]
+A=%pi*((D/2)^2) //[sq m]
+Ngr=(g*Beta*dT*(L^3))/(v^2)     //Grashoff number
+Npr=0.697
+Nra=Ngr*Npr 
+Nnu=0.10*(Nra^(1/3))    //Nusselt number
+h=Nnu*k/D   //[W/(m^2.K)]
+Q=h*A*dT    //[W]
+printf("For vertical disc,heat transferred is %f W",Q);
diff --git a/1073/CH3/EX3.38/3_38.sce b/1073/CH3/EX3.38/3_38.sce
new file mode 100755
index 000000000..b660a3be6
--- /dev/null
+++ b/1073/CH3/EX3.38/3_38.sce
@@ -0,0 +1,24 @@
+clc;
+clear;
+//Example 3.38
+v=23.13*10^-6   ; //[m^2/s]
+k=0.0321   ; //[W/m.K]
+Beta=2.68*10^-3; //[K^-1]
+Tw=443  ;//[K]
+T_inf=303  ; //[K]
+dT=Tw-T_inf;    //[K]
+g=9.81 ; //[m/s^2]
+Npr=0.688;   //Prandtl number
+D=100 ;  //Diameter [mm]
+D=D/1000    //Diameter [m]
+Nra=(g*Beta*dT*(D^3)*Npr)/(v^2)
+Nnu=0.53*(Nra^(1.0/4.0))    //Nusselt number
+h=Nnu*k/D   //[W/(m^2.K)]
+h=7.93      //Approximation
+e=0.90; //Emissivity
+sigma=5.67*10^-8    ;
+//Q=Q_conv+Q_rad  //Total heat loss
+//for total heat  loss per meter length
+Q_by_l=h*%pi*D*dT+sigma*e*%pi*D*(Tw^4-T_inf^4)  //[W/m]
+printf("Total heat loss per metre length of pipe is %f W/m",Q_by_l)
+
diff --git a/1073/CH3/EX3.39/3_39.sce b/1073/CH3/EX3.39/3_39.sce
new file mode 100755
index 000000000..972a7ad88
--- /dev/null
+++ b/1073/CH3/EX3.39/3_39.sce
@@ -0,0 +1,19 @@
+clc;
+clear;
+//Example 3.39
+k=0.035; //[W/(m.K)]
+Npr=0.684   ;//Prandtl number
+Beta=2.42*10^-3; //[K^-1]
+v=27.8*10^-6;    //[m^2/s]
+Tw=533;  //[K]
+T_inf=363  ;    //[K]
+dT=Tw-T_inf //[K]
+D=0.01  ;//[m]
+g=9.81; //[m/s^2]
+Nra=(g*Beta*dT*(D^3))/(v^2)
+//For this <10^5,we have for sphere
+A=4*%pi*(D/2)^2 //Area of sphere in [m^2]
+Nnu=(2+0.43*Nra^(1.0/4.0))//Nusslet number
+h=Nnu*k/D   //W/(m^2.K)
+Q=h*A*dT    //[W]
+printf("\nRate of heat loss is %f W",Q)
diff --git a/1073/CH3/EX3.4/3_4.sce b/1073/CH3/EX3.4/3_4.sce
new file mode 100755
index 000000000..ab0d4cdb9
--- /dev/null
+++ b/1073/CH3/EX3.4/3_4.sce
@@ -0,0 +1,32 @@
+clc;
+clear;
+//Example 3.4
+mu=1.85*10^-5           //[kg/(m.s)]
+P=101.325;               //Pressure in [kPa]
+M_avg=29;                //Avg molecular wt of air
+R=8.31451;               //Gas constant
+T=300;                   //[K]
+rho=P*M_avg/(R*T)   //[kg/m^3]
+u_inf=2             //Viscosity in [m/s]
+//At x=20 cm =0.2 m
+x=0.2;                   //[m]
+Nre_x=rho*u_inf*x/mu     //[Reynolds number]
+del_by_x=4.64/sqrt(Nre_x)   //[Boundary layer]
+del=del_by_x*x              //[m]
+//del=del*1000                //[mm]
+
+//At
+x=0.4 ;         //[m]
+Nre_x=(rho*u_inf*x)/mu      //<3*10^5
+//Boundary layer is laminar
+del_by_x=4.64/sqrt(Nre_x)   
+del1=del_by_x*x              //[m]
+//del1=del1*1000                //[mm]
+d=del1-del                      //Del 
+function m_dot=f(y),m_dot=u_inf*(1.5*(y/d)-0.5*(y/d)^3)*rho,endfunction
+m_dot=intg(0,d,f)
+printf("\nBoundary layer thickness at distance 20 cm from leading edge is %f m=%f mm\n",del,del*1000);
+printf("\nBoundary layer thickness at distance 40 cm from leading edge is %f m=%f mm\n",del1,del1*1000);
+printf("\nThus,Mass flow rate entering the boundary layer is %f kg/s",m_dot);
+
+
diff --git a/1073/CH3/EX3.40/3_40.sce b/1073/CH3/EX3.40/3_40.sce
new file mode 100755
index 000000000..81064552e
--- /dev/null
+++ b/1073/CH3/EX3.40/3_40.sce
@@ -0,0 +1,34 @@
+
+clc;
+clear;
+//Exampe 3.40
+v=17.95*10^-6   //[m^2/s]
+dT=353-293  //[K]
+k=0.0283    //[W/m.K]
+g=9.81  //[m/s^2]
+Npr=0.698   //Prandtl number
+Cp=1005 //J/(kg.K)
+Tf=323  //Film temperature in [K]
+Beta=1/Tf   //[K^-1]
+l=1 //[m]
+Nra=(g*Beta*dT*(l^3)*Npr)/(v^2)
+
+//In textbook result of above statement is wrongly calculated,So
+Nra=3.95*10^8
+//For Nra <10^9,for a vertical plate,the average nusselt number is
+Nnu=0.59*Nra^(1.0/4.0)  //Nusselt number
+h=Nnu*k/l   //[W/m^2.K]
+h=2.35      //Approx in book
+A=l^2   //Area [m^2]
+//Heat loss form 4 vertical faces of 1m*1m is 
+Q1=4*(h*A*dT)   //[W]
+//For top surface 
+P=4*l   //Perimeter in [m]
+L=A/P   //[m]
+Nra=(Npr*g*Beta*dT*(L^3))/(v^2)
+Nnu=0.15*Nra^(1.0/3.0)  //Nusselt number
+h=Nnu*k/L   //[W/m^2.K]
+h=6.7           //Approx
+Q2=h*A*dT   //[W]
+Q_total=Q1+Q2   //Total heat loss[W]
+printf("\n Therefore total heat loss is %d W",Q_total);
diff --git a/1073/CH3/EX3.41/3_41.sce b/1073/CH3/EX3.41/3_41.sce
new file mode 100755
index 000000000..b003c4cf2
--- /dev/null
+++ b/1073/CH3/EX3.41/3_41.sce
@@ -0,0 +1,31 @@
+clc;
+clear;
+//Example 3.41
+rho=0.910;           //Density in [kg/m^3]
+Cp=1.009*1000;            //[J/kg.K]
+k=0.0331;            //[W/m.K]
+mu=22.65*10^-6;      //[N.s/m^2]
+//Let a=smaller side
+//b=bigger side
+//Qa=ha*A*dT
+//Qb=hb*A*dT
+//Qa=1.14*Qb
+//Given a*b=15*10^-4
+//On solving we get:
+a=0.03;          //[m]
+b=0.05;          //[m]
+A=a*b           //Area in [sq m]
+Tf=388;          //[K]
+Beta=1/Tf       //[K^-1]
+T1=303;          //[K]
+T2=473;         //[K]
+dT=T2-T1        //[K]
+v=mu/rho        
+g=9.81      //m/s^2[acceleration due to gravity ]
+hb=0.59*(((g*Beta*dT*(b^3))/(v^2))*Cp*mu/k)^(1/4)*(k/b)       //[W/sq m.K]
+Qb=hb*A*(dT)            //[W]
+
+Qa=1.14*Qb              //[W]
+printf("\nDimensions of the plate are %fx%f m\n",a,b);
+printf("\nHeat transfer when the bigger side held vertical is %f W\n",Qb);
+printf("\nHeat transfer when the small side held vertical is %f W\n",Qa);
diff --git a/1073/CH3/EX3.42/3_42.sce b/1073/CH3/EX3.42/3_42.sce
new file mode 100755
index 000000000..f1f30d295
--- /dev/null
+++ b/1073/CH3/EX3.42/3_42.sce
@@ -0,0 +1,25 @@
+clc;
+clear;
+//Example 3.42
+Ts=373  //[K]
+rho_l=957.9    //rho*l[kg/m^3]
+Cpl=4217    //[J/kg.K]
+mu_l=27.9*10^-5 //[kg/(m.s)]
+rho_v=0.5955   //[kg/m^3]
+Csf=0.013
+sigma=5.89*10^-2    //[N/m]
+Nprl=1.76
+lambda=2257 //[kJ/kg]
+lambda=lambda*1000  //in [J/kg]
+n=1 //for water
+m_dot=30    //Mass flow rate [kg/h]
+m_dot=m_dot/3600    //[kg/s]
+D=30    //Diameter of pan [cm]
+D=D/100 //[m]
+g=9.81  //[m/s^2]
+A=%pi*(D/2)^2   //Area in [sq m]
+Q_by_A=m_dot*lambda/A   //[W/sq m]
+//For nucleate boiling point we have:
+dT=(lambda/Cpl)*Csf*(((Q_by_A)/(mu_l*lambda))*sqrt(sigma/(g*(rho_l-rho_v))))^(1.0/3.0)*(Nprl^n) //[K]
+Tw=Ts+dT    //[K]
+printf("\n Temperature of the bottom surface of the pan is %f W/(sq m)",Tw);
diff --git a/1073/CH3/EX3.43/3_43.sce b/1073/CH3/EX3.43/3_43.sce
new file mode 100755
index 000000000..f28a6d143
--- /dev/null
+++ b/1073/CH3/EX3.43/3_43.sce
@@ -0,0 +1,13 @@
+clc;
+clear;
+//Example 3.4
+lambda=2257 //[kJ/kg]
+lambda=lambda*1000  //in [J/kg]
+rho_l=957.9    //rho*l[kg/m^3]
+rho_v=0.5955   //[kg/m^3]
+sigma=5.89*10^-2    //[N/m]
+g=9.81  //[m/s^2]
+//Peak heat flux is given by
+Q_by_A_max=(%pi/24)*(lambda*rho_v^0.5*(sigma*g*(rho_l-rho_v))^(1/4))    //W/m^2
+Q_by_A_max=Q_by_A_max/(10^6)    //MW/(sq m)
+printf("\n Peak heat flux is %f MW/sq m",Q_by_A_max);
diff --git a/1073/CH3/EX3.44/3_44.sce b/1073/CH3/EX3.44/3_44.sce
new file mode 100755
index 000000000..b4036adc6
--- /dev/null
+++ b/1073/CH3/EX3.44/3_44.sce
@@ -0,0 +1,26 @@
+clc;
+clear;
+//Example 3.44
+rho_l=957.9 //[kg/m^3]
+lambda=2257 //[kJ/kg]
+lambda=lambda*10^3  //[J/kg]
+rho_v=31.54 //[kg/m^3]
+Cpv=4.64    //[kJ/kg.K]
+Cpv=Cpv*10^3    //[J/kg.K]
+kv=58.3*10^-3//[W/(m.K)]
+g=9.81  //[m/s^2]
+mu_v=18.6*10^-6 //[kg/(m.s)]
+e=1.0   //Emissivity
+sigma=5.67*10^-8;
+Ts=373  //[K]
+Tw=628  //[K]
+dT=Tw-Ts    //[K]
+D=1.6*10^-3 //[m]
+T=(Tw+Ts)/2 //[K]
+hc=0.62*((kv^3)*rho_v*(rho_l-rho_v)*g*(lambda+0.40*Cpv*dT)/(D*mu_v*dT))^(1.0/4.0)//Convective heat transfer coeff  [W/sq m.K]
+hr=e*sigma*(Tw^4-Ts^4)/(Tw-Ts)  //Radiation heat transfer coeff in [W/sq m.K]
+h=hc+(3/4)*hr   //Total heat transfer coefficient W/(sq m.K)
+Q_by_l=h*%pi*D*dT   //Heat dissipation rate per unit length in [kW/m]
+printf("\n Stable film boiling point heat transfer coefficient is %f W/(sq m.K)",h);
+Q_by_l=Q_by_l/1000  //[kW/m]
+printf("\n Heat dissipated per unit length of the heater is %f kW/m",Q_by_l);
diff --git a/1073/CH3/EX3.45/3_45.sce b/1073/CH3/EX3.45/3_45.sce
new file mode 100755
index 000000000..03362b840
--- /dev/null
+++ b/1073/CH3/EX3.45/3_45.sce
@@ -0,0 +1,13 @@
+clc;
+clear;
+//Exmaple 3.45
+dT=10   //[K]
+P=506.625   //[kPa]
+P=P/10^3    //[Mpa]
+D=25.4  //Diameter [mm]
+D=D/1000    //[m]
+h=2.54*(dT^3)*(%e^(P/1.551))   //[W/sq m.K]
+//Q=h*%pi*D*L*dT
+//Heat transfer rate per meter length of tube is 
+Q_by_l=h*%pi*D*dT   //[W/m]
+printf("\n Rate of heat transfer per 1m length of tube is %f W/m",round(Q_by_l));
diff --git a/1073/CH3/EX3.46/3_46.sce b/1073/CH3/EX3.46/3_46.sce
new file mode 100755
index 000000000..e261fd5ca
--- /dev/null
+++ b/1073/CH3/EX3.46/3_46.sce
@@ -0,0 +1,16 @@
+clc;
+clear;
+//Example 3.46
+dT=8    //[K]
+P=0.17  //[Mpa]
+P=P*1000    //[kPa]
+h1=2847 //[W/(sq m.K)]
+P1=101.325  //[kPa]
+h=5.56*(dT^3)   //[W/sq m.K]
+Q_by_A=h*dT //[W/sq m]
+hp=h*(P/P1)^(0.4)   //[W/sq m.K]
+//Correponding heat flux is :
+Q_by_A1=hp*dT //[W/sq m]
+per=(Q_by_A1-Q_by_A)*100/Q_by_A //Percent increase in heat flux
+printf("\nHeat flux when pressure  is 101.325 kPa is %f W/sq m\n",Q_by_A);
+printf("\n Percent increase in heat flux is %f  percent",round(per));
diff --git a/1073/CH3/EX3.47/3_47.sce b/1073/CH3/EX3.47/3_47.sce
new file mode 100755
index 000000000..001d8706b
--- /dev/null
+++ b/1073/CH3/EX3.47/3_47.sce
@@ -0,0 +1,22 @@
+clc;
+clear;
+//Example 3.47
+mu=306*10^-6    //[N.s/m^2]
+k=0.668 //[W/m.K]
+rho=974 //[kg/m^3]
+lambda=2225 //[kJ/kg]
+lambda=lambda*10^3  //[J/kg.K]
+g=9.81  //[m/s^2]
+Ts=373  //[K]
+Tw=357  //[K]
+dT=Ts-Tw    //[K]
+Do=25   //[mm]
+Do=Do/1000  //[m]
+h=0.725*((rho^2*g*lambda*k^3)/(mu*Do*dT))^(1.0/4.0) //[W/sq m.K]
+Q_by_l=h*%pi*Do*dT  //[W/m]
+m_dot_byl=(Q_by_l/lambda)   //[kg/s]
+m_dot_byl=m_dot_byl*3600    //[kg/h]
+
+printf("\nMean heat transfer coefficient is %f W/(sq m.K)\n",h);
+printf("\nHeat transfer per unit length is %f W/m\n",Q_by_l);
+printf("\nCondensate rate per unit length is %f kg/h",m_dot_byl);
diff --git a/1073/CH3/EX3.48/3_48.sce b/1073/CH3/EX3.48/3_48.sce
new file mode 100755
index 000000000..28c8b0738
--- /dev/null
+++ b/1073/CH3/EX3.48/3_48.sce
@@ -0,0 +1,21 @@
+clc;
+clear;
+//Example 3.48
+rho=960 //[kh/m^3]
+mu=2.82*10^-4   //[kg/(m.s)]
+k=0.68  //[W/(m.K)]
+lambda=2255 //[kJ/kg]
+lambda=lambda*10^3  //[J/kg]
+Ts=373  //Saturation temperature of steam [K]
+Tw=371  //[K]
+dT=Ts-Tw    //[K]
+L=0.3   //Dimension [m]
+g=9.81  //[m/s^2]
+h=0.943*(rho^2*g*lambda*k^3/(L*mu*dT))^(1/4)    //W/sq m.K
+A=L^2   //[sq m] 
+Q=h*A*(Ts-Tw)   //[W]=[J/s]
+m_dot=Q/lambda  //Condensate rate[kg/s]
+m_dot=m_dot*3600    //[kg/h]
+printf("\n Average heat transfer coefficient is %f W/(sq m.K)\n",h);
+printf("\nHeat transfer rate is %f J/kg\n",Q);
+printf("\n Steam condensate rate per hour is %f kg/h\n",m_dot);
diff --git a/1073/CH3/EX3.49/3_49.sce b/1073/CH3/EX3.49/3_49.sce
new file mode 100755
index 000000000..5f5094744
--- /dev/null
+++ b/1073/CH3/EX3.49/3_49.sce
@@ -0,0 +1,25 @@
+
+clc;
+clear;
+//EXample 3.49
+rho=1174    //[kg/m^3]
+k=0.069 //[W/(m.K)]
+mu=2.5*10^-4    //[N.s/m^2]
+lambda=132*10^3 //[J/kg]
+g=9.81  //[m/s^2]
+Ts=323  //[K]
+Tw=313  //[K]
+dT=Ts-Tw    //[K]
+//For square array,n=4
+n=4 //number of tubes
+Do=12   //[mm]
+Do=Do/1000  //[m]
+h=0.725*(rho^2*lambda*g*k^3/(n*Do*mu*dT))^(1/4) //W/(sq m.K) 
+//For heat transfer area calcualtion,n=16
