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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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treeab291cffc65280e58ac82470ba63fbcca7805165 /23/CH2
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diff --git a/23/CH2/EX2.1/Example_2_1.pdf b/23/CH2/EX2.1/Example_2_1.pdf
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diff --git a/23/CH2/EX2.1/Example_2_1.sce b/23/CH2/EX2.1/Example_2_1.sce
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index 000000000..4413217e9
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@@ -0,0 +1,33 @@
+clear;
+clc;
+
+//Example 2.1
+//Caption : Program to find Energy in a Waterfall
+
+
+//Given values
+H=100;//height=100m
+M=1;//Mass of water=1Kg
+g=9.8066;//Acceleration due to gravity(m/s^2)
+
+//Solution
+
+//Del(Energy of the system)=0
+//hence,del(U)+del(KE)+del(PE)=0
+
+//(a)
+PE1=M*H*g;//(J)
+disp('J',PE1,'(a)Potential energy of Water at the Top');
+
+//(b)
+del_U=0;
+KE1=0;
+PE2=0;
+KE2=PE1;//(J)
+disp('J',KE2,'(b)Kinetic energy of Water');
+
+//(c)
+del_U=KE2;
+disp('J',del_U,'(c)Change in Internal energy when 1kg Water added');
+
+//End \ No newline at end of file
diff --git a/23/CH2/EX2.10/Example_2_10.pdf b/23/CH2/EX2.10/Example_2_10.pdf
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diff --git a/23/CH2/EX2.10/Example_2_10.sce b/23/CH2/EX2.10/Example_2_10.sce
new file mode 100755
index 000000000..6da4ad349
--- /dev/null
+++ b/23/CH2/EX2.10/Example_2_10.sce
@@ -0,0 +1,44 @@
+clear;
+clc;
+
+//To find Approx Value
+function[A]=approx(V,n)
+ A=round(V*10^n)/10^n;//V-Value n-To what place
+ funcprot(0)
+endfunction
+
+//Example 2.10
+//Caption : Program to Find change in Internal Energy and Enthalpy
+
+//Given values
+
+//Initial values
+T1=277;//Temp=277K
+P1=10;//Pressure=10bar
+V1=2.28;//molar Volume=2.28m^3/Kmol
+
+//Final value
+T2=333;//Temp=333K
+P2=1;//Pressure=1atm
+
+Cv=21;//KJ/Kmol/K
+Cp=29.3;//KJ/Kmol/K
+
+//Solution
+//(a)-Cooled at const Vol to the final pressure
+//(b)-Heated at const Pressure to final temperature
+T_=T1*(1/10);//Intermediate temperature
+del_Ta=T_-T1;
+del_Tb=T2-T_;
+del_Ua=Cv*del_Ta;//KJ/Kmol
+del_Ha=del_Ua+(V1*(P2-P1)*(10^5)/(10^3));//KJ/Kmol
+V2=(V1*P1*T2)/(P2*T1);//m^3/kmol
+del_Hb=Cp*del_Tb;
+del_Ub=del_Hb-(P2*(V2-V1)*(10^5)/(10^3));//KJ/Kmol
+
+del_U=approx(del_Ua+del_Ub,0);
+del_H=approx(del_Ha+del_Hb,0);
+disp('KJ/Kmol',del_U,'Change In Internal Energy')
+disp('KJ/Kmol',del_H,'Change In Enthalpy')
+
+//End \ No newline at end of file
diff --git a/23/CH2/EX2.13/Example_2_13.pdf b/23/CH2/EX2.13/Example_2_13.pdf
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diff --git a/23/CH2/EX2.13/Example_2_13.sce b/23/CH2/EX2.13/Example_2_13.sce
new file mode 100755
index 000000000..aabb6dc5e
--- /dev/null
+++ b/23/CH2/EX2.13/Example_2_13.sce
@@ -0,0 +1,27 @@
+clear;
+clc;
+
+//To find Approx Value
+function[A]=approx(V,n)
+ A=round(V*10^n)/10^n;//V-Value n-To what place
+ funcprot(0)
+endfunction
+
+//Example 2.13
+//Caption : Program to Find the time for a certain Temperature Drop
