initial commit / add all books

5 files changed, 86 insertions, 0 deletions

diff --git a/1328/CH2/EX2.1/2_1.sce b/1328/CH2/EX2.1/2_1.sce
new file mode 100644
index 000000000..c134c4e1d
--- /dev/null
+++ b/1328/CH2/EX2.1/2_1.sce
@@ -0,0 +1,11 @@
+printf("\t Example 2.1 \n");

+printf("\t approximate values are mentioned in the book \n");

+Tavg=900; // average temperature of the wall,F

+k=0.15; // Thermal conductivity at 932 F,Btu/(hr)(ft^2)(F/ft)

+T1=1500; // hot side temperature,F

+T2=300; // cold side temperature,F

+A=192; // surface area,ft^2

+L=0.5; // thickness,ft

+Q=(k)*(A)*(T1-T2)/L; // formula for heat,Btu/hr

+printf("\t heat is : %.2e Btu/hr \n",Q);

+//end

diff --git a/1328/CH2/EX2.2/2_2.sce b/1328/CH2/EX2.2/2_2.sce
new file mode 100644
index 000000000..ca133d6b9
--- /dev/null
+++ b/1328/CH2/EX2.2/2_2.sce
@@ -0,0 +1,31 @@
+printf("\t Example 2.2 \n");

+printf("\t approximate values are mentioned in the book \n");

+La=8/12; // Thickness of firebrick wall,ft

+Lb=4/12; // Thickness of insulating brick wall,ft

+Lc=6/12; // Thickness of building brick wall,ft

+Ka=0.68; // themal conductivity of firebrick,Btu/(hr)*(ft^2)*(F/ft)

+Kb=0.15; // themal conductivity of insulating brick,Btu/(hr)*(ft^2)*(F/ft)

+Kc=0.40; // themal conductivity of building brick,Btu/(hr)*(ft^2)*(F/ft)

+A=1; // surface area,ft^2

+Ta=1600; // temperature of inner wall,F

+Tb=125; // temperature of outer wall.F

+Ra=La/(Ka)*(A); // formula for resistance,(hr)*(F)/Btu

+printf("\t resistance offered by firebrick : %.2f (hr)*(F)/Btu \n",Ra);

+Rb=Lb/(Kb)*(A); // formula for resistance,(hr)*(F)/Btu

+printf("\t resistance offered by insulating brick : %.2f (hr)*(F)/Btu \n",Rb);

+Rc=Lc/(Kc)*(A); // formula for resistance,(hr)*(F)/Btu

+printf("\t resistance offered by buildingbrick : %.2f (hr)*(F)/Btu \n",Rc);

+R=Ra+Rb+Rc; // total resistance offered by three walls,(hr)*(F)/Btu

+printf("\t total resistance offered by three walls : %.2f (hr)*(F)/Btu \n",R);

+Q=(1600-125)/4.45; // using formula for heat loss/ft^2,Btu/hr

+printf("\t heat loss/ft^2 : %.0f Btu/hr \n",Q);

+// T1,T2 are temperatures at interface of firebrick and insulating brick, and insulating brick and building brick respectively,F

+delta=(Q)*(Ra); // formula for temperature difference,F

+printf("\t delta is : %.0f F \n",delta);

+T1=Ta-((Q)*(Ra)); // temperature at interface of firebrick and insulating brick,F

+printf("\t temperature at interface of firebrick and insulating brick :%.0f F \n",T1);

+deltb=Q*(Rb);

+printf("\t deltb is : %.0f F \n",deltb);

+T2=T1-((Q)*(Rb)); //temperature at interface of insulating brick and building brick,F

+printf("\t temperature at interface of insulating brick and building brick :%.0f F \n",T2);

+//end

diff --git a/1328/CH2/EX2.3/2_3.sce b/1328/CH2/EX2.3/2_3.sce
new file mode 100644
index 000000000..c05db8979
--- /dev/null
+++ b/1328/CH2/EX2.3/2_3.sce
@@ -0,0 +1,16 @@
+printf("\t example 2.3 \n");