+A=n*%pi*Do //[sq m]
+A=0.603
+Q=h*A*dT//[W/m]
+m_dot=Q/lambda  //[kg/s]
+m_dot=0.049     //Appriximation in book
+m_dot=m_dot*3600    //[kg/h]
+printf("\n Rate of condensation per unit length is %f kg/h",m_dot);
diff --git a/1073/CH3/EX3.5/3_5.sce b/1073/CH3/EX3.5/3_5.sce
new file mode 100755
index 000000000..554a2e09a
--- /dev/null
+++ b/1073/CH3/EX3.5/3_5.sce
@@ -0,0 +1,26 @@
+
+clc;
+clear;
+//Example 3.5
+//Given
+mu=3.9*10^-4    //Kinematic viscosity in sq m/s
+k=36.4*10^-3    //Thermal conductivity in W/(m.K)
+Npr=0.69
+u_inf=8    //[m/s]
+L=1    //Lenght of plate in [m]
+Nre_l=u_inf*L/mu 
+//Since Nre_l is less than 3*10^5 ,the flow is laminar over the entire length of plate
+Nnu=0.664*sqrt(Nre_l)*Npr^(1.0/3.0)    //=hL/k
+
+h=k*Nnu/L    //w/sq m.K
+h=3.06      //Approximation   [W/sq m.K]
+T_inf=523    //[K]
+Tw=351    //[K]
+W=0.3    //Width of plate [m]
+A=W*L    //Area in [sq m]
+Q=h*A*(T_inf-Tw)    // Rate of heat removal from one side in [W]
+printf("\nRate of heat removal is %f W\n",Q)
+//from two side:
+Q=2*Q    //[W]
+printf("\n %f W heat should be removed continously from the plate",Q);
+
diff --git a/1073/CH3/EX3.50/3_50.sce b/1073/CH3/EX3.50/3_50.sce
new file mode 100755
index 000000000..68598ddbd
--- /dev/null
+++ b/1073/CH3/EX3.50/3_50.sce
@@ -0,0 +1,28 @@
+
+clc;
+clear;
+//Example 3.50
+rho=960 //[kg/m^3]
+k=0.68  //[W/m.K]
+mu=282*10^-6    //[kg/(m.s)]
+Tw=371  //Tube wall temperature [K]
+Ts=373  //Saturation temperature in [K]
+dT=Ts-Tw    //[K]
+lambda=2256.9   //[kJ/kg]
+lambda=lambda*10^3  //[J/kg]
+//Fora square array with 100tubes,n=10
+Do=0.0125 //[m]
+g=9.81  //[m/s^2]
+n=10
+h=0.725*(((rho^2)*g*lambda*(k^3)/(mu*n*Do*dT))^(1.0/4.0)) //W/(sq m.K)
+
+L=1 //[m]
+//n=100
+n=100;
+A=n*%pi*Do*L    //[m^2/m length]
+Q=h*A*dT    //Heat transfer rate in [W/m]
+ms_dot=Q/lambda //[kg/s]
+ms_dot=ms_dot*3600  //[kg/h]
+printf("\n Mass rate of steam condensation is %d kg/h\n",round(ms_dot));
+
+printf("\n NOTE:ERROR in Solution in book.Do is wrongly taken as 0.012 in lines 17 and 22 of the book,Also A is wrongly calculated\n")
diff --git a/1073/CH3/EX3.51/3_51.sce b/1073/CH3/EX3.51/3_51.sce
new file mode 100755
index 000000000..a7ecf46f4
--- /dev/null
+++ b/1073/CH3/EX3.51/3_51.sce
@@ -0,0 +1,28 @@
+
+clc;
+clear;
+//Example 3.51
+rho=975 //[kg/m^3]
+k=0.871 //[W/m.K]
+dT=10   //[K]
+mu=380.5*10^-6  //[N.s/m^2]
+lambda=2300 //[kJ/kg]
+lambda=lambda*1000  // Latent heat of condensation [J/kg]
+Do=100  //Outer diameter [mm]
+Do=Do/1000  //[m]
+g=9.81  //[m/s^2]
+//for horizontal tube
+h1=0.725*((rho^2*lambda*g*k^3)/(mu*Do*dT))^(1/4)    //Average heat transfer coefficient
+//for vertical tube
+//h2=0.943*((rho^2*lambda*g*k^3)/(mu*L*dT))^(1/4)    //Average heat transfer coefficient
+h2=h1   //For vertical tube
+//implies that
+L=(0.943*((rho^2*lambda*g*k^3)^(1/4))/(h1*((mu*dT)^(1/4))))^4   //[m]
+L=0.29      //Approximate in book
+h=0.943*((rho^2*lambda*g*k^3)/(mu*L*dT))^(1/4)  //[W/(sq m.K)]
+A=%pi*Do*L  //Area in [m^2]
+Q=h*A*dT    //Heat transfer rate [W]
+mc_dot=Q/lambda  //[Rate of condensation]in [kg/s]
+mc_dot=mc_dot*3600  //[kg/h]
+printf("\n Tube length is %f m\n",L);
+printf("\n Rate of condemsation per hour is %f kg/h",mc_dot);
diff --git a/1073/CH3/EX3.52/3_52.sce b/1073/CH3/EX3.52/3_52.sce
new file mode 100755
index 000000000..17afa555d
--- /dev/null
+++ b/1073/CH3/EX3.52/3_52.sce
@@ -0,0 +1,18 @@
+clc;
+clear;
+//Example 3.52
+m1_dot=50    // For horizontal position[kg/h]
+Do=10   //[mm]
+Do=Do/1000  //[m]
+L=1  //[m]
+//For 100 tubes n=10
+n=10;
+//We know that
+//m_dot=Q/lambda=h*A*dT/lambda
+//m_dot is proportional to  h
+//m1_dot prop to h1
+//m2_dot  propn to h2
+//m1_dot/m2_dot=h1/h2
+//or :
+m2_dot=m1_dot/((0.725/0.943)*(L/(n*Do))^(1/4))  //[kg/h]
+printf("\n For vertical position,Rate of condensationis %f kg/h",m2_dot);
diff --git a/1073/CH3/EX3.53/3_53.sce b/1073/CH3/EX3.53/3_53.sce
new file mode 100755
index 000000000..48e23b7e9
--- /dev/null
+++ b/1073/CH3/EX3.53/3_53.sce
@@ -0,0 +1,22 @@
+clc;
+clear;
+rho=975 //[kg/m^3]
+k=0.671 //[W/(m.K)]
+mu=3.8*10^-4    //[N.s/m^2]
+dT=10   //[K]
+lambda=2300*10^3    //[J/kg]
+L=1 //[m]
+g=9.81  //[m/s^2]
+h=0.943*((rho^2*lambda*g*k^3)/(mu*L*dT))^(1/4)  //W/(sq m.K)    //[W/sq m.K]
+
+printf("\n (i)- Average heat transfer coefficient is %d W/(m^2.K)\n",round(h));
+
+//Local heat transfer coefficient
+//at x=0.5  //[m]
+x=0.5   //[m]
+h=((rho^2*lambda*g*k^3)/(4*mu*dT*x))^(1/4)  //[W/sq m.K]
+printf("\n (ii)-Local heat transfer coefficient at 0.5 m height is %d W/(sq m.K)\n",round(h));
+delta=((4*mu*dT*k*x)/(lambda*rho^2*g))^(1/4)    //[m]
+delta=delta*10^3    //[mm]
+printf("\n (iii)-Film thickness is %f mm",delta);
+
diff --git a/1073/CH3/EX3.6/3_6.sce b/1073/CH3/EX3.6/3_6.sce
new file mode 100755
index 000000000..be2317b77
--- /dev/null
+++ b/1073/CH3/EX3.6/3_6.sce
@@ -0,0 +1,26 @@
+
+clc;
+clear;
+//Example 3.6
+P1=101.325    //Pressure in [kPa]
+mu1=30.8*10^-6    //Kinematic viscosity in[sq m /s]
+k=36.4*10^-3    //[W/(m.K)]
+Npr=0.69
+u_inf=8    //Velocity in [m/s]
+Cp=1.08    //kJ/(kg.K)
+L=1.5    //Length of plate in [m]
+W=0.3    //Width in [m]
+A=L*W    //Area in [sq m]
+//At constant temperature: mu1/mu2=P2/P1
+P2=8     //[kPa]
+mu2=mu1*P1/P2    //Kinematic viscosity at P2 in [sq m/s]
+Nre_l=u_inf*L/mu2    //Reynold's no.
+//Since this is less than 3*10^5
+Nnu=0.664*sqrt(Nre_l)*(Npr^(1.0/3.0))
+h=Nnu*k/L  // Heat transfer coeffficient in [W/sq m.K]
+h=2.5       //Approximation in [W/sq m.K]
+T_inf=523    //[K]
+Tw=353    //[K]
+Q=2*h*A*(T_inf-Tw)    //Heat removed from both sides in [W]
+printf("Rate of heat removed from both sides of plate is %f W",Q);    
+
diff --git a/1073/CH3/EX3.7/3_7.sce b/1073/CH3/EX3.7/3_7.sce
new file mode 100755
index 000000000..b56e80d9d
--- /dev/null
+++ b/1073/CH3/EX3.7/3_7.sce
@@ -0,0 +1,32 @@
+clc;
+clear;
+//Example 3.7
+rho=0.998    //kg/cubic m
+v=20.76*10^-6    //[sq m/s]
+Cp=1.009    //[kJ/kg.K]
+k=0.03    //[W/m.K]
+u_inf=3    //[m/s]
+x=0.4    //[m]
+w=1.5    //[m]
+Nre_x=u_inf*x/v    //Reynolds no at x=0.4 m
+//Since this is less than 3*10^5.The flow is laminar upto x=0.4 m
+mu=rho*v  //[kg/(m.s)]
+
+Cp=1.009    //[kJ/kg.K]
+Cp=Cp*1000    //[J/kg.K]
+k=0.03    //W/(m.K)
+Npr=Cp*mu/k
+Nnu_x=0.332*(sqrt(Nre_x))*(Npr^(1.0/3.0))
+hx=Nnu_x*k/x    //[W/(m.K)]
+//Average value is twice this value
+h=2*hx    //[W/(m.K)]
+h=10.6      //Approximation
+A=x*w    //Area in [sq m]
+Tw=407    //[k]
+T_inf=293    //[K]
+Q=h*A*(Tw-T_inf)     //[W]
+//From both sides of the plate:
+Q=2*Q    //[W]
+printf("The heat transferred from both sides of the plate is %d W",round(Q));
+
+
diff --git a/1073/CH3/EX3.8/3_8.sce b/1073/CH3/EX3.8/3_8.sce
new file mode 100755
index 000000000..ee2af3b9f
--- /dev/null
+++ b/1073/CH3/EX3.8/3_8.sce
@@ -0,0 +1,27 @@
+clc;
+clear;
+//Example 3.8
+rho=0.998    //[kg/cubic m]
+v=20.76*10^-6    //[sq m/s]
+k=0.03    //[W/m.K]
+Npr=0.697
+x=0.4    //[m] from leading edge of the plate
+u_inf=3    //[m/s]
+Nre_x=u_inf*x/v    //Reynold numebr at x=0.40 m
+//Since this is less than 3*10^5    
+//therefore flow is laminar and 
+Nnu_x=0.332*sqrt(Nre_x)*(Npr^(1.0/3.0));
+hx=Nnu_x*k/x    //[W/sq m.K]
+//Average heat tarnsfer coefficient is twice this value
+h=2*hx    //[W/sq m.K]
+//Given:
+Q=1450    //[W]
+Tw=407    //[K]
+T_inf=293    //[K]
+L=0.4    //[m]
+//Q=h*w*L*(Tw-T_inf)
+//L=Q/(h*w*(Tw-T_inf))
+w=Q/(h*L*(Tw-T_inf))    //[m]
+printf("\n Width of plate is %f m",w);
+
+
diff --git a/1073/CH3/EX3.9/3_9.sce b/1073/CH3/EX3.9/3_9.sce
new file mode 100755
index 000000000..e6faa8f33
--- /dev/null
+++ b/1073/CH3/EX3.9/3_9.sce
@@ -0,0 +1,27 @@
+
+clc;
+clear;
+//Example 3.9
+v=17.36*10^-6    //Viscosity for air    [sq m./s]
+k=0.0275    //for air ..[W/(m.K)]
+Cp=1.006    //[kJ/(kg.K)]
+Npr=0.7    //for air
+u_inf=2    //[m/s]
+x=0.2    //[m]
+Nre_x=u_inf*x/v    //Reynolds number at x=0.2    m
+//Since this is less than 3*10^5
+Nnu_x=0.332*sqrt(Nre_x)*(Npr^(1.0/3.0))
+hx=Nnu_x*k/x    //[W/(sq m.K]
+//Average value of heat transfer coeff is twice this value
+h=2*hx    //[W/sq m.K)]
+h=12.3      //Approximation
+w=1    //width in [m]
+A=x*w    //[sq m] Area of plate
+Tw=333    //[K]
+T_inf=300    //[K]
+Q=h*A*(Tw-T_inf)    //Heat flow in [W]
+printf("\nANSWER:\nHeat flow is :%f W\n",Q)
+//From both sides of plate:
+Q=2*Q    //[W]
+printf("\nANSWER\n Heat flow from both sides of plate is %f W",Q);
+ 
diff --git a/1073/CH4/EX4.1/4_1.sce b/1073/CH4/EX4.1/4_1.sce
new file mode 100755
index 000000000..2ed8935ed
--- /dev/null
+++ b/1073/CH4/EX4.1/4_1.sce
@@ -0,0 +1,9 @@
+clc;
+clear;
+//Example 4.1
+e=0.9   //[Emissivity]
+sigma=5.67*10^-8    //[W/m^2.K^4]
+T1=377  //[K]
+T2=283  //[K]
+Qr_by_a=e*sigma*(T1^4-T2^4) //[W/sq m]
+printf("Heat loss by radiation is %d W/sq m",round(Qr_by_a));
diff --git a/1073/CH4/EX4.10/4_10.sce b/1073/CH4/EX4.10/4_10.sce
new file mode 100755
index 000000000..ebde4488c
--- /dev/null
+++ b/1073/CH4/EX4.10/4_10.sce
@@ -0,0 +1,20 @@
+clc;
+clear;
+//Example 4.10
+e1=0.05
+e2=e1
+A1=0.6944;
+A2=1;
+T1=293  //[K]
+T2=90   //[K]
+sigma=5.67*10^-8    //[W/m^2.K^4]
+D=0.3   //Diameter in [m]
+
+F12=1/(1/e1+(A1/A2)*(1/e2-1))
+Q_by_A=sigma*F12*(T1^4-T2^4)    //[W/sq m]
+Q=Q_by_A*%pi*(D^2)  //[kJ/h]
+Q=Q*3600/1000   //[kJ/h]
+lambda=21.44    //Latent heat in [kJ/kg]
+m_dot=Q/lambda    //kg/h
+printf("\n The liquid oxygen will evaporate at %f kg/h",m_dot);
+
diff --git a/1073/CH4/EX4.11/4_11.sce b/1073/CH4/EX4.11/4_11.sce
new file mode 100755
index 000000000..51514e282
--- /dev/null
+++ b/1073/CH4/EX4.11/4_11.sce
@@ -0,0 +1,22 @@
+clc;
+clear;
+//Example4.11
+sigma=5.67*10^-8    //W/(m^2.K^4)
+e1=0.3;
+e2=e1;
+D1=0.3  //[m]
+D2=0.5  //[m]
+T1=90   //[K]
+T2=313  //[K]
+A1=%pi*D1^2 //Area in [sq m]
+A2=%pi*D2^2//Area in [sq m]
+Q1=sigma*A1*(T1^4-T2^4)/(1/e1+(A1/A2)*(1/e2-1))  //[W]
+Q1=abs(Q1); //Absolute value in [W]
+printf("\n Rate of heat flow due to radiation is %f W",Q1);
+//When Aluminium is used
+e1=0.05
+e2=0.5
+Q2=sigma*A1*(T1^4-T2^4)/(1/e1+(A1/A2)*(1/0.3-1))  //[W]
+Q2=abs(Q2) //Absolute value in [W]
+Red=(Q1-Q2)*100/Q1  //Percent reduction
+printf("\n Reduction in heat flow will be %f percent ",Red);
diff --git a/1073/CH4/EX4.12/4_12.sce b/1073/CH4/EX4.12/4_12.sce
new file mode 100755
index 000000000..0f0667341
--- /dev/null
+++ b/1073/CH4/EX4.12/4_12.sce
@@ -0,0 +1,22 @@
+
+clc;
+clear;
+//Example 4.12
+sigma=5.67*10^-8    //[W/sq m.K^4]
+T1=77   //[K]
+T2=303  //[K]
+D1=32   //cm
+D1=D1/100   //[m]
+D2=36   //[cm]
+D2=D2/100   //[m]
+A1=%pi*D1^2 //[sq m]
+A2=%pi*D2^2 //[sq m]
+e1=0.03;
+e2=e1;
+Q=sigma*A1*(T1^4-T2^4)/(1/e1+(A1/A2)*(1/e2-1))  //[W]
+Q=Q*3600/1000   //[kJ/h]
+Q=abs(Q);   //[kJ/h]
+lambda=201  //kJ/kg
+m_dot=Q/lambda  //Evaporation rate in [kg/h]
+printf("\n Nitrogen evaporates at %f kg/h",m_dot);
+
diff --git a/1073/CH4/EX4.13/4_13.sce b/1073/CH4/EX4.13/4_13.sce
new file mode 100755
index 000000000..062059f3d
--- /dev/null
+++ b/1073/CH4/EX4.13/4_13.sce
@@ -0,0 +1,22 @@
+
+clc;
+clear;
+//Example 4.13
+D1=250  //Inner sphere idameter[mm]
+D1=D1/1000  //Outer diameter [m]
+D2=350  //[mm]
+D2=D2/1000  //[m]
+sigma=5.67*10^-8    //W/(sq m.K^4)
+A1=%pi*D1^2 //[sq m]
+A2=%pi*D2^2 //[sq m]
+T1=76   //[K]
+T2=300  //[K]
