diff options
Diffstat (limited to '3751')
179 files changed, 6208 insertions, 0 deletions
diff --git a/3751/CH11/EX11.1/Ex11_1.sce b/3751/CH11/EX11.1/Ex11_1.sce new file mode 100644 index 000000000..c899a7b12 --- /dev/null +++ b/3751/CH11/EX11.1/Ex11_1.sce @@ -0,0 +1,37 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.1 +//To Find the Work done by the Impeller on the water per unit weight of water. + + clc + clear + +//Given Data:- + Di=210; //Internal diameter of Impeller, mm + Do=420; // External diameter of Impeller, mm + N=1100; //speed, rpm + beta_i=20; //Vane Angle at Inlet, degrees + beta_o=30; //Vane Angle at Outlet, degrees + //As water enters the impeller radially, + alpha_i=90; //degrees + + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + + +//Computations:- + Di=Di/1000; //m + Do=Do/1000; //m + + ui=%pi*Di*N/60; //m/s + uo=%pi*Do*N/60; //m/s + Vfi=ui*tand(beta_i); //m/s + Vfo=Vfi; + Vwo=uo-Vfo/tand(beta_o); //m/s + Work=Vwo*uo/g; //N-m/N + +//Result:- + printf(" The Work done by the Impeller on the water per unit weight of water =%.2f N-m/N \n",Work) //The answer vary due to round off error + + diff --git a/3751/CH11/EX11.10/Ex11_10.sce b/3751/CH11/EX11.10/Ex11_10.sce new file mode 100644 index 000000000..435caed65 --- /dev/null +++ b/3751/CH11/EX11.10/Ex11_10.sce @@ -0,0 +1,58 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.10 +//To Find (i)Manometric Head (ii)Manometric Efficiency (iii)Overall Efficiency of the Pump. + + clc + clear + +//Given Data:- + Do=480; //External Diameter of the Impeller, mm + Di=240; //Internal Diameter of the Impeller, mm + N=1000; //Speed, rpm + Q=0.0576; //Rate of Flow, m^3/s + Vfo=2.4; //Velocity of Flow, m/s + Vfi=Vfo; + Ds=180; //Diameter of Suction Pipe, mm + Dd=120; //Diameter of Delivery Pipe, mm + h_s=6.2; //Suction Head, m of water (abs) + h_d=30.2; //Delivery Head, m of water (abs) + P=35; //Power required to drive the pump, kW + beta_o=45; //Vane Angle at outlet, degrees + + +//Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Do=Do/1000; //m + Di=Di/1000; //m + Ds=Ds/1000; //m + Dd=Dd/1000; //m + P=P*1000; //W + + //(i)Manometric Head, Hm + + As=(%pi/4)*Ds^2; //m^2 + Ad=(%pi/4)*Dd^2; //m^2 + Vd=Q/Ad; //m/s + Vs=Q/As; //m/s + Hm=(h_d+Vd^2/(2*g))-(h_s+Vs^2/(2*g)); //m + + + uo=%pi*Do*N/60; // Tangential velocity of Impeller at Outlet, m/s + Vwo=uo-Vfo/tand(beta_o); //m/s + + //(ii) Manometric Efficiency, eta_man + eta_man=g*Hm/(Vwo*uo)*100; //In Percentage + + //(iii) The Overall Efficiency of the Pump, eta_o + eta_o=rho*Q*g*Hm/P*100; //In percentage + +//Results:- + printf(" (i)Manometric Head, Hm =%.2f m \n ",Hm) + printf(" (ii) Manometric Efficiency, eta_man =%.2f Percent \n ",eta_man) //The answer vary due to round off error + printf(" (iii) The Overall Efficiency of the Pump, eta_o =%.2f Percent \n ",eta_o) //The answer vary due to round off error + + diff --git a/3751/CH11/EX11.11/Ex11_11.sce b/3751/CH11/EX11.11/Ex11_11.sce new file mode 100644 index 000000000..508265377 --- /dev/null +++ b/3751/CH11/EX11.11/Ex11_11.sce @@ -0,0 +1,34 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.11 +//To Find Vane Angle at Outer periphery of Impeller. + + clc + clear + +//Given Data:- + Q=0.118; //discharge, m^3/s + N=1450; //Speed, rpm + Hm=25; //Manometric Head, m + Do=250; //Diameter of the Impeller at Outlet, mm + bo=50; //Width at Outlet, mm + eta_man=75/100; //Manometric Efficiency + + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Do=Do/1000; //m + bo=bo/1000; //m + + uo=%pi*Do*N/60; // Tangential velocity of Impeller at Outlet, m/s + Vwo=g*Hm/(uo*eta_man); //m/s + Vfo=Q/(%pi*Do*bo); //m/s + beta_o=atand(Vfo/(uo-Vwo)); //degrees + +//Results:- + printf(" Vane Angle at Outlet, beta_o=%.2f Degrees \n ",beta_o) //The answer vary due to round off error + + + diff --git a/3751/CH11/EX11.12/Ex11_12.sce b/3751/CH11/EX11.12/Ex11_12.sce new file mode 100644 index 000000000..0f946d77d --- /dev/null +++ b/3751/CH11/EX11.12/Ex11_12.sce @@ -0,0 +1,48 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.12 +//To Determine (i)Manometric Efficiency (ii)Vane Angle at Inlet (iii)The Least Speed at which the pump commence to work. + + clc + clear + +//Given Data:- + Do=0.5; //Outer Diameter of the Impeller, m + N=600; //Speed, rpm + Q=8000; //Discharge, litres/min. + Hm=8.5; //Manometric Head, m + Di=0.25; //Inner Diameter of Impeller, m + beta_o=45; //Vane Angle at outlet, degrees + Af=0.06; //Area of Flow, m^2 + + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Q=Q/60000; //m^3/s + + Vfo=Q/Af; //m/s + Vfi=Vfo; + ui=%pi*Di*N/60; //Tangential velocity of Impeller at Inlet,m/s + uo=%pi*Do*N/60; // Tangential velocity of Impeller at Outlet, m/s + Vwo=uo-Vfo/tand(beta_o); //m/s + + //(i) Manometric Efficiency, eta_man + eta_man=g*Hm/(Vwo*uo)*100; //In Percentage + + + //(ii) Vane Angle at Inlet, beta_i + beta_i=atand(Vfi/ui); //degrees + + //(iii) The Least Speed at which the pump commence to work, Nmin + Nmin=120*Vwo*Do*(eta_man/100)/(%pi*(Do^2-Di^2)); //rpm + + + +//Results:- + printf(" (i) Manometric Efficiency, eta_man =%.2f Percent \n ",eta_man ) //The answer vary due to round off error + printf(" (ii) Vane Angle at Inlet, beta_i=%.2f Degrees \n ",beta_i ) //The answer vary due to round off error + printf(" (iii) The Least Speed at which the pump commence to work, Nmin=%.2f rpm \n",Nmin ) //The answer vary due to round off error + + diff --git a/3751/CH11/EX11.13/Ex11_13.sce b/3751/CH11/EX11.13/Ex11_13.sce new file mode 100644 index 000000000..0dde95b26 --- /dev/null +++ b/3751/CH11/EX11.13/Ex11_13.sce @@ -0,0 +1,56 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.13 +//To Find (i)The Discharge of the Pump (ii)The Pressure at Suction and Delivery side of the Pump. + + clc + clear + +//Given Data:- + h_st=35; //Static Head, m + h_s=4; //Suction Head, m + D=150; //Diameter of Pipes, mm + Ds=D; //Diameter of Suction Pipe, mm + Dd=D; //Diameter of Delivery Pipe, mm + h_fs=1.6; //Head loss in Suction pipe, m + h_fd=6.5; //Head loss in Delivery Pipe, m + Do=380; //Diameter of Impeller at Outlet, mm + bo=25; //Width of Impeller at Outlet, mm + N=1200; //Speed, rpm + beta_o=35; //Ezxit Blade Angle, degrees + eta_man=80/100; //Manometric Efficiency + + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Do=Do/1000; //m + D=D/1000; //m + Ds=Ds/1000; //m + Dd=Dd/1000; //m + bo=bo/1000; //m + + Hm=h_st+h_fs+h_fd; // Manometric Head, m + uo=%pi*Do*N/60; // Tangential velocity of Impeller at Outlet, m/s + Vwo=g*Hm/(uo*eta_man); //m/s + Vfo=(uo-Vwo)*tand(beta_o); //m/s + + //(i)The Discharge of the Pump, Q + Q=%pi*Do*bo*Vfo*1000; //litres/s + + // (ii)The Pressure at Suction and Delivery side of the Pump + + A=(%pi/4)*D^2; //m^2 + Vd=Q*10^-3/A; //m/s + Vs=Vd; //m/s + Hs=h_s+h_fs+Vs^2/(2*g); //Pressure on Suction Side, m of water + h_d=h_st-h_s; //m + Hd=h_d+h_fd+Vd^2/(2*g); //Pressure on Delivery Side, m of water + + +//Result:- + printf(" (i)The Discharge of the Pump, Q =%.2f litres/s\n",Q) //The answer vary due to round off error + printf(" (ii) Pressure on Suction Side, Hs =-%.3f m of water \n",Hs) //The answer vary due to round off error + printf(" Pressure on Delivery Side, Hd =%.2f m of water \n",Hd) //The answer vary due to round off error + diff --git a/3751/CH11/EX11.14/Ex11_14.sce b/3751/CH11/EX11.14/Ex11_14.sce new file mode 100644 index 000000000..bc0da4a1d --- /dev/null +++ b/3751/CH11/EX11.14/Ex11_14.sce @@ -0,0 +1,59 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.14 +//To Find (a)Vane Angle of Impeller at Inlet (b) Overall Efficiency of the Pump (c) Manometric Efficiency of the Pump. + + clc + clear + +//Given Data:- + Do=400; //Diameter of the Impeller at Outlet, mm + Di=200; //Diameter of the Impeller at Inlet, mm + N=1000; //Speed, rpm + Q=39; //Discharge, litres/s + Vfo=2.2; //Velocity of Flow, m/s + Vfi=Vfo; + Ds=150; //Diameter of Suction Pipe, mm + Dd=100; //Diameter of Delivery Pipe, mm + h_s=6; //Suction Head, m of water (abs) + h_d=30; //Delivery Head, m of water (abs) + P=15.75; //Power required to drive the pump, kW + beta_o=45; //Vane Angle at outlet, degrees + + +//Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Do=Do/1000; //m + Di=Di/1000; //m + Ds=Ds/1000; //m + Dd=Dd/1000; //m + Q=Q/1000; //m^3/s + P=P*1000; //W + + //(a)Vane Angle of Impeller at Inlet, beta_i + ui=%pi*Di*N/60; //m/s + beta_i=atand(Vfi/ui); //degrees + + // (b) Overall Efficiency of the Pump + As=(%pi/4)*Ds^2; //m^2 + Ad=(%pi/4)*Dd^2; //m^2 + Vd=Q/Ad; //m/s + Vs=Q/As; //m/s + Hm=(h_d+Vd^2/(2*g))-(h_s+Vs^2/(2*g)); //m + eta_o=rho*Q*g*Hm/P*100; //In percentage + + + // (c) Manometric Efficiency of the Pump, eta_man + uo=%pi*Do*N/60; // Tangential velocity of Impeller at Outlet, m/s + Vwo=uo-Vfo/tand(beta_o); //m/s + eta_man=g*Hm/(Vwo*uo)*100; //In Percentage + + +//Results:- + printf(" (a)Vane Angle of Impeller at Inlet, beta_i=%.2f Degrees \n ",beta_i) //The answer vary due to round off error + printf(" (b) The Overall Efficiency of the Pump, eta_o =%.2f Percent \n ",eta_o) //The answer vary due to round off error + printf(" (c) Manometric Efficiency of the Pump, eta_man =%.2f Percent \n ",eta_man) //The answer vary due to round off error + diff --git a/3751/CH11/EX11.15/Ex11_15.sce b/3751/CH11/EX11.15/Ex11_15.sce new file mode 100644 index 000000000..76183f367 --- /dev/null +++ b/3751/CH11/EX11.15/Ex11_15.sce @@ -0,0 +1,31 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.15 +//To Determine Minimum Starting Speed of the Pump. + + clc + clear + +//Given Data:- + Di=300; //Diameter of Impeller at Inlet, mm + Do=600; //Diameter of the Impeller at Outlet, mm + Vfo=2.6; //Velocity of Flow at Outlet, m/s + beta_o=42; //Vane Angle at outlet, degrees + eta_man=65/100; //Manomwtric Efficiency, m^2 + + +//Computations:- + Di=Di/1000; //m + Do=Do/1000; //m + + uo_by_N=%pi*Do/60; // uo/N + + //Minimum Starting Speed of The Centrifugal Pump, Nmin + Nmin=(120*Do*eta_man*Vfo/(tand(beta_o)*%pi*(Do^2-Di^2)))/(120*eta_man*Do*uo_by_N/(%pi*(Do^2-Di^2))-1); //rpm + + + +//Results:- + printf("The Minimum Starting Speed of the Centrifugal Pump, Nmin =%.2f rpm \n",Nmin ) //The answer vary due to round off error + + diff --git a/3751/CH11/EX11.16/Ex11_16.sce b/3751/CH11/EX11.16/Ex11_16.sce new file mode 100644 index 000000000..e7435d7a1 --- /dev/null +++ b/3751/CH11/EX11.16/Ex11_16.sce @@ -0,0 +1,33 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.16 +//To Determine Minimum Starting Speed of the Pump. + + clc + clear + +//Given Data:- + Di=200; //Diameter of Impeller at Inlet, mm + Do=400; //Diameter of the Impeller at Outlet, mm + Hm=25; //Manometric Head, m + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + + +//Computations:- + Di=Di/1000; //m + Do=Do/1000; //m + + uo_by_Nmin=%pi*Do/60; // uo/Nmin + ui_by_Nmin=%pi*Di/60; // ui/Nmin + + //Minimum Starting Speed of The Centrifugal Pump, Nmin + Nmin=sqrt(2*g*Hm/(uo_by_Nmin^2-ui_by_Nmin^2)); //rpm + + + +//Results:- + printf("The Minimum Starting Speed of the Centrifugal Pump, Nmin =%.f rpm \n",Nmin ) //The answer vary due to round off error + + diff --git a/3751/CH11/EX11.17/Ex11_17.sce b/3751/CH11/EX11.17/Ex11_17.sce new file mode 100644 index 000000000..7380fdca4 --- /dev/null +++ b/3751/CH11/EX11.17/Ex11_17.sce @@ -0,0 +1,40 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.17 +//To Find (a)Manometric Efficiency. (b)Minimum Starting Speed + + clc + clear + +//Given Data:- + Di=1200; //Inner Diameter of the Impeller, mm + Do=600; //Outer Diameter of the Impeller, mm + N=200; //Speed, rpm + Hm=6; //Manometric Head, m + beta_o=26; // Vane Angle at Outlet, degrees + Vfo=2.5; // Velocity of flow at Outlet, m/s + + +//Data Used: - + g=9.81; //Acceleration due to gravity, m/s^2 + + +//Computations:- + Di=Di/1000; //m + Do=Do/1000; //m + + uo=%pi*Di*N/60; //Tangential Velocity of Impeller at Outlet, m/s + Vwo=uo-Vfo/tand(beta_o); //m/s + + //(a)Manometric Efficiency, eta_man + eta_man=g*Hm/(Vwo*uo)*100; //In Percentage + + //(b) Minimum Starting Speed, Nmin + Nmin =sqrt(2*g*Hm*60^2/(%pi^2*(Di^2-Do^2))); //rpm + + +//Results:- + printf("(a)Manometric Efficiency =%.2f Percent \n",eta_man) //The answer vary due to round off error + printf(" (b)Minimum Starting Speed, Nmin =%.f rpm",Nmin) //The answer vary due to round off error + + diff --git a/3751/CH11/EX11.18/Ex11_18.sce b/3751/CH11/EX11.18/Ex11_18.sce new file mode 100644 index 000000000..f4fdce99e --- /dev/null +++ b/3751/CH11/EX11.18/Ex11_18.sce @@ -0,0 +1,51 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.18 +//To Find (a)Manometric Efficiency. (b)Inlet Vane Angles. (c)Loss of Head at Inlet of Impeller + + clc + clear + +//Given Data:- + Q=0.21; //Discharge, m^3/s + Af=0.085; //Cross-sectional Area of Flow, m^2 + Di=300; //Inner Diameter of the Impeller, mm + Do=600; //Outer Diameter of the Impeller, mm + N=600; //Speed, rpm + Hm=19; //Manometric Head, m + beta_o=35; //degrees + Q_per=30; //Percentage by which Discharge is reduced + + +//Data Used: - + g=9.81; //Acceleration due to gravity, m/s^2 + + +//Computations:- + Di=Di/1000; //m + Do=Do/1000; //m + + ui=%pi*Di*N/60; //Tangential Velocity of Impeller at Inlet, m/s + uo=%pi*Do*N/60; //Tangential Velocity of Impeller at Outlet, m/s + Vfo=Q/Af; //Velocity of Flow, m/s + Vfi= Vfo; + Vwo=uo-Vfo/tand(beta_o); //m/s + + //(a)Manometric Efficiency, eta_man + eta_man=g*Hm/(Vwo*uo)*100; //In Percentage + + //(b)Inlet Vane Angle, beta_i + beta_i=atand(Vfi/ui); //degrees + + //(c)Loss of Head at inlet, H_L + Q_dash=Q-Q_per/100*Q; //m^3/s + Vfi_dash=Q_dash/Af; //m/s + H_L=(ui-Vfi_dash*cotd(beta_i) )^2/(2*g); // m of water + +//Results + printf("(a)Manometric Efficiency, eta_man =%.2f Percent \n",eta_man) //The answer vary due to round off error + printf ("(b)Inlet Vane Angle, beta_i =%.2f Degrees \n",beta_i) //The answer vary due to round off error + printf ("(c)Loss of Head at Inlet to the Impeller =%.3f m of water", H_L) //The answer vary due to round off error + + + diff --git a/3751/CH11/EX11.19/Ex11_19.sce b/3751/CH11/EX11.19/Ex11_19.sce new file mode 100644 index 000000000..90e4e142b --- /dev/null +++ b/3751/CH11/EX11.19/Ex11_19.sce @@ -0,0 +1,49 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.19 +//To Find (a)Head generated and (b)Power consumed + + clc + clear + +//Given Data:- + n=2; //Number of Stages + Q=100; //Discharge, litres/s + N=1000; //Speed, rpm + Do=500; //Diameter of the Impeller at Outlet, mm + bo=25; //Width of Impeller at outlet, mm + beta_o=30; //degrees + Area_per=10; //Percentage of Total Area which is covered due to blade thickness + eta_o=78/100; //Overall Efficiency + eta_man=85/100; //Manometric Efficiency + + +//Data Used: - + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + + +//Computations:- + Q=Q/1000; //m^3/s + Do=Do/1000; //m + bo=bo/1000; //m + A=%pi*Do*bo*(1-Area_per/100); //Actual Area of Flow, m^2 + + uo=%pi*Do*N/60; //Tangential Velocity of Impeller at Outlet, m/s + Vfo=Q/A; //Velocity of Flow, m/s + Vfi= Vfo; + Vwo=uo-Vfo/tand(beta_o); //m/s + + //(a)Head generated, H_Tm + Hm=eta_man*Vwo*uo/g; //m + H_Tm=n*Hm; //m + + //(b) Power consumed, P + P=rho*Q*g*H_Tm/(eta_o*1000); //kW + + +//Results:- + printf("(a)Head Generated, H_Tm=%.2f m \n",H_Tm) //The answer vary due to round off error + printf(" (b)Power consumed, P =%.2f kW \n",P) //The answer vary due to round off error + + diff --git a/3751/CH11/EX11.2/Ex11_2.sce b/3751/CH11/EX11.2/Ex11_2.sce new file mode 100644 index 000000000..570404a98 --- /dev/null +++ b/3751/CH11/EX11.2/Ex11_2.sce @@ -0,0 +1,36 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.2 +//To Find the Vane Angle at Outer Periphery of the Impeller. + + clc + clear + +//Given Data:- + N=1470; //Speed, rpm + Q=100; //Discharge, litres/s + Hm=24; //manometric Head, m + Do=240; // Diameter of Impeller