+printf("\t approximate values are mentioned in the book \n");

+Lair=0.25/12; // thickness of air film,ft

+Kair=0.0265; // thermal conductivity of air at 572F,Btu/(hr)*(ft^2)(F/ft)

+A=1; // surface area,ft^2

+Rair=Lair/(Kair*(A)); // resistance offered by air film, (hr)(F)/Btu

+printf("\t resistance offered by air film %.2f (hr)(F)/Btu \n",Rair);

+R=4.45; // resistance from previous example 2.2,(hr)(F)/Btu

+Rt=(R)+Rair; // total resistance,(hr)(F)/Btu

+printf("\t total resistance %.2f (hr)(F)/Btu \n",Rt);

+Ta=1600; // temperature of inner wall,F

+Tb=125; // temperature of outer wall,F

+Q=(1600-125)/Rt; // heat loss, Btu/hr

+printf("\t heat loss %.2f Btu/hr \n",Q);

+printf("\t It is seen that in a wall 18 in. thick a stagnant air gap only .25 in. thick reduces the heat loss by 15 percent \n");

+//end

diff --git a/1328/CH2/EX2.4/2_4.sce b/1328/CH2/EX2.4/2_4.sce
new file mode 100644
index 000000000..b0f4ef99e
--- /dev/null
+++ b/1328/CH2/EX2.4/2_4.sce
@@ -0,0 +1,10 @@
+printf("\t example 2.4 \n");

+printf("\t approximate values are mentioned in the book \n");

+k=0.63; // thermal conductivity of pipe, Btu/(hr)*(ft^2)*(F/ft)

+Do=6; // in

+Di=5; // in

+Ti=200; // inner side temperature,F

+To=175; // outer side temperature,F

+q=(2*(3.14)*(k)*(Ti-To))/(2.3*log10(Do/Di)); // formula for heat flow,Btu/(hr)*(ft)

+printf("\t heat flow is : %.0f Btu/(hr)*(ft) \n",q); // caculation mistake in book

+// end

diff --git a/1328/CH2/EX2.5/2_5.sce b/1328/CH2/EX2.5/2_5.sce
new file mode 100644
index 000000000..74052be15
--- /dev/null
+++ b/1328/CH2/EX2.5/2_5.sce
@@ -0,0 +1,18 @@
+printf("\t example 2.5 \n");

+printf("\t approximate values are mentioned in the book \n");

+t1=150; // assume temperature of outer surface of rockwool,F

+ta=70; // temperature of surrounding air,F

+ha=2.23; // surface coefficient,Btu/(hr)*(ft^2)*(F)

+q=(3.14)*(300-70)/(((2.3/(2*0.033))*log10(3.375/2.375))+(1/((2.23)*(3.375/12)))); // using formula for heat loss,Btu/(hr)*(lin ft), calculation mistake

+printf("\t heat loss for linear foot is : %.1f Btu/(hr)*(lin ft) \n",q);

+printf("\t Check between ts and t1, since delt/R = deltc/Rc \n");

+t1=300-(((104.8)*((2.3)*(log10(3.375/2.375))))/((2)*(3.14)*(.033))); // using eq 2.31,F

+printf("\t t1 is : %.1f F \n",t1);

+t1=125; // assume temperature of outer surface of rockwool,F

+ha=2.10; // surface coefficient,Btu/(hr)*(ft^2)*(F)

+q=((3.14)*(300-70))/(((2.3/(2*0.033))*log10(3.375/2.375))+(1/((2.10)*(3.375/12)))); // using formula for heat loss,Btu/(hr)*(lin ft)

+printf("\t heat loss for linear foot is : %.1f Btu/(hr)*(lin ft) \n",q);

+printf("\t Check between ts and t1, since delt/R = deltc/Rc \n");

+t1=300-(((103)*((2.3)*(log10(3.375/2.375))))/((2)*(3.14)*(.033))); // using eq 2.31,F

+printf("\t t1 is : %.1f F \n",t1);

+// end