+e1=0.04;
+e2=e1;
+Q=sigma*A1*(T1^4-T2^4)/((1/e1)+(A1/A2)*((1/e2)-1))   //[W]
+Q=-2.45     //Approximate
+Q=abs(Q)    //[W]
+Q=Q*3600/1000   //[kJ/h]
+lambda=200  //kJ/kg
+Rate=Q/lambda   //[kg/h]
+printf("\n Rate of evaporation is %f kg/h(approx)",Rate);
diff --git a/1073/CH4/EX4.15/4_15.sce b/1073/CH4/EX4.15/4_15.sce
new file mode 100755
index 000000000..062f44275
--- /dev/null
+++ b/1073/CH4/EX4.15/4_15.sce
@@ -0,0 +1,19 @@
+clc;
+clear;
+//Example 4.15
+sigma=5.67*10^-8    //[W/(m^2.K^4)]
+e1=0.4
+e3=0.2
+T1=473  //[K]
+T3=303  //[K]
+Q_by_a=sigma*(T1^4-T3^4)/((1/e1)+(1/e3)-1)  //[W/sq m]
+//Q1_by_a=sigma*(T1^4-T2^4)/((1/e1)+(1/e2)-1)=sigma*A*(T2^4-T3^4)/((1/e2)+(1/e3)-1) //[W/sq m]
+e2=0.5
+//Solving we get
+T2=((6/9.5)*((3.5/6)*T3^4+T1^4))^(1/4)  //[K]
+Q1_by_a=sigma*(T1^4-T2^4)/((1/e1)+(1/e2)-1) //[W/sq m]
+red=(Q_by_a-Q1_by_a)*100/Q_by_a
+printf("\nHeat transfer rate per unit area(WITHOUT SHIELD) due to radiation is %f W/sq m\n",Q_by_a);
+printf("\nHeat transfer rate per unit area(WITH SHIELD) due to radiation is %f W/sq m\n",Q1_by_a);
+printf("\nReduction in heat loss is %f percent",red);
+
diff --git a/1073/CH4/EX4.16/4_16.sce b/1073/CH4/EX4.16/4_16.sce
new file mode 100755
index 000000000..8c9a68af0
--- /dev/null
+++ b/1073/CH4/EX4.16/4_16.sce
@@ -0,0 +1,27 @@
+clc;
+clear;
+//Example 4.16
+//In steady state,we can write:
+//Qcd=Qdb
+//sigma(Tc^4-Td^4)*/(1/ec+1/ed-1)=sigma(Td^4-Tb^4)/(1/ed+1/eb-1)
+// i.e Td^4=0.5*(Tc^4-Tb^4)
+//Given:
+Ta=600  //[K]
+eA=0.8;
+eC=0.5;
+eD=0.4;
+sigma=5.67*10^-8        //For air
+//(600^4-Tc^4)/2.25=(Tc^4-Td^4)/3.5
+//1.56*(600^4-Tc^4)=Tc^4-Td^4
+//Putting value of Td in terms of Tc
+//1.56*(600^4-Tc^4)=Tc^4-0.5*(Tc^4-300^4)
+function y=f(Tc)
+  y=1.56*(600^4-Tc^4)-Tc^4+0.5*(Tc^4-300^4)
+endfunction
+Tc=fsolve(500,f);           //[K]
+//or
+Tc=560.94       //[K] Approximate after solving
+Td=sqrt(sqrt(0.5*(Tc^4-300^4)))         //[K]
+Q_by_a=sigma*(Ta^4-Tc^4)/(1/eA+1/eC-1)      //[W/sq m]
+printf("\nRate of heat exchange per unit area=%f W/m^2",Q_by_a);
+printf("\nSteady state temperatures,Tc=%f K,and Td=%f K",Tc,Td);
diff --git a/1073/CH4/EX4.17/4_17.sce b/1073/CH4/EX4.17/4_17.sce
new file mode 100755
index 000000000..d73d598fe
--- /dev/null
+++ b/1073/CH4/EX4.17/4_17.sce
@@ -0,0 +1,27 @@
+clc;
+clear;
+//Example 4.17
+sigma=5.67*10^-8    //[W/(sq m.K^4)]
+e=0.8
+T1=673;  //[K]
+T2=303;  //[K]
+Do=200   //[mm]
+Do=Do/1000  //[m]
+L=1      //Let [m]
+A1=%pi*Do*L //[m^2/m]
+//CAse 1: Pipe to surrundings
+
+Q1=e*A1*sigma*(T1^4-T2^4)   //[W/m]
+Q1=5600     //Approximated
+//Q1=5600         //[W/m] approximated in book for calculation purpose
+//Concentric cylinders
+e1=0.8;
+e2=0.91;
+D1=0.2  //[m]
+D2=0.4  //[m]
+Q2=sigma*0.628*(T1^4-T2^4)/((1/e1)+(D1/D2)*((1/e2)-1))   //[W/m] length
+Red=Q1-Q2   //Reduction in heat loss
+
+printf("\nDue to thermal radiaiton,Loss of heat to surrounding is %d W/m\n",round(Q1));
+printf("\nWhen pipe is enclosed in 1 400 mm diameter brick conduit,Loss of heat is %d W/m\n",round(Q2));
+printf("\n Reduction in heat loss is %d W/m\n",round(Red));
diff --git a/1073/CH4/EX4.18/4_18.sce b/1073/CH4/EX4.18/4_18.sce
new file mode 100755
index 000000000..894eb30a8
--- /dev/null
+++ b/1073/CH4/EX4.18/4_18.sce
@@ -0,0 +1,16 @@
+
+
+clc;
+clear;
+//Example 4.18
+
+
+sigma=5.67*10^-8 ;   //[W/(sq m.K^4)]
+T1=813;  //[K]
+T2=473;  //[K]
+e1=0.87;
+e2=0.26;
+D1=0.25 ;//[m]
+D2=0.3; //[m]
+Q_by_a1=sigma*(T1^4-T2^4)/(1/e1+(D1/D2)*(1/e2-1))   //[W/ sqm]
+printf("\n Heat transfer by radiaiton is %d W/sq m",Q_by_a1);
diff --git a/1073/CH4/EX4.19/4_19.sce b/1073/CH4/EX4.19/4_19.sce
new file mode 100755
index 000000000..97c6bcd66
--- /dev/null
+++ b/1073/CH4/EX4.19/4_19.sce
@@ -0,0 +1,11 @@
+
+clc;
+clear;
+//Example 4.19
+sigma=5.67*10^-8    //[W/sq m.K^4]
+A1=0.5*1    //[sq m]
+F12=0.285
+T1=1273 ///[K]
+T2=773  //[K]
+Q=sigma*A1*F12*(T1^4-T2^4)    //[W]
+printf("\n Net radiant heat exchange between plates is %d W",Q);
diff --git a/1073/CH4/EX4.2/4_2.sce b/1073/CH4/EX4.2/4_2.sce
new file mode 100755
index 000000000..c01af8bd2
--- /dev/null
+++ b/1073/CH4/EX4.2/4_2.sce
@@ -0,0 +1,9 @@
+clc;
+clear;
+//Example 4.2
+e=0.9   //Emissivity
+T1=393  //[K]
+T2=293  //[K]
+sigma=5.67*10^-8    //[W/sq m.K]
+Qr_by_a=e*sigma*(T1^4-T2^4) //W/sq m
+printf("\n Rate of heat transfer by radiation is %f W/sq m",Qr_by_a);
diff --git a/1073/CH4/EX4.20/4_20.sce b/1073/CH4/EX4.20/4_20.sce
new file mode 100755
index 000000000..4ca535724
--- /dev/null
+++ b/1073/CH4/EX4.20/4_20.sce
@@ -0,0 +1,35 @@
+clc;
+clear;
+//Example 4.20
+sigma=5.67*10^-8    //[W/sq m.K^4]
+T1=750  //[K]
+T2=500  //[K]
+e1=0.75;
+e2=0.5;
+//Heat transfer without shield :
+
+Q_by_a=sigma*(T1^4-T2^4)/((1/e1)+(1/e2)-1)  //[W/sq m]
+
+//Heat transfer with shield:
+R1=(1-e1)/e1    //Resistance 1
+
+F13=1;
+R2=1/F13        //Resistance 2
+
+e3=0.05
+R3=(1-e3)/e3    //Resistance 3
+
+R4=(1-e3)/e3    //Resistance 4
+
+F32=1;
+R5=1/F32        //Resistance 5
+
+R6=(1-e2)/e2       //Resistance 6
+
+Total_R=R1+R2+R3+R4+R5+R6   //Total resistance
+
+Q_by_as=sigma*(T1^4-T2^4)/Total_R   //[W/sq m]
+
+Red=(Q_by_a-Q_by_as)*100/Q_by_a     //Reduciton in heat tranfer due to shield 
+
+printf("\n Reduction in heat transfer rate as a result of radiaiotn shield is %f percent",Red);
diff --git a/1073/CH4/EX4.21/4_21.sce b/1073/CH4/EX4.21/4_21.sce
new file mode 100755
index 000000000..7c91c8dec
--- /dev/null
+++ b/1073/CH4/EX4.21/4_21.sce
@@ -0,0 +1,24 @@
+clc;
+clear;
+//Example 4.21
+e1=0.3
+e2=0.8
+//Let sigma*(T1^4-T2^4)=z=1(const)
+z=1;    //Let
+Q_by_A=z/(1/e1+1/e2-1)  //W/sq m
+
+//Heat transfer with radiation shield 
+e3=0.04
+F13=1;
+F32=1;
+//The resistances are:
+R1=(1-e1)/e1
+R2=1/F13
+R3=(1-e3)/e3
+R4=R3
+R5=1/F32
+R6=(1-e2)/e2
+R=R1+R2+R3+R4+R5+R6     //Total resistance
+Q_by_As=z/R //where z=sigma*(T1^4-T2^4) //W/sq m
+red=(Q_by_A-Q_by_As)*100/Q_by_A    //Percent reduction in heat transfer
+printf("\n The heat transfer is reduced by %f percent due to shield",red)
diff --git a/1073/CH4/EX4.22/4_22.sce b/1073/CH4/EX4.22/4_22.sce
new file mode 100755
index 000000000..c6cb26c81
--- /dev/null
+++ b/1073/CH4/EX4.22/4_22.sce
@@ -0,0 +1,43 @@
+
+clc;
+clear;
+//Example 4.22
+sigma=5.67*10^-8;
+T1=1273 //[K]
+T2=773  //[K]
+T3=300  //[K]
+A1=0.5  //[sq m]
+A2=A1;  //[sq m]
+F12=0.285;
+F21=F12;
+F13=1-F12;
+F23=1-F21;
+e1=0.2;
+e2=0.5;
+//Resistance in the network are calculated as:
+R1=1-e1/(e1*A1)
+R2=1-e2/(e2*A2)
+R3=1/(A1*F12)
+R4=1/(A1*F13)
+R5=1/(A2*F23)
+R6=0    //Given (1-e3)/e3*A3=0
+//Also
+Eb1=sigma*T1^4  //W/sq m
+Eb2=sigma*T2^4   //[W/sq m]
+Eb3=sigma*T3^4 //[W/sq m]
+
+//Equations are:
+//(Eb1-J1)/2+(J2-J1)/7.018+(Eb3-J1)/2.797=0
+//(J1-J2)/7.018+(Eb3-J2)/2.797+(Eb2-J2)/2=0
+
+//On solving we get:
+J1=33515    //[W/sq m]
+J2=15048    //[W/sqm]
+J3=Eb3  //[W/sq m]
+Q1=(Eb1-J1)/((1-e1)/(e1*A1))        //[W/sq m]
+Q2=(Eb2-J2)/((1-e2)/(e2*A2))        //[W/sq m]
+Q3=(J1-J3)/(1/(A1*F13))+(J2-J3)/(1/(A2*F23))    //[W/sq m]
+printf("\n Total heat lost by plate 1 is %f W/sq m\n",Q1);
+printf("\n Total heat lost by plate 2 is %f W/sq m\n",Q2);
+printf("\nThe net energy lost by both plates must be absorbed by the room,\n %f=%f",Q3,Q1+Q2)
+
diff --git a/1073/CH4/EX4.23/4_23.sce b/1073/CH4/EX4.23/4_23.sce
new file mode 100755
index 000000000..be2dd7a03
--- /dev/null
+++ b/1073/CH4/EX4.23/4_23.sce
@@ -0,0 +1,22 @@
+clc;
+clear;
+//Example 4.23
+sigma=5.67*10^-8    //[W/sq m.K^4]
+e1=0.7;
+e2=0.7;
+T1=866.5    //[K]
+T2=588.8    //[K]
+Q_by_A=sigma*(T1^4-T2^4)/((1/e1)+(1/e2)-1)  //[W/sq m]
+e1=0.7;
+e2=e1;
+e3=e1;
+e4=e1;
+e=e1;
+//Q with n shells =1/(n+1)
+n=2
+Q_shield=1/(n+1);
+es1=e1;
+es2=e1;
+Q_by_A=sigma*(T1^4-T2^4)/((1/e1)+(1/e2)+2*(1/es1+1/es2)-(n+1))  //[W/sq m]
+printf("\n New Radiaiton loss is %f W/sq m",Q_by_A);
+
diff --git a/1073/CH4/EX4.24/4_24.sce b/1073/CH4/EX4.24/4_24.sce
new file mode 100755
index 000000000..62dd800f6
--- /dev/null
+++ b/1073/CH4/EX4.24/4_24.sce
@@ -0,0 +1,38 @@
+clc;
+clear;
+//Example 4.24
+//1.WITHOUT SHIELD
+sigma=5.67*10^-8        
+e1=0.12;
+e2=0.15;
+T1=100  //[K]
+T2=300  //[K]
+r1=0.015    //[m]
+r2=0.045    //[m]
+L=1         //[m]
+A1=2*%pi*r1*L   //[sq m]
+Q_by_L=2*%pi*r1*sigma*(T1^4-T2^4)/(1/e1+(r1/r2)*(1/e2-1))   //[W/m]
+//-ve saign indicates that the net heat flow is in the radial inward direction
+//2.WITH CYLINDRICAL RADIATION SHIELD
+e3=0.10;
+e4=0.05;
+r3=0.0225   //[m]
+Qs_by_L=2*%pi*r1*sigma*(T1^4-T2^4)/(1/e1+r1/r2*(1/e2-1)+(r1/r3)*(1/e3+1/e4-1))  //[W/sq m]
+red=(abs(Q_by_L)-abs(Qs_by_L))*100/abs(Q_by_L)  //percent reduction in heat gain
+
+//Radiation network approach
+A3=2*%pi*r3     //[sq m]
+A2=2*%pi*r2     //[sq m]
+F13=1;
+F32=1;
+R1=(1-e1)/(e1*A1)
+R2=1/(A1*F13)
+R3=(1-e3)/(e3*A3)
+R4=(1-e4)/(e4*A3)
+R5=1/(A3*F32)
+R6=(1-e2)/(e2*A2)
+
+Qs=sigma*(T1^4-T2^4)/((1-e1)/(e1*A1)+1/(A1*F13)+(1-e3)/(e3*A3)+(1-e4)/(e4*A3)+1/(A3*F32)+(1-e2)/(e2*A2))        
+printf("\n With cylindrical radiaiton shield Heat gained by fluid per 1 m lengh of tube is %f W/m\n",Qs_by_L);
+printf("\nPercent reduction in heat gain is %f percent\n",red);
+printf("\nWith radiaiton network approach %f W/sqm ",Qs);
diff --git a/1073/CH4/EX4.3/4_3.sce b/1073/CH4/EX4.3/4_3.sce
new file mode 100755
index 000000000..340e0eec3
--- /dev/null
+++ b/1073/CH4/EX4.3/4_3.sce
@@ -0,0 +1,16 @@
+
+clc;
+clear;
+//Example 4.3
+L=1; //[m]
+e=0.8  ; //Emissivity
+sigma=5.67*10^-8 ;   //[m^2.K^4]
+T1=423;  //[K]
+T2=300;  //[K]
+Do=60;   //[mm]
+Do=Do/1000;  //[m]
+A=%pi*Do*L  //[sq m]
+A=0.189     //Approx in book [m^2]
+Qr=e*sigma*A*(T1^4-T2^4)    //[W/m]
+printf("\n Net radiaiton rate per 1 metre length of pipe is %d W/m",round(Qr));
+
diff --git a/1073/CH4/EX4.4/4_4.sce b/1073/CH4/EX4.4/4_4.sce
new file mode 100755
index 000000000..69d78cd61
--- /dev/null
+++ b/1073/CH4/EX4.4/4_4.sce
@@ -0,0 +1,17 @@
+clc;
+clear;
+//Example 4.4
+e=0.9   //Emissivity
+L=1 //[m]
+Do=50   //[mm]
+Do=Do/1000  //[m]
+sigma=5.67*10^-8    //[W/(m^2.K^4)]
+T1=415  //[K]
+T2=290  //[K]
+dT=T1-T2    //[K]
+hc=1.18*(dT/Do)^(0.25)  //[W/sq m.K]
+A=%pi*Do*L  //Area in [sq m]
+Qc=hc*A*dT //Heat loss by convection W/m
+Qr=e*sigma*A*(T1^4-T2^4)    //Heat loss by radiation per length W/m
+Qt=Qc+Qr    //Total heat loss in [W/m]
+printf("\n Total heat loss by convection is %f W/m",Qt);
diff --git a/1073/CH4/EX4.5/4_5.sce b/1073/CH4/EX4.5/4_5.sce
new file mode 100755
index 000000000..00ba6c238
--- /dev/null
+++ b/1073/CH4/EX4.5/4_5.sce
@@ -0,0 +1,17 @@
+clc;
+clear;
+//Example 4.5
+e=0.85
+sigma=5.67*10^-8    //[W/sq m.K]
+T1=443  //[K]
+T2=290  //[K]
+dT=T1-T2    //[K]
+hc=1.64*dT^0.25     //W/sq m.K
+Do=60   //[mm]
+Do=Do/1000  //[m]
+L=6 //Length [m]
+A=%pi*Do*L  //Surface area of pipe in [sq m]
+Qr=e*sigma*A*(T1^4-T2^4)    // Rate of heat loss by radiaiton W
+Qc=hc*A*(T1-T2) // Rate of heat loss by convection [W]
+Qt=Qr+Qc    //Total heat loss  [W]
+printf("\n Total heat loss is %d W",round(Qt))
diff --git a/1073/CH4/EX4.6/4_6.sce b/1073/CH4/EX4.6/4_6.sce
new file mode 100755
index 000000000..9f41ff8cb
--- /dev/null
+++ b/1073/CH4/EX4.6/4_6.sce
@@ -0,0 +1,18 @@
+
+clc;
+clear;
+//EXample 4.6
+sigma=5.67*10^-8    //[W/m^2.K^4]
+e1=0.79;
+e2=0.93;
+T1=500 ; //[K]
+T2=300 ; //[K]
+D=70    //[mm]
+D=D/1000    //[m]
