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-rwxr-xr-x2417/CH4/EX4.1/Ex4_1.sce31
-rwxr-xr-x2417/CH4/EX4.10/Ex4_10.sce12
-rwxr-xr-x2417/CH4/EX4.11/Ex4_11.sce21
-rwxr-xr-x2417/CH4/EX4.12/Ex4_12.sce22
-rwxr-xr-x2417/CH4/EX4.13/Ex4_13.sce21
-rwxr-xr-x2417/CH4/EX4.14/Ex4_14.sce25
-rwxr-xr-x2417/CH4/EX4.15/Ex4_15.sce33
-rwxr-xr-x2417/CH4/EX4.2/Ex4_2.sce27
-rwxr-xr-x2417/CH4/EX4.3/Ex4_3.sce22
-rwxr-xr-x2417/CH4/EX4.4/Ex4_4.sce24
-rwxr-xr-x2417/CH4/EX4.5/Ex4_5.sce22
-rwxr-xr-x2417/CH4/EX4.7/Ex4_7.sce20
-rwxr-xr-x2417/CH4/EX4.8/Ex4_8.sce15
-rwxr-xr-x2417/CH4/EX4.9/Ex4_9.sce38
14 files changed, 333 insertions, 0 deletions
diff --git a/2417/CH4/EX4.1/Ex4_1.sce b/2417/CH4/EX4.1/Ex4_1.sce
new file mode 100755
index 000000000..bf325932b
--- /dev/null
+++ b/2417/CH4/EX4.1/Ex4_1.sce
@@ -0,0 +1,31 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.1\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.1 (page no. 148)
+// Solution
+
+//given data
+t1=1000; //(unit:fahrenheit) //Source temperature
+t2=80; //(unit:fahrenheit) //Sink temperature
+//solution
+//converting temperatures to absolute temperatures;
+T1=t1+460; //Source temperature //Unit:R
+T2=t2+460; //Sink temperature //Unit:R
+
+printf("Solution for (a)\n");
+ans=((T1-T2)/T1)*100;//(ans in %) //Efficiency of the engine
+printf("Efficiency of the engine is %f percentage\n\n",ans);
+
+printf("Solution for (b)\n");
+T1=2000+460; //Source temperature //Unit:R
+T2=t2+460; //Sink temperature //Unit:R
+ans=((T1-T2)/T1)*100;//(ans in %) //Efficiency of the engine
+printf("When the upper tempretrature is increased upto certain ,Efficiency of the engine is %f percentage \n\n",ans);
+
+printf("Solution for (c)\n");
+T1=t1+460; //Source temperature //Unit:R
+T2=160+460; //Sink temperature //Unit:R
+ans=((T1-T2)/T1)*100;//(ans in %) //Efficiency of the engine
+printf("When the lower tempretrature is increased upto certain ,Efficiency of the engine is %f percentage \n\n",ans);
diff --git a/2417/CH4/EX4.10/Ex4_10.sce b/2417/CH4/EX4.10/Ex4_10.sce
new file mode 100755
index 000000000..30ba0ca22
--- /dev/null
+++ b/2417/CH4/EX4.10/Ex4_10.sce
@@ -0,0 +1,12 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.10\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.10 (page no. 159)
+// Solution
+
+hfg=1959.7; //Unit:kJ/kg //Evaporative enthalpy
+T=195.07+273; //Converted into Kelvin //Temperature
+deltaS=hfg/T; //Change in entropy //kJ/kg*K
+printf("Change in entropy at 1.4MPa for the vaporization of 1 kg is %f kJ/kg*K",deltaS); //Values compares very closely to the Steam Tables value
diff --git a/2417/CH4/EX4.11/Ex4_11.sce b/2417/CH4/EX4.11/Ex4_11.sce
new file mode 100755
index 000000000..639d3aa22
--- /dev/null
+++ b/2417/CH4/EX4.11/Ex4_11.sce
@@ -0,0 +1,21 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.11\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.11 (page no. 159)
+// Solution
+
+//Let is assume that a Carnot engine cycle operates between two temperatures in each case.
