diff options
Diffstat (limited to '172')
112 files changed, 1945 insertions, 0 deletions
diff --git a/172/CH10/EX10.1/ex1.sce b/172/CH10/EX10.1/ex1.sce new file mode 100755 index 000000000..277d38356 --- /dev/null +++ b/172/CH10/EX10.1/ex1.sce @@ -0,0 +1,17 @@ +//example 1
+//Calculating reversible work
+clear
+clc
+//Form the Steam Tables,the inlet and the exit state properties are
+hi=171.95 //initial specific heat of enthalpy in kJ/kg
+si=0.5705 //initial specific entropy in kJ/kg-K
+se=2.1341 //final specific entropy in kJ/kg-K
+he=765.34 //final specific heat of enthalpy in kJ/kg-K
+m=5 //mass flow rate of feedwater in kg/s
+q1=900/m //heat added by one of the sources in kJ/kg
+q2=he-hi-q1 //second heat transfer in kJ/kg
+To=25+273.3 //Temp. of the surroundings in K
+T1=100+273.2 //temp. of reservoir of one of the source in K
+T2=200+273.2 //temp. of reservoir of second source in K
+wrev=To*(se-si)-(he-hi)+q1*(1-To/T1)+q2*(1-To/T2) //reversible work in kJ/kg
+printf("\n Hence, the irreversibility is i=%.1f kJ/kg.\n",wrev)
diff --git a/172/CH10/EX10.2/ex2.sce b/172/CH10/EX10.2/ex2.sce new file mode 100755 index 000000000..6e3c27f3a --- /dev/null +++ b/172/CH10/EX10.2/ex2.sce @@ -0,0 +1,18 @@ +//example 2
+//Calculating reversible work
+clear
+clc
+//Form the Steam Tables,the inlet and the exit state properties are
+hi=298.6 //initial specific heat of enthalpy in kJ/kg
+si=6.8631 //initial specific entropy in kJ/kg-K
+se=7.4664 //final specific entropy in kJ/kg-K
+he=544.7 //final specific heat of enthalpy in kJ/kg-K
+q=-50 //heat lost to surroundings in kJ/kg
+w=hi-he+q //work in kJ/kg
+To=25+273.2 //Temp. of the surroundings in K
+P1=100 //Pressure of ambient air in kPa
+P2=1000 //Final pressure of air after compression in kPa
+R=0.287 //Universal gas constant in kJ/kg-K
+wrev=To*(se-si-R*log(P2/P1))-(he-hi)+q*(1-To/To)//reversible work for the given change of state in kJ/kg
+i=wrev-w //irreversibility in kJ/kg
+printf("\n Hence, the irreversibility is i=%.1f kJ/kg.\n",i)
diff --git a/172/CH10/EX10.3/ex3.sce b/172/CH10/EX10.3/ex3.sce new file mode 100755 index 000000000..28dcfac8d --- /dev/null +++ b/172/CH10/EX10.3/ex3.sce @@ -0,0 +1,20 @@ +//example 3
+//Calculating reversible work and irreversibility
+clear
+clc
+//Form the Steam Tables at state 1
+u1=1243.5 //initial specific internal energy in kJ/kg
+s1=4.4819 //initial specific entropy in kJ/kg-K
+v1=28.895 //initial specific volume in m^3/kg
+v2=2*v1 //final specific volume in kg/m^3
+u2=u1 //initial specific internal energy in kJ/kg
+//These two independent properties, v2 and u2 , fix state 2.The final temp. is calculated by interplotation using the data for T2=5C and v2,x=0/3928 and u=948.5 kJ/kg. For T2=10C and v2, x=0.5433 and u=1317 kJ/kg
+T2=9.1+273.2 //final temp. in K
+x2=0.513 //quality in final state
+s2=4.644 //final specific entropy in kJ/kg
+V1=1 //volume of part of A in m^3
+m=V1/v1 //mass flow rate in kg/s
+To=20+273.2 //Room temperature in K
+Wrev=To*m*(s2-s1) //reversible work in kJ
+I=Wrev //irreversibility of the process
+printf("\n The irreversibility is I=%.3f kJ/kg.\n",I)
\ No newline at end of file diff --git a/172/CH11/EX11.1/ex1.sce b/172/CH11/EX11.1/ex1.sce new file mode 100755 index 000000000..bae723ced --- /dev/null +++ b/172/CH11/EX11.1/ex1.sce @@ -0,0 +1,27 @@ +//Ques 1
+//To determine the efficiency of Rankine cycle
+clc
+clear
+//1-Inlet state of pump
+//2-Exit state of pump
+P2=2000;//Exit pressure in kPa
+P1=10;//Inlet pressure in kPa
+v=0.00101;//specific weight of water in m^3/kg
+wp=v*(P2-P1);//work done in pipe in kJ/kg
+h1=191.8;//Enthalpy in kJ/kg from table
+h2=h1+wp;//enthalpy in kJ/kg
+//2-Inlet state for boiler
+//3-Exit state for boiler
+h3=2799.5;//Enthalpy in kJ/kg
+//3-Inlet state for turbine
+//4-Exit state for turbine
+//s3=s4(Entropy remain same)
+s4=6.3409;//kJ/kg
+sf=0.6493;//Entropy at liquid state in kJ/kg
+sfg=7.5009;//Entropy difference for vapor and liquid state in kJ/kg
+x4=(s4-sf)/sfg;//x-factor
+hfg=2392.8;//Enthalpy difference in kJ/kg for turbine
+h4=h1+x4*hfg;//Enthalpy in kJ/kg
+
+nth=((h3-h2)-(h4-h1))/(h3-h2);
+printf('Percentage efficiency = %.1f ',nth*100);
\ No newline at end of file diff --git a/172/CH11/EX11.2/ex2.sce b/172/CH11/EX11.2/ex2.sce new file mode 100755 index 000000000..ea7fd759a --- /dev/null +++ b/172/CH11/EX11.2/ex2.sce @@ -0,0 +1,27 @@ +//Ques 2
+//To determine the efficiency of Rankine cycle
+clc
+clear
+//1-Inlet state of pump
+//2-Exit state of pump
+P2=4000;//Exit pressure in kPa
+P1=10;//Inlet pressure in kPa
+v=0.00101;//specific weight of water in m^3/kg
+wp=v*(P2-P1);//work done in pipe in kJ/kg
+h1=191.8;//Enthalpy in kJ/kg from table
+h2=h1+wp;//enthalpy in kJ/kg
+//2-Inlet state for boiler
+//3-Exit state for boiler
+h3=3213.6;//Enthalpy in kJ/kg from table
+//3-Inlet state for turbine
+//4-Exit state for turbine
+//s3=s4(Entropy remain same)
+s4=6.7690;//Entropy in kJ/kg from table
+sf=0.6493;//Entropy at liquid state in kJ/kg from table
+sfg=7.5009;//Entropy difference for vapor and liquid state in kJ/kg from table
+x4=(s4-sf)/sfg;//x-factor
+hfg=2392.8;//Enthalpy difference in kJ/kg for turbine
+h4=h1+x4*hfg;//Enthalpy in kJ/kg
+
+nth=((h3-h2)-(h4-h1))/(h3-h2);
+printf('Percentage efficiency = %.1f ',nth*100);
\ No newline at end of file diff --git a/172/CH11/EX11.3/ex3.sce b/172/CH11/EX11.3/ex3.sce new file mode 100755 index 000000000..045aaa202 --- /dev/null +++ b/172/CH11/EX11.3/ex3.sce @@ -0,0 +1,39 @@ +//Ques 3
+//To determine the efficiency of a cycle
+clc
+clear
+//1-Inlet state of pump
+//2-Exit state of pump
+P2=4000;//Exit pressure in kPa
+P1=10;//Inlet pressure in kPa
+v=0.00101;//specific weight of water in m^3/kg
+wp=v*(P2-P1);//work done in pipe in kJ/kg
+h1=191.8;//Enthalpy in kJ/kg from table
+h2=h1+wp;//enthalpy in kJ/kg
+//2-Inlet state for boiler
+//3-Exit state for Boiler
+h3=3213.6;//Enthalpy in kJ/kg from table
+//3-Inlet state for high pressure turbine
+//4-Exit state for high pressure turbine
+//s3=s4(Entropy remain same)
+s4=6.7690;//Entropy in kJ/kg from table
+sf=1.7766;//Entropy at liquid state in kJ/kg from table
+sfg=5.1193;//Entropy difference for vapor and liquid state in kJ/kg from table
+x4=(s4-sf)/sfg;//x-factor
+hf=604.7//Enthalpy of liquid state in kJ/kg
+hfg=2133.8;//Enthalpy difference in kJ/kg for turbine
+h4=hf+x4*hfg;//Enthalpy in kJ/kg
+//5-Inlet state for low pressure turbine
+//6-Exit pressure for low pressure turbine
+sf=0.6493;//Entropy in liquid state in kJ/kg for turbine
+h5=3273.4;//enthalpy in kJ/kg
+s5=7.8985;//Entropy in kJ/kg
+sfg=7.5009;//entropy diff in kJ/kg
+x6=(s5-sf)/sfg;//x-factor
+hfg=2392.8;//enthalpy difference for low pressure turbine in kj/kg
+h6=h1+x6*hfg;//entropy in kg/kg
+wt=(h3-h4)+(h5-h6);//work output in kJ/kg
+qh=(h3-h2)+(h5-h4);
+
+nth=(wt-wp)/qh;
+printf('Percentage efficiency = %.1f ',nth*100);
\ No newline at end of file diff --git a/172/CH11/EX11.4/ex4.sce b/172/CH11/EX11.4/ex4.sce new file mode 100755 index 000000000..edbe4739f --- /dev/null +++ b/172/CH11/EX11.4/ex4.sce @@ -0,0 +1,33 @@ +//ques4
+//Efficiency of Refrigeration cycle
+clc
+clear
+//from previous examples
+h1=191.8;//kJ/kg
+h5=3213.6;//kg/kg
+h6=2685.7;//kJ/kg
+h7=2144.1;//kJ/kg
+h3=604.7;//kJ/kg
+//1-Inlet state of pump
+//2-Exit state of pump
+P2=400;//Exit pressure in kPa
+P1=10; //Inlet pressure in kPa
+v=0.00101;//specific weight of water in m^3/kg
+wp1=v*(P2-P1);//work done for low pressure pump in kJ/kg
+h1=191.8;//Enthalpy in kJ/kg from table
+h2=h1+wp1;//enthalpy in kJ/kg
+//5-Inlet state for turbine
+//6,7-Exit state for turbine
+y=(h3-h2)/(h6-h2);//extraction fraction
+wt=(h5-h6)+(1-y)*(h6-h7);//turbine work in kJ/kg
+//3-Inlet for high pressure pump
+//4-Exit for high pressure pump
+P3=400;//kPa
+P4=4000;//kPa
+v=0.001084;//specific heat for 3-4 process in m^3/kg
+wp2=v*(P4-P3);//work done for high pressure pump
+h4=h3+wp2;//Enthalpy in kJ/kg
+wnet=wt-(1-y)*wp1-wp2;
+qh=h5-h4;//Heat output in kJ/kg
+nth=wnet/qh;
+printf('Refrigerator Efficiency = %.1f ',nth*100);
diff --git a/172/CH11/EX11.5/ex5.sce b/172/CH11/EX11.5/ex5.sce new file mode 100755 index 000000000..fd38837e4 --- /dev/null +++ b/172/CH11/EX11.5/ex5.sce @@ -0,0 +1,35 @@ +//ques5
+//To determine thermal efficiency of cycle
+clear
+clc
+//5-Inlet state for turbine
+//6-Exit state for turbine
+//h-Enthalpy at a state
+//s-Entropy at a state
+//from steam table
+h5=3169.1;//kJ/kg
+s5=6.7235;//kJ/kg
+s6s=s5;
+sf=0.6493;//Entropy for liquid state in kJ/kg
+sfg=7.5009;//Entropy difference in kJ/kg
+hf=191.8;//kJ/kg
+hfg=2392.8;//Enthalpy difference in kJ/kg
+x6s=(s6s-sf)/sfg;//x-factor
+h6s=hf+x6s*hfg;//kJ/Kg at state 6s
+nt=0.86;//turbine efficiency given
+wt=nt*(h5-h6s);
+//1-Inlet state for pump
+//2-Exit state for pump
+np=0.80;//pump efficiency given
+v=0.001009;//specific heat in m^3/kg
+P2=5000;//kPa
+P1=10;//kPa
+wp=v*(P2-P1)/np;//Work done in pump in kJ/kg
+wnet=wt-wp;//net work in kJ/kg
+//3-Inlet state for boiler
+//4-Exit state for boiler
+h3=171.8;//in kJ/kg from table
+h4=3213.6;//kJ/kg from table
+qh=h4-h3;
+nth=wnet/qh;
+printf('Cycle Efficiency = %.1f ',nth*100);
\ No newline at end of file diff --git a/172/CH12/EX12.1/ex1.sce b/172/CH12/EX12.1/ex1.sce new file mode 100755 index 000000000..334158848 --- /dev/null +++ b/172/CH12/EX12.1/ex1.sce @@ -0,0 +1,37 @@ +//ques1
+//Standard brayton cycle
+clc
+clear
+//1-Inlet for compressor
+//2-Exit for compressor
+//T-Temperature at a state
+//P-Pressure at a state
+T1=288.2;//K
+P2=1000;//kPa
+P1=100;//kPa
+k=1.4;
+T2=T1*(P2/P1)^(1-1/k);//K
+Cp=1.004;//Specific heat at constant pressure in kJ/kg
+wc=Cp*(T2-T1);//compressor work in kJ/kg;
+printf('Temperature T2 = %.1f K\n',T2);
+printf(' Compressor work = %.1f kJ/kg \n',wc);
+//3-Turbine Inlet
+//4-Turbine Exit
+P4=P1;
+P3=P2;
+T3=1373.2;//K
+T4=T3*(P4/P3)^(1-1/k);//K
+wt=Cp*(T3-T4);
+wnet=wt-wc;
+printf(' Temperature T3 = %.1f K\n',T3);
+printf(' Temperature T4 = %.1f K\n',T4);
+printf(' Turbine work = %.1f kJ/kg\n',wt);
+printf(' Net work = %.1f kJ/kg\n',wt-wc);
+//2-Also high temperature heat exchanger Inlet
+//3-(-do-) Exit
+qh=Cp*(T3-T2);//Heat of source in kJ/kg
+//4-high temp heat exchanger inlet
+//1-(-do-) Exit
+ql=Cp*(T4-T1);//Heat of sink in kJ/kg
+nth=wnet/qh;
+printf(' Thermal Efficiency of cycle = %.1f percent',nth*100);
\ No newline at end of file diff --git a/172/CH12/EX12.2/ex2.sce b/172/CH12/EX12.2/ex2.sce new file mode 100755 index 000000000..4c0325a98 --- /dev/null +++ b/172/CH12/EX12.2/ex2.sce @@ -0,0 +1,43 @@ +//Calculation mistake in book
+//ques2
+//Standard brayton cycle
+clc
+clear
+//Calculation mistake in book
+//1-Inlet for compressor
+//2-Exit for compressor
+//T-Temperature at a state
+//P-Pressure at a state
+T1=288.2;//K
+P2=1000;//kPa
+P1=100;//kPa
+k=1.4;
+T2s=T1*(P2/P1)^(1-1/k);//K
+nc=.80;//Compressor Efficiency
+T2=T1+(T2s-T1)/0.80;
+Cp=1.004;//Specific heat at constant pressure in kJ/kg
+wc=Cp*(T2-T1);//compressor work in kJ/kg;
+printf('Temperature T2 = %.1f K\n',T2);
+printf(' Compressor work = %.1f kJ/kg \n',wc);
+//3-Turbine Inlet
+//4-Turbine Exit
+P4=P1;
+P3=P2;
+T3=1373.2;//K
+T4s=T3*(P4/P3)^(1-1/k);//K
+nt=0.85;//turbine Efficiency
+T4=T3-(T3-T4s)*0.85;
+wt=Cp*(T3-T4);
+wnet=wt-wc;
+printf(' Temperature T3 = %.1f K\n',T3);
+printf(' Temperature T4 = %.1f K\n',T4);
+printf(' Turbine work = %.1f kJ/kg\n',wt);
+printf(' Net work = %.1f kJ/kg\n',wt-wc);