+
+//Given values
+M=190;//Mass=190Kg
+T0=333.15;//Temperature=333.15K(60`C)
+m=0.2;//Steady rate of mass(Kg/s)
+T=308.15;//Temperature=308.15K(35`C)
+T1=283.15;//Temperature=283.15K(10`C)
+
+//Solution
+//Using the Eqn (2.29)
+t=approx(-(M/m)*log((T-T1)/(T0-T1)),1);//s
+disp('s',t,'Time Taken for temperature of water to drop from 333.15K to 308.15K')
+t=round(t/60);//min
+disp('min',t,'Time Taken for temperature of water to drop from 333.15K to 308.15K')
+
+//End \ No newline at end of file
diff --git a/23/CH2/EX2.14/Example_2_14.pdf b/23/CH2/EX2.14/Example_2_14.pdf
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index 000000000..df26dbd35
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diff --git a/23/CH2/EX2.14/Example_2_14.sce b/23/CH2/EX2.14/Example_2_14.sce
new file mode 100755
index 000000000..2df15042e
--- /dev/null
+++ b/23/CH2/EX2.14/Example_2_14.sce
@@ -0,0 +1,17 @@
+clear;
+clc;
+
+//Example 2.14
+//Caption : Program to find the Enthalpy of Steam
+
+//Given values
+
+rQ=4.15;//[g/s] flow rate
+rQ2=12740;//Rate of Heat addition from resistance heater
+
+//Solution
+//del_z and del_u*2 are negligible if Ws and H1=0..then H2=Q
+H2=round(rQ2/rQ);//[J/g]
+disp('J/g',H2,'Enthalpy of Steam')
+
+//End \ No newline at end of file
diff --git a/23/CH2/EX2.15/Example_2_15.pdf b/23/CH2/EX2.15/Example_2_15.pdf
new file mode 100755
index 000000000..503fa5dec
--- /dev/null
+++ b/23/CH2/EX2.15/Example_2_15.pdf
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diff --git a/23/CH2/EX2.15/Example_2_15.sce b/23/CH2/EX2.15/Example_2_15.sce
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index 000000000..716feefed
--- /dev/null
+++ b/23/CH2/EX2.15/Example_2_15.sce
@@ -0,0 +1,18 @@
+clear;
+clc;
+
+//Example 2.15
+//Caption : Program To find the Heat to be Removed during Compression
+
+//Given values
+
+V=600;//[m/s]
+W_compression=240;//[KJ/Kg]
+
+//Solution
+//Using Eqn(2.32a)
+Q=(1/2*(V*V)/1000)-W_compression;
+
+disp('KJ/kg',-Q,'Thus Heat Removed from each KG of air compressed is')
+
+//End \ No newline at end of file
diff --git a/23/CH2/EX2.16/Example_2_16.pdf b/23/CH2/EX2.16/Example_2_16.pdf
new file mode 100755
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diff --git a/23/CH2/EX2.16/Example_2_16.sce b/23/CH2/EX2.16/Example_2_16.sce
new file mode 100755
index 000000000..f76312856
--- /dev/null
+++ b/23/CH2/EX2.16/Example_2_16.sce
@@ -0,0 +1,42 @@
+clear;
+clc;
+
+//To find Approx Value
+function[A]=approx(V,n)
+ A=round(V*10^n)/10^n;//V-Value n-To what place
+ funcprot(0)
+endfunction
+
+//Example 2.15
+//Caption : Program to Find the Temperature in the second Tank
+
+//Given values
+
+R=3.15*10^-3;//[m^3/s] Rate of pumping
+rH=-700;//[KW] Rate of Heat lost
+h=15;//[m] Height
+rW=1.5;//[KW]
+rho=958;//[Kg/m^3] at 366.65K
+g=9.805;
+gc=1000;
+del_z=h;
+
+//Solution
+
+rm=approx(R*rho,3);//[Kg/s] Mass flow rate
+Q=approx(rH/rm,1);//[KJ/Kg]
+W=approx(rW/rm,3);//[KJ/Kg] Shaft Work
+K=approx(g/gc*del_z,3);
+
+//using Eqn(2.32b)
+del_H=Q+W-K;
+
+//From Steam tables for water at 366.65K
+H1=391.6;//[KJ/Kg]
+H2=del_H+H1;