+t=1000; //(unit:fahrenheit)
+//converting temperatures to absolute temperatures;
+T1=t+460;
+//T1*deltaS=Qin;
+Qin=100; //Unit:Btu //heat added to the cycle
+deltaS=Qin/T1; //Change in entropy //Btu/R
+T2=50+460; //converting 50 F temperature to absolute temperature;
+Qr=T2*deltaS; //Heat rejected //Unit:Btu
+printf("%f Btu energy is unavailable with respect to a receiver at 50 fahrenheit \n",Qr);
+T2=0+460; //converting 0 F temperature to absolute temperature;
+Qr=T2*deltaS; //Heat rejected //unit:Btu
+printf("%f Btu energy is unavailable with respect to a receiver at 0 fahrenheit \n",Qr);
diff --git a/2417/CH4/EX4.12/Ex4_12.sce b/2417/CH4/EX4.12/Ex4_12.sce
new file mode 100755
index 000000000..f9fb0921f
--- /dev/null
+++ b/2417/CH4/EX4.12/Ex4_12.sce
@@ -0,0 +1,22 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.12\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.12 (page no. 160)
+// Solution
+
+Qin=1000; //Unit:Joule //heat entered to the system
+t=500; //(unit:Celcius) //temperature
+//converting temperature
+T1=t+273; //Unit:Kelvin
+deltaS=Qin/T1; //Change in entropy //Unit:J/K
+printf("Solution for (a),\n");
+T2=20+273; //converted 20 Celcius temperature to Kelvin;
+Qr=T2*deltaS; //Heat rejected at 20 celcius //Joule
+printf("%f Joule energy is unavailable with respect to a receiver at 20 Celcius\n\n",Qr);
+
+printf("Solution for (b),\n")
+T2=0+273; //converted 0 Celcius temperature to Kelvin
+Qr=T2*deltaS; //heat rejected at 0 celcius //Joule
+printf("%f Joule energy is unavailable with respect to a receiver at 0 Celcius\n",Qr);
diff --git a/2417/CH4/EX4.13/Ex4_13.sce b/2417/CH4/EX4.13/Ex4_13.sce
new file mode 100755
index 000000000..3c3cf2370
--- /dev/null
+++ b/2417/CH4/EX4.13/Ex4_13.sce
@@ -0,0 +1,21 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.13\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.13 (page no. 161)
+// Solution
+
+//deltas=Cp*ln(T2/T1)
+//Multiplying both the sides of equation by the mass m,
+//DeltaS=m*Cp*ln(T2/T1)
+m=6; //mass //Unit:lbm
+Cp=0.361; //Btu/lbm*R //Specific heat constant
+DeltaS=-0.7062; //Unit:Btu/R //change in entropy
+t=1440; //(unit:fahrenheit)
+//converting temperatures to absolute temperatures;
+T1=t+460; //Unit:R
+//Rearranging the equation,
+T2=T1*exp(DeltaS/(m*Cp)); //final temperature //Unit:R
+printf("Final temperature is %f R",T2);
+printf("or %f fahrenheit",T2-460);
diff --git a/2417/CH4/EX4.14/Ex4_14.sce b/2417/CH4/EX4.14/Ex4_14.sce
new file mode 100755
index 000000000..8fbdd0d09
--- /dev/null
+++ b/2417/CH4/EX4.14/Ex4_14.sce
@@ -0,0 +1,25 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.14\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.14 (page no. 162)
+// Solution
+
+//1 lbm of water at 500F is mixed with 1 lbm of water at 100F
+m1=1; //Unit:lbm //mass
+m2=1; //Unit:lbm //mass
+c1=1; //Specific heat constant
+c2=1; //Specific heat constant
+t1=500; //(unit:fahrenheit)
+t2=100; //(unit:fahrenheit)
+cmix=1; //Specific heat constant of mixture
+//now, m1*c1*t1 +m2*c2*t2 = (m1+m2)*cmix*t
+//So,
+t=((m1*c1*t1)+(m2*c2*t2))/((m1+m2)*cmix) //resulting temperature of the mixture