+//2-Also high temperature heat exchanger Inlet
+//3-(-do-) Exit
+qh=Cp*(T3-T2);//Heat of source in kJ/kg
+//4-high temp heat exchanger inlet
+//1-(-do-) Exit
+ql=Cp*(T4-T1);//Heat of sink in kJ/kg
+nth=wnet/qh;
+printf(' Thermal Efficiency of cycle = %.1f percent',nth*100);
\ No newline at end of file diff --git a/172/CH12/EX12.3/ex3.sce b/172/CH12/EX12.3/ex3.sce new file mode 100755 index 000000000..34e28fddb --- /dev/null +++ b/172/CH12/EX12.3/ex3.sce @@ -0,0 +1,12 @@ +//ques3
+//efficiency of the cycle
+clc
+clear
+wnet=395.2;//kJ/kg from example no 1
+//Tx=T4
+Tx=710.8;//K from example no 1
+T3=1373.2;//K from example no 1
+Cp=1.004;//specific heat in kJ/kg
+qh=Cp*(T3-Tx);
+nth=wnet/qh;
+printf('Thermal efficiency = %.1f percent',nth*100);
\ No newline at end of file diff --git a/172/CH12/EX12.4/ex4.sce b/172/CH12/EX12.4/ex4.sce new file mode 100755 index 000000000..58b5c9499 --- /dev/null +++ b/172/CH12/EX12.4/ex4.sce @@ -0,0 +1,13 @@ +//ques4
+//Calculation of work in the given cycle
+clear
+clc
+R=0.287;//gas constant
+T1=288.2;//compressor temperature K
+T2=1373.2;//K turbine temperature K
+//Pe/Pi=c=10, Pi/Pe=1/c from example 12.1
+c=10;
+wc=-R*T1*log(c);
+printf('Isothermal work in compressor = %.1f kJ/kg \n',wc);
+wt=-R*T2*log(1/c);
+printf(' Isothermal work in turbine = %.1f kJ/kg\n',wt);
\ No newline at end of file diff --git a/172/CH12/EX12.5/ex5.sce b/172/CH12/EX12.5/ex5.sce new file mode 100755 index 000000000..3d52c5815 --- /dev/null +++ b/172/CH12/EX12.5/ex5.sce @@ -0,0 +1,26 @@ +//ques5
+//air standard cycle for jet repulsion
+clear
+clc
+//1-compressor inlet
+//2-Compressor exit
+//P-Pressure at given point
+//T-Temperature at given point
+P1=100;//kPa
+P2=1000;//kPa
+T1=288.2;//K
+T2=556.8;//K
+wc=269.5;//from ex 12.1 work done in compressor in kJ/kg
+//2-Burner inlet
+//3-Burner exit
+P3=1000;//kPa
+T3=1373.2;//K
+//wc=wt
+Cp=1.004;//specific enthalpy of heat at constant pressure in kJ/kg
+k=1.4;
+T4=T3-wc/Cp;
+P4=P3*(T4/T3)^(1-1/k);
+//from s4=s5 and h4=h5+v2/2 we get
+T5=710.8//K, from second law
+v=sqrt(2*Cp*1000*(T4-T5));//m/s
+printf('Velocity of air leaving the nozel = %.0f m/s',v);
\ No newline at end of file diff --git a/172/CH12/EX12.6/ex6.sce b/172/CH12/EX12.6/ex6.sce new file mode 100755 index 000000000..8de46cb53 --- /dev/null +++ b/172/CH12/EX12.6/ex6.sce @@ -0,0 +1,24 @@ +//ques6
+//air standard refrigeration cycle
+clear
+clc
+//1-compressor inlet
+//2-compressor exit
+P1=100;//kPa
+P2=500;//kPa
+k=1.4;
+rp=P2/P1;
+cop=(rp^(1-1/k)-1)^-1;
+printf('Coefficient of performance = %.3f \n',cop);
+//3-Expander inlet
+//4-Expander exit
+P3=P2;
+P4=P1;
+T3=288.23;//K, given and fixed
+T4=T3/(P3/P4)^(1-1/k);
+T1=253.2;//K, given
+Cp=1.004;//Specific heat at cons pressure in kJ/kg
+ql=Cp*(T1-T4);//heat released in kJ/kg
+P=1//power required in kW
+ms=P/ql;//kg/s
+printf(' Rate at which the air enter the compressor = %.3f kg/s ',ms);
\ No newline at end of file diff --git a/172/CH12/EX12.7/ex7.sce b/172/CH12/EX12.7/ex7.sce new file mode 100755 index 000000000..91f621436 --- /dev/null +++ b/172/CH12/EX12.7/ex7.sce @@ -0,0 +1,36 @@ +//ques7
+//the otto cycle
+clear
+clc
+//1-compressor inlet
+//2-compressor exit
+P1=100;//kPa
+T1=288.2;//K
+R=0.287;//gas constant
+v1=R*T1/P1;//specific volume at inlet in m^3/kg
+rv=10;//compression ratio given
+k=1.4;//constant
+T2=T1*rv^(k-1);//K
+printf('Temperature at compressor exit, T2 = %.1f K \n',T2);
+P2=P1*rv^k;//kPa
+printf(' Pressure at compressor exit, P2 = %.3f MPa \n ',P2/1000);
+v2=v1/rv;//specific heat at exit in m^3/kg
+//23-heat addition process
+//q23=Cv*(T3-T2)=1800 kJ/kg given
+q23=1800;//kJ/kg heat addition, given
+Cv=0.717;//specific heat at constant volume in kJ/kg
+T3=T2+q23/Cv;//K
+printf('\n Initial Temperature during heat additon process, T3 = %.0f K \n',T3);
+P3=P2*(T3/T2);//kPa
+printf( ' Initial pressure during heat addition process, P3 = %.3f MPa \n',P3/1000);
+r=10;//k=V4/V3=P3/P4
+T4=T3*(1/r)^(k-1);
+printf(' \n Final temperature during heat addition process, T4 = %.1f K \n',T4);
+P4=P3/r^k;//kPa
+printf(' Final pressure during heat addition process, P4 = %.4f MPa \n',P4/1000);
+nth=1-1/r^k;//thermal efficiency
+printf('\n Thermal efficiency = %.1f percent \n',nth*100);
+q41=Cv*(T1-T4);///heat for process 4-1 in kJ/kg
+wnet=q23+q41;
+mep=wnet/(v1-v2);//effective mean pressure n kPa
+printf('\n Mean effective pressure = %.0f kPa \n',mep);
\ No newline at end of file diff --git a/172/CH12/EX12.8/ex8.sce b/172/CH12/EX12.8/ex8.sce new file mode 100755 index 000000000..861478b8a --- /dev/null +++ b/172/CH12/EX12.8/ex8.sce @@ -0,0 +1,36 @@ +//ques7
+//the diesel cycle
+clear
+clc
+//1-compressor inlet
+//2-compressor exit
+P1=100;//kPa
+T1=288.2;//K
+R=0.287;//gas constant
+v1=R*T1/P1;//specific volume at inlet in m^3/kg
+rv=20;//compression ratio given
+k=1.4;//constant
+T2=T1*rv^(k-1);//K
+printf('Temperature at compressor exit, T2 = %.1f K \n',T2);
+P2=P1*rv^k;//kPa
+printf(' Pressure at compressor exit, P2 = %.3f MPa \n ',P2/1000);
+v2=v1/rv;//specific heat at exit in m^3/kg
+//23-heat addition process
+//q23=Cv*(T3-T2)=1800 kJ/kg given
+q23=1800;//kJ/kg heat addition, given
+Cv=.717;
+Cp=1.004;//specific heat at constant pressure in kJ/kg
+T3=T2+q23/Cp;//K
+printf('\n Initial Temperature during heat addition process, T3 = %.0f K \n',T3);
+r=T3/T2;//T3/T2=V3/V2=r
+v3=r*v2;
+T4=T3/(v1/v3)^(k-1);
+printf(' Final temperature during heat addition process, T4 = %.0f K \n',T4);
+q41=Cv*(T1-T4);///heat for process 4-1 in kJ/kg
+wnet=q23+q41;
+mep=wnet/(v1-v2);//effective mean pressure in kPa
+qh=1800;//heat transfer in kJ/kg
+nth=wnet/qh;//thermal efficiency
+
+printf('\n Thermal efficiency = %.1f percent \n',nth*100);
+printf('\n Mean effective pressure = %.0f kPa \n',mep);
\ No newline at end of file diff --git a/172/CH13/EX13.3/ex3.sce b/172/CH13/EX13.3/ex3.sce new file mode 100755 index 000000000..e9a1b29a6 --- /dev/null +++ b/172/CH13/EX13.3/ex3.sce @@ -0,0 +1,16 @@ +//ques3
+//calculating humidity ratio, dew point,mass of air, mass of vapor
+clear
+clc
+r=0.70;//relative humidity
+Pg=5.628;//saturation pressure in kPa
+Pv=r*Pg;//vapour pressure in kPa
+P=100;//net pressure kPa
+Pa=P-Pv;//Partial pressure of air
+w=0.622*Pv/Pa;//humidity ratio formula
+V=100;//volume in m^3
+Ra=0.287;//gas constant for water vapour
+T=308.2;//Temperature in K
+ma=Pa*V/(Ra*T);//mass in kg
+mv=w*ma;//mass of vapour
+printf('Mass of vapour = %.2f Kg ', mv);
\ No newline at end of file diff --git a/172/CH13/EX13.4/ex4.sce b/172/CH13/EX13.4/ex4.sce new file mode 100755 index 000000000..3ab556040 --- /dev/null +++ b/172/CH13/EX13.4/ex4.sce @@ -0,0 +1,14 @@ +//ques4
+//calculating amount of water vapour condensed on cooling
+clear
+clc
+//from example 3
+w1=0.0255;//w1=w, humidity ratio at initial temperature
+ma=108.6;//mass of air in kg
+P=100;//kPa net pressure
+//at 5 C mixture is saturated so Pv2=Pg2
+Pg2=0.8721;
+Pv2=Pg2;
+w2=0.622*Pv2/(P-Pg2);
+mc=ma*(w1-w2);
+printf('Mass of vapour condense = %.3f kg \n',mc);
\ No newline at end of file diff --git a/172/CH13/EX13.5/ex5.sce b/172/CH13/EX13.5/ex5.sce new file mode 100755 index 000000000..98ac6de61 --- /dev/null +++ b/172/CH13/EX13.5/ex5.sce @@ -0,0 +1,24 @@ +//ques5
+//calculating heat transfer per kilogram of dry air
+clear
+clc
+//1-inlet state
+//2-Exit state
+r1=0.80;//realtive humidity at state 1
+Pg1=4.246;//saturation pressure of vapour in kPa
+P1=105;//net pressure at state 1 in kPa
+P2=100;//net pressure at state 2 in kPa
+Pv1=r1*Pg1;//partial pressure of vapour in kPa
+w1=0.622*Pv1/(P1-Pv1);//humidity ratio at state 1
+r2=0.95;//relative humidity at state 2
+Pg2=1.7051;//saturation pressure of vapour in kPa
+Pv2=r2*Pg2;//partial pressure of vapour in kPa
+w2=0.622*Pv2/(P2-Pv2);//humidity ratio at state 2
+T1=30;//C
+T2=15;//C
+Cp=1.004;//specific heat of water vapour in kJ/kg
+hv2=2528.9;//enthalpy of vapourisation of vapour in kJ/kg
+hv1=2556.3;//enthalpy of vapourisation of vapour in kJ/kg
+hl2=62.99;//enthalpy of
+q=Cp*(T2-T1)+w2*hv2-w1*hv1+hl2*(w1-w2);//kJ/kg
+printf('Heat transferred per unit mass = %.2f kJ/kg of dry air',q);
\ No newline at end of file diff --git a/172/CH13/EX13.6/ex6.sce b/172/CH13/EX13.6/ex6.sce new file mode 100755 index 000000000..27c050799 --- /dev/null +++ b/172/CH13/EX13.6/ex6.sce @@ -0,0 +1,24 @@ +//ques6
+//calculating heat transferred in gas vapour mixture
+clear
+clc
+//n-Nitrogen
+//v-water vapour
+Pn2=1995;//Pressure of nitrogen in kPa
+V=0.5;//Volume in m^3
+Rn2=0.2968;//Gas constant for nitrogen in kJ/kg.K
+Rv=0.4615;//gas constant for vapour
+T1=323.2;//Temperature in K
+T2=283.2;//Temperature in K
+Pv1=5;//Pressure of water vapour in kPa at state 1
+Pv2=1.2276;//Pressure of water vapour in kPa at state 2
+mn2=Pn2*V/(Rn2*T1);//mass of nitrogen
+mv1=Pv1*V/(Rv*T1);//mass of vapour in kg
+mv2=Pv2*V/(Rv*T2);//mass of vapour in kg
+ml2=mv1-mv2;//mass of liquid condensed n kg
+uv1=2443.1;//specific internal energy of vapour in kJ/kg at state 1
+uv2=2389.2;//specific internal energy of vapour in kJ/kg at state 2
+ul2=42.0;//specific internal energy of liquid water in kJ/kg
+Cv=0.745;//specific heat at constant volume in kJ/kg.K
+Q=mn2*Cv*(T2-T1)+mv2*uv2+ml2*ul2-mv1*uv1;
+printf('Heat transferred = %.1f kJ ',Q);
\ No newline at end of file diff --git a/172/CH13/EX13.7/ex7.sce b/172/CH13/EX13.7/ex7.sce new file mode 100755 index 000000000..46920ea9e --- /dev/null +++ b/172/CH13/EX13.7/ex7.sce @@ -0,0 +1,25 @@ +//ques7
+//calculating humidity ratio and relative humidity
+clear
+clc
+//1-Inlet state
+//2-Exit state
+P=100;//net pressure n kPa
+//it is steady state adiabatic process
+//water vapour leaving is saturated so Pv2=Pg2
+Pg2=2.339;//saturation pressure of vapour in kPa
+Pv2=Pg2;//partial pressure of vapour
+w2=0.622*Pv2/(P-Pg2);
+Cpa=1.004;//specific heat n kJ/kg/K
+T2=20;// final temp in C
+T1=30;// initial temp in C
+Hfg2=2454.1;//specific heat difference at state 2 in kJ/kg
+hv1=2556.3;//enthalpy of water vapour at state 1 in kJ/kg
+hl2=83.96;//enthalpy of liquid water in kJ/kg
+w1=(Cpa*(T2-T1)+w2*Hfg2)/(hv1-hl2);
+printf('Relative humidity = %.4f \n',w1);
+//also w1=0.622*Pv1/(100-Pv2)
+Pv1=100*w1/(0.622+w1);
+Pg1=4.246;//saturation pressure at state 1 in kPa
+r=Pv1/Pg1;//humidity ratio
+printf(' Humidity ratio = %.3f ',r);
\ No newline at end of file diff --git a/172/CH14/EX14.1/ex1.sce b/172/CH14/EX14.1/ex1.sce new file mode 100755 index 000000000..611e46fea --- /dev/null +++ b/172/CH14/EX14.1/ex1.sce @@ -0,0 +1,13 @@ +//ques1
+//to determine the sublimation pressure of water
+clear
+clc
+//from table in appendix B.1.5
+T1=213.2;//K, Temperature at state 1
+P2=0.0129;//kPa, pressure at state 2
+T2=233.2;//K, Temperature at state 2
+hig=2838.9;//kJ/kg, enthalpy of sublimation
+R=.46152;//Gas constant
+//using relation log(P2/P1)=(hig/R)*(1/T1-1/T2)
+P1=P2*exp(-hig/R*(1/T1-1/T2));
+printf('Sublimation Pressure = %.5f kPa \n',P1);
\ No newline at end of file diff --git a/172/CH14/EX14.4/ex4.sce b/172/CH14/EX14.4/ex4.sce new file mode 100755 index 000000000..a6ace02c8 --- /dev/null +++ b/172/CH14/EX14.4/ex4.sce @@ -0,0 +1,18 @@ +//ques4
+//Volume expansivity, Isothermal and Adiabatic compressibility
+clear
+clc
+//known data
+ap=5*10^-5;//K^-1 Volume expansivity
+bt=8.6*10^-12;//m^2/N, Isothermal compressibility
+v=0.000114;//m^3/kg, specific volume
+P2=100*10^6;//pressure at state 2 in kPa