+L=3 //[m]
+W=0.3   //Side of conduit [m]
+A1=%pi*D*L  //[sq m]
+A1=0.659        //Approximate calculation in book in [m^2]
+A2=4*(L*W)  //[sq m]
+Q=sigma*A1*(T1^4-T2^4)/(1/e1+((A1/A2)*(1/e2-1)))    //[W]
+printf("\n Heat lost by radiation is %f W",Q);
diff --git a/1073/CH4/EX4.7/4_7.sce b/1073/CH4/EX4.7/4_7.sce
new file mode 100755
index 000000000..34f1a17cc
--- /dev/null
+++ b/1073/CH4/EX4.7/4_7.sce
@@ -0,0 +1,10 @@
+clc;
+clear;
+//Example 4.7
+sigma=5.67*10^-8    //[W/sq m.K^4]
+T1=703  //[K]
+T2=513  //[K]
+e1=0.85 
+e2=0.75
+Q_by_Ar=sigma*(T1^4-T2^4)/(1/e1+1/e2-1) //[W/sq m]
+printf("\n Net radiant interchange per square metre is %d W/sq m",round(Q_by_Ar));
diff --git a/1073/CH4/EX4.8/4_8.sce b/1073/CH4/EX4.8/4_8.sce
new file mode 100755
index 000000000..10ad30c9d
--- /dev/null
+++ b/1073/CH4/EX4.8/4_8.sce
@@ -0,0 +1,13 @@
+clc;
+clear;
+//Example 4.8
+L=3 ;//[m]
+A=L^2   //Area in [sq m]
+sigma=5.67*10^-8;    //[W/sq m.K^4]
+T1=373;  //[K]
+T2=313;  //[K]
+e1=0.736;
+e2=e1;
+F12=1/((1/e1)+(1/e2)-1)
+Q=sigma*A*F12*(T1^4-T2^4)   //[W]
+printf("\n Net radiant interchange is %d W",round(Q));
diff --git a/1073/CH4/EX4.9/4_9.sce b/1073/CH4/EX4.9/4_9.sce
new file mode 100755
index 000000000..887a1f7c4
--- /dev/null
+++ b/1073/CH4/EX4.9/4_9.sce
@@ -0,0 +1,19 @@
+clc;
+clear;
+sigma=5.67*10^-8    //[W/sq m.K^4]
+e1=0.05 
+e2=0.05
+//A1=A2=1 (let)
+A1=1;
+A2=A1;
+F12=1/(1/e1+(A1/A2)*(1/e2-1))   
+T1=368  //[K]
+T2=293  //[K]
+Q_by_A=sigma*F12*(T1^4-T2^4)    //Heat loss per unit Area  [W/sq m]
+printf("\nRate of heat loss when of silvered surface is %f W/sq m",Q_by_A);
+//When both the surfaces are black
+e1=1;
+e2=1;
+F12=1/(1/e1+(A1/A2)*(1/e2-1))   
+Q_by_A=sigma*F12*(T1^4-T2^4)    //[W/sq m]
+printf("\n When both surfaces are black,Rate of heat loss  is %d W/sq m",round(Q_by_A));
diff --git a/1073/CH5/EX5.1/5_1.sce b/1073/CH5/EX5.1/5_1.sce
new file mode 100755
index 000000000..493320e62
--- /dev/null
+++ b/1073/CH5/EX5.1/5_1.sce
@@ -0,0 +1,63 @@
+
+
+clc;
+clear;
+//Example 5.1
+Di=35       //[mm]
+Di=Di/1000      //[m]
+Do=42       //[mm]
+Do=Do/1000      //[m]
+//for benzene
+mb_dot=4450     //[kg/h]
+Cpb=1.779       //[kJ/(kg.K)]
+t2=322          //[K]
+t1=300  //[K]
+Q=mb_dot*Cpb*(t2-t1)        //for benzene in [kJ/h]
+
+//For toulene
+T1=344  //[K]
+T2=311      //[K]
+Cpt=1.842       //[kJ/kg.K]
+mt_dot=Q/(Cpt*(T1-T2))      //[kg/h]
+Q=Q*1000/3600           //[W]
+//Hot fluid(toluene)
+//Cold fluid(benzene)
+dT1=22      //[K]
+dT2=11      //[K]
+dTlm=(dT1-dT2)/(log(dT1/dT2))       //[K]
+
+//Clod fluid:Inner pipe,benzene
+Di=0.035        //[m]
+Ai=(%pi/4)*Di^2     //Flow area[sq m]
+Gi=mb_dot/Ai        //Mass velocity [kg/m^2.h]
+Gi=Gi/3600          //[kg/m^2.s]
+mu=4.09*10^-4       //[kg/(m.s)]
+Nre=Di*Gi/mu        //Reynolds number
+
+Cp=Cpb*10^3         //[J/(kg.K)]
+k=0.147             //[W/m.K]
+Npr=Cp*mu/k         //Prandtl number
+hi=(k/Di)*0.023*(Nre^0.8)*(Npr^0.4)     //[W/sq m.K]
+hio=hi*Di/Do            //[W/sq m.K]
+D1=0.042                //Outside dia of inside pipe[mm]
+D2=0.0525           //Inside dia of outside pipe[m]
+De=(D2^2-D1^2)/D1           //[m]
+De=0.0236               //Approximated 
+aa=%pi*(D2^2-D1^2)/4           //Flow area[sq m]
+Ga=mt_dot/aa        //Mass velocity in [kg/m^2.h]
+Ga=Ga/3600          //[kg/m^2.s]
+mu=5.01*10^-4       //[kg/(m.s)]
+Nre=De*Ga/mu        //Reynolds number
+Npr=Cp*mu/k     //Prandtl number
+ho=(k/De)*0.023*(Nre^0.8)*(Npr^0.3)     //[W/sq m.K]
+Uc=1/(1/ho+1/hio)       //[W/sq m.K]
+Rdi=1.6*10^-4           //Fouling factor [m^2.K/W]
+Rdo=1.6*10^-4           //Fouling factor[m^2.K/W]
+Rd=Rdi+Rdo              //(m^2.K/W)
+Ud=1/(1/Uc+Rd)          //[W/sq m.K]
+A=Q/(Ud*dTlm)           //sq m
+ex=0.136                    //[sq m]
+l=A/ex                  //m
+tl=12       //Total length of one harpin of 6m [m]
+printf("b%f",l);
+printf("\n\Required surface is fulfilled by connecting %d(three) 6m harpins in series\n",round(l/tl))
diff --git a/1073/CH5/EX5.10/5_10.sce b/1073/CH5/EX5.10/5_10.sce
new file mode 100755
index 000000000..065278ec7
--- /dev/null
+++ b/1073/CH5/EX5.10/5_10.sce
@@ -0,0 +1,61 @@
+
+clc;
+clear;
+//Example 5.10
+m_dot=7250      //[kg/h] of nitrobenzene 
+Cp=2.387;            //[kJ/kg.K]
+mu=7*10^-4;          //[kg/m.s]
+k=0.151;             //[W/m.K]
+vis=1;
+Ft=0.9;             //LMTD correction factor
+T1=400              //[K]
+T2=317              //[K]
+t1=333              //[K]
+t2=300              //[K]
+dT1=T1-t1          //[K]
+dT2=T2-t2          //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)       //[K]
+//For nitrobenzene
+Q=m_dot*Cp*(T1-T2)          //[kJ/h]
+Q=Q*1000/3600                    //[W]
+n=170           //No. of tubes
+L=5     //[m]
+Do=0.019            //[m]
+Di=0.015            //[m]
+Ao=n*%pi*Do*L       //[sq m]
+Uo=Q/(Ao*Ft*dTlm)   //[W/sq m.K]
+Ud=Uo           //[W/sq m.K]
+B=0.15              //Baffle spacing [m]
+Pt=0.025            //Tube pitch in [m]
+C_dash=Pt-Do        //Clearance in [m]
+id=0.45             //[m]
+
+//Shell side cross flow area
+as=id*C_dash*B/Pt           //[sq m]
+
+//Equivalent diameter of shell
+De=4*(Pt^2-(%pi/4)*(Do^2))/(%pi*Do)        //[m]
+
+//Mass velocity on shell side
+Gs=m_dot/as             //[kg/(m.h)]
+Gs=Gs/3600              //[kg/m^2.s]
+mu=7*10^-4          //[kg/m.s]
+Cp=Cp*1000      //[J/kg.K]
+Nre=De*Gs/mu        //Reynolds number
+Npr=Cp*mu/k         //Prandtl number
+
+//From empirical eqn:
+mu_w=mu         //
+Nnu=0.36*Nre^0.55*Npr^(1/3)
+ho=Nnu*k/De         //[W/sq m.K]
+hi=1050             //Given [W/sq m.K]
+Uo=1/(1/ho+(1/hi)*(Do/Di))      //[W/sq m.K]
+Uc=Uo               //W/sq m.K
+
+//Suitability of heat exchanger
+Rd_given=9*10^-4          //[W/sq m.K]
+Rd=(Uc-Ud)/(Uc*Ud)          //[W/sq m.K]
+printf("\n Rd calculated(%f W/m^2.K) is mazimum allowable scale resistance\n",Rd);
+printf("\n\nAs Rd calculated (%f W/sq m.K)(OR 1.1*10^-3) is more than Rd given(%f W/sq m,K),the given heat exchanger is suitable\n",Rd,Rd_given);
+
+
diff --git a/1073/CH5/EX5.11/5_11.sce b/1073/CH5/EX5.11/5_11.sce
new file mode 100755
index 000000000..6bb05fb5f
--- /dev/null
+++ b/1073/CH5/EX5.11/5_11.sce
@@ -0,0 +1,41 @@
+clc;
+clear;
+//Example 5.11
+mw_dot=1720;         //water in [kg/h]
+t1=293 ;             //[K]
+t2=318 ;             //[K]
+Cpw=4.28;            //[kJ/kg.K]
+Q=mw_dot*Cpw*(t2-t1)    //[kJ/h]
+Q=Q*1000/3600           //W
+lambda=2230;         //[kJ/kg]
+dT1=90 ;             //[K]
+dT2=65 ;             //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)     //[K]
+
+//Calculation of inside heat transfer coefficient
+Di=0.0225;            //[m]
+u=1.2 ;          //[m/s]\
+rho=995.7 ;          //[kg/m^3]
+v=0.659*10^-6           //[m/s]
+mu=v*rho        //[kg/m.s]
+Nre=Di*u*rho/mu     //reynolds number
+Cp=Cpw*1000         //[J/kg.K]
+k=2.54 ;     //[kJ/h.m.K]
+k=k*1000/3600           //[W/m.K]
+Npr=Cp*mu/k             //Prandtl number
+Nnu=0.023*Nre^0.8*Npr^0.4       //Nusselt number
+hi=k*Nnu/Di             //[W/sq m.K]
+ho=19200            //[kJ/h.m^2.K]
+ho=ho*1000/3600     //[W/m^2.K]
+Do=0.025        //[m]
+Dw=(Do-Di)/log(Do/Di)       //[m]
+x=(Do-Di)/2     //[m]
+kt=460           //For tube wall material [kJ/h.m.K]
+kt=kt*1000/3600         //[W/m.K]
+Uo=1/(1/ho+(1/hi)*(Do/Di)+(x/kt)*(Do/Dw))        //[W/sq m.K]
+//Q=Uo*Ao*dTlm
+Ao=Q/(Uo*dTlm)          //[sq  m]
+L=4             //Tube length in [m]
+n=Ao/(%pi*Do*L)         //[Number of tubes]
+n=round(n)          //Approximate
+printf("\n Number of tubes reuired= %d",n);
diff --git a/1073/CH5/EX5.12/5_12.sce b/1073/CH5/EX5.12/5_12.sce
new file mode 100755
index 000000000..ce3d7dbb0
--- /dev/null
+++ b/1073/CH5/EX5.12/5_12.sce
@@ -0,0 +1,58 @@
+clc;
+clear;
+//Example 5.12
+t1=290          //Inlet temperature of cooling water [K]
+ho=2250         //Heat transfer coefficient based on inside area in [W/sq m.K]
+lambda=400      //[kJ/kg] LAtent heat of benzene 
+mb_dot=14.4     //[t/h] Condensation rate of benzene vapour
+Cpw=4.187       //Specific heat
+//With no Scale
+
+Q=mb_dot*1000*lambda     //Heat duty of condenser in [kJ/h]
+Q=(Q/3600)*1000     //[W]
+//Shell and tube type of heat exchanger is used as a single pass surface condenser
+Di=0.022        //I.D of tube[m]
+L=2.5           //Length of each tube in [m]
+n=120           //Number of tubes
+A=%pi*Di*L      //Area of heat transfer per metre length in [m^2/m]
+A=n*A           //Total area of heat transfer in [m^2]
+Ai=(%pi/4)*Di^2 //Cross-sectional area of each tube in [m^2]
+Ai=n*Ai         //Total area of flow in [m^2]
+u=0.75          //Velocty of water [ms^-1]
+V=u*Ai          //Volumetric flow of water 
+rho=1000        //[Density of water in [kg/m^3]]
+mw_dot=V*rho    //Mass flow rate of water in  [kg/s]
+
+//Heat balance
+
+//Q=mw_dot*Cpw*(t2-t1)
+t2=Q/(mw_dot*Cpw*1000)+t1        //[K]
+T=350       //Condensing benzene temperature in [K]
+dT1=T-t1    //[K]
+dT2=T-t2    //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)     //LMTD
+U=Q/(A*dTlm)                    //[W/m^2.K]
+U=round(U)
+//Neglecting resistance,we have:
+hi=1/(1/U-1/ho)         //[W/m^2.K]
+//hi is proportional to u^0.8
+C=hi/(u^0.8)        //Constant
+
+//With Scale
+
+Rd=2.5*10^-4        //[m^2 K./W]
+//1/U=1/hi+1/ho+Rd
+//U=hi/(1+3.38*u^0.8)
+//mw_dot=rho*u*Ai     //[kg/s]
+//Let t2 be the outlet temperature of water 
+//Q=mw_dot*Cpw*(t2-t1)
+//t2=Q/(mw_dot*Cpw)+t1  
+dT1=60
+//dT2=T-(t1+8.373/u)
+//dTlm=8.373/(u*log(60*u/(60*u-8.373)))
+//Q=U*A*dTlm
+//1.89=((u^-0.2)/(1+3.38*u^0.8))*(1/log((60*u)/60*u-8.373)
+//If we assume values of u greater than 0.75 m/s
+//For u=3.8     //[ms^-1]
+u=3.8           //]ms^-1]
+printf("\nWater velocity must be 3.88 ms^-1");
diff --git a/1073/CH5/EX5.13/5_13.sce b/1073/CH5/EX5.13/5_13.sce
new file mode 100755
index 000000000..86a257f4c
--- /dev/null
+++ b/1073/CH5/EX5.13/5_13.sce
@@ -0,0 +1,48 @@
+clc;
+clear;
+//Example 5.13
+mh_dot=1.25         //[kg/s]
+Cpw=4.187*10^3      //Heat capacity of water in [J/kg.K]
+lambda=315          //[kJ/kg]
+Q=mh_dot*lambda     //Rate of heat transfer from vapour     [kJ/s]
+Q=Q*10^3            //[W]
+Ts=345              //Temperature of condensing vapour[K]
+t1=290              //Inlet temperature of water [K]
+t2=310              //Outlet temperature of water[K]
+dT1=Ts-t1           //[K]
+dT2=Ts-t2           //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)         //[K]
+//Heat removed from vapour = Heat gained
+mw_dot=Q/(Cpw*(t2-t1))      //[kg/s]
+hi=2.5                      //[kW/sq m.K]
+hi=hi*1000                  //[W/sq m.K]
+Do=0.025                    //[m]
+Di=0.020                    //[m]
+hio=hi*(Di/Do)              //Inside heat transfer cosfficient referred to outside dia in [W/sq m.K]
+ho=0.8                      //Outside heat tranbsfer coefficient in [kW/sq m.K]
+ho=ho*1000                  //[W/sq m.K]
+Uo=1/(1/ho+1/hio)           //[W/sq m.K]
+//Ud is 80% of Uc
+Ud=(80/100)*Uo             //[W/sq m.K]
+Ao=Q/(Ud*dTlm)              //[sq m]
+L=1                         //[m]
+A=%pi*Do*L                  //Outside area of pipe per m length of pipe
+len=Ao/A                    //Total length of piping required.
