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-rw-r--r--555/CH6/EX6.1/1.sce19
-rw-r--r--555/CH6/EX6.2/2.sce15
-rw-r--r--555/CH6/EX6.3/3.sce34
-rw-r--r--555/CH6/EX6.5/5.sce19
-rw-r--r--555/CH6/EX6.6/6.sce17
-rw-r--r--555/CH6/EX6.7/7.sce18
6 files changed, 122 insertions, 0 deletions
diff --git a/555/CH6/EX6.1/1.sce b/555/CH6/EX6.1/1.sce
new file mode 100644
index 000000000..b8a9efb1a
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+++ b/555/CH6/EX6.1/1.sce
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+// Implementation of example 6.1
+// Basic and Applied Thermodynamics by P.K.Nag
+
+clc
+clear
+
+// T1 & T2 are source & sink temperatures respectively
+// n is efficiency of engine & W is work done
+
+t1=800
+t2=30
+T1=t1+273; // K
+T2=t2+273 // K
+W=1 // kW
+Nmax=1-(T2/T1);
+Q1=W/Nmax;
+Q2=Q1-W;
+printf("least rate of heat rejection = %.3f kW",Q2);
+// end \ No newline at end of file
diff --git a/555/CH6/EX6.2/2.sce b/555/CH6/EX6.2/2.sce
new file mode 100644
index 000000000..a9e4917f1
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+++ b/555/CH6/EX6.2/2.sce
@@ -0,0 +1,15 @@
+// Implementation of example 6.2
+// Basic and Applied Thermodynamics by P.K.Nag
+
+clc
+clear
+
+//Temperature T, Rate of heat transfer Q, Power W
+T1 = 30+273;//K
+T2 = -15+273;//K
+Q2 = 1.75;//kJ/s
+//For minimum power requirement
+Q1 = Q2*T1/T2;
+W = Q1 - Q2;
+printf('Least power necessary to pump the heat continuously is %0.2f kW',W);
+// end \ No newline at end of file
diff --git a/555/CH6/EX6.3/3.sce b/555/CH6/EX6.3/3.sce
new file mode 100644
index 000000000..feb44e1ee
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+++ b/555/CH6/EX6.3/3.sce
@@ -0,0 +1,34 @@
+// Implementation of example 6.3
+// Basic and Applied Thermodynamics by P.K.Nag
+
+clc
+clear
+
+// T temperature,Nmax max efficiency,W work done,Q heat trnsfer,COP coefficient of performance of refrigerator..
+t1=600
+T1=t1+273 // K
+t2=40
+T2=t2+273 // K
+t3=-20
+T3=t3+273 // K
+Q1=2000 // kJ
+W=360 // kJ
+
+Nmax=1-(T2/T1);
+W1=Nmax*Q1;
+COP=T3/(T2-T3);
+W2=W1-W;
+Q4=COP*W2;
+Q3=Q4+W2;
+Q2=Q1-W1;
+printf("heat rejection to 40 degree celsius reservoir = %.2f kJ \n",Q2+Q3);
+// part b
+N=0.4*Nmax;
+W1=N*Q1;
+W2=W1-W;
+COP2=0.4*COP;
+Q4=W2*COP2;
+Q3=Q4+W2;
+Q2=Q1-W1;
+printf("heat rejection to 40 degree celsius reservoir with decreased efficiency= %.2f kJ \n",Q2+Q3);
+// end \ No newline at end of file
diff --git a/555/CH6/EX6.5/5.sce b/555/CH6/EX6.5/5.sce
new file mode 100644
index 000000000..fba6315db
--- /dev/null
+++ b/555/CH6/EX6.5/5.sce
@@ -0,0 +1,19 @@
+// Implementation of example 6.5
+// Basic and Applied Thermodynamics by P.K.Nag
+
+clc
+clear
+
+T1 = 473; //K
+T2 = 293; //K
+T3 = 273; //K
+//let Q1 be any constant
+Q1 = 1;
+Q2 = Q1*T2/T1;
+W = (T1 - T2)*Q1/T1;
+//COP = T2/(T2 - T3) = Q'/W
+Q2_ = T2/(T2 - T3) * (T1 - T2)/T1 * Q1;
+
+MF = (Q2 + Q2_)/Q1; //multiplication factor
+printf("Multiplication factor, MF = %0.2f",MF);
+// end \ No newline at end of file
diff --git a/555/CH6/EX6.6/6.sce b/555/CH6/EX6.6/6.sce
new file mode 100644
index 000000000..dcb734622
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+++ b/555/CH6/EX6.6/6.sce
@@ -0,0 +1,17 @@
+// Implementation of example 6.6
+// Basic and Applied Thermodynamics by P.K.Nag
+
+clc
+clear
+
+//Energy collected by plate per unit area E
+//Temperature of plate T1, Atmospheric temperature T2
+T1 = 90+273;//K
+T2 = 20+273;//K
+W = 1;//kJ/s
+E = 1880;//kJ/m^2 h
+nmax = 1 - T2/T1;
+Qmin = W/nmax;
+A = Qmin*3600/E;
+printf('Minimum Area of collector required, A = %0.0f m^2',A);
+// end \ No newline at end of file
diff --git a/555/CH6/EX6.7/7.sce b/555/CH6/EX6.7/7.sce
new file mode 100644
index 000000000..bab5492ab
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+++ b/555/CH6/EX6.7/7.sce
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+// Implementation of example 6.7
+// Basic and Applied Thermodynamics by P.K.Nag
+
+clc
+clear
+
+// T temperature,W work done,Q heat trnsfer..
+// Radiation from panel is proportional to area and T2^4
+// so if A is area then Q2=K*A*(T2)^4
+// for minimum area we differentiate the expression A=W/[K*(T2^3)*(T1-T2)]..
+// finally the expression for minimum area is Amin=256*W/[27*K*(T1^4)]
+
+W=1 // kW
+K= 5.67*10d-9 // W/(m^2)*(K^4)
+T1=1000 // K
+Amin= (256*W*1000)/[27*K*(T1*T1*T1*T1)];
+printf("minimum area = %.4f m^2",Amin);
+// end \ No newline at end of file