+P1=100;//pressure at state 1 in kPa
+w=-v*bt*(P2^2-P1^2)/2;//work done in J/kg
+//q=T*ds and ds=-v*ap*(P2-P1)
+//so q=-T*v*ap*(P2-P1)
+T=288.2;//Temperature in K
+q=-T*v*ap*(P2-P1);//heat in J/kg
+du=q-w;//change in internal energy in J/kg
+printf('Change in internal energy = %.1f J/kg ',du);
+
\ No newline at end of file diff --git a/172/CH14/EX14.5/ex5.sce b/172/CH14/EX14.5/ex5.sce new file mode 100755 index 000000000..e27cd050e --- /dev/null +++ b/172/CH14/EX14.5/ex5.sce @@ -0,0 +1,22 @@ +//ques5
+//adiabatic steady state processes
+clear
+clc
+//from table A.2
+P1=20;//pressure at state 1 in MPa
+P2=2;//pressure at state 2 in MPa
+T1=203.2;//Temperature at state 1 in K
+Pr1=P1/3.39;//Reduced pressure at state 1
+Pr2=P2/3.39;//Reduced pressure at state 2
+Tr1=T1/126.2;//Reduced temperature
+//from compressibility chart h1*-h1=2.1*R*Tc
+//from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)
+//h2*-h2=0.5*R*Tc
+//this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.5*R*Tc
+R=0.2968;//gas constant for given substance
+Tc=126.2;//K, Constant temperature
+Cp=1.0416;//heat enthalpy at constant pressure in kJ/kg
+T2=146;//temperature at state 2
+dh=-1.6*R*Tc+Cp*(T1-T2);//
+printf('Enthalpy change = %.0f kJ/kg \n',dh);
+printf(' Since Enthalpy change is %.0f kJ/kg so Temperature = %.1f K',dh,T2);
\ No newline at end of file diff --git a/172/CH14/EX14.6/ex6.sce b/172/CH14/EX14.6/ex6.sce new file mode 100755 index 000000000..ec041b50a --- /dev/null +++ b/172/CH14/EX14.6/ex6.sce @@ -0,0 +1,29 @@ +//ques6
+//isothermal steady state processes
+clear
+clc
+//from table A.2
+P1=8;//pressure at state 1 in MPa
+P2=0.5;//pressure at state 2 in MPa
+T1=150;//Temperature at state 1 in K
+Pr1=P1/3.39;//Reduced pressure at state 1
+Pr2=P2/3.39;//Reduced pressure at state 2
+Tr1=T1/126.2;//Reduced temperature
+T2=125;//temperature at state 2
+//from compressibility chart h1*-h1=2.1*R*Tc
+//from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)
+//h2*-h2=0.5*R*Tc
+//this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.15*R*Tc
+R=0.2968;//gas constant for given substance
+Tc=126.2;//K, Constant temperature
+Cp=1.0416;//heat enthalpy at constant pressure in kJ/kg
+dh=(2.35)*R*Tc+Cp*(T2-T1);//
+printf('Enthalpy change = %.2f kJ/kg \n',dh);
+//change in entropy
+//ds= -(s2*-s2)+(s2*-s1*)+(s1*-s1)
+//s1*-s1=1.6*R
+//s2*-s2=0.1*R
+//s2*-s1*=Cp*log(T2/T1)-R*log(P2/P1)
+//so
+ds=1.6*R-0.1*R+Cp*log(T2/T1)-R*log(P2/P1);
+printf(' Entropy Change = %.4f kJ/kg.K ',ds);
\ No newline at end of file diff --git a/172/CH14/EX14.7/ex7.sce b/172/CH14/EX14.7/ex7.sce new file mode 100755 index 000000000..8875a921e --- /dev/null +++ b/172/CH14/EX14.7/ex7.sce @@ -0,0 +1,49 @@ +//ques7
+//percent deviation using specific volume calculated by kays rule and vander waals rule
+clear
+clc
+//a-denotes C02
+//b-denotes CH4
+T=310.94;//Temperature of mixture K
+P=86.19;//Pressure of mixture in MPa
+//Tc- critical Temperature
+//Pc-critical pressure
+Tca=304.1;//K
+Tcb=190.4;//K
+Pca=7.38;//MPa
+Pcb=4.60;//MPa
+Ra=0.1889;//gas constant for a in kJ/kg.K
+Rb=0.5183;//gas constant for b in kJ/kg.K
+xa=0.8;//fraction of CO2
+xb=0.2;//fraction of CH4
+Rm=xa*Ra+xb*Rb;//mean gas constant in kJ/kg.K
+Ma=44.01;//molecular mass of a
+Mb=16.043;//molecular mass of b
+//1.Kay's rule
+ya=xa/Ma/(xa/Ma+xb/Mb);//mole fraction of a
+yb=xb/Mb/(xa/Ma+xb/Mb);//mole fraction of b
+Tcm=ya*Tca+yb*Tcb;//mean critical temp in K
+Pcm=ya*Pca+yb*Tcb;//mean critical pressure n MPa
+//therefore pseudo reduced property of mixture
+Trm=T/Tcm;
+Prm=P/Pcm;
+Zm=0.7;//Compressiblity from generalised compressibility chart
+vc=Zm*Rm*T/P/1000;//specific volume calculated in m^3/kg
+ve=0.0006757;//experimental specific volume in m^3/kg
+pd1=(ve-vc)/ve*100;//percent deviation
+printf('Percentage deviation in specific volume using Kays rule = %.1f percent \n',pd1);
+
+//2. using vander waals equation
+//values of vander waals constant
+Aa=27*Ra^2*Tca^2/(64*Pca*1000);
+Ba=Ra*Tca/(8*Pca*1000);
+Ab=27*Rb^2*Tcb^2/(64*Pcb*1000);
+Bb=Rb*Tcb/(8*Pcb*1000);
+//mean vander waals constant
+Am=(xa*sqrt(Aa)+xb*sqrt(Ab))^2;
+Bm=(xa*Ba+xb*Bb);
+//using vander waals equation we get cubic equation
+//solving we get
+vc=0.0006326;//calculated specific volume in m^3/kg
+pd2=(ve-vc)/ve*100;
+printf(' Percentage deviation in specific volume using vander waals eqn = %.1f percent \n',pd2);
\ No newline at end of file diff --git a/172/CH15/EX15.1/ex1.sce b/172/CH15/EX15.1/ex1.sce new file mode 100755 index 000000000..27a21f37e --- /dev/null +++ b/172/CH15/EX15.1/ex1.sce @@ -0,0 +1,7 @@ +//ques1
+//theoratical air-fuel ratio for combustion of octane
+clear
+clc
+rm=(12.5+47.0)/1;//air fuel ratio on mole basis
+rma=rm*28.97/114.2;//air fuel ratio on mass basis;
+printf('Theoratical air fuel ratio on mass basis = %.1f kg air/kg fuel \n',rma);
diff --git a/172/CH15/EX15.15/ex15.sce b/172/CH15/EX15.15/ex15.sce new file mode 100755 index 000000000..24f7dadc2 --- /dev/null +++ b/172/CH15/EX15.15/ex15.sce @@ -0,0 +1,21 @@ +//ques15
+//calculatng reversible elecromotive force
+clear
+clc
+//1-H2O
+//2-H2
+//3-O2
+//hf-standard enthalpy
+//sf-standard entropy
+hf1=-285.830;//kJ
+hf2=0;//kJ
+hf3=0;//kJ
+sf1=69.950;//kJ/K
+sf2=130.678;//kJ/K
+sf3=205.148;//kJ/K
+dH=2*hf1-2*hf2-hf3;//change in enthalpy in kJ
+dS=2*sf1-2*sf2-sf3;//change in entropy in kJ/K
+T=298.15;//temperature in K
+dG=dH-T*dS/1000;//change in gibbs free energy in kJ
+E=-dG*1000/(96485*4);//emf in V
+printf('Reversible electromotive Force = %.3f V',E);
\ No newline at end of file diff --git a/172/CH15/EX15.17/ex17.sce b/172/CH15/EX15.17/ex17.sce new file mode 100755 index 000000000..13fa0768f --- /dev/null +++ b/172/CH15/EX15.17/ex17.sce @@ -0,0 +1,11 @@ +//ques17
+//efficiency of generator and plant
+clear
+clc
+q=325000*(3398.3-856.0);//heat transferred to H2O/kg fuel in kJ/kg
+qv=26700*33250;//higher heating value in kJ/kg
+nst=q/qv*100;//efficiency of steam generator
+w=81000*3600;//net work done in kJ/kg
+nth=w/qv*100;//thermal efficiency
+printf('Efficiency of generator = %.1f percent\n',nst);
+printf(' Thermal Efficiency = %.1f percent\n',nth);
diff --git a/172/CH15/EX15.6/ex6.sce b/172/CH15/EX15.6/ex6.sce new file mode 100755 index 000000000..de3cd3f57 --- /dev/null +++ b/172/CH15/EX15.6/ex6.sce @@ -0,0 +1,13 @@ +//ques6
+//determining heat transfer per kilomole of fuel entering combustion chamber
+clear
+clc
+//1-CH4
+//2-CO2
+//3-H2O
+//hf-standard enthalpy of given substance
+hf1=-74.873;//kJ
+hf2=-393.522;//kJ
+hf3=-285.830;//kJ
+Qcv=hf2+2*hf3-hf1;//kJ
+printf('Heat transfer per kilomole of fuel entering combustion chamber = %.3f kJ ',Qcv);
\ No newline at end of file diff --git a/172/CH15/EX15.7/ex7.sce b/172/CH15/EX15.7/ex7.sce new file mode 100755 index 000000000..58104be3b --- /dev/null +++ b/172/CH15/EX15.7/ex7.sce @@ -0,0 +1,38 @@ +//ques7
+//calculating enthalpy of water at given pressure and temperature
+clear
+clc
+//1.Assuming steam to be an ideal gas with value of Cp
+T1=298.15;//Initial temperature in K
+T2=573.15;//final temperature in K
+T=(T1+T2)/2;//average temperature in K
+Cp=1.79+0.107*T/1000+0.586*(T/1000)^2-.20*(T/1000)^3;//specific heat at constant pressure in kj/kg.K
+M=18.015;//mass in kg
+dh=M*Cp*(T2-T1);//enthalpy change in kJ/kmol
+ho=-241.826;//enthalpy at standard temperature and pressure in kJ/mol
+htp1=ho+dh/1000;//enthalpy at given temp and pressure in kJ/kmol
+printf('1. Enthalpy of water at given pressure and temperature using value of Cp = %.3f kJ/kmol \n',htp1);
+
+//2..Assuming steam to be an ideal gas with value for dh
+dh=9359;//enthalpy change from table A.9 in kJ/mol
+htp2=ho+dh/1000;//enthalpy at given temp and pressure in kJ/kmol
+printf(' 2. Enthalpy of water at given pressure and temperature assuming value od dh = %.3f kJ/kmol \n',htp2);
+
+//3. Using steam table
+dh=M*(2977.5-2547.2);//enthalpy change for gases in kJ/mol
+htp3g=dh/1000+ho;
+dh=M*(2977.5-104.9);//enthalpy change for liquid in kJ/mol
+hl=-285.830;//standard enthalpy for liquid in kJ/kmol
+htp3l=hl+dh/1000;//enthalpy at given temp and pressure in kJ/kmol
+printf(' 3.(i) enthalpy at given temp and pressure in kJ/kmol in terms of liquid = %.3f kJ/kmol \n',htp3l);
+printf(' 3.(ii) enthalpy at given temp and pressure in kJ/kmol in terms of liquid = %.3f kJ/kmol \n',htp3g);
+//4.using generalised charts
+//htp=ho-(h2*-h2)+(h2*-h1*)+(h1*-h1);
+//h2*-h2=Z*R*Tc,
+//h2*-h1*=9539 kJ/mol, from part 2
+//h1*-h1=0 ,as ideal gas
+Z=0.21;//from chart
+R=8.3145;//gas constant in SI units
+Tc=647.3;//critical temperature in K
+htp4=ho+9539/1000-Z*R*Tc/1000;//enthalpy at given temp and pressure in kJ/kmol
+printf(' 4. enthalpy at given temp and pressure in kJ/kmol using compressibility chart = %.3f kJ/kmol \n',htp4);
\ No newline at end of file diff --git a/172/CH16/EX16.2/ex2.sce b/172/CH16/EX16.2/ex2.sce new file mode 100755 index 000000000..bd10a909d --- /dev/null +++ b/172/CH16/EX16.2/ex2.sce @@ -0,0 +1,33 @@ +//ques2
+//to determine change in gibbs free energy
+clear
+clc
+//1-H2
+//2-O2
+//3-H2O
+
+//at T=298 K
+T1=298;//K
+Hf1=0;//Enthalpy of formation of H2 at 298 K
+Hf2=0;//Enthalpy of formation of O2 at 298 K
+Hf3=-241.826;//enthalpy of formation of H2O at 298 K in kJ
+dH=2*Hf1+Hf2-2*Hf3;//Change in enthalpy in kJ
+Sf1=130.678;//Entropy of H2 at 298 K n J/K
+Sf2=205.148;//Entropy of O2 at 298 K in J/K
+Sf3=188.834;//entropy of H2O at 298 K in J/K
+dS=2*Sf1+Sf2-2*Sf3;//Change in entropy in J/K
+dG1=dH-T1*dS/1000;//change n gibbs free energy in kJ
+printf(' Change in gibbs free energy at %.0f K = %.3f kJ \n' ,T1,dG1);
+
+//at T=2000 K
+T2=2000;//K
+Hf1=52.942-0;//Enthalpy of formation of H2 at 2000 K
+Hf2=59.176-0;//Enthalpy of formation of O2 at 2000 K
+Hf3=-241.826+72.788;//enthalpy of formation of H2O at 2000 K in kJ
+dH=2*Hf1+Hf2-2*Hf3;//Change in enthalpy in kJ
+Sf1=188.419;//Entropy of H2 at 2000 K n J/K
+Sf2=268.748;//Entropy of O2 at 2000 K in J/K
+Sf3=264.769;//entropy of H2O at 2000 K in J/K
+dS=2*Sf1+Sf2-2*Sf3;//Change in entropy in J/K
+dG2=dH-T2*dS/1000;//change n gibbs free energy in kJ
+printf(' Change in gibbs free energy at %.0f K = %.3f kJ ' ,T2,dG2);
\ No newline at end of file diff --git a/172/CH16/EX16.3/ex3.sce b/172/CH16/EX16.3/ex3.sce new file mode 100755 index 000000000..01cc047eb --- /dev/null +++ b/172/CH16/EX16.3/ex3.sce @@ -0,0 +1,13 @@ +//ques3
+//calculating equilibrium constant
+clear
+clc
+dG1=-457.166;//change in gibbs free energy at temp 298 K from example2 in kJ
+dG2=-271.040;;//change in gibbs free energy at temp 2000 K from example2 n kJ
+T1=298;//K
+T2=2000;//K
+R=8.3145;//gas constant
+K1=dG1*1000/(R*T1);
+K2=dG2*1000/(R*T2);
+printf('Equilibrium constant at %.0f K = %.3f \n',T1,K1);
+printf(' Equilibrium constant at %.0f K = %.3f \n',T2,K2);
\ No newline at end of file diff --git a/172/CH17/EX17.1/ex1.sce b/172/CH17/EX17.1/ex1.sce new file mode 100755 index 000000000..81693cafc --- /dev/null +++ b/172/CH17/EX17.1/ex1.sce @@ -0,0 +1,13 @@ +//ques1
+//to determine isentropic stagnation pressure and temperature
+clear
+clc
+T=300;//Temperature of air in K
+P=150;//Pressure of air in kPa
+v=200;//velocity of air flow n m/s
+Cp=1.004;//specific heat at constant pressure in kJ/kg
+To=v^2/(2000*Cp)+T;//stagnation temperature in K
+k=1.4;//constant
+Po=P*(To/T)^(k/(k-1));//stagnation pressure in kPa
+printf('Stagnation Temperature = %.1f K \n',To);