+rho=1000                    //[kg/m^3]
+V=mw_dot/rho                //[m^3/s]
+v=0.6                       //[m/s]
+a=V/v                       //Cross-sectional area for flow pass [sq m]
+a1=(%pi*Di^2)/4             //[sq m]
+//for single pass on tube side fluid(water)
+n=round(a/a1)                      //No. of tubes per pass
+l=len/n                     //Length of each tube in [m]
+//For two passes on water side:
+tn=2*n                      //Total no of tubes
+l2=len/tn                   //Length of each tube in [m]
+//For four passes on water side/tube side
+tn2=4*n                     //Total no. of tubes
+l3=len/tn2                  //Length of each tube in [m]
+
+printf("\nNo. of tubes=%d ,\nLength of tube=%f m",tn2,l3);
+
+  
diff --git a/1073/CH5/EX5.14/5_14.sce b/1073/CH5/EX5.14/5_14.sce
new file mode 100755
index 000000000..2ac1afc44
--- /dev/null
+++ b/1073/CH5/EX5.14/5_14.sce
@@ -0,0 +1,72 @@
+
+clc 
+clear 
+//Example 5.14
+//Properties of crude oil:
+Cpc=1.986   ;              //[kJ/(kg.K)]
+mu1=2.9*10^-3;             //[N.s/sq m]
+k1=0.136     ;             //[W/m.K]
+
+rho1=824      ;            //[kg/m^3]
+
+//Properties of bottom product:
+Cp2=2.202      ;          //[kJ/kg.K]
+rho2=867        ;         //[kg/m^3]
+mu2=5.2*10^-3   ;         //[N.s/sq m]
+k2=0.119        ;         //[W/sq m.K]
+
+mc_dot=135000    ;           //Basis: cruid oil flow rate in [kg/h]
+m_dot=106000     ;           //Bottom product flow rate inn [kg/h]
+t1=295           ;        //[K]
+t2=330           ;        //[K]
+T1=420           ;        //[K]
+T2=380           ;        //[K]
+dT1=T1-t2                //[K]
+dT2=T2-t1                //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)         //[K]
+Q=mc_dot*Cpc*(t2-t1)                //kJ/h
+Q=Q*1000/3600                       //[W]
+
+//Shell side calculations:
+Pt=25            ;                    //[mm]
+Pt=Pt/1000       ;                   //[m]
+B=0.23           ;                    //[m]
+Do=0.019         ;                     //[m]Outside diameter for square pitch 
+c_dash=Pt-Do                        //Clearance in [m]
+id=0.6           ;                     //[m]
+as=id*c_dash*B/Pt                     // Cross flow area of shell [sq m]
+//since there is a Calculaiton mistake ,we take:
+as=0.0353;
+Gs=m_dot/as                           //Shell side mass velocity in [kg/sq m.h]
+Gs=Gs/3600;                            //[kg/sq m.s]
+De=4*(Pt^2-(%pi/4)*Do^2)/(%pi*Do)     //[m]
+Nre=De*Gs/mu2                         //Reynolds number
+Npr=Cp2*1000*mu2/k2                        //Prandtl number
+muw=mu2                             //Since mu/muw=1
+Nnu=0.36*(Nre^0.55)*Npr^(1.0/3.0)*(mu2/muw)^(0.14)      //Nusselt number
+ho=Nnu*k2/De                        //[W/sq m.K]
+
+//Tube side heat transfer coefficient:
+n=324      ;             //No. of tubes
+n_p=324/2   ;            //No.of tubes per pass
+t=2.1         ;          //Thickness in [mm]
+t=t/1000     ;           //[m]
+Di=Do-2*t                 //I.d of tube in [m]
+A=(%pi/4)*(Di^2)        //Cross-sectional area of one tube in [sq m]
+A_p=n_p*A               //Total area for flow per pass in [sq m]
+G=mc_dot/A_p            //[kg/sq m h]
+G=G/3600                //[kg/sq m.s]
+Nre=Di*G/mu1            //Reynolods number
+Npr=42.35 ;              //Prandtl number
+Nnu=0.023*(Nre^0.8)*(Npr^0.4)       //Nusselt number
+hi=Nnu*k1/Di                         //[W/sq m.K]
+hio=hi*Di/Do                        //[W/sq m.K]
+Uo=1/(1/ho+1/hio)                   //[W/sq m.K]
+Uc=Uo
+L=4.88 ;                             //Length of tube in [m]
+Ao=n*%pi*Do*L                       //[sq m]
+Ud=Q/(Ao*dTlm)                      //[W/sq m.K]
+Rd=(Uc-Ud)/(Uc*Ud)                  //[m^2.K/W]
+printf("\n The calculation of line no.36 to calculated as is wrongly done in Book by printing 0.0353,,..which is wrong\n");
+printf("\nRd=%f K/w,or 7.34*10^-4 which is less than the provided,so this if installed will not give required temperarues without frequent cleaning\n\n",Rd); 
+
diff --git a/1073/CH5/EX5.15/5_15.sce b/1073/CH5/EX5.15/5_15.sce
new file mode 100755
index 000000000..af5b0d837
--- /dev/null
+++ b/1073/CH5/EX5.15/5_15.sce
@@ -0,0 +1,55 @@
+clc;
+clear;
+//Example 5.15
+
+//CASE I:
+Cp=4*10^3;           //[J/kg.K]
+t1=295;              //[K]
+t2=375;              //[K]
+sp=1.1;              //Specific gravity of liquid 
+v1=1.75*10^-4;        //Flow of liquid in [m^3/s]
+rho=sp*1000         //[kg/m^3]
+m_dot=v1*rho         //[kg/s]
+Q=m_dot*Cp*(t2-t1)  //[W]
+T=395;               //[K]
+dT1=T-t1            //[K]
+dT2=T-t2            //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2) //[K]
+U1A=Q/dTlm          //[W/K]
+
+//CASE-II
+v2=3.25*10^-4                   //Flow in [m^3/s]
+T2=370                          //[K]
+m_dot=v2*rho          //[kg/s]
+Q=m_dot*Cp*(T2-t1)                //[W]
+dT1=T-t1                          //[K]
+dT2=T-T2                          //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)       //[K]
+U2A=Q/dTlm                        //[W/K]
+//since u is propn to v
+//hi =C*v^0.8
+
+U2_by_U1=U2A/U1A                
+
+ho=3400                         //Heat transfer coeff for condensing steam in [W/sq m.K]
+C=poly(0,"C")
+//Let C=1 and v=v1
+//C=1;
+v=v1;                           //=1.75*10^-4 m^3/s
+hi=C*v^0.8
+U1=1/(1/ho+1/hi)                 //
+
+//When v=v2
+v=v2;
+hi=C*v^0.8
+U2=1/(1/ho+1/hi)               //
+
+//Since U2=1.6U1
+//On solving we get:
+C=142497
+v=v1
+hi=C*v^0.8
+U1=1/(1/ho+1/hi)                 //
+A=U1A/U1                //Heat transfer area in [sq m]
+printf("\n Overall heat transfer coefficient is %f W/sq m.K and\n\nHeat transfer area is %f sq m",U1,A);
+
diff --git a/1073/CH5/EX5.16/5_16.sce b/1073/CH5/EX5.16/5_16.sce
new file mode 100755
index 000000000..84582e457
--- /dev/null
+++ b/1073/CH5/EX5.16/5_16.sce
@@ -0,0 +1,22 @@
+clc;
+clear;
+//Example 5.16
+mo_dot=6*10^-2              //[kg/s]
+Cpo=2*10^3                       //Specific heat of oil in [J/kg.K]
+Cpw=4.18*10^3               //Specific heat of water in [J/kg.K]
+T1=420                      //[K]
+T2=320                      //[K]
+T=290                       //[K] Water entering temperature
+Q=mo_dot*Cpo*(T1-T2)        //[J/s]=[W]
+//Heat given out =Heat gained
+t2=Q/(mo_dot*Cpw)+T         //[K]
+dT1=T1-t2                   //[K]
+dT2=T2-T                    //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2) //[K]
+hi=1.6*1000             //[W/sq m.K]
+ho=3.6*1000           //[W/sq m.K]
+U=1/(1/ho+1/hi) //[W/sq m.K]
+A=Q/(U*dTlm)        //[sq m]
+D=0.025             //[m]
+L=A/(%pi*D)     //[m]
+printf("\n Length of tube required = %f m",L);
diff --git a/1073/CH5/EX5.17/5_17.sce b/1073/CH5/EX5.17/5_17.sce
new file mode 100755
index 000000000..fa3b43766
--- /dev/null
+++ b/1073/CH5/EX5.17/5_17.sce
@@ -0,0 +1,31 @@
+clc;
+clear;
+//Example 5.17
+mb_dot=1.25         //Benzene in [kg/s]
+Cpb=1.9*10^3             //For benzene in [J/kg.K]
+Cpw=4.187*10^3           //in [J/kg.K]
+T1=350              //[K]
+T2=300              //[K]
+Q=mb_dot*Cpb*(T1-T2)    //[W]
+t1=290                  //[K]
+t2=320                  //[K]
+dT1=T1-t2               //[K]
+dT2=T2-t1               //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2) //[K]
+mw_dot=Q/(Cpw*(t2-t1))        //Minimum flow rate of water in [kg/s]
+hi=850                      //[W/sq m.K]
+ho=1700                     //[W/sq m.K]
+Do=0.025                    //[m]
+Di=0.022                    //[m]
+x=(Do-Di) /2                    //Thickness in [m]
+hio=hi*(Di/Do)              //[W/sq m.K]
+Dw=(Do-Di)/log(Do/Di)       //[m]
+k=45                        //[W/m.K]
+Uo=1/((1/ho)+(1/hio)+(x/k)*(Do/Dw)) //[W/sq m.K]
+Ao=Q/(Uo*dTlm)                  //[sq m]
+L=1                             //Length in [m]
+area=%pi*Do*L                   // Outside surface area of tube per i m length 
+Tl=Ao/area                      //Total length of tubing required in [m]
+printf("\nTotal length of tubing required=%d m",round(Tl));      
+
+
diff --git a/1073/CH5/EX5.18/5_18.sce b/1073/CH5/EX5.18/5_18.sce
new file mode 100755
index 000000000..70294a8c0
--- /dev/null
+++ b/1073/CH5/EX5.18/5_18.sce
@@ -0,0 +1,62 @@
+clc;
+clear;
+//Example 5.18
+m_dot=4500          //Benzene condensation rate in [kg/h]
+lambda=394          //Latent heat of condensation of benzene in [kJ/kg]
+Q=m_dot*lambda      //[kJ/h]
+Q=Q*1000/3600       //[W]
+Cpw=4.18            //[kJ/kg.K]
+t1=295              //[K]
+t2=300              //[K]
+//For water :
+mw_dot=Q/(Cpw*1000*(t2-t1))      //[kg/s]
+rho=1000                    //[kg/m^3
+V=mw_dot/rho          //Volumetric flow rate in [m^3/s]
+u=1.05                //[m/s]
+A=V/u                 //Cross-sectional area  required in [sq m]
+
+//For tube:
+x=1.6               //thickness in [mm]
+x=x/1000            //[m]
+Do=0.025            //[m]
+Di=Do-2*x           //[m]
+A1=(%pi*Di^2)/4      //Of one tube [sq m]
+n=A/A1               //No. of tubes reuired   
+n=round(n)
+L=2.5               //Length of tube in [m]
+Ao=n*%pi*Do*L       //Surface area for heat transfer in [sq  m]
+Ts=353              //Condensing temp of benzene in [K]
+T1=295              //Inlet temperature in [K]
+T2=300              //Outlet temperature in [K]
+dT1=Ts-T1           //[K]
+dT2=Ts-T2           //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)     //[K]
+Uo=Q/(Ao*dTlm)          //[W/sq mK]
+Ud=Uo                   //[W/sq m.K]
+
+//OVERALL HEAT TRANSFER COEFFCIENT:
+//Inside side:
+T=(T2+T1)/2             //[K]
+
+hi=1063*((1+0.00293*T)*u^0.8)/(Di^0.2)        //[W/sq m.K]
+hio=hi*(Di/Do)                                  //[W/sq m.K]
+Dw=(Do-Di)/log(Do/Di)                           //[m]
+k=45                //For tube in [W/(m.)]
+
+//Outside of tube:
+mdot_dash=1.25/n                //[kg/s]
+M=mdot_dash/(%pi*Do)            //[kg/(m.s)]
+k=0.15                          //[W/(m.K)]
+rho=880                         //[kg/m^3]
+mu=0.35*10^-3                   //[N.s/sq m]
+g=9.81                          //[m/s^2] Acceleration due to gravity
+hm=(1.47*((4*mdot_dash)/mu)^(-1/3))/(mu^2/(k^3*rho^2*g))^(1/3)  //[W/sq m.K]
+ho=hm                           //[W/sq m.K]
+k=45                //[W/m]
+Uo=1/(1/ho+1/hio+(x*Do)/(k*Dw))
+//Uo=1/(1/ho+1/hio+(x*Do/(k*Dw)))     //Overall heat transfer coefficient in [W/sq m.K]
+Uc=Uo                               //[W/sq m.K]
+
+Rd=(Uc-Ud)/(Uc*Ud)                  //Maximum allowable sclae resistance in [K/W]
+printf("\n Uc(%f) is in excess of Ud(%f),therefore we allow for reasonable scale resistance,\nRd=%f K/W\n",Uc,Ud,Rd);
+printf("\n No.of tubes = %d ",n)
diff --git a/1073/CH5/EX5.19/5_19.sce b/1073/CH5/EX5.19/5_19.sce
new file mode 100755
index 000000000..98ecbc858
--- /dev/null
+++ b/1073/CH5/EX5.19/5_19.sce
@@ -0,0 +1,20 @@
+clc;
+clear;
+//Example 5.19
+mw_dot=5;            //Water flow rate in [kg/s]
+Cpw=4.18;            //Heat capacity of water [kJ/kg.K]
+t1=303;              //[K]
+t2=343;              //[K]
+Q=mw_dot*Cpw*(t2-t1)    //[kJ/s]
+Q=Q*1000;                //[W]
+T1=413;                  //[K]
+T2=373;                  //[K]
+dT1=T1-t2               //[K]
+dT2=T2-t1               //[K]
+dTlm=dT1                 ///[K]
+hi=1000;         //[W/sq m.K]
+ho=2500;         //[W/sq m.K]
+Rd=1/(0.714*1000)      //Fouling factor[m^2.K/KW]
+U=1/(1/hi+1/ho+Rd)      //[W/sq m.K]
+A=Q/(U*dTlm)            //[sq m]
+printf("\nHeat transfer area is %f sq m",A);
diff --git a/1073/CH5/EX5.2/5_2.sce b/1073/CH5/EX5.2/5_2.sce
new file mode 100755
index 000000000..35733abad
--- /dev/null
+++ b/1073/CH5/EX5.2/5_2.sce
@@ -0,0 +1,51 @@
+clc;
+clear;
+//Example 5.2
+ma_dot=300*1000/24          //Mass flow rate of acid in [kg/h]
+mw_dot=500*1000/24          //Mass flow rate of water in [kg/h]
+Cp1=1.465                    //[kJ/kg.K]
+T1=333                      //[K]
+T2=313                      //[K]
+Q=ma_dot*Cp1*(T1-T2)         //[kJ/h]
+Q=Q*1000/3600               //[W]
+Cp2=4.187                   //[kJ/kg.K]
+t1=288                      //[K]
+t2=(Q/(mw_dot*Cp2))+t1       //[K]
+dT1=T1-t2                   //[K]
+dT2=T2-t1                   //[K]    
+dTlm=(dT1-dT2)/log(dT1/dT2) //[K]
+dTlm=32.26              //Approximation in [K]
+//Inner pipe
+m_dot=12500                 //[kg/h]
+Di=0.075                    //[m]
+Ai=(%pi/4)*Di^2             //[sq m]
+G=ma_dot/Ai                 //[kg/m^2.h]
+G=G/3600                    //[kg/m^2.s]
+mu=0.0112                   //[kg/m.s]
+k=0.302                     //W/(m.K)
+Nre=Di*G/mu                 //Reynold number
+Npr=Cp1*10^3*mu/k                //Prandtl number
+hi=(k/Di)*0.023*(Nre^0.8)*(Npr^0.3)     //W/sq m.K
+Do=0.1                      //[m]
+hio=hi*Di/Do                //W/sq m.K
+D1=0.1                      //[m]
+D2=0.125                    //[m]
+De=(D2^2-D1^2)/D1           //[m]
+Aa=(%pi/4)*(D2^2-D1^2)      //[sq m]
+Ga=mw_dot/Aa                //[kg/m^2.h]
+Ga=Ga/3600                  //[kg/sq m.s]
+mu=0.0011                   //[kg/m.s]
+Nre=De*Ga/mu                //Reynolds number
+k=0.669                     //for water
+Npr=Cp2*10^3*mu/k                 //Prandtl number
+ho=(k/De)*0.023*(Nre^0.8)*Npr^0.4    //[W/sq m.K]
+xw=(Do-Di)/2                     //[m]
+Dw=(Do-Di)/log(Do/Di)           //[m]
+kw=46.52                    //thermal conductivity of wall in [W/m.K]
+Uc=1/(1/ho+1/hio+xw*Do/(kw*Dw))  //[W/sq m.K]
+Ud=Uc                       //As dirt factor values are not given
+Ud=195.32           //Approximation
+A=Q/(Ud*dTlm)             //[sq m]
+
+L=A/(%pi*Do)                //[sq m]
+printf("\nArea =%f m^2,\nLength fo pipe required =%f m(approx)",A,L)
diff --git a/1073/CH5/EX5.20/5_20.sce b/1073/CH5/EX5.20/5_20.sce
new file mode 100755
index 000000000..3c952a063
--- /dev/null
+++ b/1073/CH5/EX5.20/5_20.sce
@@ -0,0 +1,20 @@
+clc;
+clear;
+//Example 5.20
+Cpo=1.9          //Heat capacity for oil[kJ/kg.K]
+Cps=1.86       //Heat capacity for steam [kJ/kg.K]