+disp('KJ/Kg',H2,'Enthalpy')
+//From Steam Tables temp at this enthalpy is
+T=311.35;//[K]
+disp('K',T,'Temperature in the Second tank')
+
+//End \ No newline at end of file
diff --git a/23/CH2/EX2.3/Example_2_3.pdf b/23/CH2/EX2.3/Example_2_3.pdf
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+clear;
+clc;
+
+//Example 2.3
+//Caption : Program to find the energy change in a System
+
+//Given values
+
+P_atm=101.3;//Atm Pressure=101.3KPa
+V1=0.1;//Volume1=0.1m^3
+V2=0.2;//Volume2=0.2m^3
+
+//Solution
+
+del_V=V2-V1;
+W_by=P_atm*del_V;
+W_on=-W_by;
+Q=0;
+del_Energy=Q+W_on;//KJ
+disp('KJ',del_Energy,'Energy Change')
+
+//End
diff --git a/23/CH2/EX2.4/Example_2_4.pdf b/23/CH2/EX2.4/Example_2_4.pdf
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+clear;
+clc;
+
+//Example 2.4
+//Caption : Program to find the Heat flow in the Path
+
+//Given values
+W_acb=40;//J
+Q_acb=100;//J
+W_aeb=20;//J
+W_bda=30;//J
+
+//Solution
+
+del_U_ab=Q_acb-W_acb;
+
+//(a)
+Q_aeb=del_U_ab-W_aeb;//J
+disp('J',Q_aeb,'(a)Heat Flow in acb')
+
+//(b)
+del_U_ba=-del_U_ab;//J
+Q_bda=del_U_ba-W_bda;
+disp('J',Q_bda,'(b)Heat Flow in bda')
+
+//End \ No newline at end of file
diff --git a/23/CH2/EX2.5/Example_2_5.pdf b/23/CH2/EX2.5/Example_2_5.pdf
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+clear;
+clc;
+
+//Example 2.5
+//Caption : Program To Find The degree of freedom for the various systems
+
+//Formula To be Used F=2-#+N (Where,#(pi)-no of phases,N-number of chemical species)
+
+//(a)-Liquid Water in equllibrium with its vapour.
+N=1;
+pi=2;
+F=2-pi+N;
+disp(F,'(a)Degree Of freedom is');
+
+//(b)-Liquid Water in equllibrium with a mixture of vapour and nitrogen.
+N=2;
+pi=2;
+F=2-pi+N;
+disp(F,'(b)Degree Of freedom is');
+
+//(c)-A liquid Soln of alcohol in water in equillibrium with its vapour
+N=2;
+pi=2;
+F=2-pi+N;
+disp(F,'(c)Degree Of freedom is');
+
+//End \ No newline at end of file
diff --git a/23/CH2/EX2.6/Example_2_6.pdf b/23/CH2/EX2.6/Example_2_6.pdf
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+clear;
+clc;
+
+//To find Approx Value
+function[A]=approx(V,n)
+ A=round(V*10^n)/10^n;//V-Value n-To what place
+ funcprot(0)
+endfunction
+
+//Example 2.6
+//Caption : Program to find the work done by gas
+
+//Given values
+P=14;//Pressure=14bar
+V1=0.03;//Initial volume=0.03m^3
+V2=0.06;//Final Volume
+//Process is isothermal
+//(a)-To find the work done by gas in moving the External force
+//(b)-To find the work done by gas if external force is suddenly reduced to half its initial value
+
+//Solution
+//(a)
+K=P*V1*(10^5);//J
+W1=approx(-K*integrate('1/V','V',0.03,0.06),0);//J
+P2=K/V2;//Final Pressure(Pa)
+P2=P2/(10^5);//bar
+disp('J',W1,'(a)The work done by gas in moving the External Force is')
+
+//(b)
+W2=-P2*(10^5)*integrate('1','V',0.03,0.06)
+n=approx((W2/W1)*100,1);//Efficiency
+disp('J',W2,'(b)The work done by gas if external force is reduced to half is')
+disp('%',n,'Hence the efficiency is')
+
+//End \ No newline at end of file
diff --git a/23/CH2/EX2.7/Example_2_7.pdf b/23/CH2/EX2.7/Example_2_7.pdf