at Outlet, mm + b_o=50; //Width of Impeller at Outlet, mm + eta_man=76/100; //Manometric EEfficiency + + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + + +//Computations:- + Q=Q/1000; //m^3/s + Do=Do/1000; //m + b_o=b_o/1000; //m + + uo=%pi*Do*N/60; //m/s + Vwo=g*Hm/(uo*eta_man); //m/s + Vfo=Q/(%pi*Do*b_o); //m/s + //From Outlet Velocity Triangle (OVT), + beta_o=atand(Vfo/(uo-Vwo)); //degrees + +//Result:- + printf("The Vane Angle at Outer Periphery of Impeller, beta_o=%.2f Degrees \n",beta_o) //The answer vary due to round off error + + diff --git a/3751/CH11/EX11.20/Ex11_20.sce b/3751/CH11/EX11.20/Ex11_20.sce new file mode 100644 index 000000000..98e25858e --- /dev/null +++ b/3751/CH11/EX11.20/Ex11_20.sce @@ -0,0 +1,48 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.20 +//To Determine (i)Head generated by the Pump (ii)Shaft Power required to run the Pump. + + clc + clear + +//Given Data:- + n=3; //Number of Stages + Do=400; //Diameter of the Impeller at Outlet, mm + bo=20; //Width of Impeller at outlet, mm + beta_o=45; //degrees + Area_per=10; //Percentage of Total Area which is reduced. + eta_o=80/100; //Overall Efficiency + eta_man=90/100; //Manometric Efficiency + N=1000; //Speed, rpm + Q=0.05; //Discharge, m^3/s + + +//Data Used: - + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + + +//Computations:- + Do=Do/1000; //m + bo=bo/1000; //m + A=%pi*Do*bo*(1-Area_per/100); //Actual Area of Flow, m^2 + + uo=%pi*Do*N/60; //Tangential Velocity of Impeller at Outlet, m/s + Vfo=Q/A; //Velocity of Flow, m/s + Vfi= Vfo; + Vwo=uo-Vfo/tand(beta_o); //m/s (Value given in book is wrong due to incorrect value of beta_o is used) + + // (i)Head generated by the Pump , H_Tm + Hm=eta_man*Vwo*uo/g; //m + H_Tm=n*Hm; //m + + //(ii) Shaft Power required to run the Pump , P + P=rho*Q*g*H_Tm/(eta_o*1000); //kW + + +//Results:- + printf(" (i)Head generated by the Pump , H_Tm=%.2f m \n",H_Tm) //The answer provided in the textbook is wrong + printf(" (ii) Shaft Power required to run the Pump , P =%.2f kW \n",P) //The answer provided in the textbook is wrong + + diff --git a/3751/CH11/EX11.21/Ex11_21.sce b/3751/CH11/EX11.21/Ex11_21.sce new file mode 100644 index 000000000..4421e1019 --- /dev/null +++ b/3751/CH11/EX11.21/Ex11_21.sce @@ -0,0 +1,40 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.21 +//To Find the Manometric Efficiency + clc + clear + +//Given Data:- + n=4; //Number of Pumps + N=400; //Speed, rpm + H_Tm=40; //Total Manometric Head, m + Q=0.2; //Discharge, m^3/s + beta_o=40; //Vane Angle at Outlet, degrees + Do=600; //Diameter of the Impeller at Outlet, mm + bo=50; //Width of Impeller at outlet, mm + + +//Data Used: - + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + + +//Computations:- + Do=Do/1000; //m + bo=bo/1000; //m + A=%pi*Do*bo; //Area of Flow, m^2 + Hm=H_Tm/n; //Manometric Head of each Pump, m + + uo=%pi*Do*N/60; //Tangential Velocity of Impeller at Outlet, m/s + Vfo=Q/A; //Velocity of Flow, m/s + Vfi= Vfo; + Vwo=uo-Vfo/tand(beta_o); //m/s + + eta_man=g*Hm/(Vwo*uo)*100; //Manometric Efficiency in Percentage + + +//Results:- + printf("Manometric Efficiency, eta_man=%.2f Percent \n",eta_man) //The answer vary due to round off error + + diff --git a/3751/CH11/EX11.22/Ex11_22.sce b/3751/CH11/EX11.22/Ex11_22.sce new file mode 100644 index 000000000..9137c0ef3 --- /dev/null +++ b/3751/CH11/EX11.22/Ex11_22.sce @@ -0,0 +1,35 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11 - Centrifugal Pumps +//Example 11.22 + + clc + clear + +//Given Data:- + //For Model, + H_mM=7.5; //Manometric Head, m + Nm=1000; //Speed, rpm + Pm=25; //Shaft Power, kW + + //For Prototype, + H_mP=23; //Manometric Head, m + + Dm_by_Dp=1/6; //Scale Ratio + + +//Computations:- + + // (a)Working Speed of the Prototype, Np + Np=Nm*Dm_by_Dp*sqrt(H_mP/H_mM); //rpm + + // (b)Shaft Power of the Prototype, Pp + Pp=Pm*(Np/Nm)^3*(1/Dm_by_Dp)^5; //kW + + // (c)Ratio of Flow Rates handled by the protoytpe and the Model, Qp/Qm + Qp_by_Qm=(Np/Nm)*(1/Dm_by_Dp)^3; + +//Results:- + printf(" (a)Working Speed of the Prototype, Np =%.2f rpm\n",Np) //The answer vary due to round off error + printf(" (b)Shaft Power of the Prototype, Pp =%.2f kW\n",Pp) //The answer vary due to round off error + printf(" (c)Ratio of Flow Rates handled by the protoytpe and the Model=%.2f ",Qp_by_Qm) //The answer provided in the textbook is wrong + diff --git a/3751/CH11/EX11.23/Ex11_23.sce b/3751/CH11/EX11.23/Ex11_23.sce new file mode 100644 index 000000000..4697d140a --- /dev/null +++ b/3751/CH11/EX11.23/Ex11_23.sce @@ -0,0 +1,30 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11 - Centrifugal Pumps +//Example 11.23 +//To Find the Head and Impeller Diameter of the other Pump. + + clc + clear + +//Given Data:- + //For Pump-1, + N1=1000; //Speed, rpm + D1=320; //Impeller Diameter, mm + Hm1=16; //Manometric Head, m + Q1=0.021; //Discharge, m^3/s + + //For Pump-2, + N2=1000; //Speed, rpm + //As other Pump has to deliver half the discharge, + Q2=Q1/2; //m^3/s + + +//Computations:- + Hm2=Hm1*(N2/N1)*sqrt(Q2/Q1); //m + D2=D1*(N1/N2)*sqrt(Hm2/Hm1); //mm + +//Results:- + printf("Head of the other Pump(Pump-2), Hm2=%.2f m\n",Hm2) + printf("Impeller Diameter of the other Pump(Pump-2), D2=%.2f mm\n",D2) //The answer vary due to round off error + + diff --git a/3751/CH11/EX11.24/Ex11_24.sce b/3751/CH11/EX11.24/Ex11_24.sce new file mode 100644 index 000000000..0023dda33 --- /dev/null +++ b/3751/CH11/EX11.24/Ex11_24.sce @@ -0,0 +1,31 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11 - Centrifugal Pumps +//Example 11.24 +//To Find the the number of stages and Diameter of each Impeller of the similar multistage Pump. + + clc + clear + +//Given Data:- + //For Single Stage Pump, + N1=2000; //Speed, rpm + D1=300; //Impeller Diameter, mm + Hm1=32; //Manometric Head, m + Q1=3; //Discharge, m^3/s + + //For Multi Stage Pump, + N2=1600; //Speed, rpm + H_mT2=200; //Total Manometric Head, m + Q2=5; //Discharge, m^3/s + + +//Computations:- + Hm2=Hm1*(N2/N1)*sqrt(Q2/Q1); //m + n=round(H_mT2/Hm2); //No. of stages + D2=D1*(N1/N2)*sqrt(Hm2/Hm1); //Diameter of Each Impeller, mm + +//Results:- + printf("Number of the Stages for the Multi stage Pump, n=%.f \n",n) + printf("Diameter of each Impeller for the Multi stage Pump, D2=%.2f mm\n",D2) //The answer vary due to round off error + + diff --git a/3751/CH11/EX11.25/Ex11_25.sce b/3751/CH11/EX11.25/Ex11_25.sce new file mode 100644 index 000000000..cf6b0ffeb --- /dev/null +++ b/3751/CH11/EX11.25/Ex11_25.sce @@ -0,0 +1,32 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11 - Centrifugal Pumps +//Example 11.25 +//To Find the Discharge and Head of the Pump at Condition '2' and '3' and Compare the Power Consumed in all the cases. + + clc + clear + +//Given Data:- + //At Condition '1' + N1=750; //Speed, rpm + Q1=60; //Discharge, l/s + H1=20; //Head, m + + //At Condition '2' + N2=1200; //Speed, rpm + + //At Condition '3' + N3=4200; //Speed, rpm + +//Computations:- + Q2=Q1*(N2/N1); // l/s + H2=H1*(N2/N1)^2; //m + Q3=Q1*(N3/N1); // l/s + H3=H1*(N3/N1)^2; //m + +//Results:- + printf("At Condition -2 Discharge, Q2=%.f l/s and Head, H2=%.1f m\n",Q2,H2) + printf(" At Condition -3 Discharge, Q3=%.f l/s and Head, H3=%.1f m\n",Q3,H3) + printf(" P1: P2 : P3 = 1 : %.2f : %.2f ",Q2*H2/(Q1*H1),Q3*H3/(Q1*H1)) //The answer vary due to round off error + + diff --git a/3751/CH11/EX11.26/Ex11_26.sce b/3751/CH11/EX11.26/Ex11_26.sce new file mode 100644 index 000000000..742e16626 --- /dev/null +++ b/3751/CH11/EX11.26/Ex11_26.sce @@ -0,0 +1,37 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11 - Centrifugal Pumps +//Example 11.26 +//To Calculate the Specific Speed of Pump and the Power Input and Find the Head, Discharge and Power required at 900 rpm. + + clc + clear + +//Given Data:- + + N=1500; //Speed, rpm + Q=0.2; //Discharge, m^3/s + H=15; //Head, m + eta_o=0.68; //Overall Efficiency + N2=900; //Speed, rpm + + //Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleratio due to gravity, m/s^2 + + +//Computations:- + Ns=N*Q^(1/2)/(H^(3/4)); //Specific Speed of Pump, SI Units + P=rho*g*Q*H /eta_o; //Power Input, W + + Q1=Q; H1=H; N1=N; P1=P; + Q2=Q1*(N2/N1); // m^3/s + H2=H1*(N2/N1)^2; //m + P2=P1*(N2/N1)^3; //W + +//Results:- + printf("Specific Speed of Pump, Ns=%.2f (SI Units)\n",Ns) + printf(" Power Input, P=%.2f W\n",P) + printf(" At 900 rpm (Condition 2)\n\t ") + printf(" Head, H2=%.1f m \n\t Discharge, Q2=%.2f m^3/s,\n\t Power required, P2=%.2f W",H2,Q2,P2) + + diff --git a/3751/CH11/EX11.3/Ex11_3.sce b/3751/CH11/EX11.3/Ex11_3.sce new file mode 100644 index 000000000..338c6ec47 --- /dev/null +++ b/3751/CH11/EX11.3/Ex11_3.sce @@ -0,0 +1,27 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.3 +//To Find the Power of the Pump. + + clc + clear + +//Given Data:- + H=40; //Total Head, m + Q=50; //Discharge, litres/s + eta_o=62/100; //Overall EEfficiency + + +//Data Used:- + rho=1000; //Density of Water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + + +//Computations:- + Q=Q/1000; //m^3/s + + P=rho*Q*g*H/(eta_o*1000); //kW + +//Result:- + printf("The Power of the Pump, P=%.3f kW \n",P) + diff --git a/3751/CH11/EX11.4/Ex11_4.sce b/3751/CH11/EX11.4/Ex11_4.sce new file mode 100644 index 000000000..da9fce543 --- /dev/null +++ b/3751/CH11/EX11.4/Ex11_4.sce @@ -0,0 +1,50 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.4 +//To Find (a)Vane Angle at Inlet (b)Work done by Impeller on water per second (c)Manometric Efficiency + + clc + clear + +//Given Data:- + //As Outer Diameter equals two times Inner Diameter, + Do_by_Di=2; //Do/Di + N=980; //Speed, rpm + Hm=52; //Manometric Head, m + Vfo=2.6; //Velocity of Flow, m/s + Vfi=Vfo; + beta_o=42; //Vane Angle at outlet, degrees + Do=600; //Outer Diameter of the Impeller, mm + bo=60; //Width at Outlet, mm + + +//Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Do=Do/1000; //m + bo=bo/1000; //m + + Di=Do/Do_by_Di; //Diameter at Inlet of Impeller, m + ui=%pi*Di*N/60; //Tangential velocity of Impeller at Inlet,m/s + uo=%pi*Do*N/60; // Tangential velocity of Impeller at Outlet, m/s + Q=%pi*Do*bo*Vfo; //Discharge, m^3/s + + //(a)Vane Angle at Inlet, beta_i + beta_i=atand(Vfi/ui); //degrees + + //(b) Work done by Impeller on water per sec, W + Vwo=uo-Vfo/tand(beta_o); //m/s + W=rho*Q*Vwo*uo/1000; //kN-m/s + + //(c) Manometric Efficiency, eta_man + eta_man=g*Hm/(Vwo*uo)*100; //In Percentage + + +//Results:- + printf(" (a)Vane Angle at Inlet, beta_i=%.2f Degrees \n ",beta_i) + printf(" (b) Work done by Impeller on water per sec =%.3f kN-m/s \n ",W) //The answer provided in the textbook is wrong. + printf(" (c) Manometric Efficiency, eta_man =%.2f Percent \n ",eta_man) //The answer provided in the textbook is wrong. + + diff --git a/3751/CH11/EX11.5/Ex11_5.sce b/3751/CH11/EX11.5/Ex11_5.sce new file mode 100644 index 000000000..f927b863d --- /dev/null +++ b/3751/CH11/EX11.5/Ex11_5.sce @@ -0,0 +1,33 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.5 +//To Find the Discharge of Pump. + + clc + clear + +//Given Data:- + Hm=14.5; //Manometric Head, m + N=1000; //Speed, rpm + beta_o=30; //Vane Angle at outlet, degrees + Do=300; //Outer Diameter of the Impeller, mm + bo=50; //Width at Outlet, mm + eta_man=95/100; //Manometric Efficiency + + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Do=Do/1000; //m + bo=bo/1000; //m + + uo=%pi*Do*N/60; //m/s + Vwo=g*Hm/(uo*eta_man); //m/s + Vfo=tand(beta_o)*(uo-Vwo); //m/s + Q=%pi*Do*bo*Vfo*1000; //Discharge, litres/s + +//Results:- + printf("The Discharge of the Pump=%.2f litres/s\n",Q) //The answer vary due to round off error + + diff --git a/3751/CH11/EX11.6/Ex11_6.sce b/3751/CH11/EX11.6/Ex11_6.sce new file mode 100644 index 000000000..7c8c9871e --- /dev/null +++ b/3751/CH11/EX11.6/Ex11_6.sce @@ -0,0 +1,59 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.6 +//To Calculate the Blade angle at Outlet, Power Required and Overall Efficiency of Pump. + + clc + clear + +//Given Data:- + Do=80; //Outer Diameter of the Impeller, cm + Q=1; //Discharge, m^3/s + H=80; //Head, m + N=1000; //Speed, rpm + bo=8; //Width at Outlet, cm + Delta_Q_per=3; //Percentage of Leakage Loss(of the Discharge) + Delta_P=10; //Mechanical Loss, kW + eta_H=80/100; //Hydraulic Efficiency + + +//Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Do=Do/100; //m + bo=bo/100; //m + + uo=%pi*Do*N/60; //m/s + Vfo=Q/(%pi*Do*bo); //m/s + Vwo=g*H/(uo*eta_H); //m/s + Vrwo=uo-Vwo; //m/s + + //(a) + beta_o=atand(Vfo/Vrwo); //Blade Angle at Outlet, degrees + + //Result1 + printf(" Blade Angle at Outlet, beta_o=%.2f Degrees \n",beta_o) //The answer vary due to round off error + + //(b)Power Required + Pi=rho*(1+Delta_Q_per/100)*Q*Vwo*uo; //Power delivered by the Impeller, W + P=Pi/1000+Delta_P; //Power required, kW + + //Result2 + printf(" Power Required, P =%.3f kW \n",P) //The answer vary due to round off error + + //(c)Overall Efficiency, eta_o + eta_V=1/(1+Delta_Q_per/100); //Volumetric Efficiency + eta_m=(P-Delta_P)/P; //Mechanical Efficiency + eta_o=eta_H*eta_V*eta_m*100; //In Percentage + + //Result3 + printf(" Overall Efficiency, eta_o =%.2f Percent \n",eta_o) //The answer vary due to round off error + + //Also, Overall Efficiency + eta_o=rho*Q*g*H/(P*1000)*100; //In Percentage + + printf("Also, Overall Efficiency, eta_o=%.2f Percent\n",eta_o) + + diff --git a/3751/CH11/EX11.7/Ex11_7.sce b/3751/CH11/EX11.7/Ex11_7.sce new file mode 100644 index 000000000..f1e0d183f --- /dev/null +++ b/3751/CH11/EX11.7/Ex11_7.sce @@ -0,0 +1,44 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.7 +//To Determine the Impeller Speed and Torque produced by it. + + clc + clear + +//Given Data:- + Q=60; //Discharge, litres/s + Ri=75; //Radius of the Impeller at Inlet, mm + Ro=150; //Radius of the Impeller at Outlet, cm + beta_o=30; //Vane Angle at Outlet, degrees + beta_i=30; //Vane Angle at Inlet, degrees + Ai=0.025; //Impeller Area at Inlet, m^2 + + +//Data Used:- + rho=1000; //Density of water, kg/m^3 + + +//Computations:- + Q=Q/1000; //m^3/s + Ri=Ri/1000; //m + Ro=Ro/1000; //m + + Di=2*Ri; //m + Do=2*Ro; //m + Vfi=Q/Ai; //m/s + Vfo=Vfi; + ui=Vfi/tand(beta_i); //m/s + N=ui*60/(%pi*Di); //rpm + + uo=%pi*Do*N/60; //m/s + Vrwo=Vfo/tand(beta_o); //m/s + Vwo=uo-Vrwo; //m/s + P=rho*Q*Vwo*uo; //Impeller Power, W + Ti=P*60/(2*%pi*N); //Impeller Torque, N-m + +//Results:- + printf("Impeller Speed, N=%.2f rpm\n",N) //The answer vary due to round off error + printf("Impeller Torque, Ti=%.2f N-m\n",Ti) //The answer vary due to round off error + + diff --git a/3751/CH11/EX11.8/Ex11_8.sce b/3751/CH11/EX11.8/Ex11_8.sce new file mode 100644 index 000000000..0cae0067a --- /dev/null +++ b/3751/CH11/EX11.8/Ex11_8.sce @@ -0,0 +1,37 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.8 +//To Determine the Power Required to drive the centrifugal Pump. + + clc + clear + +//Given Data:- + Q=40; //Discharge, litres/s + Hst=20; //Static Head, m + D=150; //Diameter of Pipe, mm + L=100; //length of pipe, m + eta_o=70/100; //Overall Efficiency + f=0.015; //Coefficient of friction + + +//Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + + +//Computations:- + Q=Q/1000; //m^3/s + D=D/1000; //m + + A=(%pi/4)*D^2; //m^2 + V=Q/A; //m/s + Vd=V; + + h_f=4*f*L*V^2/(2*g*D); //Frictional Head Loss in Pipe, m + Hm=Hst+h_f+Vd^2/(2*g); //Manometric Head, m + P=rho*Q*g*Hm/(eta_o*1000); //kW + +//Result:- + printf("Power Required to drive the Centrifugal Pump=%.3f kW\n",P) //The answer vary due to round off error + diff --git a/3751/CH11/EX11.9/Ex11_9.sce b/3751/CH11/EX11.9/Ex11_9.sce new file mode 100644 index 000000000..67c4b0361 --- /dev/null +++ b/3751/CH11/EX11.9/Ex11_9.sce @@ -0,0 +1,49 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 11- Centrifugal Pumps +//Example 11.9 +//To Find (i)Vane Angle at Inlet (ii)Work done by Impeller on water per second and (iii)Manometric Efficiency. + + clc + clear + +//Given Data:- + Do=500; //Outer Diameter of the Impeller, mm + Di=250; //Inner Diameter of the Impeller, mm + N=1000; //Speed, rpm + Hm=40; //Manometric Head, m + Vfo=2.5; //Velocity of Flow, m/s + Vfi=Vfo; + beta_o=40; //Vane Angle at outlet, degrees + bo=50; //Width at Outlet, mm + + +//Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Do=Do/1000; //m + Di=Di/1000; //m + bo=bo/1000; //m + + ui=%pi*Di*N/60; //Tangential velocity of Impeller at Inlet,m/s + uo=%pi*Do*N/60; // Tangential velocity of Impeller at Outlet, m/s + Q=%pi*Do*bo*Vfo; //Discharge, m^3/s + + //(i)Vane Angle at Inlet, beta_i + beta_i=atand(Vfi/ui); //degrees + + //(ii) Work done by Impeller on water per sec, W + Vwo=uo-Vfo/tand(beta_o); //m/s + W=rho*Q*Vwo*uo/1000; //kN-m/s + + //(iii) Manometric Efficiency, eta_man + eta_man=g*Hm/(Vwo*uo)*100; //In Percentage + + +//Results:- + printf(" (i)Vane Angle at Inlet, beta_i=%.2f Degrees \n ",beta_i) //The answer vary due to round off error + printf(" (ii) Work done by Impeller on water per sec =%.3f kN-m/s \n ",W) //The answer vary due to round off error + printf(" (iii) Manometric Efficiency, eta_man =%.2f Percent \n ",eta_man) //The answer vary due to round off error + + diff --git a/3751/CH12/EX12.1/Ex12_1.sce b/3751/CH12/EX12.1/Ex12_1.sce new file mode 100644 index 000000000..7a859e0f7 --- /dev/null +++ b/3751/CH12/EX12.1/Ex12_1.sce @@ -0,0 +1,46 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 12- Reciprocating Pumps +//Example 12.1 + + clc + clear + +//Given Data:- + Hs_th=4.8; //Suction Head (Theoretical), m + Hd_th=12; //Delivery Head (Theoretical), m + N=90; //Speed of Pump, rpm + D=100; //Piston Diameter, mm + L=150; //Length of Stroke, mm + Q=102; //Actual Discharge, lit./min + eta_s=60/100; //Efficiency of Suction Stroke + eta_d=75/100; //Efficiency of Delivery Stroke + +//Data Used:- + rho=1000; //Density of Water, kg/m^3 + g=9.81; //Accelerationdue to gravity, m/s^2 + +//Computations:- + Vs=(%pi/4)*(D/1000)^2*(L/1000); //Swept volume in one revolution, m^3 + Vth=Vs*N/60; //Theoritical Volume of Water pumped per second, m^3 + m=Vth*rho; //Theoritical Mass Flow rate, kg/s + m_act=Q*1000/(60*1000); //Actual mas flow rate, kg/s + + Slip=(m-m_act)*100/m; //Slip in Percentage + Cd=m_act/m*100; //Co-efficient of Discharge in Percentage + Hs=Hs_th/eta_s; //Suction Head taking suction efficiency in account, m + Hd=Hd_th/eta_d; //Delivery Head taking delivery efficiency in account, m + H=Hs+Hd; //Total Head, m + Pth=m*g*H; //Theoritical power required to Drive the Pump, W + A=(%pi/4)*(D/1000)^2; //Cross section Area of piston, m^2 + Fs=rho*g*Hs*A; //Average Force during Suction, N + Fd=rho*g*Hd*A; //Average Force during Delivery, N + P=(Fs+Fd)*L*N/60; //Power required by Pump (Same as Pth), W + +//Results:- + printf(" 1. Slip=%.2f Percent \n",Slip) //The answer vary due to round off error + printf(" 2. The Co-efficient of Discharge =%.2f Percent \n",Cd) //The answer vary due to round off error + printf(" 3. Theoretical Power Required to Drive the Pump =%.2f W \n",Pth) //The answer vary due to round off error + printf(" 4. Force Required to Work the Piston during Suction Stroke =%.2f N \n",Fs) + printf(" 5. Force Required to Work the Piston during Delivery Stroke =%.2f N \n",Fd) + + diff --git a/3751/CH12/EX12.10/Ex12_10.sce b/3751/CH12/EX12.10/Ex12_10.sce new file mode 100644 index 000000000..b445bd4ef --- /dev/null +++ b/3751/CH12/EX12.10/Ex12_10.sce @@ -0,0 +1,50 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 12- Reciprocating Pumps +//Example 12.10 +//To Find the Power required to overcome the friction of Delivery pipe when (a)No air vessel is fitted on it , (b)A large air vessel is fitted at the centre line of the pump. + + clc + clear + +//Given Data:- + N=60; //Speed of the Pump, rpm + D=250; //Plunger Diameter, mm + L=450; //Stroke Length, mm + d_d=112.5; //Diameter of Delivery Pipe, mm + l_d=48; //Length of Delivery Pipe, m + f=0.04; //Co-efficient of friction + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + rho=1000; //Density of water, kg/m^3 + + +//Computations:- + d_d=d_d/1000; //m + D=D/1000; //m + L=L/1000; //m + + a=(%pi/4)*d_d^2; //m^2 + A=(%pi/4)*D^2; //m^2 + omega=2*%pi*N/60; //rad/s + r=L/2; //m + + //(a)Without Air Vessel + H_fd=f*(l_d/d_d)*(omega*r*A/a)^2/(2*g); //Maximum loss of head due to friction in delivery pipe, m + m=rho*A*L*N/60; //Mass of water lifted, kg/s + Power=(2/3)*H_fd*m; //W + + //Result (a) + printf("(a)Without Air Vessel\n\t") + printf("Power Required to Overcome Friction=%.2f W\n\n",Power) //The answer provided in the textbook is wrong + + //(b)With Air Vessel + Ud=A*L*N/(a*60); //m/s + H_fd=f*(l_d/d_d)*(Ud^2/(2*g)); //m + Power=m*H_fd; //W + //Result (a) + printf("(a)With Air Vessel\n\t") + printf("Power Required to Overcome Friction=%.2f W\n",Power) //The answer vary due to round off error + + + diff --git a/3751/CH12/EX12.11/Ex12_11.sce b/3751/CH12/EX12.11/Ex12_11.sce new file mode 100644 index 000000000..f8c6af6ce --- /dev/null +++ b/3751/CH12/EX12.11/Ex12_11.sce @@ -0,0 +1,43 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 12- Reciprocating Pumps +//Example 12.11 +//(a)Theoritical Question. +//(b)Theoritical Question. +//(c)To Find the Rate of flow into or from the air vessel when crank makes angle of 30, 90 and 120 degrees with inner dead centre and + //Also Determine crank angle at which there is no flow to or from the air vessel. + + clc + clear + +//Given Data:- + D=200; //Bore of the Pump, mm + L=350; //Stroke Length, mm + d_s=150; //Diameter of Suction Pipe, mm + N=120; //Speed of the Pump, rpm + + +//Computations:- + d_s=d_s/1000; //m + D=D/1000; //m + L=L/1000; //m + + A=(%pi/4)*D^2; //m^2 + omega=2*%pi*N/60; //rad/s + r=L/2; //m + + //Using the Equation 12.28 from the textbook, Rates of Flow are + Q_30=A*omega*r*(2/%pi-sind(30) )*1000; //For 30 degree angle, litres/s + Q_90=A*omega*r*(2/%pi-sind(90) )*1000; //For 90 degree angle, litres/s + Q_120=A*omega*r*(2/%pi-sind(120) )*1000; //For 120 degree angle, litres/s + + theta=asind(2/%pi); //Angle at which there is no flow, degrees + //This is NOT Calculated in the Textbook. + +//Results:- + printf("Rate of Flow from the Air Vessel=%.1f litre/s for 30 Degree Angle\n\t\t\t\t",Q_30) + printf(" =%.f litre/s for 90 Degree Angle\n\t\t\t\t",Q_90) + printf(" =%.1f litre/s for 120 Degree Angle\n",Q_120) + + printf("The angle at which there is no flow from or to the air vessel = %.2f Degrees\n",theta) + + diff --git a/3751/CH12/EX12.12/Ex12_12.sce b/3751/CH12/EX12.12/Ex12_12.sce new file mode 100644 index 000000000..443aeb95e --- /dev/null +++ b/3751/CH12/EX12.12/Ex12_12.sce @@ -0,0 +1,57 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 12- Reciprocating Pumps +//Example 12.12 +//To Find the Maximum Speed at which the Pump may run without seperation. + + clc + clear + +//Given Data:- + D=10; //Plunger Diameter, cm + L=20; //Stroke Length, cm + H_s=4; //Suction Head, m + H_d=14; //Delivery Head, m + d_s=4; //Diameter of Suction Pipe, cm + l_s=6; //Length of Suction Pipe, m + d_d=3; //Diameter of Delivery Pipe, cm + l_d=18; //Length of Delivery Pipe, m + p=7.85; //Pressure (below atm.) for seperation, N/cm^2 + H_a=10.3; //Atmospheric Pressure Head, m of water + + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + rho=1000; //Density of water, kg/m^3 + + +//Computations:- + d_s=d_s/100; //m + d_d=d_d/100; //m + D=D/100; //m + L=L/100; //m + + a_s=(%pi/4)*d_s^2; //m^2 + a_d=(%pi/4)*d_d^2; //m^2 + A=(%pi/4)*D^2; //m^2 + r=L/2; //m + + H_sp=p*100^2/(rho*g); //Pressure Head of water for seperation, m (below atmosphere) (Value given in textbook is wrong due to incorrect value of p is used) + H_abs=H_a-H_sp; //Absolute Pressure Head of water for seperation, m + H_as_by_omega2=(l_s/g)*(A/a_s)*r; //H_as/omega^2 + omega=sqrt((H_sp-H_s)/H_as_by_omega2); //rad/s + N_s=omega*60/(2*%pi); //rpm + + H_ad_by_omega2=(l_d/g)*(A/a_d)*r; //H_as/omega^2 + omega=sqrt((H_sp+H_d)/H_ad_by_omega2); //rad/s + N_d=omega*60/(2*%pi); //rpm + + //Selecting maximum speed, + if N_s>N_d then + N=N_s; + else + N=N_d; + +//Result:- + printf("Hence, The Maximum Speed at which Pump should be Run is %.2f rpm\n",N) //The answer vary due to round off error + + diff --git a/3751/CH12/EX12.13/Ex12_13.sce b/3751/CH12/EX12.13/Ex12_13.sce new file mode 100644 index 000000000..ae1c93ceb --- /dev/null +++ b/3751/CH12/EX12.13/Ex12_13.sce @@ -0,0 +1,30 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 12- Reciprocating Pumps +//Example 12.13 +//To Determine the Crank Angle, at which there is no flow of water to or from the vessel. + + clc + clear + +//Given Data:- + D=17.5; //Bore diameter, cm + L=35; //Stroke Length, cm + d_s=15; //Diameter of Suction pipe, cm + N=150; //Speed, rpm + +//Computations:- + D=D/100; //m + L=L/100; //m + d_s=d_s/100; //m + + omega=2*%pi*N/60; //rad/s + A=(%pi/4)*D^2; //m^2 + r=L/2; //m + Q_s=2*A*omega*r/%pi; //Rate of flow from sump upto air vessel, m^3/s + theta=asind(Q_s/(A*omega*r)); //degrees + + +//Result:- + printf("The Crank Angle at which there is no flow, theta=%.2f Degrees\n",theta) + + diff --git a/3751/CH12/EX12.2/Ex12_2.sce b/3751/CH12/EX12.2/Ex12_2.sce new file mode 100644 index 000000000..f611ce645 --- /dev/null +++ b/3751/CH12/EX12.2/Ex12_2.sce @@ -0,0 +1,54 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 12- Reciprocating Pumps +//Example 12.2 +//Referring to Example 12.1 +//To Determine 1.The Slip 2. The Co-efficient of Discharge 3. Theoretical Power Requied to Drive the Pump 4. Force Required to Work the Piston during Suction Stroke 5. Force Required to Work the Piston during Delivery Stroke. + + clc + clear + +//Given Data:- + //The Pump is Double Acting + //From Example 12.1 + Hs_th=4.8; //Suction Head (Theoretical), m + Hd_th=12; //Delivery Head (Theoretical), m + N=90; //Speed of Pump, rpm + D=100; //Piston Diameter, mm + L=150; //Length of Stroke, mm + eta_s=60/100; //Efficiency of Suction Stroke + eta_d=75/100; //Efficiency of Delivery Stroke + + Q=200; //Actual Discharge, lit./min + d=20; //Diameter of Piston Rod, mm + + +//Data Used:- + rho=1000; //Density of Water, kg/m^3 + g=9.81; //Accelerationdue to gravity, m/s^2 + +//Computations:- + A=(%pi/4)*(D/1000)^2; //m^2 + a= (%pi/4)*(d/1000)^2; //m^2 + L=L/1000; //m + Vs=2*A*L; //Swept volume in one revolution, m^3 + Vth=A*L*N/60+(A-a)*L*N/60; //Theoritical Volume of Water pumped per second, m^3 + m=Vth*rho; //Theoritical Mass Flow rate, kg/s + m_act=Q*1000/(60*1000); //Actual mas flow rate, kg/s + + Slip=(m-m_act)*100/m; //Slip in Percentage + Cd=m_act/m*100; //Co-efficient of Discharge in Percentage + Hs=Hs_th/eta_s; //Suction Head taking suction efficiency in account, m + Hd=Hd_th/eta_d; //Delivery Head taking delivery efficiency in account, m + H=Hs+Hd; //Total Head, m + Pth=m*g*H; //Theoritical power Required to Drive the Pump, W + Fb=rho*(Hs*A+Hd*(A-a)); //Force to be provided by Pump during Backward Stroke, kg + Ff=rho*(Hs*(A-a)+Hd*A); // Force to be provided by Pump during Forward Stroke, kg + +//Results:- + printf(" 1. Slip=%.1f Percent \n",Slip) //The answer vary due to round off error + printf(" 2. The Co-efficient of Discharge =%.1f Percent \n",Cd) //The answer vary due to round off error + printf(" 3. Theoretical Power Requied to Drive the Pump =%.2f W \n",Pth) //The answer vary due to round off error + printf(" 4. Force to be provided by Pump during Backward Stroke =%.1f kg \n",Fb) + printf(" 5. Force to be provided by Pump during Forward Stroke =%.f kg \n",Ff) + + diff --git a/3751/CH12/EX12.3/Ex12_3.sce b/3751/CH12/EX12.3/Ex12_3.sce new file mode 100644 index 000000000..8593ca45a --- /dev/null +++ b/3751/CH12/EX12.3/Ex12_3.sce @@ -0,0 +1,59 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 12- Reciprocating Pumps +//Example 12.3 +//To Calculate The Maximum Speed at which pump may be run and Determine Resultant Suction Head at Begining, Middle and End of the Stroke. + + clc + clear + +//Given Data:- + D=150; //Diameter of Plunger, mm + L=250; //Stroke length, mm + l_s=10; //Length of Suction Pipe, m + d=75; //Diameter of Suction Pipe, mm + hs=4; //Suction Head, m of water + Ha=10.34; //Atmospheric Pressure, m of water + Habs=2.44; //Absolute Pressure Head, m of water + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + + +//Computations:- + Hv=Ha-Habs; //Vaccume Pressure, m of water + //For Maximum Resultant Suction Head, + Hs=Hv; + A=(%pi/4)*(D/1000)^2; //m^2 + a_s= (%pi/4)*(d/1000)^2; //m^2 + r=L/2000; //m + omega=sqrt((Hs-hs)*g*a_s/(l_s*A*r)); //radian/sec + N=60*omega/(2*%pi); //rpm + +//Result 1 + printf(" The Maximum Speed at which pump may be run, N=%.2f rpm \n",N) //The answer vary due to round off error + + //At Begining + Has=(l_s/g)*(A/a_s)*omega^2*r*cosd(0); //m + Hs=hs+Has; //Resultant Head at Begining of Stroke, m of water + +//Result 2 + printf(" Resultant Head at Begining of Stroke, Hs=%.1f m of water \n",Hs) + + + //At Middle + Has=(l_s/g)*(A/a_s)*omega^2*r*cosd(90); //m + Hs=hs+Has; //Resultant Head at Middle of Stroke (Has=0), m of water + +//Result 3 + printf(" Resultant Head at Middle of Stroke, Hs=%.f m of water \n",Hs) + + + //At the End + Has=(l_s/g)*(A/a_s)*omega^2*r*cosd(180); //m + Hs=hs+Has; //Resultant Head at End of Stroke, m of water + // Resultant Head at End of Stroke is not calculated and displayed as result in the textbook. + +//Result 4 + printf(" Resultant Head at End of Stroke, Hs=%.1f m of water \n ",Hs) + + diff --git a/3751/CH12/EX12.4/Ex12_4.sce b/3751/CH12/EX12.4/Ex12_4.sce new file mode 100644 index 000000000..10a18d233 --- /dev/null +++ b/3751/CH12/EX12.4/Ex12_4.sce @@ -0,0 +1,42 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 12- Reciprocating Pumps +//Example 12.4 +//To Find whether seperation will take place, and if so, at which section of pipe. + + clc + clear + +//Given Data:- + ld=60.96; //Length of Delivery Pipe, m + a=1.83; //Acceleration of Plunger Pump, m/s^2 + A_by_ad=2; //ratio of Sectional Area of Plunger to that of Delivery Pipe. + //Referring to Fig 12.6 in the textbook, + ef=18.3; //m + eq=12.19; //m + dq=1.83; //m + slope=3; + + Hsp=2.44; //Pressure Head in pipe when seperation takes place, m of water + Hatm=10.36; //Atmospheric Pressure Head (Barometer Reading), m of water + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Had=-(ld/g)*A_by_ad*a; //Head at end of stroke, a=acceleration=omega^2*r, Had in m + dp=Had; // Referring to Fig 12.6 in the textbook + ed=eq+dq; + Hd=ed; //Total Delivery Head, m + Hrd=Had+Hd; //Resultant Pressure in Delivery pipe at end of Stroke, m + Habs=Hatm+Hrd; //Absolute Pressure. m of water + + Hv=Hatm-Hsp; //Vaccum pressure, m + x=-Hv-Had; //m + + if Habs<Hsp then + printf("The Seperation Will Take Place at x=%.2f m\n",x) //The answer vary due to round off error + else + printf("The Seperation Will Not Take Place \n") + end + + diff --git a/3751/CH12/EX12.5/Ex12_5.sce b/3751/CH12/EX12.5/Ex12_5.sce new file mode 100644 index 000000000..93c09f621 --- /dev/null +++ b/3751/CH12/EX12.5/Ex12_5.sce @@ -0,0 +1,62 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 12- Reciprocating Pumps +//Example 12.5 +//To Determine the Pressure Head on Piston at Begining, Middle and End of Suction Stroke. + + clc + clear + +//Given Data:- + L=150; //Length of Stroke, mm + l_s=7; //Length of Suction Pipe, m + ds_by_D=3/4; //Ratio of Suction Pipe Diameter to Piston Diameter, ds/D + hs=2.5; //Suction Head, m + ds=75; //Diameter of Suction Pipe, mm + N=75; //Crank Speed, rpm + f=0.01; //Co-efficient of Friction + + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + h_atm=10.33; //Atmospheric Pressure Head, m of water + + +//Computations:- + L=L/1000; //m + ds=ds/1000; //m + + r=L/2; //Crank radius, m + A_by_as=(1/ds_by_D)^2; + omega=2*%pi*N/60; //Angular Velocity, rad/s + +//At Begining of Suction Stroke, + theta=0; //degrees + h_as=(l_s/g)*A_by_as*omega^2*r*cosd(theta); //Acceleration Head, m of water + h_fs=(4*f*l_s/(2*g*ds))*(A_by_as*omega*r*sind(theta))^2; //Head loss due to friction, m of water + h_v=hs+h_fs+h_as; //Pressure Head on Piston, m of water Vaccum + h_abs=h_atm-h_v; //Pressure Head on Piston, m of water Absolute + //Result 1 + printf("At Begining of Suction Stroke\n Pressure Head on Piston=%.2f m of water Vaccum \n\t\t\t =%.2f m of water Absolute\n\n",h_v,h_abs) //The answer vary due to round off error + + +//At Mid of Suction Stroke, + theta=90; //degrees + h_as=(l_s/g)*A_by_as*omega^2*r*cosd(theta); //Acceleration Head, m of water + h_fs=(4*f*l_s/(2*g*ds))*(A_by_as*omega*r*sind(theta))^2; //Head loss due to friction, m of water + h_v=hs+h_fs+h_as; //Pressure Head on Piston, m of water Vaccum + h_abs=h_atm-h_v; //Pressure Head on Piston, m of water Absolute + //Result 2 + printf("At Middle of Suction Stroke\n Pressure Head on Piston=%.4f m of water Vaccum \n\t\t\t =%.3f m of water Absolute\n\n",h_v,h_abs) //The answer vary due to round off error + + +//At End of Suction Stroke, + theta=180; //degrees + h_as=(l_s/g)*A_by_as*omega^2*r*cosd(theta); //Acceleration Head, m of water + h_fs=(4*f*l_s/(2*g*ds))*(A_by_as*omega*r*sind(theta))^2; //Head loss due to friction, m of water + h_v=hs+h_fs+h_as; //Pressure Head on Piston, m of water Vaccum + h_abs=h_atm-h_v; //Pressure Head on Piston, m of water Absolute + //Result 3 + printf("At End of Suction Stroke\n Pressure Head on Piston=%.2f m of water Vaccum \n\t\t\t =%.2f m of water Absolute\n\n",h_v,h_abs) //The answer vary due to round off error + + + diff --git a/3751/CH12/EX12.6/Ex12_6.sce b/3751/CH12/EX12.6/Ex12_6.sce new file mode 100644 index 000000000..05840dcb1 --- /dev/null +++ b/3751/CH12/EX12.6/Ex12_6.sce @@ -0,0 +1,81 @@ +//Fluid Systems - By- Shiv Kumar +//Chapter 12- Reciprocating Pumps +//Example 12.6 + + clc + clear + +//Given Data:- + D=200; //Piston Diameter, mm + L=300; //Stroke length, mm + H_s=4; //Suction Head, m + H_d=35; //Delivery Head, m + d_s=100; //Diameter of Suction Pipe, mm + d_d=d_s; //Diameter of Delivery Head + l_d=50; //Length of Delivery Pipe, m + l_s=10; //Length of Suction Pipe, m + f_s=0.04; //Co-efficient of friction for Suction Pipe + f_d=f_s; //Co-efficient of friction for Delivery Pipe + N=30; //Speed of Pump, rpm + + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + rho=1000; //Density of water, kg/m^3 + + +//Computations:- + D=D/1000; //m + L=L/1000; //m + d_s=d_s/1000; //m + d_d=d_d/1000; //m + + a_s=(%pi/4)*d_s^2; //m^2 + a_d=(%pi/4)*d_d^2; //m^2 + A=(%pi/4)*D^2; //m^2 + omega=2*%pi*N/60; //rad/s + r=L/2; //m + + // (1) Suction Stroke + //At end of Stroke, + H_as=(l_s/g)*(A/a_s)*omega^2*r; //m of water + + //At middle of Stroke, + h_fs=f_s*(l_s/d_s)*(1/(2*g))*((A/a_s)*omega*r)^2; //m of water + + H_sb=H_s+H_as; //Pressure at begining of suction stroke, m of water (vaccum) + H_se=H_s-H_as; //Pressure at end of suction stroke, m of water + H_se=abs(H_se); //m above atmosphere + H_sm=H_s+h_fs; //Pressure at middle of suction stroke, m of water (vaccum) + + // (1) Delivery Stroke + //At end of Stroke, + H_ad=(l_d/g)*(A/a_d)*omega^2*r; //m of water + + //At middle of Stroke, + h_fd=f_d*(l_d/d_d)*(1/(2*g))*((A/a_d)*omega*r)^2; //m of water + + H_db=H_d+H_ad; //Pressure at begining of delivery stroke, m of water (above atmosphere) + H_de=H_d-H_ad; //Pressure at end of delivery stroke, m of water (above atm.) + H_dm=H_d+h_fd; //Pressure at middle of delivery stroke, m of water (above atm.) + + m=rho*A*L*N/60; //Mass of Water Discharge, kg/s + //Referring to Equation 12.18 in the textbook, + Work= m*g*(H_s+H_d+(2/3)*h_fs+(2/3)*h_fd); //Total Work done by Pump, W + +//Results:- + printf("(1)Suction Stroke\n\t") + printf("Pressure at Begining of the Stroke=%.2f m of water (vaccum)\n\t",H_sb) //The answer vary due to round off error + printf("Pressure at End of the Stroke=%.1f m of water (above atmosphere\n\t",H_se) //The answer vary due to round off error + printf("Pressure at Middle of the Stroke=%.3f m of water (vaccum)\n\n",H_sm) //The answer vary due to round off error + + printf("(2)Delivery Stroke\n\t") + printf("Pressure at Begining of the Stroke=%.2f m of water ( above atmosphere )\n\t",H_db) //The answer vary due to round off error + printf("Pressure at End of the Stroke=%.2f m of water (above atm.)\n\t",H_de) //The answer vary due to round off error + printf("Pressure at Middle of the Stroke=%.2f m of water ( above atm. )\n",H_dm) //The answer vary due to round off error + + printf(" Power Required to drive the Pump=%.2f W",Work) //The answer vary due to round off error + + + + diff --git a/3751/CH12/EX12.7/Ex12_7.sce b/3751/CH12/EX12.7/Ex12_7.sce new file mode 100644 index 000000000..b7503f450 --- /dev/null +++ b/3751/CH12/EX12.7/Ex12_7.sce @@ -0,0 +1,67 @@ +//Fluid Systems - By- Shiv Kumar +//Chapter 12- Reciprocating Pumps +//Example 12.7 +//To Calculate (a)The Absolute Head in the Pump corresponding to the four corners of the cord and also the mid strokes (b)The Work done/minute. + + clc + clear + +//Given Data:- + D=125; //Bore of the Pump, mm + L=125; //Stroke length, mm + N=30; //Speed of Pump, rpm + H_s=3; //Suction Head, m + H_d=15; //Delivery Head, m + d_s=62.5; //Diameter of Suction Pipe, mm + d_d=d_s; //Diameter of Delivery Head + l_d=18; //Length of Delivery Pipe, m + l_s=l_d; //Length of Suction Pipe + f=0.032; //Co-efficient of friction for both Pipes + + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + rho=1000; //Density of water, kg/m^3 + H_a=10.2; //Atmospheric Pressure Head, m of water + +//Computations:- + D=D/1000; //m + L=L/1000; //m + d_s=d_s/1000; //m + d_d=d_d/1000; //m + + a=(%pi/4)*d_s^2; //m^2 + A=(%pi/4)*D^2; //m^2 + omega=2*%pi*N/60; //rad/s + r=L/2; //m + + H_as=(l_s/g)*(A/a)*omega^2*r; //m + h_fs_max=f*(l_s/d_s)*(1/(2*g))*((A/a)*omega*r)^2; //m + //As Pipes are of same diameter and length, + H_ad=H_as; + h_fd_max=h_fs_max; + + H_m=H_a-H_s-H_as; //Pressure Head at 'm' , m of water + H_r= H_a-H_s-h_fs_max ; //Pressure Head at 'r' , m of water + H_n=H_a-H_s+H_as ; //Pressure Head at 'n' , m of water + H_at_s= H_a+H_s+H_as ; //Pressure Head at 's' , m of water + H_o=H_a+H_d+h_fd_max ; //Pressure Head at 'o' , m of water + H_q=H_a+H_d+H_ad ; //Pressure Head at 'q' , m of water + + m=rho*A*L*N*2/60; //mass of water/s, kg/s + Work_s=m*g*(H_s+H_d+(2/3)*h_fs_max+(2/3)*h_fd_max); //Word done/s, W + Work_m=Work_s*60; //Work done/min. , J/min + +//Results:- + printf("Pressure Head at m =%.2f m of water\n ",H_m) //The answer vary due to round off error + printf("Pressure Head at r =%.3f m of water\n ",H_r) //The answer vary due to round off error + printf("Pressure Head at n =%.2f m of water\n ",H_n) //The answer provided in the textbook is wrong + printf("Pressure Head at s =%.2f m of water\n ",H_at_s ) //The answer provided in the textbook is wrong + printf("Pressure Head at o =%.3f m of water\n ",H_o) //The answer vary due to round off error + printf("Pressure Head at q =%.2f m of water\n\n ",H_q) //The answer provided in the textbook is wrong + + printf("Work done/s=%.1f W \n Work done/minute=%.2f J/min. \n",Work_s,Work_m) //The answer provided in the textbook is wrong + + + + diff --git a/3751/CH12/EX12.7/Ex12_7_Indicator_Diagram.jpg b/3751/CH12/EX12.7/Ex12_7_Indicator_Diagram.jpg Binary files differnew file mode 100644 index 000000000..efa8ec78c --- /dev/null +++ b/3751/CH12/EX12.7/Ex12_7_Indicator_Diagram.jpg diff --git a/3751/CH12/EX12.8/Ex12_8.sce b/3751/CH12/EX12.8/Ex12_8.sce new file mode 100644 index 000000000..3f1975680 --- /dev/null +++ b/3751/CH12/EX12.8/Ex12_8.sce @@ -0,0 +1,52 @@ +//Fluid Systems - By- Shiv Kumar +//Chapter 12- Reciprocating Pumps +//Example 12.8 +//To (a) Find the Speed at which seperation may take place at commencement of suction stroke, (b)Find the change in Speed of Pump if an air vessel is fitted in the suction side. + + clc + clear + +//Given Data:- + H_s=3.60; //Suction Head, m + d_s=225; //Diameter of Suction Pipe, mm + l_s=9.6; //Length of Suction Pipe, m + D=300; //Pump cylinder diameter, mm + L=450; //Stroke length, mm + + H_a=9.6; //Barometric Head, m of water + H_sp=2.4; //Head (m of water) for seperation + f=0.04; + + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + + +//Computations:- + D=D/1000; //m + L=L/1000; //m + d_s=d_s/1000; //m + + a_s=(%pi/4)*d_s^2; //m^2 + A=(%pi/4)*D^2; //m^2 + r=L/2; //m + + //Without Air Vessel + H_as_by_omega2=(l_s/g)*(A/a_s)*r; //H_as/omega^2 + omega=sqrt((H_a-H_s-H_sp)/H_as_by_omega2); //rad/s + N=omega*60/(2*%pi); //rpm + + //With Air Vessel + Us_by_N=(A/a_s)*L/60; //Us/N + l_v=H_sp/2; //m + H_as_by_N2=(l_v/g)*(A/a_s)*(2*%pi/60)^2*r; //H_as/N^2 + h_fs_by_N2=f*(l_s-l_v)*Us_by_N^2/(r*2*g); + N1=sqrt((H_a-H_sp-H_s)/(H_as_by_N2+h_fs_by_N2)); //Speed of Pump if air vessel is fitted, rpm + Change_In_Speed=N1-N; //rpm + + +//Results:- + printf("(a)Speed at which Seperation may take place, N=%.f rpm\n",N) + printf("(b)Change in Speed with air vessel=%.f rpm\n",Change_In_Speed) //The answer provided in the textbook is wrong + + diff --git a/3751/CH12/EX12.9/Ex12_9.sce b/3751/CH12/EX12.9/Ex12_9.sce new file mode 100644 index 000000000..905a7ea51 --- /dev/null +++ b/3751/CH12/EX12.9/Ex12_9.sce @@ -0,0 +1,71 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 12- Reciprocating Pumps +//Example 12.9 +//To Determine the Pressure on the Cylinder at the Begining of the Stroke (a)When no air vessel is fitted, (b)When air vessel is fitted at the cylinder level. + + clc + clear + +//Given Data:- + d_s=150; //Diameter of Suction Pipe, mm + l_s=12; //Length of Suction pipe, m + H_s=3; //Suction Head, m + D=225; //Cylinder Diameter, mm + L_s=375; //Stroke Length, mm + L=1.5; //Length of Connecting Rod, m + N=20; //Crank Speed, rpm + l_v=1.5; //m + f=0.04; //Co-efficient of friction + +//Data Used:- + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + d_s=d_s/1000; //m + D=D/1000; //m + L_s=L_s/1000; //m + + a_s=(%pi/4)*d_s^2; //m^2 + A=(%pi/4)*D^2; //m^2 + omega=2*%pi*N/60; //rad/s + r=L_s/2; //m + + printf("Without Air Vessel : \n\t") + //(i) Assuming Simple Harmonic Motion : + printf("(i) Assuming Simple Harmonic Motion\n\t\t") + H_as=(l_s/g)*(A/a_s)*omega^2*r; //m of water + H=H_s+H_as; //Pressure at the begining of stroke, m of water (vaccum) + //Result (a) (i) + printf(" Pressure at the begining of stroke=%.2f m of water (vaccum) \n\t",H) //The answer vary due to round off error + + //(ii) If Simple Harmonic Motion is not assumed : + printf(" (ii) If Simple Harmonic Motion is not assumed : \n\t\t") + H_as=H_as*(1+r/L); + H=H_s+H_as; //Pressure at the begining of stroke, m of water (vaccum) + //Result (a) (ii) + printf(" Pressure at the begining of stroke=%.3f m of water (vaccum) \n\n",H) //The answer vary due to round off error + + + //(b) When Air Vessel is fitted + printf(" When Air Vessel is fitted : \n\t") + + Us=(A/a_s)*L_s*N/60; //m/s + h_fs=(f*(l_s-l_v)/d_s)*(Us^2/(2*g)); //m of water + + //(i) Assuming Simple Harmonic Motion : + printf("(i) Assuming Simple Harmonic Motion\n\t\t") + H_as=(l_v/g)*(A/a_s)*omega^2*r; //m of water (vaccum) + H=H_s+H_as+h_fs; //Total Pressure Head in the Cylinder, m of water below atmospheric + //Result (b) (i) + printf(" Total Pressure Head in the Cylinder =%.4f m of water below atmospheric or vaccum \n\t",H) //The answer vary due to round off error + + //(ii) If Simple Harmonic Motion is not assumed : + printf(" (ii) If Simple Harmonic Motion is not assumed : \n\t\t") + H_as=H_as*(1+r/L); + H=H_s+H_as+h_fs; // Total Pressure Head in the Cylinder , m of water below atmospheric + //Result (b) (ii) + printf(" Total Pressure Head in the Cylinder =%.4f m of water below atmospheric \n",H) //The answer vary due to round off error + + + + diff --git a/3751/CH16/EX16.1/Ex16_1.sce b/3751/CH16/EX16.1/Ex16_1.sce new file mode 100644 index 000000000..2c55c1fe5 --- /dev/null +++ b/3751/CH16/EX16.1/Ex16_1.sce @@ -0,0 +1,28 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.1 +//To Find the Maximum Power Available at the Outlet of Pipe. + clc + clear + +//Given Data:- + d=300; //Diameter of the Pipe, mm + l=3000; //Length of the Pipe, m + H=400; //Total Head at Inlet, m + f=0.005; + +//Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + //Condition for Maximum Power transmission + hf=H/3; //m + V=sqrt(hf*(2*g*d/1000)/(4*f*l)); //m/s + Q=(%pi/4)*(d/1000)^2*V; //Discharge, m^3/s + Pmax=rho*g*Q*(H-hf)/1000; //Maximum Power Available at Outlet of Pipe, kW + + +//Results:- + printf("The Maximum Power Available at Outlet of Pipe=%.3f kW",Pmax) //The answer vary due to round off error + diff --git a/3751/CH16/EX16.10/Ex16_10.sce b/3751/CH16/EX16.10/Ex16_10.sce new file mode 100644 index 000000000..5539e21dc --- /dev/null +++ b/3751/CH16/EX16.10/Ex16_10.sce @@ -0,0 +1,33 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.10 +//To Find (i)The Weight of Loaded Cylinder and energy stored by the Cylinder (ii)Ther Power supplied by the Accumulator (iii)The Diameter of ram of an ordinary Accumulator. + clc + clear + +//Given Data:- + D=180; // mm + d=150; //mm + L=1.25; //Stroke length, m + p=100; //Pressure of Water, bar + +//Computations:- + D=D/1000; //m + d=d/1000; //m + p=p*10^5; //N/m^2 + + A=(%pi/4)*(D^2-d^2); //Annular area of Ram, m^2 + //(i) + W=p*A/1000; //Weight of Loaded Cylinder, kN + Energy=W*L; //Energy stored in the Accumulator, kNm + //(ii) + t=90; //Time taken by ram to complete the stroke, seconds + P=W*L/t; //kW + //(iii) + D=(W*1000/(p*%pi/4))^(1/2)*1000; //mm + +//Results:- + printf("(i)Weight of Loaded Cylinder, W=%.2f kN\n",W) //The answer vary due to round off error + printf(" Energy stored in the Accumulator=%.3f kNm\n",Energy) //The answer vary due to round off error + printf("(ii)Power Supplied by the Accumulator=%.3f kW\n",P) //The answer vary due to round off error + printf("(iii)Ram Diameter (In case of Ordinary Accumulator) = %.2f mm\n",D) //The answer vary due to round off error diff --git a/3751/CH16/EX16.11/Ex16_11.sce b/3751/CH16/EX16.11/Ex16_11.sce new file mode 100644 index 000000000..207696186 --- /dev/null +++ b/3751/CH16/EX16.11/Ex16_11.sce @@ -0,0 +1,24 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.11 +//To Find the Diameters of Fixed ram and Sliding Cylinder. + clc + clear + +//Given Data:- + p1=50; //Pressure Intensity of Low Pressure Liquid, bar + p2=150; // Pressure Intensity of High Pressure Liquid, bar + Capacity=32; //Capacity of Intensifier, Litres + l=1.5; //Stroke Length, m + +//Computations:- + Capacity=Capacity/1000; //m^3 + + D2=sqrt(Capacity/((%pi/4)*l))*1000; //mm + D1=sqrt((p2/p1)*D2^2); //mm + +//Results:- + printf("Diameter of Fixed Cylinder, D2=%.2f mm\n",D2) //The answer vary due to round off error + printf("Diameter of Sliding Ram, D1=%.2f mm\n",D1) //The answer vary due to round off error + + diff --git a/3751/CH16/EX16.12/Ex16_12.sce b/3751/CH16/EX16.12/Ex16_12.sce new file mode 100644 index 000000000..5d76ed585 --- /dev/null +++ b/3751/CH16/EX16.12/Ex16_12.sce @@ -0,0 +1,23 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.12 +//To Calculate the Diameters of Fixed ram and Sliding Cylinder. + clc + clear + +//Given Data:- + p1=50; //Pressure Intensity of Low Pressure Liquid, bar + p2=150; // Pressure Intensity of High Pressure Liquid, bar + Capacity=0.025; //Capacity of Intensifier, m^3 + l=1.25; //Stroke Length, m + +//Computations:- + + D2=sqrt(Capacity/((%pi/4)*l))*1000; //mm + D1=sqrt((p2/p1)*D2^2); //mm + +//Results:- + printf("Diameter of Fixed Cylinder, D2=%.2f mm\n",D2) //The answer vary due to round off error + printf("Diameter of Sliding Ram, D1=%.2f