+printf("The resulting temperature of the mixture is %f fahrenheit\n",t);
+//For this problem,the hot steam is cooled
+deltas=cmix*log((t+460)/(t1+460)); //temperatures converted to absolute temperatures; //deltas=change in entropy //Unit:Btu/(lbm*R)
+//The cold steam is heated
+deltaS=cmix*log((t+460)/(t2+460)); //temperatures converted to absolute temperatures; //deltaS=change in entropy //Unit:Btu/(lbm*R)
+printf("The net change in entropy is %f Btu/(lbm*R)\n",deltaS+deltas);
diff --git a/2417/CH4/EX4.15/Ex4_15.sce b/2417/CH4/EX4.15/Ex4_15.sce
new file mode 100755
index 000000000..82fca6503
--- /dev/null
+++ b/2417/CH4/EX4.15/Ex4_15.sce
@@ -0,0 +1,33 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.15\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.15 (page no. 163)
+// Solution
+
+//In problem 4.15,
+//1 lbm of water at 500F is mixed with 1 lbm of water at 100F
+m1=1; //Unit:lbm //mass
+m2=1; //Unit:lbm //mass
+c1=1; //Specific heat constant
+c2=1; //Specific heat constant
+t1=500; //(unit:fahrenheit)
+t2=100; //(unit:fahrenheit)
+cmix=1; //Specific heat constant of mixture
+//now, m1*c1*t1 +m2*c2*t2 = (m1+m2)*cmix*t //So,
+t=((m1*c1*t1)+(m2*c2*t2))/((m1+m2)*cmix) //resulting temperature of the mixture
+printf("In problem 4.14,The resulting temperature of the mixture is %f fahrenheit\n",t);
+
+//Now,in problem 4.15,taking 0F as a reference temperature,
+//For hot fluid,
+deltas=cmix*log((t1+460)/(0+460)); //temperatures converted to absolute temperatures; //deltas=change in entropy //Unit:Btu/(lbm*R)
+//For cold fluid,
+s=cmix*log((t2+460)/(0+460)); //temperatures converted to absolute temperatures; //s=change in entropy //Unit:Btu/(lbm*R)
+//At final mixture temperature of t F,the entropy of each system above 0F is,for the hot fluid
+s1=cmix*log((t+460)/(0+460)); //temperatures converted to absolute temperatures; //s1=change in entropy //Unit:Btu/(lbm*R)
+//and for the cold fluid,
+s2=cmix*log((t+460)/(0+460)); //temperatures converted to absolute temperatures; //s2=change in entropy //Unit:Btu/(lbm*R)
+printf("The change in the entropy for hot fluid is %f Btu/(lbm*R)\n",s1-deltas);
+printf("The change in the entropy for cold fluid is %f Btu/(lbm*R)\n",s2-s);
+printf("The total change in entropy if %f Btu/(lbm*R",s1-deltas+s2-s);
diff --git a/2417/CH4/EX4.2/Ex4_2.sce b/2417/CH4/EX4.2/Ex4_2.sce
new file mode 100755
index 000000000..204476d2d
--- /dev/null
+++ b/2417/CH4/EX4.2/Ex4_2.sce
@@ -0,0 +1,27 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.2\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.2 (page no. 149)
+// Solution
+
+//given data
+Qin=100; //heat added to the cycle
+
+printf("In problem 4.1,\n")
+//given data
+t1=1000; //(unit:fahrenheit) //Source temperature
+t2=80; //(unit:fahrenheit) //Sink temperature
+//solution
+//converting temperatures to absolute temperatures;
+T1=t1+460; //Source temperature //Unit:R
+T2=t2+460; //Sink temperature //Unit:R
+printf("Solution for (a)\n");
+printf("Efficiency of the engine is %f percentage\n\n",((T1-T2)/T1)*100);
+
+printf("Now in problem 4.2,\n")
+W=0.63*Qin; //W=W/J; //Efficiency in problem 4.1