+printf(' Stagnation Pressure = %.1f kPa \n',Po);
\ No newline at end of file diff --git a/172/CH17/EX17.3/ex3.sce b/172/CH17/EX17.3/ex3.sce new file mode 100755 index 000000000..c55647364 --- /dev/null +++ b/172/CH17/EX17.3/ex3.sce @@ -0,0 +1,18 @@ +//ques3
+//determining the thrust acting on a control surface
+clear
+clc
+//i-inlet
+//e-exit
+//using momentum equation on control surface in x direction
+me=20.4;//mass exiting in kg
+mi=20;//mass entering in kg
+ve=450;//exit velocity in m/s
+vi=100;//exit velocity in m/s
+Pi=95;//Pressure at inlet in kPa
+Pe=125;//Pressure at exit in kPa
+Po=100;//surrounding pressure in kPa
+Ai=0.2;//inlet area in m^2
+Ae=0.1;//exit area in m^2
+Rx=(me*ve-mi*vi)/1000-(Pi-Po)*Ai+(Pe-Po)*Ae;//thrust in x direction in kN
+printf('Thrust acting in x direction = %.2f kN',Rx);
\ No newline at end of file diff --git a/172/CH17/EX17.5/ex5.sce b/172/CH17/EX17.5/ex5.sce new file mode 100755 index 000000000..d655a00f7 --- /dev/null +++ b/172/CH17/EX17.5/ex5.sce @@ -0,0 +1,13 @@ +//ques5
+//determining velocity of sound in air
+clear
+clc
+k=1.4;//constant
+R=0.287;//gas constant
+//at 300K
+T1=300;//K
+c1=sqrt(k*R*T1*1000);
+printf('Speed of sound at %.0f K = %.1f m/s \n',T1,c1);
+T2=1000;//K
+c2=sqrt(k*R*T2*1000);
+printf(' Speed of sound at %.0f K = %.1f m/s \n',T2,c2);
\ No newline at end of file diff --git a/172/CH17/EX17.6/ex6.sce b/172/CH17/EX17.6/ex6.sce new file mode 100755 index 000000000..b1b91e1c4 --- /dev/null +++ b/172/CH17/EX17.6/ex6.sce @@ -0,0 +1,23 @@ +//ques6
+//determining mass flow rate through control volume
+clear
+clc
+k=1.4;//constant
+R=0.287;//gas constant
+To=360;//stagnation Temperature in K
+T=To*0.8333;//Temperature of air in K, 0.8333 stagnation ratio from table
+v=sqrt(k*R*T*1000);//velocity in m/s
+P=528;//stagnation pressure in kPa
+d=P/(R*T);//stagnation density in kg/m^3
+A=500*10^-6;//area in m^2
+ms=d*A*v;//mass flow rate in kg/s
+printf('Mass flow rate at the throat section = %.4f kg/s \n',ms);
+//e-exit state
+Te=To*0.9381;//exit temperature in K, ratio from table
+ce=sqrt(k*R*Te*1000);//exit velocity of sound in m/s
+Me=0.573;//Mach number
+ve=Me*ce;
+Pe=800;//exit pressure in kPa
+de=Pe/R/Te;
+mse=de*A*ve;
+printf(' Mass flow rate at the exit section = %.4f kg/s \n',mse);
\ No newline at end of file diff --git a/172/CH17/EX17.7/ex7.sce b/172/CH17/EX17.7/ex7.sce new file mode 100755 index 000000000..37ad710db --- /dev/null +++ b/172/CH17/EX17.7/ex7.sce @@ -0,0 +1,31 @@ +//ques7
+//determining exit properties in a control volume
+clear
+clc
+Po=1000;//stagnation pressure in kPa
+To=360;//stagnation temperature in K
+
+//when diverging section acting as nozzle
+Pe1=0.0939*Po;//exit pressure of air in kPa
+Te1=0.5089*To;//exit temperature in K
+k=1.4;//constant
+R=0.287;//gas constant for air
+ce=sqrt(k*R*Te1*1000);//velocity of sound in exit section in m/s
+Me=2.197;//mach number from table
+ve1=Me*ce;//velocity of air at exit section in m/s
+disp(" When diverging section act as a nozzle :-");
+printf('Exit pressure = %.1f kPa \n',Pe1);
+printf('Exit Temperature = %.1f K \n',Te1);
+printf('Exit velocity = %.1f m/s \n',ve1);
+
+//when diverging section act as diffuser
+Me=0.308;
+Pe2=0.0936*Po;//exit pressure of air in kPa
+Te2=0.9812*To;//exit temperature in K
+ce=sqrt(k*R*Te2*1000);//velocity of sound in exit section in m/s
+ve2=Me*ce;
+disp(" When diverging section act as a diffuser :-");
+printf('Exit pressure = %.1f kPa \n',Pe2);
+printf('Exit Temperature = %.1f K \n',Te2);
+printf('Exit velocity = %.1f m/s \n',ve2);
+
diff --git a/172/CH17/EX17.9/ex9.sce b/172/CH17/EX17.9/ex9.sce new file mode 100755 index 000000000..4da76a754 --- /dev/null +++ b/172/CH17/EX17.9/ex9.sce @@ -0,0 +1,18 @@ +//ques9
+//determining exit plane properties in control volume
+clear
+clc
+//x-inlet
+//y-exit
+Mx=1.5;//mach number for inlet
+My=0.7011;//mach number for exit
+Px=272.4;//inlet pressure in kPa
+Tx=248.3;//inlet temperature in K
+Pox=1000;//stagnation pressure for inlet
+Py=2.4583*Px;//Exit Pressure in kPa
+Ty=1.320*Tx;//Exit temperature in K
+Poy=0.9298*Pox;//Exit pressure in kPa
+
+printf('Exit pressure = %.1f kPa \n',Py);
+printf(' Exit temperature = %.1f K \n',Ty);
+printf(' Exit stagnation pressure = %.1f kPa \n',Poy);
\ No newline at end of file diff --git a/172/CH2/EX2.1/ex1.sce b/172/CH2/EX2.1/ex1.sce new file mode 100755 index 000000000..c6f730a04 --- /dev/null +++ b/172/CH2/EX2.1/ex1.sce @@ -0,0 +1,8 @@ +//example 1
+//weight of a person
+clear
+clc
+m=1 //kg
+g=9.75 //acc.due to gravity in m/s^2
+F=m*g //weight of 1 kg mass in N
+printf("\n hence,weight of person is F = %.3f N. \n",F)
\ No newline at end of file diff --git a/172/CH2/EX2.2/ex2.sce b/172/CH2/EX2.2/ex2.sce new file mode 100755 index 000000000..a41434acc --- /dev/null +++ b/172/CH2/EX2.2/ex2.sce @@ -0,0 +1,22 @@ +//example 2
+//average volume and density
+clear
+clc
+Vliq=0.2 //volume of liquid in m^3
+dliq=997 //density of liquid in kg/m^3
+Vstone=0.12 //volume of stone in m^3
+Vsand=0.15 //volume of sand in m^3
+Vair=0.53 //vo;ume of air in m^3
+mliq=Vliq*dliq //mass of liquid in kg
+dstone=2750 //density of stone in kg/m^3
+dsand=1500 //density of sand in kg/m^3
+mstone=Vstone*dstone //volume of stone in m^3
+msand=Vsand*dsand //volume of sand in m^3
+Vtot=1 //total volume in m^3
+dair=1.1 //density of air in kg/m^3
+mair=Vair*dair //mass of air
+mtot=mair+msand+mliq+mstone //total mass in kg
+v=Vtot/mtot //specific volume in m^3/kg
+d=1/v //overall density in kg/m^3
+printf("\n hence,average specific volume is v=%.6f m^3/kg. \n",v )
+printf("\n and overall density is d=%.0f kg/m^3. \n",d)
\ No newline at end of file diff --git a/172/CH2/EX2.3/ex3.sce b/172/CH2/EX2.3/ex3.sce new file mode 100755 index 000000000..5f17aadba --- /dev/null +++ b/172/CH2/EX2.3/ex3.sce @@ -0,0 +1,14 @@ +//example 3
+//calculating the required force
+clear
+clc
+Dcyl=0.1 //cylinder diameter in m
+Drod=0.01 //rod diameter in m
+Acyl=%pi*Dcyl^2/4 //cross sectional area of cylinder in m^2
+Arod=%pi*Drod^2/4 //cross sectional area of rod in m^2
+Pcyl=250000 //inside hydaulic pressure in Pa
+Po=101000 //outside atmospheric pressure in kPa
+g=9.81 //acc. due to gravity in m/s^2
+mp=25 //mass of (rod+piston) in kg
+F=Pcyl*Acyl-Po*(Acyl-Arod)-mp*g //the force that rod can push within the upward direction in N
+printf("\n hence,the force that rod can push within the upward direction is F = %.3f N. \n",F)
\ No newline at end of file diff --git a/172/CH2/EX2.4/ex4.sce b/172/CH2/EX2.4/ex4.sce new file mode 100755 index 000000000..6e4c3266e --- /dev/null +++ b/172/CH2/EX2.4/ex4.sce @@ -0,0 +1,9 @@ +//example 4
+//Calculating atmospheric pressure
+clear
+clc
+dm=13534 //density of mercury in kg/m^3
+H=0.750 //height difference between two columns in metres
+g=9.80665 //acc. due to gravity in m/s^2
+Patm=dm*H*g/1000 //atmospheric pressure in kPa
+printf("\n hence, atmospheric pressure is Patm = %.2f kPa. \n",Patm)
\ No newline at end of file diff --git a/172/CH2/EX2.5/ex5.sce b/172/CH2/EX2.5/ex5.sce new file mode 100755 index 000000000..4688e7a4c --- /dev/null +++ b/172/CH2/EX2.5/ex5.sce @@ -0,0 +1,12 @@ +//example 5
+//pressure inside vessel
+clear
+clc
+dm=13590 //density of mercury in kg/m^3
+H=0.24 //height difference between two columns in metres
+g=9.80665 //acc. due to gravity in m/s^2
+dP=dm*H*g //pressure difference in Pa
+Patm=13590*0.750*9.80665 //Atmospheric Pressure in Pa
+Pvessel=dP+Patm //Absolute Pressure inside vessel in Pa
+Pvessel=Pvessel/101325//Absolute Pressure inside vessel in atm
+printf("\n hence, the absolute pressure inside vessel is Pvessel = %.3f atm. \n",Pvessel)
\ No newline at end of file diff --git a/172/CH2/EX2.6/ex6.sce b/172/CH2/EX2.6/ex6.sce new file mode 100755 index 000000000..3d74b243c --- /dev/null +++ b/172/CH2/EX2.6/ex6.sce @@ -0,0 +1,16 @@ +//example 6
+//calculating pressure
+clear
+clc
+dg=750 //density of gaasoline in kg/m^3
+dR=1206 //density of R-134a in kg/m^3
+H=7.5 //height of storage tank in metres
+g=9.807 //acc. due to gravity in m/s^2
+dP1=dg*g*H/1000 //in kPa
+Ptop1=101 //atmospheric pressure in kPa
+P1=dP1+Ptop1
+disp('hence,pressure at the bottom of storage tank if fluid is gasoline is 156.2 kPa')
+dP2=dR*g*H/1000 //in kPa
+Ptop2=1000 //top surface pressure in kPa
+P2=dP2+Ptop2
+printf("\n hence, pressure at the bottom of storage tank if liquid is R-134a is P2 = %.0f kPa. \n",P2)
\ No newline at end of file diff --git a/172/CH2/EX2.7/ex7.sce b/172/CH2/EX2.7/ex7.sce new file mode 100755 index 000000000..d3681adae --- /dev/null +++ b/172/CH2/EX2.7/ex7.sce @@ -0,0 +1,15 @@ +//example 6
+//calculating balancing force
+clear
+clc
+Po=100//Outside atmospheric pressure in kPa
+F1=25 //net force on the smallest piston in kN
+A1=0.01 //cross sectional area of lower piston in m^2
+P1=Po+F1/A1 //fluid pressure in kPa
+d=900 //density of fluid in kg/m^3
+g=9.81 //acc. due to gravity in m/s^2
+H=6 //height of second piston in comparison to first one in m
+P2=P1-d*g*H/1000 //pressure at higher elevation on piston 2 in kPa
+A2=0.05 // cross sectional area of higher piston in m^3
+F2=(P2-Po)*A2 //balancing force on second piston in kN
+printf("\n hence, balancing force on second larger piston is F2 = %.1f N. \n",F2)
\ No newline at end of file diff --git a/172/CH3/EX3.1/ex1.sce b/172/CH3/EX3.1/ex1.sce new file mode 100755 index 000000000..0e37d18a8 --- /dev/null +++ b/172/CH3/EX3.1/ex1.sce @@ -0,0 +1,6 @@ +//example 1
+//determinig the phase of water
+clear
+clc
+disp('from the table,we find that at 120C,saturation pressure of water is 198.5 kPa.But here we have pressure of 500 kPa.hence,water exists as a compressed liquid here.')
+disp('also at 120C,vf=0.00106 kg/m^3 and vg=0.89186 kg/m^3.given v=0.5 m^3/kg i.e. vf<v<vg,so we have two phase mixture of liquid and vapor.')
\ No newline at end of file diff --git a/172/CH3/EX3.10/ex10.sce b/172/CH3/EX3.10/ex10.sce new file mode 100755 index 000000000..61ac81d16 --- /dev/null +++ b/172/CH3/EX3.10/ex10.sce @@ -0,0 +1,13 @@ +//example 10
+//mass flow rate
+clear
+clc
+dt=185 //time period in seconds over which there is incrrease in volume
+dV=0.75 //increase in volume in 0.75 in m^3
+V=dV/dt //volume flow rate in m^3/s
+P=105 //pressure inside gas bell kPa
+T=21 //temperature in celsius
+R=0.1889 //ideal gas constant in kJ/kg-K
+m=P*V/(R*(T+273.15)) //mass flow rate of the flow in kg/s
+printf("\n hence,mass flow rate is m = %.3f kg/s. \n",m)
+printf("\n and volume flow rate is V = %.3f m^3/s. \n",V)
\ No newline at end of file diff --git a/172/CH3/EX3.11/ex11.sce b/172/CH3/EX3.11/ex11.sce new file mode 100755 index 000000000..a95abc399 --- /dev/null +++ b/172/CH3/EX3.11/ex11.sce @@ -0,0 +1,7 @@ +//example 11
+//predicting the nature of given state
+clear
+clc
+disp('For Nitrogn, the critical properties are 126.2 K, 3.39 MPa. Given T=20+273.2 K, P=1.0 MPa. Since, given temperature is more than twice Tc and the reduced pressure is less than 0.3, ideal gas behaviour is a very good assumption.')
+disp('For Carbon Dioxide, the critical properties are 304.1 K,7.38 MPa.Given T=20+273.2 K, P=1.0 MPa Therefore, reduced properties are 0.96(T/Tc) and 0.136 (P/Pc). CO2 is a gas with a Z of about 0.95, os the ideal gas is accurate to within about 5% in this case.')
+disp('Given P=1.0MPa, T=20+273.2 K. For Ammonia, at T=293.2 K, Pg=858 kPa. Since, P>Pg, this state is compressible liquid and not a gas.')