mm\n",D1) //The answer vary due to round off error + + diff --git a/3751/CH16/EX16.13/Ex16_13.sce b/3751/CH16/EX16.13/Ex16_13.sce new file mode 100644 index 000000000..046769591 --- /dev/null +++ b/3751/CH16/EX16.13/Ex16_13.sce @@ -0,0 +1,18 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.13 +//To Find the Diameter of Cylinder. + clc + clear + +//Given Data:- + F=400; //Force, N + p=4000; //Pressure, kPa + +//Computations:- + + d=sqrt(4*F/(%pi*p*1000))*1000; //mm + +//Results;- + printf("Cylinder Diameter, d=%.2f mm\n",d) + diff --git a/3751/CH16/EX16.14/Ex16_14.sce b/3751/CH16/EX16.14/Ex16_14.sce new file mode 100644 index 000000000..b4699d3b0 --- /dev/null +++ b/3751/CH16/EX16.14/Ex16_14.sce @@ -0,0 +1,45 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.14 +//To Find (i)The Force applied in Plunger (ii) The Number of Strokes performed by Plunger (iii) Work done by the Press Ram and (iv) Power required to drive the Plunger. + + clc + clear + +//Given Data:- + D=180; //Diameter of ram, mm + d=36; //Diameter of Plunger, mm + W=7 ; //Weight exerted by Press ram, kN + L=300; //Stroke Length of Plunger, mm + l=0.9; //Distance moved by ram, m + t=15; //Time, minutes + +//Computations:- + D=D/1000; //m + A=(%pi/4)*D^2; //m^2 + d=d/1000; //m + a= (%pi/4)*d^2; //m^2 + W=W*1000; //N + L=L/1000; //m + t=t*60; //seconds(s) + + // (i)The Force applied in Plunger, F1 + F1=(a/A)*W; //N + + //(ii) The Number of Strokes performed by Plunger, n + n=(A/a)*(l/L); + + // (iii) Work done by the Press Ram + Work=W*l; //N-m + + // (iv) Power required to drive the Plunger, P + P=Work/t; //W + + +//Results:- + printf(" (i) The Force applied in Plunger, F1=%.2f N \n",F1) //The answer vary due to round off error + printf(" (ii) The Number of Strokes performed by Plunger, n =%.f \n",n) + printf(" (iii) Work done by the Press Ram =%.f N.m \n",Work) + printf(" (iv) Power required to drive the Plunger, P =%.f W \n",P) + + diff --git a/3751/CH16/EX16.15/Ex16_15.sce b/3751/CH16/EX16.15/Ex16_15.sce new file mode 100644 index 000000000..8e670453c --- /dev/null +++ b/3751/CH16/EX16.15/Ex16_15.sce @@ -0,0 +1,45 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.15 +//To Find (i)The Force applied in Plunger (ii) The Number of Strokes performed by Plunger (iii) Work done by the Press Ram and (iv) Power required to drive the Plunger. + + clc + clear + +//Given Data:- + D=165; //Diameter of ram, mm + d=33; //Diameter of Plunger, mm + W=5.5; //Weight exerted by Press ram, kN + L=250; //Stroke Length of Plunger, mm + l=1.2; //Distance moved by ram, m + t=20; //Time, minutes + +//Computations:- + D=D/1000; //m + A=(%pi/4)*D^2; //m^2 + d=d/1000; //m + a= (%pi/4)*d^2; //m^2 + W=W*1000; //N + L=L/1000; //m + t=t*60; //seconds(s) + + // (i)The Force applied in Plunger, F1 + F1=(a/A)*W; //N + + //(ii) The Number of Strokes performed by Plunger, n + n=(A/a)*(l/L); + + // (iii) Work done by the Press Ram + Work=W*l; //N-m + + // (iv) Power required to drive the Plunger, P + P=Work/t; //W + + +//Results:- + printf(" (i) The Force applied in Plunger, F1=%.f N \n",F1) + printf(" (ii) The Number of Strokes performed by Plunger, n =%.f \n",n) + printf(" (iii) Work done by the Press Ram =%.f N.m \n",Work) + printf(" (iv) Power required to drive the Plunger, P =%.1f W \n",P) + + diff --git a/3751/CH16/EX16.16/Ex16_16.sce b/3751/CH16/EX16.16/Ex16_16.sce new file mode 100644 index 000000000..ad9444410 --- /dev/null +++ b/3751/CH16/EX16.16/Ex16_16.sce @@ -0,0 +1,30 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.16 +//To Find (i) Power required to drive the Lift (ii) Working Period of Lift and (iii) Ideal Period of Lift. + + clc + clear + +//Given Data:- + W=60; //Load lifted by Lift, kN + H=14; //Height, m + V=0.5; //Speed of Lift, m/s + t=60; //Time for one operation, s + +//Computations:- + + // (i) Power required to drive the Lift, P + P=W*H/t; //kJ/s + + // (ii) Working Period of Lift , tw + tw=H/V; //s + + // (iii) Ideal Period of Lift, ti + ti=t-tw; //s + +//Results + printf(" (i) Power required to drive the Lift, P=%.f kW \n",P) + printf(" (ii) Working Period of Lift , tw =%.f s \n",tw) + printf(" (iii) Ideal Period of Lift, ti =%.f s \n",ti) + diff --git a/3751/CH16/EX16.17/Ex16_17.sce b/3751/CH16/EX16.17/Ex16_17.sce new file mode 100644 index 000000000..02b3707fb --- /dev/null +++ b/3751/CH16/EX16.17/Ex16_17.sce @@ -0,0 +1,24 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.17 +//To Find the Efficiency of Hydraulic Crane. + + clc + clear + +//Given Data:- + V=340; //Volume of water utilized, litres + p=50; //Pressure Intensity, bar + W=125; //Load Lift, kN + l=10; //Displacement of Weight, m + + +//Computations:- + Energy=p*10^5*V/1000; //Energy Supplied to Crane, J + Work=W*1000*l; //Work done by crane in lifting load, J + eta=Work/Energy*100; //Efficiency In Percentage + +//Result:- + printf("Efficiency of Hydraulic Crane, eta=%.2f Percent\n",eta) //The answer vary due to round off error + + diff --git a/3751/CH16/EX16.18/Ex16_18.sce b/3751/CH16/EX16.18/Ex16_18.sce new file mode 100644 index 000000000..c2a88b031 --- /dev/null +++ b/3751/CH16/EX16.18/Ex16_18.sce @@ -0,0 +1,28 @@ +//Fluid Systyems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmission +//Example 16.18 +//To Find (i)The Load Lifted by Crane (ii)The Quantity of Water needed to Lift the Load. + + clc + clear + +//Given Data:- + d=200; //Diameter of Ram, mm + p=7.5; //Pressure of Water Supplied, MPa + VR=6; //Velocity Ratio + eta=50/100; //Efficiency of Crane + h=10; //Height through which water is to be lifted, m] + +//Computations:- + d=d/1000; //m + p=p*10^6; //Pa + + Fp=(%pi/4)*d^2*p; //Pressure Force Exerted on Ram, N (answer vary due to value of %pi) + W=Fp*eta/VR; //Load Lifted by Crane, N + Vw=(%pi/4)*d^2*h/VR*1000; //Quantity of Water needed, Litres + +//Results:- + printf(" (i)The Load Lifted by Crane, W=%.f N \n",W) //The answer provided in textbook is wrong + printf(" (ii)The Quantity of Water needed to Lift the Load by 10 m =%.2f Litres \n",Vw) //The answer vary due to round off error + + diff --git a/3751/CH16/EX16.2/Ex16_2.sce b/3751/CH16/EX16.2/Ex16_2.sce new file mode 100644 index 000000000..fd716311d --- /dev/null +++ b/3751/CH16/EX16.2/Ex16_2.sce @@ -0,0 +1,28 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.2 +//To Determine the Flow Rate and the Minimum Diameter of Pipe. + clc + clear + +//Given Data:- + P=1000; //Power Transmitted, kW + eta=85/100; //Efficiency + l=500; //Length of the Pipe, m + H=150; //Head of Water at Inlet, m + f=0.006; + +//Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + hf=H*(1-eta); //m + Q=P*10^3/(rho*g*(H-hf)); //m^3/s + d=(64*f*l*Q^2/(2*g*%pi^2*hf))^(1/5); //m + +//Results:- + printf("The Required Flow Rate, Q=%.4f m^3/s\n",Q) + printf("The Minimum Diameter, d=%.4f m\n",d) //The answer vary due to round off error + + diff --git a/3751/CH16/EX16.3/Ex16_3.sce b/3751/CH16/EX16.3/Ex16_3.sce new file mode 100644 index 000000000..1c382287e --- /dev/null +++ b/3751/CH16/EX16.3/Ex16_3.sce @@ -0,0 +1,32 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.3 +//To Determine the Minimum Number of Pipes. + clc + clear + +//Given Data:- + l=7500; //Length of each Pipe, m + d=125; //Diameter of each Pipe, mm + Pr=6000; //Pressure at Discharge End, kPa + eta=85/100; //Efficiency + P=156; //Power Delivered, kW + f=0.006; + +//Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + H_minus_hf=Pr*10^3/(rho*g); //H-hf, m + H=H_minus_hf/eta; //m + hf=H-H_minus_hf; //m + Q=P*1000/(rho*g*(H-hf)); //m^3/s + q=sqrt((hf*2*g*%pi^2*(d/1000)^5)/(64*f*l)); //Discharge in each Pipe, m^3/s + n=Q/q; //Number of Pipes + + +//Results:- + + printf("The Minimum Number of Pipes Required=%.f\n",n) + diff --git a/3751/CH16/EX16.4/Ex16_4.sce b/3751/CH16/EX16.4/Ex16_4.sce new file mode 100644 index 000000000..756a5b755 --- /dev/null +++ b/3751/CH16/EX16.4/Ex16_4.sce @@ -0,0 +1,29 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.4 +//To Find the Diameter of Pipe. + clc + clear + +//Given Data:- + l=2100; //Length of the Pipe, m + P=103; //Power Transmitted, kW + pi=392.4; //Pressure at Inlet of Pipe, N/cm^2 + eta=80/100; //Efficiency + f=0.005; + +//Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + H=pi*10^4/(rho*g); //m + hf=H*(1-eta); //m + Q=P*1000/(rho*g*(H-hf)); //m^3/s + d=((64*f*l*Q^2)/(hf*2*g*%pi^2))^(1/5)*1000; //mm + + +//Results:- + + printf("The Diameter of Pipe=%.2f mm\n",d) //The answer vary due to round off error + diff --git a/3751/CH16/EX16.5/Ex16_5.sce b/3751/CH16/EX16.5/Ex16_5.sce new file mode 100644 index 000000000..d20aa02d4 --- /dev/null +++ b/3751/CH16/EX16.5/Ex16_5.sce @@ -0,0 +1,26 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.5 +//To Calculate the Increase in Pressure Intensity. + clc + clear + +//Given Data:- + d=200; //diameter of Pipe, mm + Q=40; //Discharge, Litres/s + l=600; //Length of Pipe, m + t=1.5; //Time taken to close the Valve gradually, s + +//Data Required:- + rho=1000; //Density of Water, Kg/m^3 + +//Computations:- + A=(%pi/4)*(d/1000)^2; //m^2 + V=(Q/1000)/A; //m/s + p=rho*l*V/(t*1000); //Pressure Rise, kPa + + +//Results:- + + printf("The Pressure Rise due to Gradual Closure of Valve=%.f kPa\n",p) //The answer vary due to round off error + diff --git a/3751/CH16/EX16.6/Ex16_6.sce b/3751/CH16/EX16.6/Ex16_6.sce new file mode 100644 index 000000000..6c14d6b67 --- /dev/null +++ b/3751/CH16/EX16.6/Ex16_6.sce @@ -0,0 +1,37 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.6 +//To Calculate the Rise in Pressure due to Valve Closure in (i)10 seconds, (ii)2.5 seconds. + clc + clear + +//Given Data:- + l=2500; //Lenfth of Pipe, m + V=1.2 ; //Velocity of Flow, m/s + K=20*10^8; //Bulk Modulus of Water, N/m^2 + +//Data Used:- + rho=1000; //Density of Water, Kg/m^3 + +//Computations:- + a=sqrt(K/rho); //Velocity of Pressure Wave, m/s + t_c=2*l /a; //Critical time, s + + // (i) + t=10; // s + //t>t_c. so, This is a case of Gradual valve closure. + p=rho*l*V/(t*1000); //Pressure Rise, kPa + + //Result (i) + printf("(i)Pressure Rise, p=%.f kPa\n",p) + + //(ii) + t=2.5; // s + // t<t_c. This is a case of Instantaneous Valve Closure. + p=rho*V*a/1000; // Pressure Rise, kPa + + //Result (ii) + printf("(ii)Pressure Rise, p=%.2f kPa\n",p) //The answer vary due to round off error + + + diff --git a/3751/CH16/EX16.7/Ex16_7.sce b/3751/CH16/EX16.7/Ex16_7.sce new file mode 100644 index 000000000..4fb8b3bc0 --- /dev/null +++ b/3751/CH16/EX16.7/Ex16_7.sce @@ -0,0 +1,36 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.7 +//To Determine the Increasse in Pressure. + clc + clear + +//Given Data:- + d=800; //Diameter of pipe, mm + Q=0.75; //Discharge, m^3/s + t=10; //Thickness of Pipe, nmnm + Es=20*10^10; //Elastic Modulus of Steel, N/m^2 + E=2*10^9; //Elastic Modulus of Water, N/m^2 + l=3500; //Lenfth of Pipe, m + T=5; //Time of Valve Closure, s + + +//Data Used:- + rho=1000; //Density of Water, Kg/m^3 + +//Computations:- + K=E/(1+(d/t)*(E/Es)); //Combined Modulus of Elasticity, N/m^2 + a=sqrt(K/rho); //Velocity of Pressure Wave, m/s + Tc=2*l /a; //Critical time, s + + //t<t_c. So, valve closure is rapid. + A=(%pi/4)*(d/1000)^2; //m^2 + V=Q/A; //Average Velocity of Flow, m/s + p=rho*V*a/1000; //Pressure Rise, kPa + + +//Result + printf("The Rise of Pressure=%.2f kPa\n",p) //The answer provided in the textbook is wrong + + + diff --git a/3751/CH16/EX16.8/Ex16_8.sce b/3751/CH16/EX16.8/Ex16_8.sce new file mode 100644 index 000000000..9485eabdb --- /dev/null +++ b/3751/CH16/EX16.8/Ex16_8.sce @@ -0,0 +1,25 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.8 +//To Find (i)Displacement of Accumulator (ii)Capacity of Accumulator (iii)Total weight placed on the ram. + clc + clear + +//Given Data:- + p=200; //Pressure of oil, kPa + D=1.5; //Diameter of Ram, m + L=6; //Stroke or Lift of Ram, m + +//Computations:- + A=(%pi/4)*D^2; //m^2 + Disp=A*L; //Displacenmenmt of Accumulator, m^3 + Capacity=p*Disp; //Capacity of Accumulator, kNm + W=p*A; //Total Weight on the Ram, kN + +//Results:- + printf("(i) Displacenmenmt of Accumulator=%.2f m^3\n ",Disp) //The answer vary due to round off error + printf("(ii) Capacity of Accumulator =%.f kNm \n ",Capacity) //The answer given in the textbook is wrong + printf("(iii) Total Weight on the Ram, W =%.1f kN \n ",W) //The answer vary due to round off error + + + diff --git a/3751/CH16/EX16.9/Ex16_9.sce b/3751/CH16/EX16.9/Ex16_9.sce new file mode 100644 index 000000000..b1562f85b --- /dev/null +++ b/3751/CH16/EX16.9/Ex16_9.sce @@ -0,0 +1,34 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 16- Hydraulic Power and Its Transmissions +//Example 16.9 +//To Deternmine the Diameter of the ram. + clc + clear + +//Given Data:- + d=125; //Diameter of Pipe, mm + l=2; //Lenght of Pipe, km + P=35; //Power Transmitted, kW + W=1250; //Load on ram, kN + loss_per=3; //Percentage of Power Loss due to friction + f_dash=0.04; //Pipe Friction Factor + +//Data Used:- + rho=1000; //Density of Water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Delta_P=loss_per/100*P*1000; //Power Loss due to friction , W + //By Darcy's Formula, + hf_by_V2=f_dash*(l*1000)/(2*g*d/1000); //hf/V^2 + + QbyV=(%pi/4)*(d/1000)^2; //Q/V + V=( Delta_P/(rho*g*QbyV*hf_by_V2))^(1/3); //m/s + Q=QbyV*V; //m^3/s + p=P*1000/Q; //N/m^2 + D=sqrt(W*1000/(p*%pi/4))*1000; //mm + +//Result:- + printf("The Diameter of ram, D=%.2f mm",D) //The answer vary due to round off error + + diff --git a/3751/CH17/EX17.1/Ex17_1.sce b/3751/CH17/EX17.1/Ex17_1.sce new file mode 100644 index 000000000..bb6e98d7d --- /dev/null +++ b/3751/CH17/EX17.1/Ex17_1.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.1 + //Theoritical Problem to Check Dimensions.
\ No newline at end of file diff --git a/3751/CH17/EX17.10/Ex17_10.sce b/3751/CH17/EX17.10/Ex17_10.sce new file mode 100644 index 000000000..ef1257964 --- /dev/null +++ b/3751/CH17/EX17.10/Ex17_10.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.10 + //Theoritical Problem (Same as 17.9).
\ No newline at end of file diff --git a/3751/CH17/EX17.11/Ex17_11.sce b/3751/CH17/EX17.11/Ex17_11.sce new file mode 100644 index 000000000..db5fadeae --- /dev/null +++ b/3751/CH17/EX17.11/Ex17_11.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.11 + //Theoritical Problem .
\ No newline at end of file diff --git a/3751/CH17/EX17.12/Ex17_12.sce b/3751/CH17/EX17.12/Ex17_12.sce new file mode 100644 index 000000000..0c2aacd61 --- /dev/null +++ b/3751/CH17/EX17.12/Ex17_12.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.12 + //Theoritical Problem to Find Expression for Delta(p).
\ No newline at end of file diff --git a/3751/CH17/EX17.13/Ex17_13.sce b/3751/CH17/EX17.13/Ex17_13.sce new file mode 100644 index 000000000..0b8709010 --- /dev/null +++ b/3751/CH17/EX17.13/Ex17_13.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.13 + //Theoritical Problem .
\ No newline at end of file diff --git a/3751/CH17/EX17.14/Ex17_14.sce b/3751/CH17/EX17.14/Ex17_14.sce new file mode 100644 index 000000000..60964aead --- /dev/null +++ b/3751/CH17/EX17.14/Ex17_14.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.14 + //Theoritical Problem to obtain Expression for Drag Force.
\ No newline at end of file diff --git a/3751/CH17/EX17.15/Ex17_15.sce b/3751/CH17/EX17.15/Ex17_15.sce new file mode 100644 index 000000000..20f602f42 --- /dev/null +++ b/3751/CH17/EX17.15/Ex17_15.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.15 + //Theoritical Problem to Develop an Expression for the Thrust developed by a Propeller.
\ No newline at end of file diff --git a/3751/CH17/EX17.16/Ex17_16.sce b/3751/CH17/EX17.16/Ex17_16.sce new file mode 100644 index 000000000..5077bcf43 --- /dev/null +++ b/3751/CH17/EX17.16/Ex17_16.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.16 + //Theoritical Problem.