+W=Qin*(W/Qin); //amount of work
+Qr=Qin-W; //Qin-Qr=W/J //Qr=heat rejected by the cycle
+printf("The heat removed from the reservoir %f units",Qr);
diff --git a/2417/CH4/EX4.3/Ex4_3.sce b/2417/CH4/EX4.3/Ex4_3.sce
new file mode 100755
index 000000000..c57a6582e
--- /dev/null
+++ b/2417/CH4/EX4.3/Ex4_3.sce
@@ -0,0 +1,22 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.3\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.3 (page no. 149)
+// Solution
+
+//given data
+t1=70; //(unit:fahrenheit) //Source temperature
+t2=15; //(unit:fahrenheit) //Sink temperature
+Qin=125000; //(unit=Btu/hr) //Qin=heat added to the cycle
+//converting temperatures to absolute temperatures;
+T1=t1+460; //Source temperature //Unit:R
+T2=t2+460; //Sink temperature //Unit:R
+Qr=Qin*(T2/T1); //Qr=heat rejected by the cycle
+printf("Qr is %f in Btu/hr\n",Qr);
+work=Qin-Qr; //reversed cycle requires atleast input //work //btu/hr
+printf("Work is %f in Btu/hr\n",work);
+// 1 hp = 33000 ft*LBf/min
+// 1 Btu = 778 ft*LBf //1 hr = 60 min
+printf("Minimum horsepower input required is %f hp",work*778/(60*33000));
diff --git a/2417/CH4/EX4.4/Ex4_4.sce b/2417/CH4/EX4.4/Ex4_4.sce
new file mode 100755
index 000000000..a6f672526
--- /dev/null
+++ b/2417/CH4/EX4.4/Ex4_4.sce
@@ -0,0 +1,24 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.4\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.4 (page no. 150)
+// Solution
+
+W=(50*33000)/778;//output //W=W/J
+// 1 hp = 33000 ft*LBf/min
+// 1 Btu = 778 ft*LBf
+printf("Output is %f in Btu/min\n",W);
+t1=1000; //Source temperature //(unit:fahrenheit)
+t2=100; //Sink temperature //(unit:fahrenheit)
+//converting temperatures to absolute temperatures;
+T1=t1+460; //Source temperature //Unit:R
+T2=t2+460; //Sink temperature //Unit:R
+n=(1-(T2/T1))*100; //efficiency
+printf("Efficiency is %f percentage\n",n);//(in %)
+//n=(W/J)/Qin
+Qin=W/(n/100);//(unit Btu/hr) //Qin=heat added to the cycle
+printf("Heat added to the cycle is %f in Btu/min\n",Qin);
+Qr=Qin*(1-(n/100));//(unit Btu/hr) //Qr=heat rejected by the cycle
+printf("Heat rejected by the cycle is %f in Btu/min \n",Qr);
diff --git a/2417/CH4/EX4.5/Ex4_5.sce b/2417/CH4/EX4.5/Ex4_5.sce
new file mode 100755
index 000000000..43e9cc1dd
--- /dev/null
+++ b/2417/CH4/EX4.5/Ex4_5.sce
@@ -0,0 +1,22 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.5\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.5 (page no. 151)
+// Solution
+
+t1=700; //Source temperature //Unit:Celcius
+t2=20; //Sink temperature //Unit:Celcius
+//converting in F
+T1=t1+273; //Source temperature //Unit:R
+T2=t2+273; //Sink temperature //Unit:R
+n=(T1-T2)/T1*100; //Efficiency
+printf("Efficiency is %f percentage\n",n);//(in %)
+output=65;//in hp //Given
+work=output*0.746;//(unit kJ/s) // 1 hp = 746 W
+printf("Work is %f kJ/s\n",work);
+Qin=work/(n/100);//(unit kJ/s) //Qin=heat added to the cycle
+printf("Heat added to the cycle is %f kJ/s \n",Qin);
+Qr=Qin*(1-(n/100));//(unit kJ/s) //Qr=heat rejected by the cycle
+printf("Heat rejected by the cycle is %f kJ/s \n",Qr);