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+clear;
+clc;
+
+//Example 2.7
+//Caption : Program to find the Enegy Changes in the Process
+
+//Given values
+P=7;//pressure=7bar
+m=45;//Mass of cube
+mt=23;//mass of piston,piston rod,pan
+x=0.5;//Distance moved=0.5m
+g=9.8;//Acceleration Due to gravity(m/s^2)
+
+//Solution
+
+//Acc to Eqn del_U_sys+del_U_surr+del_PE_surr=0
+del_PE_surr=(m+mt)*g*x;
+//ans=del_U_sys+del_U_surr
+disp('J',-del_PE_surr,'Energy Changes in the Process')
+
+//End \ No newline at end of file
diff --git a/23/CH2/EX2.8/Example_2_8.pdf b/23/CH2/EX2.8/Example_2_8.pdf
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+clear;
+clc;
+
+//To find Approx Value
+function[A]=approx(V,n)
+ A=round(V*10^n)/10^n;//V-Value n-To what place
+ funcprot(0)
+endfunction
+
+//Example 2.8
+//Caption : Program to find Change in Enthalpy and Internal Energy
+
+//Given values
+m=1;//1kg of water
+T=373.15;//Temp=373.15K(100`C)
+P=101.325;//Pressure=101.325KPa
+V2=1.673;//Final Volume[m^3]
+V1=0.00104;//Initial Volume[m^3]
+Sv_liqiud=0.00104;//Specific Volume of Liqiud
+Sv_vapour=1.673;//Specific Volume of Vapour
+del_H=2256.9;//Heat Added(KJ)
+
+//Solution
+Q=del_H;
+del_V=V2-V1;
+W=P*del_V;//KJ
+del_U=approx(del_H-(P*del_V),1);
+disp('KJ',del_H,'Change in Enthalpy');
+disp('KJ',del_U,'Change in Internal energy');
+//End \ No newline at end of file
diff --git a/23/CH2/EX2.9/Example_2_9.pdf b/23/CH2/EX2.9/Example_2_9.pdf
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+clear;
+clc;
+
+//Example 2.9
+//Caption : Program To Find Work,Heat,del U and del H
+
+//Given values
+//Initial
+P1=1;//Pressure=1bar
+T1=298.15;//Temp=298.15K(25`C)
+V1=0.02479;//Molar Volume=0.02479m^3/mol
+//Final
+P2=5;//Pressure=5bar
+Cv=20.78;//J/mol/K
+Cp=29.10;//J/mol/K
+
+//to Find del_U,del_H by two processes
+V2=V1*(P1/P2);//m^3(1 mol)
+disp('m^3',V2,'Final Volume')
+
+//Solution
+
+//(a)-Cooling at const pressure followed by heating at const Volume
+T2=T1*(V2/V1);//K
+disp('K',T2,'Final Temperature')
+del_H=round(Cp*(T2-T1));//J
+Q1=del_H;//J
+del_U1=round(del_H-(P1*(10^5)*(V2-V1)));//J
+//Second Step
+del_U2=round(Cv*(T1-T2));//J
+Q2=del_U2;
+Q=Q1+Q2;
+del_U=0;
+W=del_U-Q;//J
+del_H=0;//const Temperature
+
+disp('(a) Cooling at const Pressure Followed by Heating at const Volume')
+disp('J',Q,'Heat Required')
+disp('J',W,'Work Required')
+disp('J',del_H,'Change in enthalpy')
+disp('J',del_U,'Change in Energy')
+//(b)-heating at Const Volume Followed by cooling at const Pressure
+T2=T1*(P2/P1);//K
+del_U1=round(Cv*(T2-T1));//J
+Q1=del_U1;
+del_H=round(Cp*(T1-T2));//J
+Q2=del_H;
+del_U2=round(del_H-(P2*(10^5)*(V2-V1)));//J
+Q=Q1+Q2;
+del_U=0;
+W=del_U-Q;//J
+del_H=0;//const Temperature
+disp('(b) Heating at const Volume Followed by Cooling at const Pressure')
+disp('J',Q,'Heat Required')
+disp('J',W,'Work Required')
+disp('J',del_H,'Change in enthalpy')
+disp('J',del_U,'Change in Energy')
+
+
+//Note
+disp('Note : The Answer varies From That in the book because in Book 4956.44 has been rounded to 4958 which is absurd')
+//End \ No newline at end of file