+ms_dot=5.2      //Mass flow rate in [kg/s]
+T1=403          //[K]
+T2=383          //[K]
+
+Q=ms_dot*Cps*(T1-T2)        //[kJ/s]
+Q=Q*1000                    //[W]
+t1=288;                      //[K]
+t2=358;                      //[K]
+dT2=T1-t2                   //[K]
+dT1=T2-t1                   //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)     //LMTD in [K]
+U=275 ;                          //Overall heat transfer coeffcient in [W//sq m.K]
+Ft=0.97                         //LMTD correction factor
+A=Q/(U*Ft*dTlm)                 //[sq m]
+printf("\nHeat exchanger surface area is %f sq m",A);
diff --git a/1073/CH5/EX5.21/5_21.sce b/1073/CH5/EX5.21/5_21.sce
new file mode 100755
index 000000000..c8b7ff5c5
--- /dev/null
+++ b/1073/CH5/EX5.21/5_21.sce
@@ -0,0 +1,38 @@
+clc;
+clear;
+//Example 5.21
+mc_dot=3.783;        //Cold water flow rate[kg/s]
+mh_dot=1.892;       //Hot water flow rate [kg/s]
+Cpc=4.18;           //Sp heat of cold water [kJ/(kg.K)]
+T1=367;             //[K]
+t2=328;             //[K]
+t1=311;             //[K]
+Cph=4.18;           //Specific heat of hot water [kJ/(kg.K)]
+rho=1000;            //Density [kg/m^3]
+D=0.019;             //Diameter of tube in [m]
+U=1450 ;             //Overal heat transfer coefficient in [W/sq m.K]   
+T2=T1-mc_dot*Cpc*(t2-t1)/(mh_dot*Cph)   //[K]
+Q=mc_dot*Cpc*(t2-t1)        //[kJ/s]
+Q=Q*1000                    //[W]
+//For counterflow heat exchanger
+dT1=T1-t2                   //[K]
+dT2=17;                     //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2) //[K]
+lmtd=dTlm                   //LMTD 
+Ft=0.88                     //LMTD correction factor
+A=Q/(U*dTlm)                //[sq m]
+u=0.366;                    //Velocity through tubes[ms^-1]
+Ai=mc_dot/(rho*u)           //Total flow Area in [sq m]
+n=Ai/((%pi/4)*(D^2))        //No. of tubes 
+L=1                         //Per m length[m]
+sa=%pi*D*L                  //S.S per tube per 1 m length
+L=A/(n*%pi*D)               //Length of tubes in [m]
+printf("\nThe length is more than allowable 2.44 m length,so we must use more than one tube \n");
+
+//For 2passes on the tube side
+A=Q/(U*Ft*lmtd)     //[sq m]
+L=A/(2*n*%pi*D)     //Length in [m]
+printf("\n This length is within 2.44 m requirement,so the design choice is \n\n");
+printf("\nType of heat exchanger : 1-2 Shell and tube heat exchanger\n")
+printf("\nNo of tubes per pass= %d\n",round(n));
+printf("\nLength of tube per pass=%f m\n ",L);
diff --git a/1073/CH5/EX5.22/5_22.sce b/1073/CH5/EX5.22/5_22.sce
new file mode 100755
index 000000000..e7c741b85
--- /dev/null
+++ b/1073/CH5/EX5.22/5_22.sce
@@ -0,0 +1,35 @@
+clc;
+clear;
+//Example 5.22
+mh_dot=16.67;       //Mass flow rate of hot fluid in [kg/s]
+mc_dot=20;         //Mass flow rate of cold fluid in [kg/s]
+Cph=3.6;            //Sp heat of hot fluid in [kJ/kg.K]
+Cph=Cph*1000;        //Sp heat of hot fluid in [J/kg.K]
+Cpc=4.2;            //Sp heat of cold fluid in [kJ/(kg.K)]
+Cpc=Cpc*1000;        //Sp heat of cold fluid in [J/(kg.K)]
+U=400;              //Overall heat transfer coefficient in [W/sq m.K]
+A=100;              //Surface area in [sq m]
+mCp_h=mh_dot*Cph     //[J/s] or [W/K]
+mCp_c=mc_dot*Cpc     //[J/s] or [W/K]
+mCp_small=mCp_h     //[W/K]
+C=mCp_small/mCp_c   //Capacity ratio           
+ntu=U*A/mCp_small   //NTU
+T1=973;             //Hot fluid inlet temperature in [K]
+t1=373;             //Cold fluid inlet temperature in [K]
+//Case 1:Countercurrent flow arrangement
+E=(1-%e^(-(1-C)*ntu))/(1-C*%e^(-(1-C)*ntu)) //Effectiveness
+//W=T1-T2/(T1-t1)  therefore:
+T2=T1-E*(T1-t1)     //[K]
+printf("\nExit temperature of hot fluid is %d K",round(T2));
+t2=mCp_h*(T1-T2)/(mCp_c)+t1     //[From energy balance eqn in ][K]
+printf("\nExit temperature of cold fluid is %d K(%d C)\n",round(t2),round(t2-273));
+
+//Case 2:Parallel flow arrangement
+E1=(1-%e^(-(1+C)*ntu))/(1+C)
+//In the textbok here is a calculation mistake,and the value of E is takne as E=0.97
+
+T2=T1-E1*(T1-t1)     //[K]
+t2=mCp_h*(T1-T2)/(mCp_c)+t1     //[From energy balance eqn in ][K]
+printf("\nExit temperature of Hot water=%f K\n",T2);
+printf("\nExit temperature of cold water=%f K\n",t2);
+
diff --git a/1073/CH5/EX5.23/5_23.sce b/1073/CH5/EX5.23/5_23.sce
new file mode 100755
index 000000000..5a9823fc7
--- /dev/null
+++ b/1073/CH5/EX5.23/5_23.sce
@@ -0,0 +1,38 @@
+clc;
+clear;
+//Example 5.23
+Cpo=2131;       //Sp heat of oil in [J/kg.K]
+Cpw=4187;       //Sp heat of water in [J/kg.K]
+mo_dot=0.10;    //Oil flow rate in [kg/s]
+mw_dot=0.20;     //Water flow rate in [kg/s]
+U=380;          //Overall heat transfer coeff in [W/sq m.K]
+T1=373;         //Initial temp of oil [K]
+T2=333;         //Final temperature of oil [K]
+t1=303;         //Water enter temperature in [K]
+t2=t1+mo_dot*Cpo*(T1-T2)/(mw_dot*Cpw)   //[K]   
+//1.LMTD method
+dT1=T1-t2           //[K]
+dT2=T2-t1           //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)     //[K]
+lmtd=dTlm;           //[K]
+Q=mo_dot*Cpo*(T1-T2)        //[J/s]
+A=Q/(U*dTlm)                //[sq m]
+Do=0.025;                   //Inner tubve diameter [m]
+L=A/(%pi*Do)                //Length in [m]
+
+//2.NTU method
+mCp_c=mw_dot*Cpw               //[W/K]
+mCp_h=mo_dot*Cpo               //[W/K]
+printf("\n In textbook this value of mCp_h is wrongly calculated as 231.1  so we will take this only for calculation\n");
+mCp_h=231.1;                //[W/K]
+//mCp_h is smaller
+C=mCp_h/mCp_c
+E=(T1-T2)/(T1-t1)               //Effeciency
+//For countercurrent flow
+deff('[x]=f(ntu)','x=E-(1-%e^(-(1-C)*ntu))/(1-C*%e^(-(1-C)*ntu))')
+ntu=fsolve(1,f)
+A=ntu*mCp_h/U       //[sq m]
+A=0.56      //Approximately
+L1=A/(%pi*Do)        //Length in [m]
+printf("\nFrom LMTD approach:\n length=%f m\n",L);
+printf("\nFrom NTU method:\n length=%f m\n",L1);
diff --git a/1073/CH5/EX5.24/5_24.sce b/1073/CH5/EX5.24/5_24.sce
new file mode 100755
index 000000000..9edcdea6e
--- /dev/null
+++ b/1073/CH5/EX5.24/5_24.sce
@@ -0,0 +1,37 @@
+
+clc;
+clear;
+//Example 5.24
+ho=200;     //[W/sq m.K]
+hi=1500;    //[W/sq m.K]
+Cpw=4.2;     //Sp heat of Water in [kJ/(kg.K)]
+Cpo=2.1;     //Sp heat of Oil in [kJ/(kg.K)]
+E=0.8;      //Effectiveness
+k=46;       //[W/m.K]
+m_dot=0.167;    //[kg/s]
+
+mCp_oil=2*m_dot*Cpo*1000      //For oil [W/K]
+//mCp_oil is wrongly calculated as 710.4
+mCp_water=m_dot*Cpw*1000      //For water [W/K]
+//mCp_oil is wrongly calculated as 710.4
+//NOTE:The above two values are wrongly calculated in book as 710.4
+//so we take here:
+mCp_small=710.4     //[W/K]
+//Since both mCp_water and mCp_oil are equal ,therefore:
+C=1;
+
+deff('[x]=f(ntu)','x=E-(ntu/(1+ntu))');
+ntu=fsolve(1,f)
+id=20;      //Internal diameter in [mm]
+od=25;      //External diameter in [mm]
+hio=hi*id/od        //[W/sq m.K]
+Dw=(od-id)/log(od/id)   //[mm]
+Dw=Dw/1000              //[m]
+x=(od-id)/2             //[mm]
+x=x/1000            //[m]
+Do=0.025            //External dia in [m]
+L=2.5;              //Length of tube in [m]
+Uo=1/(1/ho+1/hio+(x/k)*(Dw/Do))     //[W/sq m.K]
+A=ntu*mCp_small/Uo   //Heat transfer area in [sq m]
+n=A/(%pi*Do*L)      //No of tubes
+printf("\nNo. of tubes required = %d",round(n+1));
diff --git a/1073/CH5/EX5.25/5_25.sce b/1073/CH5/EX5.25/5_25.sce
new file mode 100755
index 000000000..ede38ee5d
--- /dev/null
+++ b/1073/CH5/EX5.25/5_25.sce
@@ -0,0 +1,25 @@
+clc;
+clear;
+//Example 5.25
+
+//(i)Parallel flow
+T1=633;     //[K]
+t2=303;     //[K]
+T2=573;     //[K]
+t1=400;     //[K]
+dT1=T1-t2;  //[K]
+dT2=T2-t1;  //[K]
+mh_dot=1.2; //[kg/s]
+U=500;      //Overall heat transfer coefficient in [W/sqm.K]
+Cp=2083;        //Sp.heat of oil J/kg.K
+dTlm=(dT1-dT2)/log(dT1/dT2) //[K]
+Q=mh_dot*Cp*(T1-T2)         //[W]
+A=Q/(U*dTlm)                //[sq m]
+
+//(ii)Counter current flow
+dT1=T1-t1;           //[K]
+dT2=T2-t2;          //[K]
+dTlm=(dT2-dT1)/log(dT2/dT1)     //[K]
+A1=Q/(U*dTlm)        //[sq m]
+printf("\nFor parallel flow,Area = %f sq m \n For countercurrent flow,Area=%f sq m\n",A,A1);
+printf("\n\nFor the same terminal temperatures of the fluid ,the surface area for the counterflow arrangement\n is less than the required for the parallel flow\n")
diff --git a/1073/CH5/EX5.3/5_3.sce b/1073/CH5/EX5.3/5_3.sce
new file mode 100755
index 000000000..d6d0a6ed4
--- /dev/null
+++ b/1073/CH5/EX5.3/5_3.sce
@@ -0,0 +1,47 @@
+clc;
+clear;
+//Example 5.3
+me_dot=5500 ;        //[kg/h]
+me_dot1=me_dot/3600          //[kg/s]
+Di=0.037     ;   //I.D of inner pipe in [m]
+Ai=(%pi/4)*Di^2         //[sq m]
+G=me_dot1/Ai             //[kg/sq m.s]
+mu=3.4*10^-3   ;         //[Pa.s] or [kg/(m.s)]
+Nre=Di*G/mu             //Reynolds number
+Cp=2.68       ;  //[kJ/kg.K]
+Cp1=Cp*10^3          //[J/kgK]
+k=0.248        ; //[W/m.K]
+Npr=Cp1*mu/k     //Prandtl number
+//Nre is greater than 10,000,Use Dittus-Boelter eqn:
+Nnu=0.023*(Nre^0.8)*(Npr^0.3)       //Nusselt number
+hi=k*Nnu/Di                 //[W/sq m.K]
+T2=358          //[K]
+T1=341          //[K]
+Cp2=1.80            //[kJ/kg.K]
+t2=335              //[K]
+t1=303              //[K]
+mt_dot=me_dot*Cp*(T2-T1)/(Cp2*(t2-t1))  //[kg/h]
+mt_dot=mt_dot/3600                      //[kg/s]
+D1=0.043                //[m]
+D2=0.064            //Inside dia of outer pipe
+De=(D2^2-D1^2)/D1           //Equivalent diameter [m]
+Aa=%pi/4*(D2^2-D1^2)         //[sq m]
+Ga=mt_dot/Aa                    //kg/(sq m.s)
+mu2=4.4*10^-4                    // Viscosity of toluene Pa.s
+k2=0.146                    //For toluene [W/m.K]
+Cp2=1.8*10^3            //J/kg.K
+Nre=De*Ga/mu2            //Reynolds number
+Npr=Cp2*mu2/k2          //Prandtl number
+Nnu=0.023*Nre^0.8*Npr^0.4       //Nusselt number
+ho=k2*Nnu/De                     //W/(sq m.K)
+Dw=(D1-Di)/log(D1/Di)             //[m]
+x=0.003                     //Wall thickness in [m]
+Uo=1/(1/ho+(1/hi)*(D1/Di)+(x*D1/(46.52*Dw)))     //[W/sq m.K]
+dT1=38      //[K]
+dT2=23      //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)                 //[K]
+Q=me_dot1*Cp*(T2-T1)       //[kJ/s]
+Q=Q*1000                //[J/s]
+L=Q/(Uo*%pi*D1*dTlm)    //[m]
+printf("\nTotal lenggth of double pipe heat exchanger is %f m",L)
+
diff --git a/1073/CH5/EX5.4/5_4.sce b/1073/CH5/EX5.4/5_4.sce
new file mode 100755
index 000000000..488b312d9
--- /dev/null
+++ b/1073/CH5/EX5.4/5_4.sce
@@ -0,0 +1,24 @@
+clc;
+clear;
+//Example 5.4
+mc_dot=1000         //[kg/h]
+mc_dot=mc_dot/3600      //[kg/s]
+mh_dot=250          //[kg/h]
+mh_dot=mh_dot/3600  //[kg/s]
+Cpc=4187          //[J/(kg.K)]
+Cph=3350        //[W/K]
+w=mc_dot*Cpc        //[W/K]
+l=mh_dot*Cph        //[W/K]
+C=mh_dot*Cph/(mc_dot*Cpc)
+U=1160      //[W/sq m.K]
+A=0.25      //Heat transfer surface for exchanger in [sq m]
+ntu=U*A/(mh_dot*Cph)    //
+E=(1-%e^(-ntu*(1+C)))/(1+C)     //Effectiveness of heat exchanger
+T1=393      //Inlet temperature in [K]
+t1=283      //Cooling water [K]
+T2=T1-E*(T1-t1)     //Outlet T of hot liquid 
+
+t2=C*(T1-T2)+t1     //[K]
+printf("\n\nEffectiveness of heat exchanger is %f\n",E);
+printf("\nOutlet temperature of hot liquid is %f\n",T2);
+printf("\nOutlet temperature of water is %f\n",t2)
diff --git a/1073/CH5/EX5.5/5_5.sce b/1073/CH5/EX5.5/5_5.sce
new file mode 100755
index 000000000..ec084f87f
--- /dev/null
+++ b/1073/CH5/EX5.5/5_5.sce
@@ -0,0 +1,26 @@
+
+clc;
+clear;
+//Example 5.5
+Cpc=4187                //Specific heat of water in [J/(kg.K)]
+Cph=2000                //Sp heat of oil in [J/(kg.K)]
+mc_dot=1300/3600        //[kg/s]
+mh_dot=550/3600         //[kg/s]
+w=mc_dot*Cpc        //[W/K]
+o=mh_dot*Cph        //[W/K]
+//Heat capacity of rate of hot fluid is smaller than water
+U=1075          //[W/sq m.K]
+A=1       //[sq m]
+ntu=(U*A)/(mh_dot*Cph)    
+C=mh_dot*Cph/(mc_dot*Cpc)   
+E=(1-%e^(-ntu*(1-C)))/(1-C*%e^(-ntu*(1-C)))     //Effeciency
+T1=367          //[K]
+t1=288          //[K]
+T2=T1-E*(T1-t1)                 //Outlet temperature [K]
+T2=291.83                   //Approximated in book without precise calculation
+t2=C*(T1-T2)+t1                 //[K]
+Q=mh_dot*Cph*(T1-T2)            //[W]
+printf("\n\nEffectiveness of exchanger is %f\n",E);
+printf("\nOutlet temperature of oil is %f K\n",T2);
+printf("\nOutlet temperature of water is %f K\n",t2);
+printf("\nRate of heat transfer is %f W",Q);
diff --git a/1073/CH5/EX5.6/5_6.sce b/1073/CH5/EX5.6/5_6.sce
new file mode 100755
index 000000000..5d37cab8e
--- /dev/null
+++ b/1073/CH5/EX5.6/5_6.sce
@@ -0,0 +1,37 @@
+clc;
+clear;
+//Example 5.6
+printf("\nLMTD Approach\n")
+
+Cph=4187            //[J/(kg.K)]
+mh_dot=600/3600     //Hot side flow rate [kg/s]
+mc_dot=1500/3600        //[kg/s]
+Cpc=Cph             //[J/kg.K]
+T1=343              //[K]
+T2=323              //[K]
+Q=mh_dot*Cph*(T1-T2)    //[W]
+t1=298                  //[K]
+t2=(mh_dot*Cph*(T1-T2))/(mc_dot*Cpc)+t1 //[K]
+dT1=45      //[K]
+dT2=17          //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)         //[K]
+hi=1600         //Heat transfer coeff in [W/sq m.K]
+ho=hi           //[W/sq m.K]
+U=1/(1/hi+1/ho)     //[W/sq m.K]
+A=Q/(U*dTlm)        //[sq m]
+
+printf("\nEffectiveness-NTU approach\n");
+
+//hot water:
+h=mh_dot*Cph            //[W/K]
+c=mc_dot*Cpc            //[W/K]
+//Heat capacity rate of hot fluid is small
+C=mh_dot*Cph/(mc_dot*Cpc)       //
+E=(T1-T2)/(T1-t1)       //Effectiveness
+//for paralell flow:
+ntu=-log(1-E*(1+C))/(1+C)       
+A2=(ntu*mh_dot*Cph)/U    //[sq m]
+t2=C*(T1-T2)+t1         //[K]
+printf("\n By LMTD approach area of heat exchanger is %f sq m\n",A);
+printf("\nBy Ntu approach Area of heat exchanger is %f sq  m\n",A);
+printf("\n Outlet temperature of cold water=%f K\n",t2)
diff --git a/1073/CH5/EX5.7/5_7.sce b/1073/CH5/EX5.7/5_7.sce
new file mode 100755
index 000000000..1520e3856
--- /dev/null
+++ b/1073/CH5/EX5.7/5_7.sce
@@ -0,0 +1,22 @@
+clc;
+clear;
+//Example 5.7
+mw_dot=10           //[kg/s]
+Cpw=4.187           //[kJ/(kg.K)]
+t2=318              //[K]
+t1=295              //[K]
+Q=mw_dot*Cpw*(t2-t1)    //[kJ/s]
+Q=Q*1000                //W