\ No newline at end of file diff --git a/3751/CH17/EX17.17/Ex17_17.sce b/3751/CH17/EX17.17/Ex17_17.sce new file mode 100644 index 000000000..e0b4dca33 --- /dev/null +++ b/3751/CH17/EX17.17/Ex17_17.sce @@ -0,0 +1,26 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.17 +//To Find the Velocity of Oil flowing in the Pipe for the condition of Dynamic Similarity. + + clc + clear + +//Given:- + //For Model ( Working Fluid is Oil), + Dm=80; //Diameter, mm + nu_m=0.03; //Kinematic Viscosity of Oil, Stoke + + //For Prototype ( Working Fluid is Water), + Dp=200; //Diameter, mm + Vp=3.5; //Velocity of Water in the Pipe, m/s + nu_p=0.01; //Kinematic Viscosity of Water, Stoke + +//Computations:- + //From Reynold's Law of Similarity, + Vm=Vp*(Dp/Dm)*(nu_m/nu_p); //Velocity of Oil, m/s + +//Results:- + printf("The Velocity of Oil , Vm=%.2f m/s\n",Vm) + + diff --git a/3751/CH17/EX17.18/Ex17_18.sce b/3751/CH17/EX17.18/Ex17_18.sce new file mode 100644 index 000000000..9a53a4acf --- /dev/null +++ b/3751/CH17/EX17.18/Ex17_18.sce @@ -0,0 +1,40 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.18 +//To Find the Velocity and Rate of Flow in the Model. + + clc + clear + +//Given:- + //For Model ( Working Fluid is Water), + Dm=150; //Diameter of Pipe, mm + mu_m=0.01; //Viscosity of Water, Poise + + //For Prototype ( Working Fluid is Oil), + Dp=1.5; //Diameter, m + Sp=0.9; //Specific Gravity of Oil + mu_p=3*10^-2; //Viscosity of Oil, Poise + Qp=3000; //Discharge of Oil, Litres/sec + +//Data Required:- + rho_m=1000; //Density of Water, Kg/m^3 + + +//Computations:- + Dm=Dm/1000; //m + Qp=Qp/1000; //m^3/s + Ap=(%pi/4)*Dp^2; //m^2 + Am= (%pi/4)*Dm^2 ; //m^2 + Vp=Qp/Ap; //m/s + rho_p=Sp*rho_m; //Kg/m^3 + + //From Reynold's Law of Similarity, + Vm=Vp*(Dp/Dm)*(mu_m/mu_p)*(rho_p/rho_m); //Velocity of Water in Model, m/s + Qm=Am*Vm*1000; //Rate of Flow in Model, Litres/sec + +//Results:- + printf("The Velocity in the Model, Vm=%.3f m/s\n",Vm) //The Answer Vary due to Round off Error + printf("The Rate of Flow in the Model, Qm=%.2f Litres/sec\n",Qm) //The Answer Vary due to Round off Error + + diff --git a/3751/CH17/EX17.19/Ex17_19.sce b/3751/CH17/EX17.19/Ex17_19.sce new file mode 100644 index 000000000..123d0f320 --- /dev/null +++ b/3751/CH17/EX17.19/Ex17_19.sce @@ -0,0 +1,30 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.19 +//To Find the Speed of Air in Wind Tunnel and Determine the The Ratio of Drag(Resistance) between the Model and its Prototype. + + clc + clear + +//Given:- + Lr=30; //Scale Ratio (Lp/Lm) + //For Model ( Working Fluid is Air), + nu_m=0.016; //Kinematic Viscosity of Air, Stoke + rho_m=1.24; //Density of Air, Kg/m^3 + + //For Prototype ( Working Fluid is Sea Water), + Vp=10; //Speed of Sub-marine (Prototype), m/s + nu_p=0.012; //Kinematic Viscosity of Sea Water, Stoke + rho_p=1030; //Density of Sea Water, Kg/m^3 + +//Computations:- + //From Reynold's Law of Similarity, + Vm=Vp*(nu_m/nu_p)*Lr; //Velocity of Air, m/s + + Fp_by_Fm=(rho_p/rho_m)*Lr^2*(Vp/Vm)^2; //Ratio of Drag Force (Resistance) + + +//Results:- + printf("The Speed of Air in Wind Tunnel, Vm=%.f m/s\n",Vm) + printf("The Ratio of Drag Force (Resistance), Fp/Fm=%.3f \n",Fp_by_Fm) + diff --git a/3751/CH17/EX17.2/Ex17_2.sce b/3751/CH17/EX17.2/Ex17_2.sce new file mode 100644 index 000000000..a08652290 --- /dev/null +++ b/3751/CH17/EX17.2/Ex17_2.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.2 + //Theoritical Problem to Obtain Expression for Velocity of Propagation of Wave diff --git a/3751/CH17/EX17.20/Ex17_20.sce b/3751/CH17/EX17.20/Ex17_20.sce new file mode 100644 index 000000000..162c05c42 --- /dev/null +++ b/3751/CH17/EX17.20/Ex17_20.sce @@ -0,0 +1,26 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.20 +//To Find the Discharge and the Velocity over the Prototype. + + clc + clear + +//Given Data:- + //For Model, + Qm=3; //Discharge over the Model, m^3/s + Vm=1.5; //Velocity of Flow over the Model, m/s + Lr=40; //Scale Ratio (Lp/Lm) + + +//Computations:- + //By Froude's Law of Similarity, + Vp=Vm*Lr^(1/2); //Velocity of Flow ovre the Prototype, m/s + Qp=Lr^2*(Vp/Vm)*Qm; //Discharge over the Prototype, m^/s + + +//Results:- + printf("The Velocity over the Prototype, Vp=%.2f m/s \n",Vp) //The Answer Vary due to Round off Error. + printf("The Discharge over the Prototype, Qp=%.f m^3/s \n",Qp) //The Answer provided in the Textbook is Wrong. + + diff --git a/3751/CH17/EX17.21/Ex17_21.sce b/3751/CH17/EX17.21/Ex17_21.sce new file mode 100644 index 000000000..db67ae8a4 --- /dev/null +++ b/3751/CH17/EX17.21/Ex17_21.sce @@ -0,0 +1,29 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.21 +//To Find the Ship Velocity and Propulsive Force in the Prototype. + + clc + clear + +//Given Data:- + //For Model, + Lm=1; //Length of Model, m + Vm=0.7; //Speed in the Model, m/s + Fm=5; //Force in the Model, N + + //For Prototype, + Lp=50; //Length of Prototype, m + + +//Computations:- + //By Froude's Law of Similarity, + Vp=Vm*(Lp/Lm)^(1/2); //Velocity of the Prototype(Ship), m/s + Fp=Fm*(Lp/Lm)^3/1000; // Propulsive Force in the Prototype, kN + + +//Results:- + printf("The Ship(Prototype) Velocity , Vp=%.2f m/s \n",Vp) + printf("The Propulsive Force in the Prototype , Fp=%.f kN \n",Fp) + + diff --git a/3751/CH17/EX17.22/Ex17_22.sce b/3751/CH17/EX17.22/Ex17_22.sce new file mode 100644 index 000000000..2953cbb55 --- /dev/null +++ b/3751/CH17/EX17.22/Ex17_22.sce @@ -0,0 +1,27 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.22 +//To Determine for the Model: (i) The Size and the Velocity of Waves (ii)The Tidal Period + + clc + clear + +//Given Data:- + Lr=200; //Scale Ratio + //For Prototype, + Ap=20; //Amplititude of Waves, m + Vp=10; //Velocity, m/s + Tp=12; //Time Period, hrs. + +//Computations:- + //By Froude's Law of Similarity, + Vm=Vp/Lr^(1/2); //Velocity of Wave in the Model, m/s + Am=Ap/Lr; //Amplitude of waves in the Model, m + Tm=Tp*60/Lr^(1/2); //Tidal Period in the Model, min. + + +//Results:- + printf("(i)For the Model:\n Velocity of Wave, Vm=%.3f m/s\n Size(Amplitude) of Wave, Am=%.1f m\n",Vm,Am) + printf("(ii)The Tidal Period in the Model, Tm=%.2f min. \n",Tm) + + diff --git a/3751/CH17/EX17.23/Ex17_23.sce b/3751/CH17/EX17.23/Ex17_23.sce new file mode 100644 index 000000000..5503e692e --- /dev/null +++ b/3751/CH17/EX17.23/Ex17_23.sce @@ -0,0 +1,28 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.23 +//To Find the Velocity of the Prototype and Force Required to Propel the Prototytpe. + + clc + clear + +//Given Data:- + Lr=40; //Scale Ratio (Lp/Lm) + //For Model, + Vm=2; //Velocity for the Model, m/s + Fm=0.5; //Propulsive Force in Model, N + //For Prototype, + Lp=45; //m + + +//Computations:- + //By Froude's Law of Similarity, + Vp=Vm*Lr^(1/2); //Velocity for the Prototype, m/s + Fp=Fm*Lr^3; //Force Required to Propel the Prototytpe, N + + +//Results:- + printf("The Velocity of the Prototype, Vp=%.2f m/s \n",Vp) //The Answer vary due to Round off Error + printf("The Force Required to Propel the Prototytpe , Fp=%.f N \n",Fp) + + diff --git a/3751/CH17/EX17.24/Ex17_24.sce b/3751/CH17/EX17.24/Ex17_24.sce new file mode 100644 index 000000000..c85521e49 --- /dev/null +++ b/3751/CH17/EX17.24/Ex17_24.sce @@ -0,0 +1,28 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.24 +//To Find the Prototype to Model Scale Ratios for : (i)Velocity (ii)Time (iii)Acceleration (iv)Force . and Wave Height and Time taken for Model. + + clc + clear + +//Given Data:- + Lr=50; //Scale Ratio (Lp/Lm) + //For Prototype, + Hp=1.5; //m + Tp=25; //Seconds(s) + +//Computations:- + Vr=Lr^(1/2); //Velocity Ratio + Tr=Lr^(1/2); //Time Ratio + ar=Vr/Tr; //Acceleration Ratio + Fr=Lr^3; //Force Ratio + + Hm=Hp/Lr; //m + Tm=Tp/Tr; //Seconds(s) + + +//Results:- + printf("(i)Velocity Ratio, Vr=%.2f\n (ii)Time Ratio, Tr=%.2f \n (iii)Acceleration Ratio, ar=%.f \n(iv)Force Ratio, Fr=%.f\n\n ",Vr,Tr,ar,Fr) + printf("Wave Height in the Model, Hm=%.2f m\n",Hm) + printf("Time taken in the Model, Tm=%.2f s",Tm) //The Answer Vary due to Round off Error diff --git a/3751/CH17/EX17.3/Ex17_3.sce b/3751/CH17/EX17.3/Ex17_3.sce new file mode 100644 index 000000000..0cb22a2c4 --- /dev/null +++ b/3751/CH17/EX17.3/Ex17_3.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.3 + //Theoritical Problem.
\ No newline at end of file diff --git a/3751/CH17/EX17.4/Ex17_4.sce b/3751/CH17/EX17.4/Ex17_4.sce new file mode 100644 index 000000000..2f3020acf --- /dev/null +++ b/3751/CH17/EX17.4/Ex17_4.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.4 + //Theoritical Problem to Find expression for Drag Force.
\ No newline at end of file diff --git a/3751/CH17/EX17.5/Ex17_5.sce b/3751/CH17/EX17.5/Ex17_5.sce new file mode 100644 index 000000000..172ed5a47 --- /dev/null +++ b/3751/CH17/EX17.5/Ex17_5.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.5 + //Theoritical Problem .
\ No newline at end of file diff --git a/3751/CH17/EX17.6/Ex17_6.sce b/3751/CH17/EX17.6/Ex17_6.sce new file mode 100644 index 000000000..41a8c0f56 --- /dev/null +++ b/3751/CH17/EX17.6/Ex17_6.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.6 + //Theoritical Problem .
\ No newline at end of file diff --git a/3751/CH17/EX17.7/Ex17_7.sce b/3751/CH17/EX17.7/Ex17_7.sce new file mode 100644 index 000000000..660d3472b --- /dev/null +++ b/3751/CH17/EX17.7/Ex17_7.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.7 + //Theoritical Problem.
\ No newline at end of file diff --git a/3751/CH17/EX17.8/Ex17_8.sce b/3751/CH17/EX17.8/Ex17_8.sce new file mode 100644 index 000000000..4a6289106 --- /dev/null +++ b/3751/CH17/EX17.8/Ex17_8.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.8 + //Theoritical Problem.
\ No newline at end of file diff --git a/3751/CH17/EX17.9/Ex17_9.sce b/3751/CH17/EX17.9/Ex17_9.sce new file mode 100644 index 000000000..1fcaba0cb --- /dev/null +++ b/3751/CH17/EX17.9/Ex17_9.sce @@ -0,0 +1,4 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 17- Dimensional and Model Analysis +//Example 17.9 + //Theoritical Problem to Find Expression for Force exerted by a Flowing Liquid.
\ No newline at end of file diff --git a/3751/CH2/EX2.1/Ex2_1.sce b/3751/CH2/EX2.1/Ex2_1.sce new file mode 100644 index 000000000..7dbb44435 --- /dev/null +++ b/3751/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,21 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.1
+ clc
+ clear
+
+//Given Data:-
+ V=32; //Velocity of the Jet, m/s
+ d=5; //Diameter of the Jet, cm
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ d=d/100; //cm
+ a=(%pi/4)*d^2; //cross-sectional area of Jet, m^2
+ m=rho*a*V; //Mass Flow Rate, kg/s
+ F=m*V/1000; //Force Exerted by the Jet on the flat plate, kN
+//Result:-
+ printf("The Force exerted by the Jet on the plate=%.3f kN \n", F) //The answer vary due to round off error
+
diff --git a/3751/CH2/EX2.10/Ex2_10.sce b/3751/CH2/EX2.10/Ex2_10.sce new file mode 100644 index 000000000..025d48f38 --- /dev/null +++ b/3751/CH2/EX2.10/Ex2_10.sce @@ -0,0 +1,21 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.10
+ clc
+ clear
+//Given Data:-
+ V1=40; //Velocity of the Jet at Inlet, m/s
+ V2=32; //Velocity of the Jet at Outlet, m/s
+ theta=65; //Angle of Deflection from original direction, degrees
+ m=0.9; //Mass flow rate, kg/s
+
+//Computations:-
+ Fx=m*(V1-V2*cosd(theta)); //N (Answer in textbook is wrong)
+ Fy=m*V2*sind(theta); //N
+ F_R=sqrt(Fx^2+Fy^2); //Resultant Force, N
+ phi=atand(Fy/Fx); //Angle made by resultant with X-axis, degrees
+
+//Results:-
+ printf("Resultant Force, F_R=%.2f N at an angle, phi=%.2f Degrees to X-axis", F_R, phi) //The answer provided in the textbook is wrong
+
+
diff --git a/3751/CH2/EX2.11/Ex2_11.sce b/3751/CH2/EX2.11/Ex2_11.sce new file mode 100644 index 000000000..9c2663172 --- /dev/null +++ b/3751/CH2/EX2.11/Ex2_11.sce @@ -0,0 +1,40 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.11
+ clc
+ clear
+//Given Data:-
+ //(a)
+ V=60; //Velocity of the Jet, m/s
+ theta=30; //Angle of Outlet, degrees
+ //(b)
+ u=25; //Velocity of vane, m/s
+
+
+//Data Used:-
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ //(a)
+ Fx=(V/g)*(1+cosd(theta)); //Force exerted by Unit weight of water in direction of Jet, N/N of Water
+ Fy=V*sind(theta)/g; //Force exerted by Unit weight of water in direction perpendicular to direction of Jet, N/N of Water
+ F_R=sqrt(Fx^2+Fy^2); //Resultant for per unit weight of water, N/N of Water
+ phi=atand(Fy/Fx); //Angle made by resultant with X-axis, degrees
+
+//Results(a):-
+ printf("(a)\nForce exerted by Unit weight of water in direction of Jet, Fx=%.2f N/N of Water \n", Fx)
+ printf("Force exerted by Unit weight of water in direction perpendicular to direction of Jet, Fy=%.2f N/N of water \n", Fy) //The answer vary due to round off error
+ printf("Resulatant Force, F_R=%.2f N/N of Water at angle, phi=%.2f degrees \n\n", F_R, phi) //The answer vary due to round off error
+ //(b)
+ Fx=(V-u)*(1+cosd(theta))/g; //Force exerted by Unit weight of water in direction of Jet, N/N of Water
+ Fy=(V-u)*sind(theta)/g; //Force exerted by Unit weight of water in direction perpendicular to direction of Jet, N/N of Water
+ F_R=sqrt(Fx^2+Fy^2); //Resultant force per unit weight of water, N/N of Water
+ phi=atand(Fy/Fx); //Angle made by resultant with X-axis, degrees
+ W=Fx*u; //N-m/s/N of Water
+ P=Fx*u/1000; //Power developed per unit weight of water, KW/N of Water
+ //Result(b)
+ printf("(b)\nForce exerted by Unit weight of water in direction of Jet, Fx=%.2f N/N of Water \n", Fx) //The answer vary due to round off error
+ printf("Force exerted by Unit weight of water in direction perpendicular to direction of Jet, Fy=%.2f N/N of water \n", Fy)
+ printf("Resulatant Force, F_R=%.2f N/N of Water at angle, phi=%.2f degrees \n\n", F_R, phi) //The answer vary due to round off error
+ printf("Work done per unit weight of water=%.2f N-m/s/N of Water \n", W) //The answer vary due to round off error
+ printf("Power developed per unit weight of water=%.4f KW/N of Water", P) //The answer vary due to round off error
diff --git a/3751/CH2/EX2.12/Ex2_12.sce b/3751/CH2/EX2.12/Ex2_12.sce new file mode 100644 index 000000000..4b8fe3e4d --- /dev/null +++ b/3751/CH2/EX2.12/Ex2_12.sce @@ -0,0 +1,37 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.12
+ clc
+ clear
+//Given Data:-
+ Vi=22; //Absolute velocity of Jet at Inlet of Vane, m/s
+ u=11; //Velocity of Vane, m/s
+ ui=u;
+ uo=u;
+ alpha_i=25; //Angle made by Jet at Inlet, degrees
+ alpha_l=135; //Angle made by Jet at leaving, degrees
+ alpha_o=180-alpha_l; //degrees
+
+//Data Used:-
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ //(a)
+ Vwi=Vi*cosd(alpha_i); //m/s
+ Vfi=Vi*sind(alpha_i); //m/s
+ Vrwi=Vwi-ui; //m/s
+ beta_i=atand(Vfi/Vrwi); //degrees
+ Vri=Vfi/sind(beta_i); //m/s
+ Vro=Vri;
+ beta_o=alpha_o-asind(uo*sind(180-alpha_o)/Vro); //degrees
+ Vwo=Vro*cosd(beta_o)-uo; //degrees
+ //(b)
+ W=(Vwi+Vwo)*u/g; //N-m/N
+
+//Results:-
+ printf("(a)Vane angle at Inlet, beta_i=%.2f degrees \n", beta_i) //The answer vary due to round off error
+ printf(" Vane angle at Outlet, beta_o=%.2f degrees \n", beta_o) //The answer vary due to round off error
+ printf("(b)Work done per second per unit weight of water striking the vane per second=%.2f N-m/N", W) //The answer vary due to round off error
+
+
+
diff --git a/3751/CH2/EX2.13/Ex2_13.sce b/3751/CH2/EX2.13/Ex2_13.sce new file mode 100644 index 000000000..874d95aac --- /dev/null +++ b/3751/CH2/EX2.13/Ex2_13.sce @@ -0,0 +1,30 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.13
