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diff --git a/479/CH9/EX9.1/Example_9_1.sce b/479/CH9/EX9.1/Example_9_1.sce
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+//Chemical Engineering Thermodynamics

+//Chapter 9

+//Fluid Flow in Pipes and Nozzles

+//Example 9.1

+clear;

+clc;

+

+//Given

+R = 848;//gas constant in m Kgf/Kgmole K

+M = 29;//molecular weight of air

+g = 9.81;

+T1 = 90+273;//initial temperature in K

+y = 1.4;//gamma = Cp/Cv

+W = 800/3600;//Mass rate of air in Kg/sec

+P1 = 3.5;//initial pressure in atm

+d = 2.5;//diameter of the pipe in cm

+

+//To find out the pressure at the final point

+v1 = (R*T1)/(M*P1*1.033*10^4);//specific volume in cubic meter/Kg

+u1 = (W*v1)/(%pi*(d^2*(10^-4))/4);//inital velocity in m/sec

+//Assume final temperature as

+T2 = [300 310];

+//Assume specific heat capacity in J/KgK corresponding to the above temperature as

+Cp = [2987.56 2983.56];

+for i = 1:2

+    us(i) = (g*y*R*T2(i)/M)^(1/2);//sonic velocity attained in m/sec

+    u2(i) = ((u1^2)-((2*g*Cp(i)/M)*(T2(i)-T1)))^(1/2);//From equation 9.18 & 9.19 (page no 170)

+end

+if us(i)-u2(i) <= 1 

+    u2 = u2(i);

+    T2 = T2(i);

+else

+end

+v2 = u2*(%pi/4)*(d^2/10^4)*(1/W);

+P2 = (P1*v1*T2)/(T1*v2);

+mprintf('The pressure at the final point is %f atm',P2);

+//end
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diff --git a/479/CH9/EX9.2/Exampe_9_2.sce b/479/CH9/EX9.2/Exampe_9_2.sce
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+//Chemical Engineering Thermodynamics

+//Chapter 9

+//Fluid Flow in Pipes and Nozzles

+//Example 9.2

+clear;

+clc;

+

+//Given

+A1 = 0.002;//inlet area in sq meter

+A2 = 0.0005;//throat area in sq meter

+P1 = 1.3*10^4;//inlet pressure in Kgf/sq m

+P2 = 0.7*10^4;//throat pressure in Kgf/sq m

+g = 9.81;

+v = 12*10^-4;//specific volume in cubic m /Kg

+

+//To find out the mass rate of alcohol

+u2 = ((v*2*g*(P1-P2))/(1-((A2/A1)^2)))^(1/2);//throat velocity in m/sec

+W = (u2*A2)/v;

+mprintf('The mass rate of alcohol is %f Kg/sec',W);
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diff --git a/479/CH9/EX9.3/Example_9_3.sce b/479/CH9/EX9.3/Example_9_3.sce
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+//Chemical Engineering Thermodynamics

+//Chapter 9

+//Fluid Flow in Pipes and Nozzles

+//Example 9.3

+clear;

+clc;

+

+//Given

+P1 = 50;//initial pressure in Kgf/sq m

+T1 = 45+273;//initial temperature in K

+g = 9.81;

+y = 1.35;//gamma

+R = 848;//gas constant in m Kgf/Kgmole K

+M = 29;//molecular weight of air

+d = 1;//pipe diameter in cm

+

+//(i)To plot velocity,specific volume,mass velocity against P2/P1

+//(ii)To calculate the critical pressure,critical mass velocity and mass rate of flow

+//(i)Plotting of graph

+V1 = (R*T1)/(M*P1*1.033*10^4);//initial volume of the gas in cubic m/Kg

+//P3 = P2/P1 (say)

+//Assume P3 values as

+P3 = [1.0 0.8 0.6 0.4 0.2 0.1 0];

+G = [0 0 0 0 0 0 0];

+for i = 1:7

+    u2(i) = (((2*g*y*R*T1)/((y-1)*M))*(1-(P3(i)^((y-1)/y))))^(1/2);//final velocity in m/sec

+end

+for i = 1:6

+     v2(i) = V1/(P3(i)^(1/y));//final specific volume in cubic meter/Kg

+end

+for i = 1:6

+    G(i) = u2(i)/v2(i);//Mass velocity in Kg/sq m sec

+end

+

+clf;

+xset('window',4);

+plot(P3,u2,"o-");

+xtitle("Velocity vs P2/P1","P2/P1","Velocity");

+xset('window',5);

+plot(P3,G,"+-");

+xtitle("Mass velocity vs P2/P1","P2/P1","Mass velocity");

+xset('window',6);

+P_3 = [1.0 0.8 0.6 0.4 0.2 0.1];

+plot(P_3,v2,"*-");

+xtitle("Sp. volume vs P2/P1","P2/P1","Specific volume");

+

+//(ii)Calculation of critical pressure,critical mass velocity and mass rate of flow

+//From equation 9.37(page no 181)

+P2 = P1*(2/(y+1))^(y/(y-1));

+mprintf('The critical pressure is %f atm',P2);

+//From equation a (page no 183)

+u2 = (((2*g*y*R*T1)/((y-1)*M))*(1-((P2/P1)^((y-1)/y))))^(1/2);

+mprintf('\n The critical velocity is %f m/sec',u2);

+//From equation b (page no 183)

+v2 = ((R*T1)/(M*P1*1.033*10^4))/((P2/P1)^(1/y));

+mprintf('\n The critical specific volume is %f cubic meter/Kg',v2);

+//From relation c (page no 183)

+G = u2/v2;

+mprintf('\n The critical mass velocity is %f Kg/sq meter sec',G);

+W = G*(%pi/4)*(d/(100))^2;

+mprintf('\n Mass rate of flow through nozzle is %f Kg/sec',W);

+//end
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diff --git a/479/CH9/EX9.4/Example_9_4.sce b/479/CH9/EX9.4/Example_9_4.sce
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+//Chemical Engineering Thermodynamics

+//Chapter 9

+//Fluid Flow in Pipes and Nozzles

+//Example 9.4

+clear;

+clc;

+

+//Given

+A1 = 0.1;//Inlet area in sq meter

+u1 = 60;//inlet velocity in m/sec

+v1 = 0.185;//inlet specific volume in cubic meter/Kg

+H1 = 715;//inlet enthalpy in Kcal/Kg

+H2 = 660;//exit enthalpy in Kcal/Kg

+v2 = 0.495;//exit specific volume in cubic meter/Kg

+g = 9.81

+

+//To calculate the area at exit of nozzle and hence decide the type of the nozzle

+//From the first law

+u2 = ((u1^2)-(2*g*(H2-H1)*427))^(1/2);

+W = (u1*A1)/v1;//Mass rate of gas in Kg/sec

+A2 = (W*v2)/u2;//Area at exit of nozzle

+if(A2 < A1)

+    mprintf('The nozzle is convergent');

+else

+    mprintf('The nozzle is divergent');

+end

+//end
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