\ No newline at end of file diff --git a/172/CH3/EX3.12/ex12.sce b/172/CH3/EX3.12/ex12.sce new file mode 100755 index 000000000..5c690ee24 --- /dev/null +++ b/172/CH3/EX3.12/ex12.sce @@ -0,0 +1,18 @@ +//example 12
+//determining specific using diffenet laws
+clear
+clc
+T=100 //given temp.in 100 celsius
+P=3 //given pressure in MPa
+v1=0.0065 //specific volume in m^3/kg using table
+printf("\n hence,the specific volume for R-134a using R-134a tables is v1 = %.3f m^3/kg. \n",v1)
+M=102.3 //molecular mass in kg
+R=8.3145 //in kJ/K
+Ru=R/M //in kJ/K-kg
+v2=Ru*(T+273)/(P*1000) //specific volume assuming R-134a to be ideal gas in m^3/kg
+printf("\n hence,the specific volume for R-134a using R-134a the ideal gas laws is v2 = %.3f m^3/kg. \n",v2)
+Tr=373.2/374.2 //reduced temperature using generalized chart
+Pr=3/4.06 //reduced pressure using generalized chart
+Z=0.67 //compressibility factor
+v3=Z*v2 // specific volume using generalized chart in m^3/kg
+printf("\n hence,the specific volume for R-134a using the generalized chart is v3 = %.3f m^3/kg. \n",v3)
diff --git a/172/CH3/EX3.13/ex13.sce b/172/CH3/EX3.13/ex13.sce new file mode 100755 index 000000000..bbe7e546f --- /dev/null +++ b/172/CH3/EX3.13/ex13.sce @@ -0,0 +1,19 @@ +//example 13
+//calculating mass of gas
+clear
+clc
+Pc=4250 //critical pressure of propane in kPa
+Tc=369.8 //critical temperature in K
+T=15 //temperature of propane in celsius
+Tr=T/Tc //reduced temperature
+Prsat=0.2 // reduced pressure
+P=Prsat*Pc //pressure in kPa
+x=0.1 //given quality
+Zf=0.035 //from graph
+Zg=0.83 //from graph
+Z=(1-x)*Zf+x*Zg //overall compressibility factor
+V=0.1 //volume of steel bottle in m^3
+R=0.1887 //in kPa-m^3/kg-K
+m=P*V/(Z*R*(T+273)) //total propane mass in kg
+printf("\n hence,the total propane mass is m = %.3f kg. \n",m)
+printf("\n and pressure is P = %.3f kPa. \n",P)
\ No newline at end of file diff --git a/172/CH3/EX3.2/ex2.sce b/172/CH3/EX3.2/ex2.sce new file mode 100755 index 000000000..ec32c2216 --- /dev/null +++ b/172/CH3/EX3.2/ex2.sce @@ -0,0 +1,6 @@ +//example 2
+//determinig the phase
+clear
+clc
+disp('from the table,we find that at 30C,saturation pressure of ammonia is 1167 kPa.But here we have pressure of 1000 kPa.hence,ammonia exists in superheated vapor state.')
+disp('for R-22 at 200 kPa,vg=0.1119 kg/m^3.given v=0.15 m^3/kg i.e. v>vg,so the state is superheated vapor')
\ No newline at end of file diff --git a/172/CH3/EX3.3/ex3.sce b/172/CH3/EX3.3/ex3.sce new file mode 100755 index 000000000..58c035041 --- /dev/null +++ b/172/CH3/EX3.3/ex3.sce @@ -0,0 +1,12 @@ +//example 3
+//determining the quality and specific volume
+clear
+clc
+v1=0.5 //given specific volume in m^3/kg
+vf=0.001073 //specific volume when only liquid phase is present in m^3/kg
+vfg=0.60475 //in m^3/kg
+disp('For water at a pressure of 300 kPa,the state at which v1 is 0.5 m^3/kg is seen to be in the liquid-vapor two-phase region,at which T=133.6 C and the quality x is ')
+x=(v1-vf)/vfg //quality
+v2=1 //given specific volume in m^3/kg
+disp('By comparing with the values given in the table,this state is seen to be in the superheated vapor region.temperature will be calculated using the method of interplotation.')
+T=((400-300)*(1.0-0.8753))/(1.0315-0.8753)+300 //temperature of the water
\ No newline at end of file diff --git a/172/CH3/EX3.4/ex4.sce b/172/CH3/EX3.4/ex4.sce new file mode 100755 index 000000000..3ecbb5a5c --- /dev/null +++ b/172/CH3/EX3.4/ex4.sce @@ -0,0 +1,13 @@ + //example 4
+//percentage of vapor
+clear
+clc
+vliq=0.1 //volume of saturated liquid in m^3
+vf=0.000843 //in m^3/kg
+vvap=0.9 //volume of saturated vapor R-134a in equilbrium
+vg=0.02671 //in m^3/kg
+mliq=vliq/vf //mass of liquid in kg
+mvap=vvap/vg //mass of vapor in kg
+m=mliq+mvap //total mass in kg
+x=mvap/m //percentage of vapor on mass basis
+disp('hence,% vapor on mass basis is 22.1')
\ No newline at end of file diff --git a/172/CH3/EX3.5/ex5.sce b/172/CH3/EX3.5/ex5.sce new file mode 100755 index 000000000..729970968 --- /dev/null +++ b/172/CH3/EX3.5/ex5.sce @@ -0,0 +1,10 @@ +//example 5
+//calculating pressure after heat addition
+clear
+clc
+v1=0.14922 //specific volume of sautrated ammonia in m^3/kg
+disp('Since the volume does not change during the process,the specific volume remains constant.therefore, ')
+v2=v1 //in m^3/kg
+disp('Since vg at 40C is less than v2,it is evident that in the final state the Ammonia is superheated vapor.By interplotation,we find that ')
+P2=945 //final pressure in kPa
+disp('hence,the final pressure is 945 kPa')
\ No newline at end of file diff --git a/172/CH3/EX3.6/ex6.sce b/172/CH3/EX3.6/ex6.sce new file mode 100755 index 000000000..b58260b51 --- /dev/null +++ b/172/CH3/EX3.6/ex6.sce @@ -0,0 +1,16 @@ +//example 6
+//Determinig the missing property
+clear
+clc
+T1=273-53.2 //given temperature in K
+P1=600 //given pressure in kPa
+disp('This temperature is higher than the critical temperature (critical temp. at P=600 kPa) is 96.37 K.Hence,v=0.10788 m^3/kg')
+T2=100 //given temp. in K
+v2=0.008 //given specific volume in m^3/kg
+vf=0.001452 //in m^3/kg
+vg=0.0312 //in m^3/kg
+Psat=779.2 //saturation pressure in kPa
+vfg=vg-vf //in m^3/kg
+x=(v2-vf)/vfg //quality
+printf("\n hence, the pressure is Psat = %.1f kPa. \n",Psat)
+printf("\n and quality is x = %.4f . \n",x)
\ No newline at end of file diff --git a/172/CH3/EX3.7/ex7.sce b/172/CH3/EX3.7/ex7.sce new file mode 100755 index 000000000..3fc0f72ff --- /dev/null +++ b/172/CH3/EX3.7/ex7.sce @@ -0,0 +1,12 @@ +//example 7
+//determining the pressure of water
+clear
+clc
+vg=0.12736 //specific volume in m^3/kg for water at 200C
+v=0.4 //specific volume in m^3/kg
+P1=500 //in kPa
+v1=0.42492 //specific volume at P1 in m^3/kg
+P2=600 //in kPa
+v2=0.35202 //specific volume at P2 in m^3/kg
+P=P1+(P2-P1)*(v-v1)/(v2-v1) //calculating pressure by interplotation
+disp('hence,the pressure of water is 534.2 kPa')
\ No newline at end of file diff --git a/172/CH3/EX3.8/ex8.sce b/172/CH3/EX3.8/ex8.sce new file mode 100755 index 000000000..565eb5815 --- /dev/null +++ b/172/CH3/EX3.8/ex8.sce @@ -0,0 +1,10 @@ +//example 8
+//calculating mass of air
+clear
+clc
+P=100 //pressure in kPa
+V=6*10*4 //volume of room in m^3
+R=0.287 //in kN-m/kg-K
+T=25 //temperature in Celsius
+m=P*V/(R*(T+273.2)) //mass of air contained in room
+printf("\n hence, mass of air contained in room is m = %.3f kg. \n",m)
\ No newline at end of file diff --git a/172/CH3/EX3.9/ex9.sce b/172/CH3/EX3.9/ex9.sce new file mode 100755 index 000000000..d1b676a4e --- /dev/null +++ b/172/CH3/EX3.9/ex9.sce @@ -0,0 +1,12 @@ +//example 9
+//calculating pressure inside tank
+clear
+clc
+V=0.5 //volumr of tank in m^3
+m=10 //mass of ideal gas in kg
+T=25 //temperature of tank in Celsius
+M=24 //molecular mass of gas in kg/kmol
+Ru=8.3145 //universal gas constant in kN-m/kmol-K
+R=Ru/M //gas constant for given ideal gas in kN-m/kg-K
+P=m*R*(T+273.2)/V //pressure inside tank
+printf("\n hence,pressure inside tank is P = %.0f kPa. \n",P)
\ No newline at end of file diff --git a/172/CH4/EX4.1/ex1.sce b/172/CH4/EX4.1/ex1.sce new file mode 100755 index 000000000..70ebd2285 --- /dev/null +++ b/172/CH4/EX4.1/ex1.sce @@ -0,0 +1,16 @@ +//example 1
+//work done during different processes
+clear
+clc
+P1=200 //initial pressure inside cylinder in kPa
+V2=0.1 //in m^3
+V1=0.04 //initial volume of gas in m^3
+W1=P1*(V2-V1) //work done in isobaric process in kJ
+printf("\n hence,the work done during the isobaric process is W1 = %.3f kJ. \n",W1)
+W2=P1*V1*log(V2/V1) //work done in isothermal process in kJ
+printf("\n hence,the work done in isothermal process is W2 = %.3f kJ. \n",W2)
+P2=P1*(V1/V2)^(1.3) //final pressure according to the given process
+W3=(P2*V2-P1*V1)/(1-1.3)
+printf("\n hence,the work done during the described process is W3 = %.3f kJ. \n",W3)
+W4=0 //work done in isovolumic process
+printf("\n hence,the work done in the isovolumic process is W4 = %.3f kJ. \n",W4)
\ No newline at end of file diff --git a/172/CH4/EX4.3/ex3.sce b/172/CH4/EX4.3/ex3.sce new file mode 100755 index 000000000..8c7f5faf8 --- /dev/null +++ b/172/CH4/EX4.3/ex3.sce @@ -0,0 +1,15 @@ +//example 3
+//work produced
+clear
+clc
+Psat=190.2 //in kPa
+P1=Psat //saturation pressure in state 1
+vf=0.001504 //in m^3/kg
+vfg=0.62184 //in m^3/kg
+x1=0.25 //quality
+v1=vf+x1*vfg //specific volume at state 1 in m^3/kg
+v2=1.41*v1 //specific volume at state 2 in m^3/kg
+P2=600 //pressure in state 2 in kPa
+m=0.5 //mass of ammonia in kg
+W=m*(P1+P2)*(v2-v1)/2 //woork produced by ammonia in kJ
+disp('hence,work produced by ammonia is 12.71 kJ')
\ No newline at end of file diff --git a/172/CH4/EX4.4/ex4.sce b/172/CH4/EX4.4/ex4.sce new file mode 100755 index 000000000..dde73edae --- /dev/null +++ b/172/CH4/EX4.4/ex4.sce @@ -0,0 +1,10 @@ +//example 4
+//calculating work done
+clear
+clc
+v1=0.35411 //specific volume at state 1 in m^3/kg
+v2=v1/2
+m=0.1 //mass of water in kg
+P1=1000 //pressure inside cylinder in kPa
+W=m*P1*(v2-v1) //in kJ
+disp('hence,the work in the overall process is -17.7 kJ')
\ No newline at end of file diff --git a/172/CH4/EX4.7/ex7.sce b/172/CH4/EX4.7/ex7.sce new file mode 100755 index 000000000..726322b83 --- /dev/null +++ b/172/CH4/EX4.7/ex7.sce @@ -0,0 +1,13 @@ +//example 7
+//heat transfer
+clear
+clc
+k=1.4 //conductivity of glass pane in W/m-K
+A=0.5 //total surface area of glass pane
+dx=0.005 //thickness of glasspane in m
+dT1=20-12.1 //temperature difference between room air and outer glass surface temperature in celsius
+Q=-k*A*dT1/dx //conduction through glass slab in W
+h=100 //convective heat transfer coefficient in W/m^2-K
+dT=12.1-(-10) //temperature difference between warm room and colder ambient in celsius
+Q2=h*A*dT //heat transfer in convective layer in W
+printf("\n hence,the rate of heat transfer in the glass and convective layer is Q2 = %.0f kW. \n",Q2)
\ No newline at end of file diff --git a/172/CH5/EX5.1/ex1.sce b/172/CH5/EX5.1/ex1.sce new file mode 100755 index 000000000..ff728fe3b --- /dev/null +++ b/172/CH5/EX5.1/ex1.sce @@ -0,0 +1,10 @@ +//example 1
+//calculating height
+clear
+clc
+m=1100 //mass of car in kg
+ke=400 //kinetic energy of car in kJ
+V=(2*ke*1000/m)^0.5 //velocity of car in m/s
+g=9.807 //acc. due to gravity in m/s^2
+H=ke*1000/(m*g) //height to which the car should be lifted so that its potential energy equals its kinetic energy
+disp('hence,the car should be raised to a height of 37.1 m to make its potential energy equal to kinetic energy')
\ No newline at end of file diff --git a/172/CH5/EX5.10/ex10.sce b/172/CH5/EX5.10/ex10.sce new file mode 100755 index 000000000..8cf87ce92 --- /dev/null +++ b/172/CH5/EX5.10/ex10.sce @@ -0,0 +1,8 @@ +//example 10
+//calculating rate of increase of internal energy
+clear
+clc
+W=-12.8*20 //power consumed in J/s
+Q=-10 //heat transfer rate from battery in J/s
+r=Q-W //rate of increase of internal energy
+printf("\n hence,the rate of increase of internal energy is r=%.0f J/s. \n", r)
\ No newline at end of file diff --git a/172/CH5/EX5.11/ex11.sce b/172/CH5/EX5.11/ex11.sce new file mode 100755 index 000000000..0648afff6 --- /dev/null +++ b/172/CH5/EX5.11/ex11.sce @@ -0,0 +1,16 @@ +//example 11
+//rate of change of temperature
+clear
+clc
+Q=1500 //power produced by burning wood in J/s
+mair=1 //mass of air in kg
+mwood=5 //mass of soft pine wood in kg
+miron=25 //mass of cast iron in kg
+Cvair=0.717 //constant volume specific heat for air in kJ/kg
+Cwood=1.38 //constant volume specific heat for wood in kJ/kg
+Ciron=0.42 //constant volume specific heat for iron in kJ/kg
+dT=75-20 //increase in temperature in Celsius