+ clc
+ clear
+
+//Given Data:-
+ d=25; //Diameter of the Jet, mm
+ V=27; //Velocity of the Jet, m/s
+ AoD=140; //Angle of Deflection, degrees
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ d=d/1000; //m
+ a=(%pi/4)*d^2; //cross-sectional area of Jet, m^2
+ m=rho*a*V; //Mass Flow Rate, kg/s
+ //For condition of Maximum work done,
+ u=V/3; //Velocity of Vane, m/s
+ theta=180-AoD; //degrees
+ //(a)Maximum work done/second
+ W=rho*a*(V-u)^2*(1+cosd(theta))*u/1000; //kJ/s
+ //(b)Efficiency of the Jet,
+ KE=(1/2)*rho*a*V^3; //kinetic energy supplied by jet per second, J
+ eta=W*1000/KE*100; //In percentage
+
+//Result:-
+ printf("(a)Maximum work done/sec=%.3f kJ/s \n", W) //The answer vary due to round off error
+ printf("(b)Effeciency of the Jet, eta=%.2f percent \n", eta) //The answer vary due to round off error
diff --git a/3751/CH2/EX2.14/Ex2_14.sce b/3751/CH2/EX2.14/Ex2_14.sce new file mode 100644 index 000000000..add2f224c --- /dev/null +++ b/3751/CH2/EX2.14/Ex2_14.sce @@ -0,0 +1,27 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.14
+ clc
+ clear
+//Given Data:-
+ Vi=50; //Absolute velocity of Jet at inlet, m/s
+ u=25; //velocity of vane, m/s
+ ui=u;
+ uo=u;
+ alpha_i=32; //Angle made by Vi at inlet, degrees
+ alpha_l=90; //Angle made by Vi at outlet, degrees
+ alpha_o=180-alpha_l; //degrees
+
+//Computations:-
+ Vfi=Vi*sind(alpha_i); //m/s
+ Vwi=Vi*cosd(alpha_i); //m/s
+ Vwi=Vwi-ui; //m/s
+ beta_i=atand(Vfi/Vwi); //degrees
+ Vri=Vfi/sind(beta_i); //m/s
+ Vro=Vri;
+ beta_o=acosd(uo/Vro); //degrees
+
+//Result:-
+ printf("Vane Angle at Inlet, beta_i=%.2f degrees \n", beta_i)
+ printf("Vane angle at outlet, beta_o=%.2f degrees \n", beta_o) //The answer vary due to round off error
+
diff --git a/3751/CH2/EX2.14/Ex2_14_Velocity_Triangles.jpg b/3751/CH2/EX2.14/Ex2_14_Velocity_Triangles.jpg Binary files differnew file mode 100644 index 000000000..ae30c0c1c --- /dev/null +++ b/3751/CH2/EX2.14/Ex2_14_Velocity_Triangles.jpg diff --git a/3751/CH2/EX2.15/Ex2_15.sce b/3751/CH2/EX2.15/Ex2_15.sce new file mode 100644 index 000000000..7ecb4ff7b --- /dev/null +++ b/3751/CH2/EX2.15/Ex2_15.sce @@ -0,0 +1,38 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.15
+ clc
+ clear
+
+//Given Data:-
+ Vi=20; //Absolute velocity of Jet at Inlet of Vane, m/s
+ u=10; //Velocity of Vane, m/s
+ ui=u;
+ uo=u;
+ alpha_i=20; //Angle made by Jet at Inlet, degrees
+ alpha_l=130; //Angle made by Jet at leaving, degrees
+ alpha_o=180-alpha_l; //degrees
+
+//Data Used:-
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ //(a)
+ Vwi=Vi*cosd(alpha_i); //m/s
+ Vfi=Vi*sind(alpha_i); //m/s
+ Vrwi=Vwi-ui; //m/s
+ beta_i=atand(Vfi/Vrwi); //degrees
+ Vri=Vfi/sind(beta_i); //m/s
+ Vro=Vri;
+ beta_o=alpha_o-asind(uo*sind(180-alpha_o)/Vro); //degrees
+ Vwo=Vro*cosd(beta_o)-uo; //degrees
+ //(b)
+ W=(Vwi+Vwo)*u/g; //N-m/N
+
+//Results:-
+ printf("(a)Vane angle at Inlet, beta_i=%.2f degrees \n", beta_i)
+ printf(" Vane angle at Outlet, beta_o=%.2f degrees \n\n", beta_o) //The answer vary due to round off error
+ printf("(b)Work done per second per unit weight of water striking the vane per second=%.2f N-m/N", W) //The answer vary due to round off error
+
+
+
diff --git a/3751/CH2/EX2.16/Ex2_16.sce b/3751/CH2/EX2.16/Ex2_16.sce new file mode 100644 index 000000000..2c4fe48d9 --- /dev/null +++ b/3751/CH2/EX2.16/Ex2_16.sce @@ -0,0 +1,39 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.16
+ clc
+ clear
+
+//Given Data:-
+ Vi=18; //velocity of Jet at Inlet of, m/s
+ u=6; //Velocity of Vane, m/s
+ ui=u;
+ uo=u;
+ AoD=110; //Angle of deflection of the Jet, degrees
+
+//Data Used:-
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ beta_i=(180-AoD)/2;
+ beta_o=beta_i;
+
+ //(a)
+ alpha_i=beta_i-asind(ui*sind(180-beta_i)/Vi); //degrees
+ Vwi=Vi*cosd(alpha_i); //m/s
+ Vfi=Vi*sind(alpha_i); //m/s
+ Vri=Vfi/sind(beta_i); //m/s
+ Vro=Vri;
+ Vfo=Vro*sind(beta_o); //m/s
+ Vrwo=Vro*cosd(beta_o); //m/s
+ Vwo=Vrwo-uo; //m/s
+ alpha_o=atand(Vfo/Vwo); //degrees
+ //(b)
+ alpha_o_dash=180-alpha_o; //degrees
+ //(c)
+ W=(Vwi+Vwo)*u/g; //N-m/N
+//Results:-
+ printf("(a)Angle of Jet at Inlet of Vane, alpha_i=%.2f Degrees \n", alpha_i) //The answer vary due to round off error
+ printf(" Angle at Outlet of Vane, alpha_o=%.2f Degrees \n", alpha_o) //The answer vary due to round off error
+ printf("(b)Angle made by leaving Jet to the direction of motion of Vane, alpha_o_dash=%.2f Degrees \n", alpha_o_dash) //The answer vary due to round off error
+ printf("(c)Work done per second per unit weight of water striking the vane per second=%.2f N-m/N", W) //The answer vary due to round off error
diff --git a/3751/CH2/EX2.17/Ex2_17.sce b/3751/CH2/EX2.17/Ex2_17.sce new file mode 100644 index 000000000..e42604cde --- /dev/null +++ b/3751/CH2/EX2.17/Ex2_17.sce @@ -0,0 +1,38 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.17
+ clc
+ clear
+//Given Data:-
+ d=60; //Diameter of Jet, mm
+ Vi=22; //Absolute Velocity of Jet at Inlet, m/s
+ u=11; //Velocity of vane, m/s
+ ui=u;
+ uo=u;
+ alpha_i=0; //degrees
+ alpha_l=65; //degrees
+ alpha_o=180-alpha_l; //degrees
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+
+ d=d/1000; //m
+ a=(%pi/4)*d^2; //cross-sectional area of Jet, m^2
+ Vwi=Vi; //m/s
+ Vri=Vi-ui; //m/s
+ Vro=Vri;
+ beta_o=alpha_o-asind(uo*sind(alpha_l)/Vro); //degrees
+ Vwo=uo-Vro*cosd(beta_o); //m/s
+ //(a)The Force exerted by Jet on Vane in direction of motion, Fx
+ Fx=rho*a*Vri*(Vwi-Vwo); //N
+ //(b)Power developed by vane,
+ P=Fx*u/1000; //kW
+ //(c)Efficiency of Vane,
+ eta=2*Fx*u/(rho*a*Vi^3)*100; //in Percentage
+
+//Results:-
+ printf("(a)The Force exerted by Jet on Vane in direction of motion, Fx=%.2f N \n", Fx) //The answer vary due to round off error
+ printf("(b)Power developed by vane=%.3f kW \n", P) //The answer vary due to round off error
+ printf("(c)Efficiency of vane, eta=%.2f percent \n", eta) //The answer vary due to round off error
diff --git a/3751/CH2/EX2.18/Ex2_18.sce b/3751/CH2/EX2.18/Ex2_18.sce new file mode 100644 index 000000000..4b9f5a90b --- /dev/null +++ b/3751/CH2/EX2.18/Ex2_18.sce @@ -0,0 +1,36 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.18
+ clc
+ clear
+
+//Given Data:-
+ Vi=18; //velocity of Jet at Inlet of, m/s
+ u=6; //Velocity of Vane, m/s
+ ui=u;
+ uo=u;
+ AoD=120; //Angle of deflection of the Jet, degrees
+
+//Data Used:-
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ beta_i=(180-AoD)/2; //degrees
+ beta_o=beta_i;
+ //(i)
+ alpha_i=beta_i-asind(ui*sind(180-beta_i)/Vi); //degrees
+ //(ii)
+ Vrwi=Vi*cosd(alpha_i)-ui; //m/s
+ Vfi=Vi*sind(alpha_i); //m/s
+ Vri=Vfi/sind(beta_i); //m/s
+ Vro=Vri;
+ Vfo=Vro*sind(beta_o); //m/s
+ Vwo=Vro*cosd(beta_o)-uo; //m/s
+ alpha_o=atand(Vfo/Vwo); //degrees
+ Vo=Vfo/sind(alpha_o); //m/s
+ //(iii)
+ W=(Vi*cosd(alpha_i)+Vwo)*u/g; //N-m/N
+//Results(a):-
+ printf("(i)Angle of Jet at Inlet, alpha_i=%.2f Degrees \n", alpha_i)
+ printf("(ii)Absolute velocity of Jet at Outlet, Vo=%.2f m/s with angle alpha_o=%.2f Degrees \n", Vo,alpha_o) //The answer vary due to round off error
+ printf("(iii)Work done per N of Water=%.2f N-m/N", W) //The answer vary due to round off error
diff --git a/3751/CH2/EX2.19/Ex2_19.sce b/3751/CH2/EX2.19/Ex2_19.sce new file mode 100644 index 000000000..41ebffe4b --- /dev/null +++ b/3751/CH2/EX2.19/Ex2_19.sce @@ -0,0 +1,58 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.19
+ clc
+ clear
+
+//Given Data:-
+ d=40; //Diameter of the Jet, mm
+ V=24; //Velocity of the Jet, m/s
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ d=d/1000; //m
+ a=(%pi/4)*d^2; //cross-sectional area of Jet, m^2
+ //CaseI - Jet strikes normal to a fixed plate
+ //(a)
+ Fx=rho*a*V^2; //N
+ //(b)Work done, W
+ W=0; //As there is no motion of flat plate
+ //(c)
+ eta=0; //Hydraulic efficiency is zero
+
+//Result I:-
+ printf("Case I: \n\t")
+ printf("(a)Force exerted by Jet on the Plate in direction of Jet, Fx=%.2f N \n\t", Fx) //The answer vary due to round off error
+ printf("(b)Work done by Jet per second=%.2f N \n\t", W) //The answer vary due to round off error
+ printf("(c)Hydraulic efficiency of the Jet, eta_H=%.2f percent \n\n", eta) //The answer vary due to round off error
+ //Case II - Jet strikes the moving plate
+ u=10; //Velocity of moving flat plate, m/s
+ //(a)
+ Fx=rho*a*(V-u)^2; //N
+ //(b)
+ W=Fx*u; //N-m/s
+ //(c)
+ eta=2*W/(rho*a*V^3)*100; //In percentage
+ //Result II
+ printf("Case II: \n\t")
+ printf("(a)Force exerted by Jet on the Plate in direction of Jet, Fx=%.2f N \n\t", Fx) //The answer vary due to round off error
+ printf("(b)Work done by Jet per second=%.2f N \n\t", W) //The answer vary due to round off error
+ printf("(c)Hydraulic efficiency of the Jet, eta_H=%.2f percent \n\n", eta) //The answer vary due to round off error
+ //Case III - Jet strikes a series of flat moving plate
+ u=10; //velocity of flat plate, m/s
+ //(a)
+ Fx=rho*a*V*(V-u); //N
+ //((b)
+ W=Fx*u; //N-m/s
+ //(c)
+ eta=W*2/(rho*a*V^3)*100; //In percentage
+ //Result III
+
+ printf("Case III: \n\t")
+ printf("(a)Force exerted by Jet on the Plate in direction of Jet, Fx=%.3f N \n\t", Fx) //The answer vary due to round off error
+ printf("(b)Work done by Jet per second=%.2f N \n\t", W) //The answer vary due to round off error
+ printf("(c)Hydraulic efficiency of the Jet, eta_H=%.2f percent \n\n", eta) //The answer vary due to round off error
+
+
diff --git a/3751/CH2/EX2.2/Ex2_2.sce b/3751/CH2/EX2.2/Ex2_2.sce new file mode 100644 index 000000000..10322f83d --- /dev/null +++ b/3751/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,28 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.2
+ clc
+ clear
+
+//Given Data:-
+ V=25; //Velocity of the Jet, m/s
+ theta=45; //Inclination of the plate with Jet axis, degrees
+ a=30; //cross-sectional area of the Jet, cm^2
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ a=a*10^-4; //m^2
+ //(a) Force normal to the plate is the maximum force of Jet on the plate Fn
+ Fn=rho*a*V^2*sind(theta); //N
+ //(b) Components of the force Fn,
+ Fx=Fn*sind(theta); //N
+ Fy=Fn*cosd(theta); //N
+ //(c) Ratio in which the discharge gets divided
+ Q1_by_Q2=(1+cosd(theta))/(1-cosd(theta));
+//Results:-
+ printf("(a)The Maximum force of the Jet on the plate, Fn=%.2f N \n", Fn) //The answer vary due to round off error
+ printf("(b)Components of the Normal force, Fn are: \n\t")
+ printf("Fx=%.2f N , Fy=%.2f N \n", Fx, Fy) //The answer vary due to round off error
+ printf("(C)The Ratio in which discharge gets divided, Q1/Q2=%.2f \n", Q1_by_Q2) //The answer vary due to round off error
diff --git a/3751/CH2/EX2.20/Ex2_20.sce b/3751/CH2/EX2.20/Ex2_20.sce new file mode 100644 index 000000000..9a06903b5 --- /dev/null +++ b/3751/CH2/EX2.20/Ex2_20.sce @@ -0,0 +1,31 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.20
+ clc
+ clear
+
+//Given Data:-
+ d=40; //Diameter of the Jet, mm
+ V=35; //Absolute Velocity of the Jet, m/s
+ u=18; //Velocity of the curved plate, m/s
+ AoD=165; //Angle of deflection of the Jet, degrees
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+
+//Computations:-
+ d=d/1000; //m
+ a=(%pi/4)*d^2; //cross-sectional area of Jet, m^2
+ theta=180-AoD; //degrees
+ //(a)
+ Fx=rho*a*V*(V-u)*(1+cosd(theta)); //N
+ //(b)Work done by Jet per second, W
+ W=Fx*u; //N-m/s
+ //(c)
+ eta=W*2/(rho*a*V^3)*100; //In percentage
+
+//Results:-
+ printf("(a)Force exerted on the series of curved plates in direction of Jet, Fx=%.2f N \n", Fx) //The answer vary due to round off error
+ printf("(b)Work done by Jet per second=%.2f N-m/s \n", W) //The answer vary due to round off error
+ printf("(c)Efficiency of the Jet, eta=%.2f percent", eta) //The answer vary due to round off error
diff --git a/3751/CH2/EX2.21/Ex2_21.sce b/3751/CH2/EX2.21/Ex2_21.sce new file mode 100644 index 000000000..b868145d7 --- /dev/null +++ b/3751/CH2/EX2.21/Ex2_21.sce @@ -0,0 +1,39 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.21
+ clc
+ clear
+//Given Data:-
+ Vi=30; //velocity of Jet at Inlet of, m/s
+ u=15; //Velocity of Vane, m/s
+ ui=u;
+ uo=u;
+ alpha_i=32; //Angle of Jet at Inlet, degrees
+ alpha=125; //Angle made by Jet at Outlet with direction fo motion of Vanes, degrees
+ alpha_o=180-alpha; //degrees
+
+//Data Used:-
+ g=9.81; //Acceleration due to gravity, m/s^2
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ Vwi=Vi*cosd(alpha_i); //m/s
+ Vfi=Vi*sind(alpha_i); //m/s
+ Vrwi=Vwi-ui; //m/s
+ beta_i=atand(Vfi/Vrwi); //degrees
+ Vri=Vfi/sind(beta_i); //m/s
+ Vro=Vri;
+ beta_o=alpha_o-asind(uo*sind(180-alpha_o)/Vro); //degrees
+
+ Vrwo=Vri*cosd(beta_o); //m/s
+ Vwo=Vrwo-uo; //m/s
+
+ //(a)
+ W=(Vwi+Vwo)*u/g; //N-m/N (Answer in textbook is wrong due to wrong value of Vwi used)
+ //(b)Work done by Jet per second, W
+ eta=2*(Vwi+Vwo)*u/(Vi^2)*100; //In percentage
+
+//Results:-
+ printf("(a)Work done per unit weight of water=%.2f N-m/N \n", W) //The answer provided in the textbook is wrong
+ printf("(b)Efficiency of the vane, eta=%.2f percent", eta) //The answer provided in the textbook is wrong
+
diff --git a/3751/CH2/EX2.21/Ex2_21_Velocity_Triangles.jpg b/3751/CH2/EX2.21/Ex2_21_Velocity_Triangles.jpg Binary files differnew file mode 100644 index 000000000..f827cd0a8 --- /dev/null +++ b/3751/CH2/EX2.21/Ex2_21_Velocity_Triangles.jpg diff --git a/3751/CH2/EX2.22/Ex2_22.sce b/3751/CH2/EX2.22/Ex2_22.sce new file mode 100644 index 000000000..95e334c50 --- /dev/null +++ b/3751/CH2/EX2.22/Ex2_22.sce @@ -0,0 +1,36 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.22
+ clc
+ clear
+//Given Data:-
+ Vi=32; //velocity of Jet at Inlet, m/s
+ u=16; //Velocity of Vane, m/s
+ ui=u;
+ uo=u;
+ alpha_i=22; //Angle of Jet at Inlet, degrees
+ K=0.92; //Co-efficient of Vane
+
+//Data Used:-
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ Vwi=Vi*cosd(alpha_i); //m/s
+ Vfi=Vi*sind(alpha_i); //m/s
+ Vrwi=Vwi-ui; //m/s
+ //(a)
+ beta_i=atand(Vfi/Vrwi); //degrees
+ Vri=Vfi/sind(beta_i); //m/s
+ Vro=K*Vri; //m/s
+ beta_o=acosd(uo/Vro); //degrees
+ //(b)
+ Vwo=0; //m/s(as alpha_o=90 degrees)
+ W=(Vwi+Vwo)*u/g; //N-m/N
+ //(c)
+ eta=2*Vwi*u/Vi^2*100; //In percentage
+
+//Results:-
+ printf("(a)Vane angle at Entrance, beta_i=%.2f degrees \n", beta_i) //The answer vary due to round off error
+ printf(" Vane angle at exit, beta_o=%.2f degrees \n", beta_o) //The answer vary due to round off error
+ printf("(b)Work done on vanes per unit weight of water=%.2f N-m/N \n", W)
+ printf("(c)Efficiency of the system, eta=%.2f percent", eta) //The answer vary due to round off error
diff --git a/3751/CH2/EX2.23/Ex2_23.sce b/3751/CH2/EX2.23/Ex2_23.sce new file mode 100644 index 000000000..70d7bee9f --- /dev/null +++ b/3751/CH2/EX2.23/Ex2_23.sce @@ -0,0 +1,47 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.23
+ clc
+ clear
+//Given Data:-
+ Vi=65; //Absolut velocity of Jet at Inlet, m/s
+ Ri=400; //Inner radius of wheel, mm
+ Ro=800; //outer radius of wheel, mm
+ alpha_i=24; //degrees
+ Vfo=12; //Flow velocity at outlet, m/s
+ beta_i=40; //blade angle at Inlet, degrees
+ beta_o=30; //Blade angle at outlet, degrees
+
+//Data Used:-
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ Ri=Ri/1000; //m
+ Di=2*Ri; //m
+ Ro=Ro/1000; //m
+ Do=2*Ro; //m
+ Vfi=Vi*sind(alpha_i); //m/s
+ Vwi=Vi*cosd(alpha_i); //m/s
+ Vrwi=Vfi/tand(beta_i); //m/s
+ //(a)
+ ui=Vwi-Vrwi; //m/s
+ N=ui*60/(%pi*Do); //rpm
+ omega=2*%pi*N/60; //rad/s