diff --git a/2417/CH4/EX4.7/Ex4_7.sce b/2417/CH4/EX4.7/Ex4_7.sce
new file mode 100755
index 000000000..60fde9f89
--- /dev/null
+++ b/2417/CH4/EX4.7/Ex4_7.sce
@@ -0,0 +1,20 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.7\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.7 (page no. 152)
+// Solution
+
+t1=700; //(unit:fahrenheit) //Source temperature
+t2=200; //(unit:fahrenheit) //Sink temperature
+//converting temperatures to absolute temperatures;
+T1=t1+460; //Source temperature //Unit:R
+T2=t2+460; //Sink temperature //Unit:R
+//n1=(T1-Ti)/T1 and n2=(Ti-T2)/Ti //n1 & n2 are efficiency
+//(T1-Ti)/T1=(Ti-T2)/Ti;
+Ti=sqrt(T1*T2); //Exhaust temperature //Unit:R
+printf("Exhaust temperature of first engine is %f in R\n",Ti);
+//converting absolute temperature to normal F temperature
+//Ti(fahrenheit)=Ti(R)-460;
+printf("Exhaust temperature of first engine is %f fahrenheit\n",Ti-460);
diff --git a/2417/CH4/EX4.8/Ex4_8.sce b/2417/CH4/EX4.8/Ex4_8.sce
new file mode 100755
index 000000000..c83682752
--- /dev/null
+++ b/2417/CH4/EX4.8/Ex4_8.sce
@@ -0,0 +1,15 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.8\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.8 (page no. 157)
+// Solution
+
+//For reversible isothermal process,
+q=843.7; //Heat //Unit:Btu //at 200 psia
+t=381.86; //(unit:fahrenheit) //temperature
+////converting temperatures to absolute temperatures;
+T=t+460; //temperature //unit:R
+deltaS=(q/T); //Change in entropy //Unit:Btu/lbm*R
+printf("Change in entropy is %f Btu/lbm*R\n",deltaS); //1 LBm of saturated water
diff --git a/2417/CH4/EX4.9/Ex4_9.sce b/2417/CH4/EX4.9/Ex4_9.sce
new file mode 100755
index 000000000..37c8350b6
--- /dev/null
+++ b/2417/CH4/EX4.9/Ex4_9.sce
@@ -0,0 +1,38 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.9\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.9 (page no. 158)
+// Solution
+
+//For reversible isothermal process,
+//In problem 4.8,
+q=843.7; //Heat //Unit:Btu //at 200 psia
+t=381.86; //(unit:fahrenheit)
+//converting temperatures to absolute temperatures;
+T=t+460; //Unit:R"
+deltaS=(q/T); //Change in entropy //Btu/lbm
+printf("Change in entropy is %f Btu/lbm*R\n",deltaS); //1 LBm of saturated water
+
+//In problem 4.9
+t1=381.86; //(unit:fahrenheit) //Source temperature
+t2=50; //(unit:fahrenheit) //Sink temperature
+//converting temperatures to absolute temperatures;
+T1=t1+460; //Source temperature //Unit:R
+T2=t2+460; //Sink temperature //Unit:R
+qin=q;//heat added to the cycle
+n=(1-(T2/T1))*100; //Efficiency
+printf("Efficiency is %f percentage\n",n);
+wbyJ=qin*n*0.01;//work output
+printf("Work output is %f Btu/lbm\n",wbyJ);
+Qr=qin-wbyJ; //heat rejected
+printf("Heat rejected is %f Btu/lbm\n\n",Qr);
+printf("As an alternative solution and refering to figure 4.12,\n")
+qin=T1*deltaS; //heat added //btu/lbm
+Qr=T2*deltaS; //Heat rejected //btu/lbm
+printf("Heat rejected is %f Btu/lbm\n",Qr);
+wbyJ=qin-Qr; //Work output //Btu/lbm
+printf("Work output is %f Btu/lbm\n",wbyJ);
+n=(wbyJ/qin)*100; //Efficiency
+printf("Efficiency is %f percentage\n",n);
generated by cgit (git 2.34.1) at 2024-12-20 17:53:31 +0000