+T=(Q/1000)/(mair*Cvair+mwood*Cwood+miron*Ciron) //rate of change of temperature in K/s
+dt=(dT/T)/60 //in minutes
+printf(" hence,the rate of change of temperature is dt=%.4f K/s.\n", T)
+printf(" and time taken to reach a temperature of T=%.0f min.\n", dt)
\ No newline at end of file diff --git a/172/CH5/EX5.2/ex2.sce b/172/CH5/EX5.2/ex2.sce new file mode 100755 index 000000000..2cc7f9fef --- /dev/null +++ b/172/CH5/EX5.2/ex2.sce @@ -0,0 +1,8 @@ +//example 2
+//change in internal energy
+clear
+clc
+W=-5090 //work input to paddle wheel in kJ
+Q=-1500 //heat transfer from tank in kJ
+dU=Q-W //change in internal energy in kJ
+disp('hence,change in internal energy is 3590 kJ')
\ No newline at end of file diff --git a/172/CH5/EX5.3/ex3.sce b/172/CH5/EX5.3/ex3.sce new file mode 100755 index 000000000..54e2fb70c --- /dev/null +++ b/172/CH5/EX5.3/ex3.sce @@ -0,0 +1,25 @@ +//example 3
+//analysis of energy transfer
+clear
+clc
+g=9.806 //acceleration due to gravity in m/s^2
+m=10 //mass of stone in kg
+H1=10.2 //initial height of stone above water in metres
+H2=0 //final height in metres
+dKE1=-m*g*(H2-H1) //change in kinetic energy when stone enters state 2 in J
+dPE1=-1 //change in potential energy when stone enters state 2 in J
+printf("\n hence,when stone is about to enter state 2, dKE = %.3f J. \n",dKE1)
+printf("\n and dPE = %.3f J. \n",dPE1)
+dPE2=0 //change in potential energy when stone enters state 3 in JQ2=0 //no heat transfer when stone enters state 3 in J
+W2=0 //no work done when stone enters state 3 in J
+dKE2=-1 //change in kinetic energy when stone enters state 3
+dU2=-dKE2 //change in internal energy when stone enters state 3 in J
+printf("\n hence,when stone has just come to rest in the bucket i.e. state 3, W=0, dPE=0, dKE1 = %.3f J. \n",dKE2)
+printf("\n and dU = %.3f J. \n",dU2)
+dKE3=0 //change in kinetic energy when stone enters state 4
+dPE=0 //change in potential energy when stone enters state 4 in J
+W3=0 //no work done when stone enters state 4 in J
+dU3=-1 //change in internal energy when stone enters state 4 in J
+Q3=dU3 //heat transfer when stone enters state 4 in J
+printf("\n hence,when stone has entered state 4, dPE=0, W3=0,dKE=0, dU= %.3f J. \n",dU3)
+printf("\n and Q3= %.3f J. \n",Q3)
\ No newline at end of file diff --git a/172/CH5/EX5.4/ex4.sce b/172/CH5/EX5.4/ex4.sce new file mode 100755 index 000000000..2f8e48dec --- /dev/null +++ b/172/CH5/EX5.4/ex4.sce @@ -0,0 +1,19 @@ +//example 4
+//Determinig the missing properties
+clear
+clc
+T1=300 //given temp. in Celsius
+u1=2780 //given specific internal enrgy in kJ/kg
+disp('From steam table, at T=300 C,ug=2563.0 kJ/kg.So,u1>ug ,it means the state is in the superheated vapor region.So, by interplotation,we find P=1648 kPa and v=0.1542 m^3/kg')
+P2=2000 //hiven pressure in kPa
+u2=2000 //given specific intrernal energy in kJ/kg
+disp('at P=2000 kPa,')
+uf=906.4 //in kJ/kg
+ug=2600.3 //in kJ/kg
+x2=(u2-906.4)/(ug-uf)
+disp('Also, under the given conditions')
+vf=0.001177 //in m^3/kg
+vg=0.099627 //in m^3/kg
+v2=vf+x2*(vg-vf)//Specific volume for water in m^3/kg
+printf("\n hence,specific volume for water is v2 = %.5f m^3/kg. \n",v2)
+printf("\n Therefore ,this state is in the two phase region with quality x2=%.4f . \n",x2)
\ No newline at end of file diff --git a/172/CH5/EX5.5/ex5.sce b/172/CH5/EX5.5/ex5.sce new file mode 100755 index 000000000..d2af49083 --- /dev/null +++ b/172/CH5/EX5.5/ex5.sce @@ -0,0 +1,21 @@ +//example 5
+//calculating heat transfer for the given process
+clear
+clc
+Vliq=0.05 //volume of saturated liquid in m^3
+vf=0.001043 //in m^3/kg
+Vvap=4.95 //volume of saturated water vapour in m^3
+vg=1.6940 //in m^3/kg
+m1liq=Vliq/vf //mass of liquid in kg
+m1vap=Vvap/vg //mass of vapors in kg
+u1liq=417.36 //specific internal energy of liquid in kJ/kg
+u1vap=2506.1 //specific internal energy of vapors in kJ/kg
+U1=m1liq*u1liq+m1vap*u1vap //total internal energy in kJ
+m=m1liq+m1vap //total mass in kg
+V=5 //total volume in m^3
+v2=V/m //final specific volume in m^3/kg
+disp('by interplotation we find that for steam, if vg=0.09831 m^3/kg then pressure is 2.03 MPa')
+u2=2600.5 //specific internal energy at final state in kJ/kg
+U2=m*u2 //internal energy at final state in kJ
+Q=U2-U1 //heat transfer for the process in kJ
+printf("\n hence,heat transfer for the process is Q = %.0f kJ. \n",Q)
\ No newline at end of file diff --git a/172/CH5/EX5.6/ex6.sce b/172/CH5/EX5.6/ex6.sce new file mode 100755 index 000000000..4adda4d24 --- /dev/null +++ b/172/CH5/EX5.6/ex6.sce @@ -0,0 +1,20 @@ +//example 6
+//calculating work and heat transfer for the process
+clear
+clc
+V1=0.1 //volume of cylinder in m^3
+m=0.5 //mass of steam in kg
+v1=V1/m //specific volume of steam in m^3/kg
+vf=0.001084 //m^3/kg
+vfg=0.4614 //m^3/kg
+x1=(v1-vf)/vfg //quality
+hf=604.74 //kJ/kg
+hfg=2133.8//kJ/kg
+h2=3066.8 //final specific heat enthalpy in kJ/kg
+h1=hf+x1*hfg //initial specific enthalpy in kJ/kg
+Q=m*(h2-h1) //heat transfer for this process in kJ
+P=400 //pressure inside cylinder in kPa
+v2=0.6548 //specific enthalpy in m^3/kg
+W=m*P*(v2-v1) //work done for the process in kJ
+printf("\n hence, work done for the process, W = %.3f kJ. \n",W)
+printf("\n and heat transfer, Q=%.3f kJ.\n",Q)
\ No newline at end of file diff --git a/172/CH5/EX5.8/ex8.sce b/172/CH5/EX5.8/ex8.sce new file mode 100755 index 000000000..e9eafd090 --- /dev/null +++ b/172/CH5/EX5.8/ex8.sce @@ -0,0 +1,18 @@ +//example 8
+//calculating change in enthalpy
+clear
+clc
+h1=273.2 //specific heat enthalpy for oxygen at 300 K
+h2=1540.2 //specific heat enthalpy for oxygen at 1500 K
+T1=300 //initial temperature in K
+T2=1500 //final temparature in K
+x=poly([0],'x');
+Cp=0.88-0.00001*x+0.54*x^2-0.33*x^3 //expression for constant pressure specific heat enthalpy for oxygen
+dh1=h2-h1 //this change in specific heat enthalpy is calculated using ideal gas tables
+dh2=1000*integrate('0.88-0.00001*x+0.54*x^2-0.33*x^3','x',T1/1000,T2/1000) //using empirical equation
+dh3=0.922*(T2-T1) //it is claculated if we assume specific heat enthalpy to be constant and uses its value at 300K
+dh4=1.0767*(T2-T1) //it is claculated if we assume specific heat enthalpy to be constant and uses its value at 900K i.e mean of initial and final temperature
+printf("\n Hence,change in specific heat enthalpy if ideal gas tables are used is dh1=%.1f kJ/kg. \n", dh1)
+printf("\n if empirical equations are used, dh2=%.1f kJ/kg. \n", dh2)
+printf("\n if specific heat is assumed to be constant and using its value at T1, dh3=%.1f kJ/kg. \n", dh3)
+printf("\n if specific heat is assumed to be constant at its value at (T1+T2)/2, dh4=%.1f kJ/kg. \n", dh4)
\ No newline at end of file diff --git a/172/CH5/EX5.9/ex9.sce b/172/CH5/EX5.9/ex9.sce new file mode 100755 index 000000000..e8574c903 --- /dev/null +++ b/172/CH5/EX5.9/ex9.sce @@ -0,0 +1,14 @@ +//example 9
+//determining amount of heat transfer
+clear
+clc
+P=150 //pressure of nitrogen in cylinder in kPa
+V=0.1 //initial volume of cylinder in m^3
+T1=25 //initial temperature of nitrogen in celsius
+T2=150 //final tempareture of nitrogen in celsius
+R=0.2968 //in kJ/kg-K
+m=P*V/(R*(T1+273)) //mass of nitrogen in kg
+Cv=0.745 //constant volume specific heat for nitrogen in kJ/kg-K
+W=-20 //work done on nitrogen gas in kJ
+Q=m*Cv*(T2-T1)+W //heat transfer during the process in kJ
+printf("\n hence,the heat transfer for the above process is Q=%.1f kJ. \n", Q)
\ No newline at end of file diff --git a/172/CH6/EX6.1/ex1.sce b/172/CH6/EX6.1/ex1.sce new file mode 100755 index 000000000..369bc8b7e --- /dev/null +++ b/172/CH6/EX6.1/ex1.sce @@ -0,0 +1,13 @@ +//example 1
+//calculating mass flow rate in kg/s
+clear
+clc
+R=0.287 //in kJ/kg-K
+T=25 //temperature in celsius
+P=150 //pressure in kPa
+v=R*(T+273.2)/P //specific volume in m^3/kg
+D=0.2 //diameter of pipe in metre
+A=%pi*D^2/4 //cross sectional area in m^2
+V=0.1 //velocity of air in m/s
+m=V*A/v //mass flow rate in kg/s
+printf("\n hence,the mass flow rate is m=%.4f kg/s.\n",m)
\ No newline at end of file diff --git a/172/CH6/EX6.10/ex10.sce b/172/CH6/EX6.10/ex10.sce new file mode 100755 index 000000000..64067f988 --- /dev/null +++ b/172/CH6/EX6.10/ex10.sce @@ -0,0 +1,18 @@ +//example 10
+//analysis of refrigerator
+clear
+clc
+hf4=167.4 //in kJ/kg
+hfg4=215.6 //in kJ/kg
+h3=241.8 //specific heat of enthalpy of R-134a entering expansion valve
+h4=h3 //specific heat of enthalpy of R-134a leaving expansion valve
+h1=387.2 //in kJ/kg
+h2=435.1 //in kJ/kg
+x4=(h3-hf4)/hfg4 //quality of R-134a at evaporator inlet
+m=0.1 //mass flow rate in kg/s
+Qevap=m*(h1-h4) //rate of heat transfer to the evaporator
+Wcomp=-5 //power input to compressor in kW
+Qcomp=m*(h2-h1)+Wcomp //rate of heat transfer from compressor
+printf("\n hence, the quality at the evaporator inlet is x4=%.3f. \n",x4')
+printf("\n hence, the rate of heat transfer to the evaporator is Qevap=%.2f kW. \n",Qevap')
+printf("\n hence, rate of heat transfer from the compressor is Qcomp=%.2f kW. \n",Qcomp')
\ No newline at end of file diff --git a/172/CH6/EX6.11/ex11.sce b/172/CH6/EX6.11/ex11.sce new file mode 100755 index 000000000..748c5dd58 --- /dev/null +++ b/172/CH6/EX6.11/ex11.sce @@ -0,0 +1,9 @@ +//example 11
+//Determining the final temperature of steam
+clear
+clc
+u2=3040.4 //final internal energy in kJ/kg
+hi=u2 //in kJ/kg
+P2=1.4 //final Pressure in MPa
+disp('Since, the final pressure is given as 1.4 MPa,we know two properties at the final state and hence,final state can be determined.The temperature corresponding to a pressure of 1.4 MPa and an internal energy of 3040.4 kJ/kg is found to be ')
+T2=452 //final temperature in Celsius
\ No newline at end of file diff --git a/172/CH6/EX6.12/ex12.sce b/172/CH6/EX6.12/ex12.sce new file mode 100755 index 000000000..9e1d5283e --- /dev/null +++ b/172/CH6/EX6.12/ex12.sce @@ -0,0 +1,16 @@ +//example 12
+//Calculating mass flow of steam in tank
+clear
+clc
+V1=0.4 //initial volume fo tank in m^3
+v1=0.5243 //initial specific volume in m^3/kg
+h1=3040.4 //initial specific enthalpy in kJ/kg
+u1=2548.9 //initial specific internal energy in kJ/kg
+m1=V1/v1 //initial mass of steam in tank in kg
+V2=0.4 //final volume in m^3
+disp('let x=V*(h1-u2)/v2-m1*(h1-u1).If we assume T2=300C, then v2=0.1823m^3/kg ,u2=2785.2kJ/kg and x=+ve.If we assume T2=350C,then v2=0.2003 m^3/kg, u2=2869.1kJ/kg and x=-ve.Hence,actualt T2 must be between these two assumed values in order that x=0.By interplotation,')
+T2=342 //final temperature in Celsius
+v2=0.1974 //final specific volume in m^3/kg
+m2=V2/v2 //final mass of the steam in the tank in kg
+m=m2-m1 //mass of steam that flowsinto the tank
+printf(" \n Hence,mass of the steam that flows into the tank is m=%.3f kg. \n",m)
\ No newline at end of file diff --git a/172/CH6/EX6.13/ex13.sce b/172/CH6/EX6.13/ex13.sce new file mode 100755 index 000000000..b28e7e13e --- /dev/null +++ b/172/CH6/EX6.13/ex13.sce @@ -0,0 +1,26 @@ +//example 13
+//Calculating mass flow of steam in tank
+clear
+clc
+vf1=0.001725 //in m^3/kg
+vf2=0.0016 //in m^3/kg
+uf1=368.7 //in kJ/kg
+uf2=226 //in kJ/kg
+vg1=0.08313 //in m^3/kg
+vfg2=0.20381
+ug1=1341 //in kJ/kg
+ufg2=1099.7 //in kJ/kg
+Vf=1 //initial volume of liquid in m^3
+Vg=1 //initial volume of vapor in m^3
+mf1=Vf/vf1 //initial mass of liquid in kg
+mg1=Vg/vg1 //initial mass of vapor in kg
+m1=mf1+mg1 //initial mass of liquid in kg
+he=1461.1 //in kJ/kg
+V=2 //volume of tank in m^3
+disp('m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.00160+0.20381*x2) and u2=uf2+x2*ufg2=226.0+1099.7*x2.')
+disp('Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.')