+dT1=98              //[K]
+dT2=75              //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)         //[K]
+hi=850          //[W/sq m.K]
+id=0.027            //Inside dia[m]
+od=0.031                 //Outside dia [m]
+hio=hi*id/od            //[W/sq  m.K]
+ho=6000     //Heat transfer coefficients[W/sq m.K]
+Uo=1/(1/ho+1/hio)       //[W/sq m.K]
+Ao=Q/(Uo*dTlm)          //[sq m]
+L=4         //Length [m]
+n=Ao/(%pi*od*L)         //[No. of tubes]
+printf("\n Number of tubes required = %d\n",round(n));
diff --git a/1073/CH5/EX5.8/5_8.sce b/1073/CH5/EX5.8/5_8.sce
new file mode 100755
index 000000000..e90dd9eb5
--- /dev/null
+++ b/1073/CH5/EX5.8/5_8.sce
@@ -0,0 +1,51 @@
+clc;
+clear;
+//Example 5.8
+mdot=7250;           //Nitrobenzene in shell in [kg/h]
+Cp=2.387;            //[kJ/(kg.K)]
+mu=7*10^-4 ;         //Pa.s
+k=0.151;             //[W/m.K]
+T1=400;              //[K]
+T2=317;              //[K]
+t1=305;              //[K]
+t2=345;              //[K]
+dT1=T1-t2              //[K]
+dT2=T2-t1              //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)     //[K]
+Q=mdot*Cp*(T1-T2)   //[kJ/h]
+Q=Q*1000/3600           //[W]
+n=166;                   //no of tubes
+L=5;                     //[m]
+Do=0.019;                //[m]
+Di=0.015                 //[m]
+Ao=n*%pi*Do*L           //[sq m]
+Uo=Q/(Ao*dTlm)          //[W/sq m.K]
+Ud=Uo
+//Shell side heat transfer coefficient
+Pt=0.025                //[m]
+C_dash=Pt-(0.5*Do+0.5*Do)
+
+//Shell side crossflow area
+B=0.15              //[m]
+id=0.45     //[m]
+as=id*C_dash*B/Pt       //[sq m]
+//As there are two shell passes,area per pass is :
+as_dash=as/2            //[sq m]
+
+//Equivalent diameter of shell 
+De=4*(Pt^2-(%pi/4)*Do^2)/(%pi*Do)       //[m]
+
+//Mass velocity on shell side
+Gs=mdot/as_dash        //[kg/m^2.h]
+Gs=Gs/3600              //[kg/m^2.s]
+mu=7*10^-4              //Pa.s
+Cp=Cp*1000              //J/kg.K
+Nre=De*Gs/mu    //Reynold number
+Npr=Cp*mu/k     //Prandtls number
+Nnu=0.36*Nre^0.55*Npr^(1.0/3.0) //Nusselts number
+hi=1050             //[W/sq m .K]
+ho=Nnu*k/De         //[W/sq m.K]
+Uo=1/(1/ho+(1/hi*(Do/Di)))          //[W/sq m K]
+Uc=Uo
+Rd=(Uc-Ud)/(Uc*Ud)                  //m^2.K/W
+printf("\n Fouling factor=Sclae resistance=%f m^2.K/W\n",Rd);
diff --git a/1073/CH5/EX5.9/5_9.sce b/1073/CH5/EX5.9/5_9.sce
new file mode 100755
index 000000000..dec92f0d8
--- /dev/null
+++ b/1073/CH5/EX5.9/5_9.sce
@@ -0,0 +1,38 @@
+clc;
+clear;
+//Example 5.9
+k=0.628         //W/(m.K)
+rho=980     //[kg/m^3]
+mu=6*10^-4      //kg/(m.s)
+Cpw=4.187        //kJ/(kg.K)
+Cp=Cpw*10^3      //J/(kg.K)
+Di=25           //[mm]
+Di=Di/1000      //[m]
+mw_dot=1200*10^-3*rho        //Mass flow rate of water [kg/h]
+mw_dot=mw_dot/3600            //[kg/s]
+Ai=(%pi*Di^2)/4              //Inside area of tube in sq m
+G=mw_dot/Ai                 //kg/m^2.s
+Nre=Di*G/mu                 //Reynolds number
+Npr=Cp*mu/k                 //Pranddtl number
+//Inside heat transfer coefficient
+Nnu=0.023*Nre^0.8*Npr^0.4       //Nusselt number
+hi=Nnu*k/Di             //[W/sq m.K]
+ho=6000     //[W//sq m.K]
+Do=0.028        //[m]
+Dw=(Do-Di)/log(Do/Di)       //[m]
+x=(Do-Di)/2                  //[m]
+k2=348.9            //thermal conductivity of metal in [W/m.K]
+Uo=1/((1/ho)+(1/hi)*(Do/Di)+(x/k2)*(Do/Dw))  //[W/sq m.K]
+t1=303          //[K]
+t2=343          //[K]
+Q=mw_dot*Cpw*(t2-t1)        //[kJ/h]
+Q=Q*1000              //[W]
+Ts=393                      //[K]
+
+dT1=Ts-t1                   //[K]
+dT2=Ts-t2                   //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)     //[K]
+Ao=Q/(Uo*dTlm)          //[sq m]
+L=Ao/(%pi*Do)           //Length
+printf("\n therefore length of tube required is %f m\n",L);
+
diff --git a/1073/CH6/EX6.1/6_1.sce b/1073/CH6/EX6.1/6_1.sce
new file mode 100755
index 000000000..bf0daa28a
--- /dev/null
+++ b/1073/CH6/EX6.1/6_1.sce
@@ -0,0 +1,10 @@
+clc;
+clear;
+//Example6.1
+T=380       //B.P of solution[K]
+T_dash=373      //B.P of water [K]
+BPE=T-T_dash        //Boiling point elevation in [K]
+Ts=399      //Saturating temperature in [K]
+DF=Ts-T     //Driving force in [K]
+printf("\nBoiling point of elevation of the solution is %d K \n",BPE);
+printf("\nDriving forve for heat transfer is %d K \n",DF)
diff --git a/1073/CH6/EX6.10/6_10.sce b/1073/CH6/EX6.10/6_10.sce
new file mode 100755
index 000000000..265cc9524
--- /dev/null
+++ b/1073/CH6/EX6.10/6_10.sce
@@ -0,0 +1,18 @@
+clc;
+clear;
+//Example 6.10
+Ts=381.3        //[K]
+dT=56.6;     //[K]
+U1=2800; //Overall heat transfer coeff in first effect
+U2=2200; //Overall heat transfer coeff in first effect
+U3=1100; //Overall heat transfer coeff in first effect
+dT1=dT/(1+(U1/U2)+(U1/U3))  ///[K]
+dT2=dT/(1+(U2/U1)+(U2/U3))  ///[K]
+dT3=dT-(dT1+dT2)             //[K]
+//dT1=Ts-T1_dash      //[K]
+T1dash=Ts-dT1
+//dT2=T1_dash-T2_dash         //[K]
+T2_dash=T1dash-dT2             //[K]
+printf("\n\nBoiling point of solution in first effect =%f K\n\n",T1dash);
+printf("\n\nBoiling point of solution in second effect =%f K\n\n",T2_dash);
+
diff --git a/1073/CH6/EX6.11/6_11.sce b/1073/CH6/EX6.11/6_11.sce
new file mode 100755
index 000000000..656d600aa
--- /dev/null
+++ b/1073/CH6/EX6.11/6_11.sce
@@ -0,0 +1,97 @@
+clc;
+clear;
+//Example 6.11
+mf_dot=10000        //[kg/h] of feed
+ic=0.09     //Initial concentration 
+fc=0.47     //Final concentration
+m1dot_dash=ic*mf_dot/fc     //[kg/h]
+Ps=686.616      //Steam pressure [kPa.g]
+Ps=Ps+101.325       //[kPa]
+Ts=442.7        //Saturation temperature in [K]
+P2=86.660       //Vacuum in second effect in [kPa]
+U1=2326     //Overall heat transfer in first effect [W/sq m.K]
+U2=1744.5   //Overall heat transfer in 2nd effect [W/sqm.K]
+P2_abs=101.325-P2   //Absolute pressure in second effect[kPa]
+T2=326.3        //Temperature in 2nd effect in [K]
+dT=Ts-T2        //[K]
+Tf=309      //Feed temperature in[K]
+T=273       //[K]
+Cpf=3.77        //kJ/kg.K  Specific heat for all caustic streams
+//Q1=Q2
+//U1*A1*dT1=U2*A2*dT2
+dT2=dT/1.75     //[K]
+dT1=(U2/U1)*dT2     //[K]
+//Since there is no B.P.R
+Tv1=Ts-dT1      //Temperature in vapor space of first effect in [K]
+Tv2=Tv1-dT2     //Second effect [K]
+Hf=Cpf*(Tf-T)       //Feed enthalpy[kJ/kg]
+H1dash=Cpf*(Tv1-T)      //Enthalpy of final product[kJ/kg]
+H2dash=Cpf*(Tv2-T)      //kJ/kg
+//For steam at 442.7 K
+lambda_s=2048.7     //[kJ/kg]
+//For vapour at 392.8 K
+Hv1=2705.22     //[kJ/kg]
+lambda_v1=2202.8        //[kJ/kg]
+//for vapour at 326.3 K:
+Hv2=2597.61     //[kJ/kg]
+lambda_v2=2377.8        //[kJ/kg]
+
+//Overall material balance:
+mv_dot=mf_dot-m1dot_dash        //[kg/h]
+
+//Equation 4 becomes:
+//mv1_dot*lambda_v1+mf_dot*Hf=(mv_dot-mv1_dot)*Hv2+(mf_dot-mv2_dot)*H2_dash
+mv1_dot=(H2dash*(mf_dot-mv_dot)-mf_dot*Hf+mv_dot*Hv2)/(Hv2+lambda_v1-H2dash)  
+mv2_dot=mv_dot-mv1_dot              //[kg/h]
+
+//From equation 2
+
+m2dot_dash=m1dot_dash+mv1_dot           //First effect material balance[kg/h]
+ms_dot=(mv1_dot*Hv1+m1dot_dash*H1dash-m2dot_dash*H2dash)/lambda_s     //[kg/h]
+
+
+//Heat transfer Area
+//First effect
+A1=ms_dot*lambda_s*(10^3)/(3600*U1*dT1)     //[sq m]
+
+//Second effect
+lambda_v1=lambda_v1*(10^3/3600)
+A2=mv1_dot*lambda_v1/(U2*dT2)       //[sq m]
+
+//Since A1 not= A2
+
+//SECOND TRIAL
+Aavg=(A1+A2)/2          //[sq m]
+dT1_dash=dT1*A1/Aavg        //[K]
+dT2_dash=dT-dT1         ///[K]
+
+//Temperature distribution
+Tv1=Ts-dT1_dash         //[K]
+Tv2=Tv1-dT2_dash            //[K]
+Hf=135.66       //[kJ/kg]
+H1dash=Cpf*(Tv1-T)      //[kJ/kg]
+H2dash=200.83          //[kJ/kg]
+
+//Vapour at 388.5 K
+Hv1=2699.8      //[kJ/kg]
+lambda_v1=2214.92       //[kJ/kg]
+mv1_dot=(H2dash*(mf_dot-mv_dot)-mf_dot*Hf+mv_dot*Hv2)/(Hv2+lambda_v1-H2dash)  
+mv2_dot=mv_dot-mv1_dot  //[kg/h]
+
+//First effect Energy balance
+ms_dot=((mv1_dot*Hv1+m1dot_dash*H1dash)-(mf_dot-mv2_dot)*H2dash)/lambda_s   //[kg/h]
+
+//Area of heat transfer
+lambda_s=lambda_s*1000/3600     
+A1=ms_dot*lambda_s/(U1*dT1_dash)        //[sq m]
+
+//Second effect:
+A2=mv1_dot*lambda_v1*1000/(3600*U2*dT2_dash)        //[sq m]
+
+printf("\nA1(%f)=A2(%f),So the area in each effect can be %f sq m\n",A1,A2,A2);
+printf("\nHeat transfer surface in each effect is %f sq m\n",A2);
+printf("\nSteam consumption=%d kg/h\n",round(ms_dot));
+printf("\nEvaporation in the first effect is %d kg/h\n",round(mv1_dot));
+printf("\nEvaporation in  2nd effect is %d kg/h\n",round(mv2_dot));
+
+
diff --git a/1073/CH6/EX6.12/6_12.sce b/1073/CH6/EX6.12/6_12.sce
new file mode 100755
index 000000000..301b1b7af
--- /dev/null
+++ b/1073/CH6/EX6.12/6_12.sce
@@ -0,0 +1,87 @@
+
+clc;
+clear;
+//Example 6.12
+Tf=353;          //[K]
+T=273           //[K]
+mf_dot=10000;         //Feed [kg/h]
+ic=0.07;         //Initial conc of glycerine 
+fc=0.4;          //FinaL CONC OF GLYCERINE
+//Overall glycerine balance
+m3dot_dash=(ic/fc)*mf_dot           //[kg/h]
+mv_dot=mf_dot-m3dot_dash            ///[kg/h]
+P=313;           //Steam pressure[kPa]
+Ts=408;      //[from steam table][K]
+P1=15.74;        //[Pressure in last effect][kPa]
+Tv3=328;         //[Vapour temperature]
+dT=Ts-Tv3      //Overall apparent [K]
+bpr1=10  ;       //[K]
+bpr2=bpr1;
+bpr3=bpr2;
+sum_bpr=bpr1+bpr2+bpr3      //[K]
+dT=dT-sum_bpr               //True_Overall
+dT1=14.5;            //[K]
+dT2=16;              //[K]
+dT3=19.5;            //[K]
+Cpf=3.768            //[kJ/(kg.K)]
+//Enthalpies of various streams
+Hf=Cpf*(Tf-T)           //[kJ/kg]
+H1=Cpf*(393.5-T)           //[kJ/kg]
+H2=Cpf*(367.5-T)           //[kJ/kg]
+H3=Cpf*(338-T)           //[kJ/kg]
+//For steam at 40K
+lambda_s=2160       //[kJ/kg]
+Hv1=2692        //[kJ/kg]
+lambda_v1=2228.3        //[kJ/kg]
+Hv2=2650.8          //[kJ/kg]
+lambda_v2=2297.4        //[kJ/kg]
+Hv3=2600.5              //[kJ/kg]
+lambda_v3=2370      //[kJ/kg]
+
+//MATERIAL AND EBERGY BALANCES
+//First effect
+//Material balance
+
+//m1dot_dash=mf_dot-mv1_dot
+//m1dot_dash=1750+mv2_dot+mv3_dot           
+
+//Energy balance
+//ms_dot*lambda_s+mf_Dot*hf=mv1_dot*Hv1+m1dot_dash*H1
+//2160*ms_dot+2238*(mv2_dot+mv3_dot)=19800500
+
+//Second effect
+//Energy balance:
+//mv3_dot=8709.54-2.076*mv2_dot
+
+//Third effect:
+//m2dot_dash=mv3_dot+m3dot_dash
+//m2dot_dash=mv3_dot+1750
+//From eqn 8 we get
+mv2_dot=(8709.54*2600.5+1750*244.92-8790.54*356.1-356.1*1750)/(-2.076*356.1+2297.4+2600.5*2.076)
+//From eqn 8:
+mv3_dot=8709.54-2.076*mv2_dot           //[kg/h]
+mv1_dot=mv_dot-(mv2_dot+mv3_dot)        //[kg/h]
+//From equation 4:
+//m1dot_dash=mf_dot-mv1_dot
+//ms_dot=(mv1_dot*Hv1+m1dot_dash*H1-mf_dot*Hf)/lambda_s   //[kg/h]
+ms_dot=(19800500-2238*(mv2_dot+mv3_dot))/2160           //[kg/h]
+
+//Heat transfer Area is
+U1=710          //[W/sq m.K]
+U2=490          //[W/sq m.K]
+U3=454          //[W/sq m.K]
+A1=ms_dot*lambda_s*1000/(3600*U1*dT1)     //[sq m]
+A2=mv1_dot*lambda_v1*1000/(3600*U2*dT2)   //[sq m]
+A3=mv2_dot*lambda_v2*1000/(3600*U3*dT3)   //[sq m]
+//The deviaiton is within +-10%
+//Hence maximum A1 area can be recommended
+
+eco=(mv_dot/ms_dot)     //[Steam economy]
+
+Qc=mv3_dot*lambda_v3        //[kJ/h]
+dT=25           //Rise in water temperature
+Cp=4.187
+mw_dot=Qc/(Cp*dT)
+printf("\nANSWER\n Area in each effect%f sq m\n",A1);
+printf("\nANSWER \n Steam economy is%f\n",eco);
+printf("\nANSWER Cooling water rate is %f t/h",mw_dot/1000)
diff --git a/1073/CH6/EX6.13/6_13.sce b/1073/CH6/EX6.13/6_13.sce
new file mode 100755
index 000000000..5982e7d1a
--- /dev/null
+++ b/1073/CH6/EX6.13/6_13.sce
@@ -0,0 +1,60 @@
+clc;
+clear;
+//Example 6.13
+Cpf=4.18        //[kJ/kg.K]
+dT1=18          //[K]
+dT2=17          //[K]
+dT3=34          //[K]
+mf_dot=4        //[kg/s]
+Ts=394          //[K]
+bp=325      //Bp of water at 13.172 kPa [K]
+dT=Ts-bp        //[K]
+lambda_s=2200           //[kJ/kg]
+T1=Ts-dT1           //[K]
+lambda1=2249        //[kJ/kg]
+lambda_v1=lambda1       //[kJ/kg]
+
+T2=T1-dT2           //[K]
+lambda2=2293        //[kJ/kg]
+lambda_v2=lambda2       //[kJ/kg]
+
+T3=T2-dT3           //[K]
+lambda3=2377        //[kJ/kg]
+lambda_v3=lambda3       //[kJ/kg]
+
+ic=0.1      //Initial conc of solids
+fc=0.5      //Final conc of solids
+m3dot_dash=(ic/fc)*mf_dot       //[kg/s]
+mv_dot=mf_dot-m3dot_dash        //Total evaporation in [kg/s]
+//Material balance over first effect
+//mf_dot=mv1_dot_m1dot_dash
+//Energy balance:
+//ms_dot*lambda_s=mf_dot*(Cpf*(T1-Tf)+mv1_dot*lambda_v1)
+
+//Material balance over second effect
+//m1dot_dash=mv2_dot+m2dot_dash
+//Enthalpy balance:
+//mv1_dot*lambda_v1+m1dot_dash(cp*(T1-T2)=mv2_dot*lambda_v2)
+
+//Material balance over third effect
+//m2dot_dash=mv3_dot+m3dot+dash
+
+//Enthalpy balance:
+//mv2_lambda_v2+m2dot_dash*cp*(T2-T3)=mv3_dot*lambda_v3
+294
+mv2_dot=3.2795/3.079        //[kg/s]
+mv1_dot=1.053*mv2_dot-0.1305     //[kg/s]
+mv3_dot=1.026*mv2_dot+0.051     //[kg/s]
+ms_dot=(mf_dot*Cpf*(T1-294)+mv1_dot*lambda_v1)/lambda_s      //[kg/s]
+eco=mv_dot/ms_dot           //Steam economy      
+eco=round(eco)
+printf("\nSteam economy is %d\n",eco);
+U1=3.10     //[kW/sq m.K]
+U2=2     //[kW/sq m.K]
+U3=1.10     //[kW/sq m.K]
+//First effect:
+A1=ms_dot*lambda_s/(U1*dT1)         //[sq m]
+A2=mv1_dot*lambda_v1/(U2*dT2)        //[sq m]
+A3=mv2_dot*lambda_v2/(U3*dT3)        //[sq m]
+//Areas are calculated witha  deviation of +-10%
+printf("\nArea pf heat transfer in each effect is %f sq m\n",A3)
diff --git a/1073/CH6/EX6.14/6_14.sce b/1073/CH6/EX6.14/6_14.sce
new file mode 100755
index 000000000..c02240541
--- /dev/null
+++ b/1073/CH6/EX6.14/6_14.sce
@@ -0,0 +1,59 @@
+clc;
+clear;
+//Example 6.14
+mf_dot=1060     //[kg/h]
+ic=0.04     //Initial concentration
+fc=0.25         //Final concentration
+m4dot_dash=(ic/fc)*mf_dot       //[kg/h]
+//Total evaporation=
+mv_dot=mf_dot-m4dot_dash        //[kg/h]
+
+//Fromsteam table:
+P1=370      //[kPa.g]
+T1=422.6        //[K]
+lambda1=2114.4      //[kJ/kg]
+
+P2=235      //[kPa.g]
+T2=410.5        //[K]
+lambda2=2151.5      //[kJ/kg]
+
+P3=80      //[kPa.g]
+T3=390.2        //[K]
