diff options
Diffstat (limited to '1073/CH2')
47 files changed, 1197 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);
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