+x2=((2*1461.1)-(2*226)-(0.00160*634706))/((634706*0.20381)+(2*1099.7)) //quality of ammonia
+v2=0.00160+(0.20381*x2) //final specific volume in m^3/kg
+m2=V/v2 //final mass of ammonia in kg
+m=m1-m2 //mass of ammonia withdrawn
+printf(" \n Hence,mass of ammonia withdrawn is m=%.1f kg. \n",m)
\ No newline at end of file diff --git a/172/CH6/EX6.2/ex2.sce b/172/CH6/EX6.2/ex2.sce new file mode 100755 index 000000000..29798c1aa --- /dev/null +++ b/172/CH6/EX6.2/ex2.sce @@ -0,0 +1,9 @@ +//example 2
+//work done for adding the fluid
+clear
+clc
+P=600 //pressure in kPa
+m=1 //in kg
+v=0.001 //specific volume in m^3/kg
+W=P*m*v //necessary work in kJ for adding the fluid
+printf(" \n hence,the work involved in this process is W=%.3f kJ. \n",W)
\ No newline at end of file diff --git a/172/CH6/EX6.3/ex3.sce b/172/CH6/EX6.3/ex3.sce new file mode 100755 index 000000000..8c193fc60 --- /dev/null +++ b/172/CH6/EX6.3/ex3.sce @@ -0,0 +1,11 @@ +//example 3
+//rate of flow of water
+clear
+clc
+hir=441.89 //in kJ/kg for refrigerant using steam table
+her=249.10 //in kJ/kg for refrigerant using steam table
+hiw=42 //in kJ/kg for water using steam table
+hew=83.95 //in kJ/kg for water using steam table
+mr=0.2 //the rate at which refrigerant enters the condenser in kg/s
+mw=mr*(hir-her)/(hew-hiw) //rate of flow of water in kg/s
+printf("\n hence,the rate at which cooling water flows thorugh the condenser is mw=%.3f kg/s. \n", mw)
\ No newline at end of file diff --git a/172/CH6/EX6.4/ex4.sce b/172/CH6/EX6.4/ex4.sce new file mode 100755 index 000000000..f149c0a87 --- /dev/null +++ b/172/CH6/EX6.4/ex4.sce @@ -0,0 +1,12 @@ +//example 4
+//determining quality of steam
+clear
+clc
+hi=2850.1 //initial specific heat enthalpy for steam in kJ/kg
+Vi=50 //initial velocity of steam in m/s
+Ve=600 //final velocity of steam in m/s
+he=hi+Vi^2/(2*1000)-Ve^2/(2*1000) //final specific heat enthalpy for steam in kJ/kg
+hf=467.1 //at final state in kJ/kg
+hfg=2226.5 //at final state in kJ/kg
+xe=(he-hf)/hfg //quality of steam in final state
+printf(" \n hence, the quality is xe=%.3f. \n",xe)
\ No newline at end of file diff --git a/172/CH6/EX6.5/ex5.sce b/172/CH6/EX6.5/ex5.sce new file mode 100755 index 000000000..b93f21942 --- /dev/null +++ b/172/CH6/EX6.5/ex5.sce @@ -0,0 +1,10 @@ +//example 5
+//quality of ammonia leaving expansion valve
+clear
+clc
+hi=346.8 //specific heat enthalpy for ammonia at initial state in kJ/kg
+he=hi //specific heat enthalpy for ammonia at final state will be equal that at initial state because it is a throttling process
+hf=134.4 //at final state in kJ/kg
+hfg=1296.4//at final state in kJ/kg
+xe=(he-hf)/hfg //quality at final state
+printf("\n hence,quality of the ammonia leaving the expansion valve is xe=%.4f. \n",xe')
\ No newline at end of file diff --git a/172/CH6/EX6.6/ex6.sce b/172/CH6/EX6.6/ex6.sce new file mode 100755 index 000000000..cd8cadd97 --- /dev/null +++ b/172/CH6/EX6.6/ex6.sce @@ -0,0 +1,15 @@ +//example 6
+//power output of turbine in kW
+clear
+clc
+hi=3137 //initial specific heat of enthalpy in kJ/kg
+he=2675.5 //final specific heat of enthalpy in kJ/kg
+Vi=50 //initial velocity of steam in m/s
+Ve=100 //final velocity of steam in m/s
+Zi=6 //height of inlet conditions in metres
+Ze=3 //height of exit conditions in metres
+m=1.5 //mass flow rate of steam in kg/s
+g=9.8066 //acc. due to gravity in m/s^2
+Qcv=-8.5 //heat transfer rate from turbine in kW
+Wcv=Qcv+m*(hi+Vi^2/(2*1000)+g*Zi/1000)-m*(he+Ve^2/(2*1000)+g*Ze/1000) //power output of turbine in kW
+printf("\n hence,the power output of the turbine is Wcv=%.3f kW. \n",Wcv)
\ No newline at end of file diff --git a/172/CH6/EX6.7/ex7.sce b/172/CH6/EX6.7/ex7.sce new file mode 100755 index 000000000..3eb6e1a06 --- /dev/null +++ b/172/CH6/EX6.7/ex7.sce @@ -0,0 +1,14 @@ +//example 7
+//heat transfer rate in aftercooler
+clear
+clc
+V1=0 //we assume initial velocity to be zero because its given that it enters with a low velocity
+V2=25 //final velocity with which carbon dioxide exits in m/s
+h2=401.52 //final specific enthalpy of heat when carbon dioxide exits in kJ/kg
+h1=198 //initial specific enthalpy of heat in kJ/kg
+w=h1-h2-V2^2/(2*1000) //in kJ/kg
+Wc=-50 //power input to the compressor in kW
+m=Wc/w //mass flow rate of carbon dioxide in kg/s
+h3=257.9 //final specific enthalpy of heat when carbon dioxide flows into a constant pressure aftercooler
+Qcool=-m*(h3-h2) //heat transfer rate in the aftercooler in kW
+printf(" \n hence,heat transfer rate in the aftercooler is Qcool=%.3f kW. \n",Qcool)
diff --git a/172/CH6/EX6.8/ex8.sce b/172/CH6/EX6.8/ex8.sce new file mode 100755 index 000000000..49e52f5b6 --- /dev/null +++ b/172/CH6/EX6.8/ex8.sce @@ -0,0 +1,13 @@ +//example 8
+//Required pump work
+clear
+clc
+m=1.5 //mass flow rate of water in kg/s
+g=9.807 //acceleration due to gravity in m/s^2
+Zin=-15 //depth of water pump in well in metres
+Zex=0 //in metres
+v=0.001001 //specific volume in m^3/kg
+Pex=400+101.3 //exit pressure in kPa
+Pin=90 //in kPa
+W=m*(g*(Zin-Zex)*0.001-(Pex-Pin)*v) //power input in kW
+printf(" \n Hence, the pump requires power input of W=%.0f W. \n",W*1000)
\ No newline at end of file diff --git a/172/CH6/EX6.9/ex9.sce b/172/CH6/EX6.9/ex9.sce new file mode 100755 index 000000000..c74622274 --- /dev/null +++ b/172/CH6/EX6.9/ex9.sce @@ -0,0 +1,21 @@ +//example 9
+//heat tranfer in simple steam power plant
+clear
+clc
+h1=3023.5 //specific heat of enthalpy of steam leaving boiler in kJ/kg
+h2=3002.5 //specific heat of enthalpy of steam entering turbine in kJ/kg
+x=0.9 //quality of steam entering condenser
+hf=226 //in kJ/kg
+hfg=2373.1 //in kJ/kg
+h3=hf+x*hfg //specific heat of enthalpy of steam entering condenser in kJ/kg
+h4=188.5 //specific heat of enthalpy of steam entering pump in kJ/kg
+q12=h2-h1 //heat transfer in line between boiler and turbine in kJ/kg
+w23=h2-h3 //turbine work in kJ/kg
+q34=h4-h3 //heat transfer in condenser
+w45=-4 //pump work in kJ/kg
+h5=h4-w45 //in kJ/kg
+q51=h1-h5 //heat transfer in boiler in kJ/kg
+printf("\n hence, heat transfer in line between boiler and turbine is q12=%.1f kJ/kg. \n",q12')
+printf("\n hence, turbine work is w23=%.1f kJ/kg. \n",w23')
+printf("\n hence, heat transfer in condenser is q34=%.1f kJ/kg. \n",q34')
+printf("\n hence, heat transfer in boiler is q51=%.0f kJ/kg. \n",q51')
\ No newline at end of file diff --git a/172/CH7/EX7.1/ex1.sce b/172/CH7/EX7.1/ex1.sce new file mode 100755 index 000000000..b750cc0a9 --- /dev/null +++ b/172/CH7/EX7.1/ex1.sce @@ -0,0 +1,12 @@ +//example 1
+//rate of fuel consumption
+clear
+clc
+W=136*0.7355 //output of automobile engine in kW
+neng=0.3 //thermal efficiency of automobile engine
+Qh=W/neng //energy output of fuel in kW
+Ql=Qh-W //total rate of energy rejected to the ambient
+qh=35000 //energy output of fuel in kJ/kg
+m=Qh/qh //rate of fuel consumption in kg/s
+printf("\n hence,total rate of energy rejected is Ql=%.0f kW.\n",Ql)
+printf("\n and rate of fuel consumption is m=%.4f kg/s.\n",m)
\ No newline at end of file diff --git a/172/CH7/EX7.2/ex2.sce b/172/CH7/EX7.2/ex2.sce new file mode 100755 index 000000000..d12cedf23 --- /dev/null +++ b/172/CH7/EX7.2/ex2.sce @@ -0,0 +1,10 @@ +//example 2
+//coefficient of performance of refrigerator
+clear
+clc
+Qh=400 //heat rejected to kitchen air in W
+W=150 //electrical input power in W
+Ql=Qh-W //rate of energy taken out to cold space in W
+B=Ql/W //coefficicent of performnace of refrigerator
+printf("\n hence,rate of energy taken out of the cold space is Ql=%.3f W.\n",Ql)
+printf("\n and coefficient of performance of the refrigerator is B=%.3f .\n",B)
\ No newline at end of file diff --git a/172/CH7/EX7.4/ex4.sce b/172/CH7/EX7.4/ex4.sce new file mode 100755 index 000000000..4693cab22 --- /dev/null +++ b/172/CH7/EX7.4/ex4.sce @@ -0,0 +1,15 @@ +//example 4
+//comparison of ideal carnot heat engine with actual heat engine
+clear
+clc
+Qh=1000 //rate of heat transfer to heat engine in kW
+W=450 //rate of production of work in kW
+Ql=Qh-W //rate of heat rejected by heat engine in kW
+nthermal=W/Qh //efficiency from the definition of efficiency
+Tl=300 //temperature of surroundings in K
+Th=550 //temperature of heat source in Celsius
+ncarnot=1-Tl/(Th+273) //efficiency if heat engine is considered to be ideal carnot heat engine
+W2=ncarnot*Qh //rate of work production if heat engine is assumed to be ideal carnot heat engine in kW
+Ql2=Qh-W2 //rate of heat rejected by heat engine in kW if heat engine is assumed to be ideal carnot heat engine
+printf("\n hence,energy discarded to the ambient surroundings is Ql2=%.0fkW.\n",Ql2)
+printf("\n and the engine efficiency is ncarnot=%.3f.\n",ncarnot)
\ No newline at end of file diff --git a/172/CH7/EX7.5/ex5.sce b/172/CH7/EX7.5/ex5.sce new file mode 100755 index 000000000..4008f1006 --- /dev/null +++ b/172/CH7/EX7.5/ex5.sce @@ -0,0 +1,10 @@ +//example 5
+//calculating required work
+clear
+clc
+Tl=24+273 //room temperature in Kelvins
+Th=35+273 //atmospheric temperature in Kelvins
+Ql=4 //rate of heat rejection from room
+B=Tl/(Th-Tl) //coefficient of performance of air conditioner
+W=Ql/B //required work in kW
+printf("\n hence,the magnitude of reqiured work is W=%.2f kW.\n",W)
\ No newline at end of file diff --git a/172/CH8/EX8.1/ex1.sce b/172/CH8/EX8.1/ex1.sce new file mode 100755 index 000000000..b5c2504e1 --- /dev/null +++ b/172/CH8/EX8.1/ex1.sce @@ -0,0 +1,15 @@ +//example 1
+//coefficient of performance of refrigerator
+clear
+clc
+Th=60 //temperature at which heat is rejected from R-134a
+Tl=0 //temperature at which heat is absorbed into the R-134a
+s1=1.7262 //specific entropy at 0 Celsius
+s2=s1 //process of state change from 1-2 is isentropic
+s3=1.2857 //specific entropy at 60 celsius
+s4=s3 //process of state change from 3-4 is isentropic
+disp('if Pressure is 1400 kPa,then s=1.7360 kJ/kg-K and if P=1600 kPa,then s=1.7135 kJ/kg-K.Therefore')
+P2=1400+(1600-1400)*(1.7262-1.736)/(1.7135-1.736) //pressure after compression in kPa
+B=(Th+273)/(Th-Tl) //coefficient of performance of refrigerator
+printf(" \n hence,pressure after compression is P2=%.3f kPa.\n",P2)
+printf("\n and coefficient of performance of refrigerator is B=%.3f .\n",B)
\ No newline at end of file diff --git a/172/CH8/EX8.10/ex10.sce b/172/CH8/EX8.10/ex10.sce new file mode 100755 index 000000000..018a8b3ce --- /dev/null +++ b/172/CH8/EX8.10/ex10.sce @@ -0,0 +1,17 @@ +//example 10
+//Determiining the entropy generated
+clear
+clc
+B=4 //COP of air conditioner
+W=10 //power input of air conditioner in kW
+Qh=B*W //in kW
+Ql=Qh-W //in kW
+Thigh=323 //in Kelvin
+Tlow=263 //in Kelvin
+SgenHP=(Qh*1000/Thigh)-(Ql*1000/Tlow) //in W/K
+Tl=281 // in K
+Th=294 //in K
+SgenCV1=Ql*1000/Tlow-Ql*1000/Tl //in W/K
+SgenCV2=Qh*1000/Th-Qh*1000/Thigh //in W/K
+SgenTOT=SgenCV1+SgenCV2+SgenHP //in W/K
+printf(" \n Hence,Total entropy generated is SgenTOT=%.1f W/K. \n",SgenTOT)
\ No newline at end of file diff --git a/172/CH8/EX8.2/ex2.sce b/172/CH8/EX8.2/ex2.sce new file mode 100755 index 000000000..36431002e --- /dev/null +++ b/172/CH8/EX8.2/ex2.sce @@ -0,0 +1,16 @@ +//example 2
+//heat transfer in a given process
+clear
+clc
+u1=87.94 //specific internal energy of R-12 at state 1 in kJ/kg
+u2=276.44 //specific internal energy of R-12 at state 2 in kJ/kg
+s1=0.3357 //specific entropy at state 1 in kJ/kg-K
+s2=1.2108 //specific entropy at state 2 in kJ/kg-K
+V=0.001 //volume of saturated liquid in m^3
+v1=0.000923 //specific volume in m^3/kg
+m=V/v1 //mass of saturated liquid in kg
+T=20 //temperature of liquid in celsius
+Q12=m*(T+273.15)*(s2-s1) //heat transfer in kJ to accomplish the process
+W12=m*(u1-u2)+Q12 //work required to accomplish the process
+printf(" \n hence,work required to accomplish the process is W12=%.1f kJ.\n",W12)
+printf(" \n and heat transfer is Q12=%.1f kJ.\n",Q12)
\ No newline at end of file diff --git a/172/CH8/EX8.3/ex3.sce b/172/CH8/EX8.3/ex3.sce new file mode 100755 index 000000000..15ab12073 --- /dev/null +++ b/172/CH8/EX8.3/ex3.sce @@ -0,0 +1,11 @@ +//example 3
+//entropy change
+clear
+clc
+C=4.184 // specific heat of water in kJ/kg-K
+T1=20 //initial temperature of water in celsius
+T2=90 //final temperature of water in celsius
+dS1=C*log((T2+273.2)/(T1+273.2)) //change in entropy in kJ/kg-K
+dS2=1.1925-0.2966 //in kJ/kg-K using steam tables
+printf("\n hence,change in entropy assuming constant specific heat is dS1=%.4f kJ/kg-K.\n",dS1)
+printf("\n using steam table is dS2=%.4f kJ/kg-K.\n",dS2)
\ No newline at end of file diff --git a/172/CH8/EX8.4/ex4.sce b/172/CH8/EX8.4/ex4.sce new file mode 100755 index 000000000..9399182cc --- /dev/null +++ b/172/CH8/EX8.4/ex4.sce @@ -0,0 +1,20 @@ +//example 4
+//entropy change with different assumptions
+clear
+clc
+T1=300 //initial temperature in kelvins
+T2=1500 //final temperature in kelvins
+P1=200 //initial pressure in kPa
+P2=150 //final pressure in kPa
+R=0.2598 // in kJ/kg-K
+Cp=0.922 //specific heat in kJ/kg-K at constant pressure
+dsT2=8.0649 //in kJ/kg-K
+dsT1=6.4168 //in kJ/kg-K
+dS1=dsT2-dsT1-R*log(P2/P1) //entropy change calculated using ideal gas tables
+dS2=integrate('0.88/x-0.0001+0.54*x-0.33*x^2','x',0.3,1.5)-R*log(P2/P1) //entropy change calculated using empirical equation