+lambda3=2210.2      //[kJ/kg]
+
+P4=50.66      //[kPa.g]
+T4=354.7        //[K]
+lambda4=2304.6      //[kJ/kg]
+
+P=700       //Latent heat of steam[kPa .g]
+lambda_s=2046.3         //[kJ/kg]
+
+//FIRST EFFECT
+//Enthalpy balance:
+//ms_dot=mf_dot*Cpf*(T1-Tf)+mv1_dot*lambda1
+//ms_dot=1345.3-1.033*m1dot_dash
+
+//SECOND EFFECT
+//m1dot_dash=m2dot_dash+mdot_v2
+//Enthalpy balance:
+//m1dot_dash=531.38+0.510*m2dot_dash
+
+//THIRD EFFECT
+//Material balance:
+//m2dot_dash-m3dot_dash+mv3_dot
+
+//FOURTH EFFECT
+//m3dot_dash=m4dot_dash+mv4_dot
+mv4dot_dash=169.6           //[kg/h]
+m3dot_dash=416.7        //[kg/h]
+
+//From eq n 4:
+m2dot_dash=-176.84+1.98*m3dot_dash      //[kg/h]
+
+//From eqn 2:
+m1dot_dash=531.38+0.510*m2dot_dash      //[kg/h]
+
+//From eqn 1:
+ms_dot=1345.3-1.033*m1dot_dash
+eco=mv_dot/ms_dot           //[kg evaporation /kg steam]
+printf("\nSteam economy is %f evaporation/kg steam",eco);
diff --git a/1073/CH6/EX6.15/6_15.sce b/1073/CH6/EX6.15/6_15.sce
new file mode 100755
index 000000000..87bb75401
--- /dev/null
+++ b/1073/CH6/EX6.15/6_15.sce
@@ -0,0 +1,44 @@
+clc;
+clear;
+//Example 6.15
+m1_dot=5000     //[kg/h]
+ic=0.1      //Initial concentration
+fc=0.5      //Final concentration
+mf_dot=(fc/ic)*m1_dot       //[kg/h]
+mv_dot=mf_dot-m1_dot        //Water evaporated[kg/h]
+P=357       //Steam pressure[kN/sq m]
+Ts=412       //[K]
+H=2732      //[kJ/kg]
+lambda=2143     //[kJ/kg]
+bpr=18.5            //[K]
+T_dash=352+bpr      //[K]
+Hf=138      //[kJ/kg]
+lambda_s=2143       //[kJ/kg]
+Hv=2659     //[kJ/kg]
+H1=568      //[kJ/kg]
+ms_dot=(mv_dot*Hv+m1_dot*H1-mf_dot*Hf)/lambda_s         //Steam consumption in kg/h
+printf("\nSteam consumption is %f kg/h\n",ms_dot);
+printf("\nCapacity is %f kg/h\n",mv_dot);
+eco=mv_dot/ms_dot       //Economy
+printf("\nSteam economy is %f\n",eco);
+dT=Ts-T_dash        //[K]
+hi=4500     //[W/sq m.K]
+ho=9000     //[W/sq m.K]
+Do=0.032        //[m]
+Di=0.028        //[m]
+x1=(Do-Di)/2        //[m]
+Dw=(Do-Di)/log(32/28)   //[m]
+x2=0.25*10^-3      //[m]
+L=2.5       //Length [m]
+hio=hi*(Di/Do)        //[W/sq m.K]
+printf("\n NOTE:In textbook this value of hio is wrongly calculated as 3975.5..So we will take this\n\n");
+hio=3975.5
+k1=45       //Tube material in [W/sq m.K]
+k2=2.25     //For scale[W/m.K]
+Uo=1/(1/ho+1/hio+(x1*Dw)/(k1*Do)+(x2/k2))     //Overall heat transfer coeff in W/sq m.K
+Q=ms_dot*lambda_s       //[kJ/h]
+Q=Q*1000/3600           //[W]
+
+A=Q/(Uo*dT)             //[sq m]
+n=A/(%pi*Do*L)          //from A=n*%pi*Do*L 
+printf("\n No. of tubes required is %d",round(n));
diff --git a/1073/CH6/EX6.16/6_16.sce b/1073/CH6/EX6.16/6_16.sce
new file mode 100755
index 000000000..98b1bcb50
--- /dev/null
+++ b/1073/CH6/EX6.16/6_16.sce
@@ -0,0 +1,41 @@
+clc;
+clear;
+//Example 6.16
+bpr=40.6;        //[K]
+Cpf=1.88;       //[kJ/kg.K]
+Hf=214;     //[kJ/kg]
+H1=505;     //[kJ/kg]
+mf_dot=4536;    //[kg/h] of feed solution
+ic=0.2;      //Initial conc
+fc=0.5;      //Final concentration
+m1dot_dash=(ic/fc)*mf_dot       //Thisck liquor flow arte[kg/h]
+mv_dot=mf_dot-m1dot_dash        //[kg/H]
+Ts=388.5;        //Saturation temperature of steam in [K]
+bp=362.5        //b.P of solution in [K]
+lambda_s=2214;     //[kJ/kg]
+P=21.7;      //Vapor space in [kPa]
+Hv=2590.3;       //[kJ/kg]
+
+//Enthalpy balance over evaporator
+ms_dot=(m1dot_dash*H1+mv_dot*Hv-mf_dot*Hf)/lambda_s     //[kg/h
+printf("\nSteam consumption is %f kg/h\n",ms_dot);
+dT=Ts-bp        //[K]
+U=1560      //[W/sq m.K]
+Q=ms_dot*lambda_s           //[kJ/h]
+Q=Q*1000/3600               //[W]
+A=Q/(U*dT)      //[sq m]
+printf("\nHeat transfer area is %f sq m\n",A);
+
+//Calculations considering enthalpy of superheated vapour
+
+Hv=Hv+Cpf*bpr   //[kJ/kg]
+ms_dot=(m1dot_dash*H1+mv_dot*Hv-mf_dot*Hf)/lambda_s     //[kg/h]
+printf("\n Now,Steam consumption is %f kg/h\n",ms_dot);
+eco=mv_dot/ms_dot       //Steam economy
+printf("\nEconomy of evaporator %f\n",eco);
+Q=ms_dot*lambda_s       //[kJ/h]
+Q=Q*1000/3600           //[w]
+A2=Q/(U*dT)              //Area
+printf("\nNow,Area is %f\n",A);
+perc=(A2-A)*100/A           //%error in the heat transfer area 
+printf("\n If enthalpy of water vapour Hv were based on the saturated vapour at the pressure\nthe error introduced is only %f percent\n",perc);
diff --git a/1073/CH6/EX6.2/6_2.sce b/1073/CH6/EX6.2/6_2.sce
new file mode 100755
index 000000000..bc91d98fd
--- /dev/null
+++ b/1073/CH6/EX6.2/6_2.sce
@@ -0,0 +1,15 @@
+clc;
+clear;
+//Example 6.2
+m_dot=10000     //Weak liquor entering in [kg/h]
+fr_in=0.04       //Fraciton of caustic soda IN i.e 4%
+fr_out=0.25    //Fraciton of caustic soda OUT i.e 25%
+//Let mdash_dot be the kg/h of thick liquor leaving
+mdash_dot=fr_in*m_dot/fr_out        //[kg/h]
+
+//Overall material balance
+//kg/h of feed=kg/h of water evaporated +kg/h of thick liquor
+//we=water evaporated in kg/h
+//Therefore
+we=m_dot-mdash_dot      //[kg/h]
+printf("\n Capacity of evaporator is %d kg/h",we);
diff --git a/1073/CH6/EX6.3/6_3.sce b/1073/CH6/EX6.3/6_3.sce
new file mode 100755
index 000000000..d534222a1
--- /dev/null
+++ b/1073/CH6/EX6.3/6_3.sce
@@ -0,0 +1,29 @@
+clc;
+clear;
+//Exmaple 6.3
+ic=0.05     //Initial concentration (5%)
+fc=0.2      //Final concentration   (20%)
+T_dash=373      //B.P of water in [K]
+bpe=5       //Boiling point elevation[K]
+mf_dot=5000       //[Basis] feed to evaporator in [kg/h]
+//Material balance of solute
+mdash_dot=ic*mf_dot/fc       //[kg/h]
+//Overall material balance
+mv_dot=mf_dot-mdash_dot      //Water evaporated [kg/h]
+lambda_s=2185        //Latent heat of condensation of steam[kJ/kg]
+lambda_v=2257       //Latent heat of vaporisation of water [kJ/kg]
+lambda=lambda_v     //[kJ/kg]
+T=T_dash+bpe        //Temperature of thick liquor[K]
+Tf=298      //Temperature of feed [K]
+Cpf=4.187       //Sp. heat of feed in [kJ/kg.K]
+//Heat balance over evaporator=ms_dot
+ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s   //Steam consumption [kg/h]
+Eco=mv_dot/ms_dot       //Economy of evaporator
+Ts=399      //Saturation temperature of steam in [K]
+dT=Ts-T     //Temperature driving force [K]        
+U=2350      //[W/sq m.K]
+Q=ms_dot*lambda_s       //Rate of heat transfer in [kJ/kg]
+Q=Q*1000/3600           //[J/s]=[W]
+A=Q/(U*dT)              //Heat transfer area in [sq m]
+printf("\nANSWER Economoy pf evaporator is %f \n",Eco);
+printf("\nHeat tarnsfer area to be provided = %f sq m\n",A);
diff --git a/1073/CH6/EX6.4/6_4.sce b/1073/CH6/EX6.4/6_4.sce
new file mode 100755
index 000000000..aecc7518a
--- /dev/null
+++ b/1073/CH6/EX6.4/6_4.sce
@@ -0,0 +1,42 @@
+
+clc;
+clear;
+//Example 6.4
+Cpf=3.98        //Specific heat of feed in kJ/(kg.K)
+lambda_s=2202    //Latent heat of conds of heat at 0.2MPa in [kJ/kg]
+lambda=2383     //Latent heat of vaporisation of water aty 323 [kJ/kg
+ic=0.1          //Initial concentration of soilds in [%]
+fc=0.5          //Final concentration
+m_dot=30000     //Feed to evaporator in [kg/h]
+mdash_dot=ic* m_dot/fc  //Mass flow rate of thick liquor in [kg/h]
+mv_dot=m_dot-mdash_dot      //Water evaporated in [kg/h]
+
+//Case 1: Feed at 293K
+mf_dot=30000        //[kg/h]
+mv_dot=24000        //[kg/h]
+Cpf=3.98        //[kJ/(kg.K)]
+Ts=393      //Saturation temperature of steam in [K]
+T=323       //Boiling point of solution [K]
+lambda_s=2202       //Latent heat of condensation [kJ/kg]
+lambda=2383     //Latent heat of vaporisation[kJ/kg]
+Tf=293          //Feed temperature
+//Enthalpy balance over the evaporator:
+ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s       //Steam consumption[kg/h]
+eco=(mv_dot/ms_dot)         //Steam economy
+printf("\nWhen Feed introduced at 293 K ,Steam economy is %f\n",eco);
+dT=Ts-T                     //[K]
+U=2900          //[W/sq m.K]
+Q=ms_dot*lambda_s           //Heat load =Rate of heat transfer in [kJ/h]
+Q=Q*1000/3600               //[J/s]
+A=Q/(U*dT)              //Heat transfer area required [sq m]
+printf("\n ANSWER-(i)\n\n At 293 K,Heat transfer area required is %f sq m\n",A);
+
+//Case2: Feed at 308K
+Tf=308      //[Feed temperature][K]
+ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s           //Steam consumption in [kg/h]
+eco=mv_dot/ms_dot               //Economy of evaporator
+printf("\n ANSWER-(ii)\n\n When T=308 K \nEconomy of evaporator is %f\n",eco);
+Q=ms_dot*lambda_s               //[kJ/h]
+Q=Q*1000/3600                   //[J/s]
+A=Q/(U*dT)                      //Heat transfer area required [sq m]
+printf('\nANSWER-(iii) \n When T=308 K,\nHeat transfer Area required is %f sq m\n",A);
diff --git a/1073/CH6/EX6.5/6_5.sce b/1073/CH6/EX6.5/6_5.sce
new file mode 100755
index 000000000..61894636a
--- /dev/null
+++ b/1073/CH6/EX6.5/6_5.sce
@@ -0,0 +1,23 @@
+clc;
+clear;
+//Example 6.5
+m_dot=5000  //Feed to the evaporator [kg/h]
+Cpf=4.187           //Cp of feed in [kJ/kg.K]
+ic=0.10             //Initial concentration
+fc=0.4              //Final concentration
+mdash_dot=m_dot*ic/fc           //[kg/h] of thick liquor
+mv_dot=m_dot-mdash_dot          //Water evaporated in[kg/h]
+lambda_s=2162           //Latent heat of condensing steam [kJ/kg]
+P=101.325       //Pressure in the evaporator[kPa]
+bp=373          //[K]
+Hv=2676     //Enthalpy of water vapor [kJ/kg]
+H_dash=419          //[kJ/kg]
+Hf=170          //[kJ/kg]
+ms_dot=(mv_dot*Hv+mdash_dot*H_dash-m_dot*Hf)/lambda_s          //Steam consumption in [kg/h]
+eco=mv_dot/ms_dot           //Steam economy of evaporator
+Q=ms_dot*lambda_s           //[kJ/h]
+U=1750          //[W/sq m.K]
+dT=34           //[K]
+Q=Q*1000/3600               //[J/s]
+A=Q/(U*dT)                  //[sq m]
+printf("\n Heat transfer area to be provided is %f sq m",A);
diff --git a/1073/CH6/EX6.6/6_6.sce b/1073/CH6/EX6.6/6_6.sce
new file mode 100755
index 000000000..ea2d07a65
--- /dev/null
+++ b/1073/CH6/EX6.6/6_6.sce
@@ -0,0 +1,27 @@
+clc;
+clear;
+//Example 6.6
+mf_dot=5000           //[kg/h]
+ic=0.01             //Initial concentration [kg/h]
+fc=0.02             //Final concentration  [kg/h]
+T=373               //Boiling pt of saturation in [K]
+Ts=383              //Saturation temperature of steam in [K]  
+mdash_dot=ic*mf_dot/fc   //[kg/h]
+mv_dot=mf_dot-mdash_dot      //Water evaporated in [kg/h]
+Hf=125.79           //[kJ/kg]
+Hdash=419.04            //[kJ/kg]
+Hv=2676.1           //[kJ/kg]
+lambda_s=2230.2     //[kJ/kg]
+ms_dot=(mdash_dot*Hdash+mv_dot*Hv-mf_dot*Hf)/lambda_s   //Steam flow rate in [kg/h]
+eco=mv_dot/ms_dot           //Steam economy
+Q=ms_dot*lambda_s           //Rate of heat transfer in [kJ/h]
+Q=Q*1000/3600               //[J/s]
+dT=Ts-T                     //[K]
+
+A=69            //Heating area of evaporator in [sq m]
+U=Q/(A*dT)      //Overall heat transfer coeff in [W/sq m.K]
+printf("\nSteam economy is %f\n",eco);
+printf("\n\nOverall heat transfer coefficient is %d  W/sq m.K",round(U));
+    
+
+
diff --git a/1073/CH6/EX6.7/6_7.sce b/1073/CH6/EX6.7/6_7.sce
new file mode 100755
index 000000000..c440f4496
--- /dev/null
+++ b/1073/CH6/EX6.7/6_7.sce
@@ -0,0 +1,19 @@
+clc;
+clear;
+//Example 6.7
+//From previous example:
+mf_dot=5000         //[kg/h]
+Hf=125.79           //[kJ/kg]
+lambda_s=2230.2     //[kJ/kg]
+mdash_dot=2500   //[kg/h]
+Hdash=313.93         //[kJ/kg]
+mv_dot=2500         //[kg/h]
+Hv=2635.3           //[kJ/kg]
+ms_dot=(mdash_dot*Hdash+mv_dot*Hv-mf_dot*Hf)/lambda_s   //Steam flow rate in [kg/h]
+Q=ms_dot*lambda_s           //[kJ/h]
+Q=Q*1000/3600           //[W]
+U=2862              //[W/sq m.K]
+dT=35       //[K]
+A=Q/(U*dT)      //[sq m]
+printf("\n The heat transfer area in this case is %f sq m\n",A);
+printf("\n\nNOTE :There is a calculation mistake in the book at the line12 of this code,ms_dot value is written as 2320.18,which is wrong\n\n");
diff --git a/1073/CH6/EX6.8/6_8.sce b/1073/CH6/EX6.8/6_8.sce
new file mode 100755
index 000000000..e37a6e123
--- /dev/null
+++ b/1073/CH6/EX6.8/6_8.sce
@@ -0,0 +1,17 @@
+clc;
+clear;
+//Example 6.8
+mf_dot=6000         //Feed rate in [kg/h]
+//Taking the given values from previous example(6.6)
+Hf=125.79           //[kJ/kg]
+ms_dot=3187.56      //[kg/h]
+lambda_s=2230.2         //[kJ/kg]
+Hdash=419.04        //[kJ/kg]
+Hv=2676.1           //[kJ/kg]
+mv_dot=(mf_dot*Hf+ms_dot*lambda_s-6000*Hdash)/(Hv-Hdash)  //Water evaporated in [kg/h]
+mdash_dot=6000-mv_dot       //Mass flow rate of product [kg/h]
+x=(0.01*mf_dot)*100/mdash_dot       //Wt % of solute in products
+printf("\nMass flow rate of product is %f kg/h\n\n",mdash_dot);
+printf("\n\nThe product concentration is %f percent by weight \n\n",x);
+
+
diff --git a/1073/CH6/EX6.9/6_9.sce b/1073/CH6/EX6.9/6_9.sce
new file mode 100755
index 000000000..99b624aa5
--- /dev/null
+++ b/1073/CH6/EX6.9/6_9.sce
@@ -0,0 +1,24 @@
+clc;
+clear;
+//Example 6.9
+Tf=298      //Feed temperature in [K]
+T_dash=373      //[K]
+Cpf=4       //[kJ/kg.K]
+fc=0.2      //Final concentration of salt
+ic=0.05     //Initial concentration
+mf_dot=20000    //[kg/h] Feed to evaporator
+mdash_dot=ic*mf_dot/fc      //Thick liquor [kg/h]
+mv_dot=mf_dot-mdash_dot     //Water evaporated in [kg/h]
+lambda_s=2185       //[kJ/kg]
+lambda=2257       //[kJ/kg]
+bpr=7       //Boiling point rise[K]
+T=T_dash+bpr     //Boiling point of solution in[K]
+Ts=39       //Temperature of condensing steam in [K]
+ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s   //Steam consumption in [kg/h]
+eco=mv_dot/ms_dot           //Economy of evaporator 
+Q=ms_dot*lambda_s               //[kJ/h]
+Q=Q*1000/3600           //[J/s]
+printf("\nHeat load is %d W or J/s",round(Q));
+printf("\n\nEconomy of evaporator is %f ",eco);
+
+printf("\n\nNOTE:Again there is a calcualtion mistake in book at line 19 of code,it is written as 4041507.1 instead of 40415071 \n\n");
author	priyanka	2015-06-24 15:03:17 +0530
committer	priyanka	2015-06-24 15:03:17 +0530
commit	b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch)
tree	ab291cffc65280e58ac82470ba63fbcca7805165 /1073
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