+dS3=Cp*log(T2/T1)-R*log(P2/P1) //entropy change assuming constant specific heat in kJ/kg-K
+dS4=1.0767*log(T2/T1)+0.0747 //entropy change assuming specific heat is constant at its value at 990K
+printf("\n hence,change in entropy using ideal gas tables is dS1=%.4f kJ/kg-K.\n",dS1)
+printf("\n hence,change in entropy using empirical equation is dS2=%.4f kJ/kg-K.\n",dS2)
+printf("\n hence,change in entropy using the value of specific heat at 300K is dS3=%.4f kJ/kg-K.\n",dS3)
+printf("\n hence,change in entropy assuming specific heat is constant at its value at 900K is dS4=%.4f kJ/kg-K.\n",dS4)
\ No newline at end of file diff --git a/172/CH8/EX8.5/ex5.sce b/172/CH8/EX8.5/ex5.sce new file mode 100755 index 000000000..f23cb3ac9 --- /dev/null +++ b/172/CH8/EX8.5/ex5.sce @@ -0,0 +1,16 @@ +//example 5
+//entropy change
+clear
+clc
+Cp=1.004 //specific heat at constant pressure in kJ/kg-K
+R=0.287 //gas constant in kJ/kg-K
+P1=400 //initial pressure in kPa
+P2=300 //final pressure in kPa
+T1=300 //initial temperature in K
+T2=600 //final temperature in K
+dS1=Cp*log(T2/T1)-R*log(P2/P1) //entropy change assuming constant specific heat
+s1=6.8693 //specific entropy at T1
+s2=7.5764 //specific entropy at T2
+dS2=s2-s1-R*log(P2/P1) //entropy change assuming variable specific heat
+printf("\n hence,entropy change assuming constant specific heat is dS1=%.4f kJ/kg-K.\n",dS1)
+printf("\n and assuming variable specific heat is dS2=%.4f kJ/kg-K.\n",dS2)
\ No newline at end of file diff --git a/172/CH8/EX8.6/ex6.sce b/172/CH8/EX8.6/ex6.sce new file mode 100755 index 000000000..ebfbe454d --- /dev/null +++ b/172/CH8/EX8.6/ex6.sce @@ -0,0 +1,17 @@ +//example 6
+//work done by air
+clear
+clc
+T1=600 //initial temperature of air in K
+P1=400 //intial pressure of air in kPa
+P2=150 //final pressure in kPa
+u1=435.10 //specific internal energy at temperature T1 in kJ/kg
+sT1=7.5764 //specific entropy at temperature T1 in kJ/kg-K
+R=0.287 //gas constant in kJ/kg-K
+ds=0
+sT2=ds+sT1+R*log(P2/P1) //specific entropy at temperature T2 in kJ/kg-K
+disp('we know the values of s and P for state 2.So,in order to fully determine the state,we will use steam table')
+T2=457 //final temperature in K
+u2=328.14 //specific internal energy at temperature T2 in kJ/kg
+w=u1-u2 //work done by air in kJ/kg
+printf("\n hence,work done by air is w=%.2f kJ/kg.\n",w)
\ No newline at end of file diff --git a/172/CH8/EX8.7/ex7.sce b/172/CH8/EX8.7/ex7.sce new file mode 100755 index 000000000..18fd02c90 --- /dev/null +++ b/172/CH8/EX8.7/ex7.sce @@ -0,0 +1,15 @@ +//example 7
+//work and heat transfer
+clear
+clc
+P2=500 //final pressure in cylinder in kPa
+P1=100 //initial pressure in cylinder in kPa
+T1=20+273.2 //initial temperature inside cylinder in Kelvins
+n=1.3
+T2=(T1)*(P2/P1)^((n-1)/n) //final temperature inside cylinder in K
+R=0.2968 //gas constant in kJ/kg-K
+w12=R*(T2-T1)/(1-n) //work in kJ/kg
+Cvo=0.745 //specific heat at constant volume in kJ/kg-K
+q12=Cvo*(T2-T1)+w12 //heat transfer in kJ/kg
+printf(" \n hence,work done is w12=%.1f kJ/kg.\n",w12)
+printf("\n and heat transfer are q12=%.1f kJ/kg.\n",q12)
\ No newline at end of file diff --git a/172/CH8/EX8.8/ex8.sce b/172/CH8/EX8.8/ex8.sce new file mode 100755 index 000000000..d5a5dbad8 --- /dev/null +++ b/172/CH8/EX8.8/ex8.sce @@ -0,0 +1,13 @@ +//example 3
+//calculating increase in entropy
+clear
+clc
+m=1 //mass of saturated water vapour
+sfg=6.0480 //in kJ/K
+T=25 //temperature of surrounding air in celsius
+dScm=-m*sfg //change in entropy of control mass in kJ/K
+hfg=2257.0 //in kJ/kg
+Qtosurroundings=m*hfg //heat transferred to surroundings in kJ
+dSsurroundings=Qtosurroundings/(T+273.15) //in kJ/K
+dSnet=dScm+dSsurroundings //net increase in entropy in kJ/K
+printf(" hence,net increase in entropy of water plus surroundings is dSnet=%.4f kJ/K.\n",dSnet)
\ No newline at end of file diff --git a/172/CH8/EX8.9/ex9.sce b/172/CH8/EX8.9/ex9.sce new file mode 100755 index 000000000..4065db4e6 --- /dev/null +++ b/172/CH8/EX8.9/ex9.sce @@ -0,0 +1,8 @@ +//example 9
+//entropy generation
+clear
+clc
+Qout=1 //value of heat flux generated by 1kW of electric power
+T=600 //temperature of hot wire surface in K
+Sgen=Qout/T //entropy generation in kW/K
+printf(" \n hence,entropy generation is Sgen=%.5f kW/K.\n",Sgen)
\ No newline at end of file diff --git a/172/CH9/EX9.1/ex1.sce b/172/CH9/EX9.1/ex1.sce new file mode 100755 index 000000000..e4d1538c7 --- /dev/null +++ b/172/CH9/EX9.1/ex1.sce @@ -0,0 +1,18 @@ +//example 1
+//work done by steam
+clear
+clc
+hi=3051.2 //initial specific heat of enthalpy of steam in kJ/kg
+si=7.1228 //initial specific entropy of steam in kJ/kg-K
+Pe=0.15 //final pressure in MPa
+se=si //specific entropy in final state in kJ/kg-K
+sf=1.4335 //in kJ/kg-K
+sfg=5.7897 //in kJ/kg-K
+vi=50 //velocity with which steam enters turbine in m/s
+ve=200 //velocity with which steam leaves the turbine in m/s
+xe=(se-sf)/sfg //quality of steam in final state
+hf=467.1 //in kJ/kg
+hfg=2226.5 //in kJ/kg
+he=hf+xe*hfg //final specific heat of enthalpy of steam in kJ/kg
+w=hi+vi^2/(2*1000)-he-ve^2/(2*1000) //work of steam for isentropic process in kJ/kg
+printf("\n hence, work per kilogram of steam for this isentropic process is w=%.1f kJ/kg-K.\n",w)
\ No newline at end of file diff --git a/172/CH9/EX9.10/ex10.sce b/172/CH9/EX9.10/ex10.sce new file mode 100755 index 000000000..eda8e8d4d --- /dev/null +++ b/172/CH9/EX9.10/ex10.sce @@ -0,0 +1,17 @@ +//example 10
+//turbine efficiency
+clear
+clc
+hi=3051.2 //initial specific heat of enthalpy in kJ/kg
+si=7.1228 //initial specific entropy in kJ/kg-K
+sf=0.7548 //in kJ/kg-K
+sfg=7.2536 //in kJ/kg-K
+ses=si //final specific entropy is same as the initial
+xes=(si-sf)/sfg //quality of steam when it leaves the turbine
+hf=225.9 //in kJ/kg
+hfg=2373.1 //in kJ/kg
+hes=hf+xes*hfg //final specific heat of enthalpy in kJ/kg
+ws=hi-hes //work output of turbine calculated ideally in kJ/kg
+wa=600 //actual work output of turbine in kJ/kg
+nturbine=wa/ws //efiiciency of turbine
+printf("\n hence,efficiency of the turbine is nturbine=%.1f.\n",nturbine*100)
\ No newline at end of file diff --git a/172/CH9/EX9.11/ex11.sce b/172/CH9/EX9.11/ex11.sce new file mode 100755 index 000000000..b95bb560d --- /dev/null +++ b/172/CH9/EX9.11/ex11.sce @@ -0,0 +1,16 @@ +//example 11
+//turbine inlet pressure
+clear
+clc
+hi=1757.3 //initial specific heat of enthalpy of air in kJ/kg
+si=8.6905 //initial specifc entropy of airin kJ/kg-K
+he=855.3 //final specific heat of enthalpy of air in kJ/kg
+w=hi-he //actual work done by turbine in kJ/kg
+n=0.85 //efficiency of turbine
+ws=w/n //ideal work done by turbine in kJ/kg
+hes=hi-ws //from first law of isentropic process
+Tes=683.7 //final temperature in kelvins from air tables
+ses=7.7148 //in kJ/kg-K
+R=0.287 //gas constant in kJ/kg-K
+Pi=100/%e^((si-ses)/-R) //turbine inlet pressure in kPa
+printf("\n hence,turbine inlet pressure is Pi=%.0f kPa.\n",Pi)
\ No newline at end of file diff --git a/172/CH9/EX9.12/ex12.sce b/172/CH9/EX9.12/ex12.sce new file mode 100755 index 000000000..7da4a7e60 --- /dev/null +++ b/172/CH9/EX9.12/ex12.sce @@ -0,0 +1,15 @@ +//example 12
+//required work input
+clear
+clc
+Pe=150 //final pressure of air in kPa
+Pi=100 //initial presure of air in kPa
+k=1.4
+Ti=300 //initial temperature of air in kelvis
+Tes=Ti*(Pe/Pi)^((k-1)/k) //from second law
+ws=1.004*(Ti-Tes) //from first law of isentropic process
+n=0.7 //efficiency of automotive supercharger
+w=ws/n //real work input in kJ/kg
+Te=Ti-w/1.004 //temperature at supercharger exit in K
+printf("\n hence,required work input is w=%.1f kJ/kg.\n",w)
+printf("\n and exit temperature is Te=%.1f K.\n",Te)
\ No newline at end of file diff --git a/172/CH9/EX9.2/ex2.sce b/172/CH9/EX9.2/ex2.sce new file mode 100755 index 000000000..5df3d2f35 --- /dev/null +++ b/172/CH9/EX9.2/ex2.sce @@ -0,0 +1,14 @@ +//example 2
+//exit velocity of steam from nozzle
+clear
+clc
+hi=3051.2 //initial specific heat of enthalpy in kJ/kg
+si=7.1228 //initial specific entropy in kJ/kg-K
+se=si //final specific entropy
+Pe=0.3 //final pressure in MPa
+disp('from steam table,various properties at final state are ')
+he=2780.2 //final specific heat of enthalpy in kJ/kg-K
+Te=159.1 //final temperature in celsius
+vi=30 //velocity with which steam enters the nozzle in m/s
+ve=((2*(hi-he)+(vi^2/1000))*1000)^0.5 //final velocity of steam with which it exits in m/s
+printf("\n hence,exit velocity of the steam from the nozzle is ve=%.0f m/s.\n",ve)
\ No newline at end of file diff --git a/172/CH9/EX9.3/ex3.sce b/172/CH9/EX9.3/ex3.sce new file mode 100755 index 000000000..2b6c3f410 --- /dev/null +++ b/172/CH9/EX9.3/ex3.sce @@ -0,0 +1,8 @@ +//example 3
+//violation of second law
+clear
+clc
+disp('from R-134a tables')
+se=1.7148 //specific entropy in final state in kJ/kg-K
+si=1.7395 //initial specific entropy in kJ/kg-K
+disp('therefore,se<si,whereas for this process the second law requires that se>=si.The process described involves a violation of the second law and thus would be impossible.')
\ No newline at end of file diff --git a/172/CH9/EX9.4/ex4.sce b/172/CH9/EX9.4/ex4.sce new file mode 100755 index 000000000..64f83d947 --- /dev/null +++ b/172/CH9/EX9.4/ex4.sce @@ -0,0 +1,12 @@ +//example 4
+//calculating required specific work
+clear
+clc
+Cp=1.004 //specific heat of air at constant pressure in kJ/kg-K
+Ti=290 //initial temperature in kelvins
+Pi=100 //initial pressure in kPa
+Pe=1000 //final pressure in kPa
+k=1.4
+Te=Ti*(Pe/Pi)^((k-1)/k) //final temperature in kelvins
+we=Cp*(Ti-Te) //required specific work in kJ/kg
+printf("\n hence,specific work required is we=%.3f kJ/kg.\n",we)
\ No newline at end of file diff --git a/172/CH9/EX9.5/ex5.sce b/172/CH9/EX9.5/ex5.sce new file mode 100755 index 000000000..c1494ff1f --- /dev/null +++ b/172/CH9/EX9.5/ex5.sce @@ -0,0 +1,15 @@ +//example 5
+//entropy generation
+clear
+clc
+h1=2865.54 //specific heat of enthalpy at state 1 in kJ/kg
+h2=83.94 //specific heat of enthalpy at state 2 in kJ/kg
+h3=2725.3 //specific heat of enthalpy at state 3 in kJ?kg
+s1=7.3115 //specific entropy at state 1 in kJ/kg-K
+s2=0.2966 //specific entropy at state 2 in kJ/kg-K
+s3=6.9918 //specific entropy at state 3in kJ/kg-K
+m1=2 //mass flow rate at state 1 in kg/s
+m2=m1*(h1-h3)/(h3-h2) //mass flow rate at state 2 in kg/s
+m3=m1+m2 //mass flow rate at state 3 in kg/s
+Sgen=m3*s3-m1*s1-m2*s2 //entropy generation in the process
+printf("\n hence,entropy generated in this process is Sgen=%.3f kW/K.\n",Sgen)
\ No newline at end of file diff --git a/172/CH9/EX9.6/ex6.sce b/172/CH9/EX9.6/ex6.sce new file mode 100755 index 000000000..58240e4c7 --- /dev/null +++ b/172/CH9/EX9.6/ex6.sce @@ -0,0 +1,21 @@ +//example 6
+//work required to fill the tank
+clear
+clc
+T1=17+273 //initial temperature of tank in Kelvins
+sT1=6.83521 //specific entropy in kJ/kg-K
+R=0.287 //gas constant in kJ/kg-K
+P1=100 //initial pressure in kPa
+P2=1000 //final pressure in kPa
+sT2=sT1+R*log(P2/P1) //specific entropy at temperature T2 in kJ/kg-K
+T2=555.7 //from interplotation
+V1=0.04 //volume of tank in m^3
+V2=V1 //final volume is equal to initial volume
+m1=P1*V1/(R*T1) //initial mass of air in tank in kg
+m2=P2*V2/(R*T2) //final mass of air in tank in kg
+Min=m2-m1 //in kg
+u1=207.19 //initial specific heat of enthalpy in kJ/kg
+u2=401.49 //final specific heat of enthalpy in kJ/kg
+hin=290.43 //in kJ/kg
+W12=Min*hin+m1*u1-m2*u2 //work required to fill the tank in kJ
+printf("\n hence,the total amount of work required to fill the tank is W12=%.1f m/s.\n",W12)
\ No newline at end of file diff --git a/172/CH9/EX9.7/ex7.sce b/172/CH9/EX9.7/ex7.sce new file mode 100755 index 000000000..a06f05b91 --- /dev/null +++ b/172/CH9/EX9.7/ex7.sce @@ -0,0 +1,9 @@ +//example 7
+//work required to pump water isentropically
+clear
+clc
+P1=100 //initial pressure in kPa
+P2=5000 //final pressure in kPa
+v=0.001004 //specific volume in m^3/kg
+w=v*(P2-P1) //work required to pump water isentropically
+printf("\n hence,work required to pump water isentropically is w=%.2f kJ/kg.\n",w)
\ No newline at end of file diff --git a/172/CH9/EX9.8/ex8.sce b/172/CH9/EX9.8/ex8.sce new file mode 100755 index 000000000..52f626ab7 --- /dev/null +++ b/172/CH9/EX9.8/ex8.sce @@ -0,0 +1,11 @@ +//example 8
+//Velocity in exit flow
+clear
+clc
+disp('From Steam Tables, for liquid water at 20 C')
+vf=0.001002 //in m^3/kg
+v=vf
+Pi=300 //Line pressure in kPa
+Po=100 //in kPa
+Ve=(2*v*(Pi-Po)*1000)^0.5 //velocity in the exit flow
+printf(" \n Hence, an ideal nozzle can generate upto Ve=%.0f m/s in the exit flow. \n",Ve)
\ No newline at end of file diff --git a/172/CH9/EX9.9/ex9.sce b/172/CH9/EX9.9/ex9.sce new file mode 100755 index 000000000..b58e59bba --- /dev/null +++ b/172/CH9/EX9.9/ex9.sce @@ -0,0 +1,15 @@ +//example 9
+//Rate of Entropy Generation
+clear
+clc
+disp('From R-410a tables,we get')
+hi=280.6 //in kJ/kg
+he=307.8 //in kJ/kg
+si=1.0272 //in kJ/kg
+se=1.0140 //in kJ/kg
+m=0.08 //flow rate of refrigerant in kg/s
+P=3 //electrical power input in kW
+Qcv=m*(he-hi)-P //in kW
+To=30 //in Celsius
+Sgen=m*(se-si)-Qcv/(To+273.2) //rate of entropy generation
+printf("\n Hence,the rate of entropy generation for this process is Sgen=%.5f kW/K. \n",Sgen)
\ No newline at end of file |