+ uo=%pi*Di*N/60; //m/s
+ Vro=Vfo/sind(beta_o); //m/s
+ Vrwo=Vro*cosd(beta_o); //m/s
+ Vwo=Vrwo-uo; //m/s
+ //(b)
+ W=(Vwi*ui+Vwo*uo)/g; //Work done per unit weight of water, N-m/N
+ //(c)
+ eta=(Vwi*ui+Vwo*uo)*2/Vi^2*100; //In percentage
+
+//Results:-
+ printf("(a)For the speed of wheel: \n\t")
+ printf("N=%.2f rpm \n\t", N) //The answer vary due to round off error
+ printf("Angular velocity, omega=%.2f rad/s \n\t", omega) //The answer vary due to round off error
+ printf("Peripheral velocity of wheel at outlet, uo=%.2f m/s \n\t", uo)
+ printf("Vwo=%.2f m/s \n\n", Vwo) //The answer vary due to round off error
+ printf("(b)Work done per unit weight of water=%.2f N-m/N \n", W) //The answer vary due to round off error
+ printf("(c)Efficiency of the system, eta=%.2f percent", eta) //The answer vary due to round off error
+
diff --git a/3751/CH2/EX2.24/Ex2_24.sce b/3751/CH2/EX2.24/Ex2_24.sce new file mode 100644 index 000000000..c93728a82 --- /dev/null +++ b/3751/CH2/EX2.24/Ex2_24.sce @@ -0,0 +1,38 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.24
+ clc
+ clear
+
+//Given Data:-
+ Do=1.5; //Diameter of rotor at inlet of vane, m
+ Di=1; //Diameter of rotor at outlet of vane, m
+ N=400; //Speed of the rotor, rpm
+ Vi=15; //m/s
+ alpha_i=12; //Nozzle angle at inlet, degrees
+ Vo=5; //m/s
+ VFo=Vo;
+//Data Used:-
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ ui=%pi*Do*N/60; //m/s
+ uo=%pi*Di*N/60; //m/s
+ Vfi=Vi*sind(alpha_i); //m/s
+ Vfo=Vo; //m/s
+ Vwi=Vi*cosd(alpha_i); //m/s
+ //(a)
+ Vrwi=ui-Vwi; //m/s
+ beta_i=180-atand(Vfi/Vrwi); //Blade angle at inlet, degrees
+ beta_o=atand(Vfo/uo); //Blade angle at outlet, degrees
+ //(b)
+ Vro=uo/cosd(beta_o); //m/s
+ //(c)
+ W=Vwi*ui/g; //N-m/N
+
+//Results:-
+ printf("(a)Blade angle at entry and exit are: \n\t")
+ printf("beta_i=%.2f degrees \t beta_o=%.2f degrees \n\n", beta_i,beta_o) //The answer vary due to round off error
+ printf("(b)Velocity of water relative to Vanes at exit, Vro=%.2f m/s \n", Vro) //The answer vary due to round off error
+ printf("(c)Work done per second per unit weight of water strikes on Vane per second=%.2f N-m/N", W) //The answer vary due to round off error
+
diff --git a/3751/CH2/EX2.25/Ex2_25.sce b/3751/CH2/EX2.25/Ex2_25.sce new file mode 100644 index 000000000..58169444a --- /dev/null +++ b/3751/CH2/EX2.25/Ex2_25.sce @@ -0,0 +1,42 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.25
+ clc
+ clear
+
+//Given Data:-
+ Vi=32; //Absolute velocity of Jet at inlet, m/s
+ N=250; //Speed of the wheel, rpm
+ alpha_i=20; //angle of Jet at inlet, degrees
+ Vo=6; //Absolute velocity of Jet at outlet, m/s
+ alpha=132; //Angle made by Jet at outlet with tangent to wheel, degrees
+ alpha_o=180-alpha; //degrees
+ Do=1.2; //outer Diameter of wheel, m
+ Di=0.75; //Inner diameter of wheel, m
+
+//Data Used:-
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ ui=%pi*Do*N/60; //m/s
+ uo=%pi*Di*N/60; //m/s
+ //(a)
+ Vfi=Vi*sind(alpha_i); //m/s
+ Vwi=Vi*cosd(alpha_i); //m/s
+ Vrwi=Vwi-ui; //m/s
+ Vwo=Vo*cosd(alpha_o); //m/s
+ Vrwo=uo+Vwo; //m/s
+ Vfo=Vo*sind(alpha_o); //m/s
+ beta_i=atand(Vfi/Vrwi); //degrees
+ beta_o=atand(Vfo/Vrwo); //degrees
+ //(b)
+ W=(Vwi*ui+Vwo*uo)/g; //N-m/N
+ //(c)
+ eta=2*(Vwi*ui+Vwo*uo)/Vi^2*100; //in percentage
+
+//Results:-
+ printf("(a)Vane angle at Inlet, beta_i=%.2f degrees \n", beta_i) //The answer vary due to round off error
+ printf(" Vane angle at Outlet, beta_o=%.2f degrees \n", beta_o) //The answer vary due to round off error
+ printf("(b)Work done per second per unit weight of water strikes on vane per second=%.2f N-m/N \n", W) //The answer vary due to round off error
+ printf("(c)Efficiency of the wheel, eta=%.2f percent",eta) //The answer vary due to round off error
+
diff --git a/3751/CH2/EX2.26/Ex2_26.sce b/3751/CH2/EX2.26/Ex2_26.sce new file mode 100644 index 000000000..5cc327b8e --- /dev/null +++ b/3751/CH2/EX2.26/Ex2_26.sce @@ -0,0 +1,24 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.26
+ clc
+ clear
+
+//Given Data:-
+ H=4; //Head of water in tank, m
+ d=150; //Diameter of orfice, mm
+ Cv=0.96;
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+
+//Computations:-
+ d=d/1000; //m
+ a=(%pi/4)*d^2; //cross-sectional are of orifice, m^2
+ V=Cv*sqrt(2*g*H); //Velocity of Jet, m/s
+ F=rho*a*V^2; //Force exerted on tank, N
+
+//Results:-
+ printf("The force exerted on the tank=%.2f n", F) //The answer vary due to round off error
diff --git a/3751/CH2/EX2.27/Ex2_27.sce b/3751/CH2/EX2.27/Ex2_27.sce new file mode 100644 index 000000000..cfe8de27e --- /dev/null +++ b/3751/CH2/EX2.27/Ex2_27.sce @@ -0,0 +1,31 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.27
+ clc
+ clear
+
+//Given Data:-
+ H=3.8; //Head of water in tank, m
+ d=200; //Diameter of orfice, mm
+ Cv=0.97;
+ u=2; //Velocity of tank, m/s
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ d=d/1000; //m
+ a=(%pi/4)*d^2; //cross-sectional are of orifice, m^2
+ V=Cv*sqrt(2*g*H); //Velocity of Jet, m/s
+ //(a)
+ F=rho*a*(V+u)*V; //N
+ //(b)
+ W=F*u; //N-m/s
+ //(c)
+ eta=2*V*u/(V+u)^2*100; //in Percentage
+
+//Results:-
+ printf("(a)Propelling Force on tank, F=%.2f N \n", F) //The answer provided in the textbook is wrong
+ printf("(b)Work done by propelling force per second=%.2f N-m/s \n", W) //The answer provided in the textbook is wrong
+ printf("(c)Efficiency of propulsion, eta=%.2f percent", eta) //The answer vary due to round off error
diff --git a/3751/CH2/EX2.28/Ex2_28.sce b/3751/CH2/EX2.28/Ex2_28.sce new file mode 100644 index 000000000..63f77a979 --- /dev/null +++ b/3751/CH2/EX2.28/Ex2_28.sce @@ -0,0 +1,28 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.28
+ clc
+ clear
+
+//Given Data:-
+ V=20; //Absolute Velocity of Jet of Water, m/s
+ a=0.02; //Cross-sectional area of Jet, m^2
+ u=30; //Speed of boat, km/hr
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ u=u*1000/3600; //m/s
+ //(a)
+ Fx=rho*a*(V+u)*V; //N
+ //(b)
+ P=Fx*u/1000; //kW
+ //(c)
+ eta=2*V*u/(V+u)^2*100; //in Percentage
+
+//Results:-
+ printf("(a)Propelling Force on the boat, Fx=%.f N \n", Fx) //The answer vary due to round off error
+ printf("(b)power required to drive the pump=%.2f kW \n", P) //The answer vary due to round off error
+ printf("(c)Efficiency of the Jet propulsion, eta=%.2f percent", eta) //The answer vary due to round off error
+
diff --git a/3751/CH2/EX2.29/Ex2_29.sce b/3751/CH2/EX2.29/Ex2_29.sce new file mode 100644 index 000000000..891bf0407 --- /dev/null +++ b/3751/CH2/EX2.29/Ex2_29.sce @@ -0,0 +1,28 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.29
+ clc
+ clear
+
+//Given Data:-
+ V=20; //Absolute Velocity of Jet of Water, m/s
+ a=0.18; //Cross-sectional area of Jet, m^2
+ u=30; //Speed of boat, km/hr
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ u=u*1000/3600; //m/s
+ //(a)
+ Fx=rho*a*(V+u)*V/1000; //kN
+ //(b)
+ P=Fx*u; //kW
+ //(c)
+ eta=2*V*u/(V+u)^2*100; //in Percentage
+
+//Results:-
+ printf("(a)Propelling Force on the boat, Fx=%.4f kN \n", Fx)
+ printf("(b)power required to drive the pump=%.2f kW \n", P)
+ printf("(c)Efficiency of the Jet propulsion, eta=%.2f percent", eta)
+
diff --git a/3751/CH2/EX2.3/Ex2_3.sce b/3751/CH2/EX2.3/Ex2_3.sce new file mode 100644 index 000000000..5d6d20b51 --- /dev/null +++ b/3751/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,21 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.3
+ clc
+ clear
+//Given Data:-
+ d=40; //Diameter of the Jet, mm
+ V=60; //Velocity of the Jet, m/s
+ AoD=125; //Angle of Deflection, degrees
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ d=d/1000; //m
+ a=(%pi/4)*d^2; //cross-sectional area of Jet, m^2
+ theta=180-AoD; //degrees
+ Fx=rho*a*V^2*(1+cosd(theta)); //N
+//Results:-
+ printf("The Force exerted by the Jet of water in the direction of Jet, Fx=%.2f N \n",Fx) //The answer provided in the textbook is wrong.
+
diff --git a/3751/CH2/EX2.30/Ex2_30.sce b/3751/CH2/EX2.30/Ex2_30.sce new file mode 100644 index 000000000..6c9f79839 --- /dev/null +++ b/3751/CH2/EX2.30/Ex2_30.sce @@ -0,0 +1,23 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.30
+ clc
+ clear
+
+//Given Data:-
+ V=40; //Absolute Velocity of Jet, m/s
+ a=0.04; //Cross-sectional area of Jet, m^2
+ u=40; //Speed of boat, km/hr
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ u=u*1000/3600; //m/s
+ F=rho*a*(V+u)*V; //N
+ eta=2*u/(V+2*u)*100; //in Percentage
+
+//Results:-
+ printf("(a)Propelling Force, F=%.f N \n", F) //The answer vary due to round off error
+ printf("(b)Efficiency of propulsion, eta=%.2f percent", eta)
+
diff --git a/3751/CH2/EX2.31/Ex2_31.sce b/3751/CH2/EX2.31/Ex2_31.sce new file mode 100644 index 000000000..dc27dab4f --- /dev/null +++ b/3751/CH2/EX2.31/Ex2_31.sce @@ -0,0 +1,30 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.31
+//To Find (i)The volume of water drawn by the pump per second (ii)The Efficiency of Jet propulsion.
+ clc
+ clear
+
+//Given Data:-
+ F=5890; //Total resistance offered to motion, N
+ a=424; //Total area of Jet, cm^2
+ u=6; //Speed of boat, m/s
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computaions:-
+ a=a/10000; //m^2
+ //For solving Quadratic in V
+ A=rho*a;
+ B=rho*a*u;
+ C=-F;
+ V=(-B+sqrt(B^2-4*C*A))/(2*A); //m/s
+ //(i)
+ Q=a*(V+u); //m^3/s
+ //(ii)
+ eta=2*V*u/(V+u)^2*100; //In percentage
+
+//Results:-
+ printf("(i)The Volume of water drawn by the pump per second=%.4f m^3/s \n", Q) //The answer vary due to round off error
+ printf("(ii)The Efficiency of Jet propulsion, eta=%.2f percent", eta) //The answer vary due to round off error
diff --git a/3751/CH2/EX2.32/Ex2_32.sce b/3751/CH2/EX2.32/Ex2_32.sce new file mode 100644 index 000000000..be962cb5c --- /dev/null +++ b/3751/CH2/EX2.32/Ex2_32.sce @@ -0,0 +1,24 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.32
+ clc
+ clear
+
+//Given Data:-
+ V=18; //Absolute Velocity of the Jet, m/s
+ a=0.04; //cross-sectional area of Jet, m^2
+ u=28; //Speed of the ship, km/hr
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ u=u*1000/3600; //m/s
+ //(a)
+ F=rho*V*a*(V+u); //N
+ //(b)
+ eta=2*u/(V+2*u)*100; //In percentage
+
+//Results:-
+ printf("(a)Propelling Force, F=%.f N \n", F)
+ printf("(b)The Efficiency of propulsion, eta=%.2f percent \n", eta) //The answer vary due to round off error
diff --git a/3751/CH2/EX2.33/Ex2_33.sce b/3751/CH2/EX2.33/Ex2_33.sce new file mode 100644 index 000000000..f231711da --- /dev/null +++ b/3751/CH2/EX2.33/Ex2_33.sce @@ -0,0 +1,33 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.33
+ clc
+ clear
+
+//Given Data:-
+ a=0.72; //Total cross-sectional area of Jets, m^2
+ Vr=12; //Velocity through the Jet relative to ship, m/s
+ u=6; //Speed of ship, m/s
+ eta_E=85/100; //Efficiency of I.C. engine
+ eta_P=70/100; //Efficiency of Centrifugal Pump
+ Pipe_Loss_per=8; //Percentage of pipe losses (of the kinetic energy of Jet per sec)
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ Pipe_Loss=(Pipe_Loss_per/100)*(rho*a*Vr^3/2); //Pipe Losses, N-m/s
+ V=Vr-u; //Absolute Velocity of the Jet, m/s
+ //(a)
+ F=rho*V*a*(V+u); //N
+ //(b)
+ W=F*u; //Work done by Jet per second, N-m/s
+ OE_P=rho*a*Vr^3/2+Pipe_Loss; //Output energy of pump per sec, N-m/s
+ IP_P=OE_P/eta_P; //Input Energy of pump per sec, N-m/s
+ OE_E=IP_P; //Output of Engine is equal to Input to the pump
+ IE_E=OE_E/eta_E; //Input Energy of Engine per sec, N-m/s
+ eta_o=W/IE_E*100; //Overall Efficiency in percentage
+
+//Results:-
+ printf("(a)Propelling Force=%.f N \n", F)
+ printf("(b)Overall Efficiency, eta_o=%.2f percent", eta_o) //The answer vary due to round off error
diff --git a/3751/CH2/EX2.34/Ex2_34.sce b/3751/CH2/EX2.34/Ex2_34.sce new file mode 100644 index 000000000..9cd9e0867 --- /dev/null +++ b/3751/CH2/EX2.34/Ex2_34.sce @@ -0,0 +1,30 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.34
+ clc
+ clear
+
+//Given Data:-
+ F=100800; //Total resistance offered to motion, N
+ a=0.8; //Total area of Jet, m^2
+ u=5; //Speed of boat, m/s
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ //For solving Quadratic in V
+ A=rho*a;
+ B=rho*a*u;
+ C=-F;
+ V=(-B+sqrt(B^2-4*C*A))/(2*A); //m/s
+ //(a)
+ Q=a*(V+u); //m^3/s
+ //(b)
+ eta=2*V*u/(V+u)^2*100; //In percentage
+
+//Results:-
+ printf("(a)The Volume of water drawn by the pump per second=%.1f m^3/s \n", Q)
+ printf("(b)The Efficiency of Jet propulsion, eta=%.2f percent", eta) //The answer vary due to round off error
+
+
diff --git a/3751/CH2/EX2.35/Ex2_35.sce b/3751/CH2/EX2.35/Ex2_35.sce new file mode 100644 index 000000000..bc8783746 --- /dev/null +++ b/3751/CH2/EX2.35/Ex2_35.sce @@ -0,0 +1,38 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.35
+ clc
+ clear
+
+//Given Data:-
+ Vr=14; //Relative Velocity of ship, m/s
+ a=0.025; //cross-sectional area of Jet, m^2
+ u=32; //Speed of ship, km/hr
+ eta_P=80/100; //Efficiency of pump
+ h_f=2.5; //Frictional Losses, m of water
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ u=u*1000/3600; //m/s
+ //(i)
+ //(a)
+ V=Vr-u; //m/s
+ F=rho*V*a*(V+u); //N
+ //(b)
+ W=F*u; //N-m/s, Value in textbook is wrong due to incorrect value of u ia used.
+ //(ii)
+ E=rho*a*Vr*((Vr^2-u^2)/2+g*h_f); //Actual energy supplied to water per second, N-m/s
+ OE_P=E; //Output fluid energy per second of pump
+ //(a)
+ P=OE_P/eta_P; //Power required to drive the pump, W
+ //(b)
+ eta_o=W/P*100; //In percentage
+
+//Results:-
+ printf("(i) (a)Resistance to the motion of ship, F=%.f N \n", F) //The answer provided in the textbook is wrong
+ printf(" (b)Propulsive work per second=%.2f N-m/s \n\n", W) //The answer provided in the textbook is wrong
+ printf("(ii) (a)Power required to drive the pump=%.2f W \n", P) //The answer provided in the textbook is wrong
+ printf(" (b)Overall Efficiency of propulsion, eta_o=%.2f percent", eta_o) //The answer vary due to round off error
diff --git a/3751/CH2/EX2.4/Ex2_4.sce b/3751/CH2/EX2.4/Ex2_4.sce new file mode 100644 index 000000000..a3fb84a20 --- /dev/null +++ b/3751/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,26 @@ +
+//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.4
+ clc
+ clear
+
+//Given Data:-
+ d=65; //Diameter of the Jet, mm
+ V=45; //Velocity of the Jet, m/s
+ theta_i=35; //Entry angle with horizontal, degrees
+ theta_o=25; //Exit angle with horizontal, degrees
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ d=d/1000; //m
+ a=(%pi/4)*d^2; //cross-sectional area of Jet, m^2
+ Fx=rho*a*V^2*(cosd(theta_i)+cosd(theta_o)); //N
+ Fy=rho*a*V^2*(sind(theta_i)-sind(theta_o)); //N
+//Results:-
+ printf("Force exerted by Jet in horizontal direction, Fx=%.2f N \n", Fx) //The answer provided in the textbook is wrong
+ printf("Force exerted by Jet in vertial direction, Fy=%.3f N(Fy acts upward) or Fy=-%.3f N(Fy acts downward) \n", Fy, Fy) //The answer vary due to round off error
+
+
diff --git a/3751/CH2/EX2.5/Ex2_5.sce b/3751/CH2/EX2.5/Ex2_5.sce new file mode 100644 index 000000000..ed3294af1 --- /dev/null +++ b/3751/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,21 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.5
+ clc
+ clear
+
+//Given Data:-
+ d=30; //Diameter of the Jet, mm
+ V=15; //Velocity of the Jet, m/s
+ W=245.25; //Weight of plate, N
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ d=d/1000; //m
+ a=(%pi/4)*d^2; //cross-sectional area of Jet, m^2
+ theta=asind(rho*a*V^2/W); //degrees
+//Results:-
+ printf("The Angle through which the plate will swing, theta=%.2f degrees \n", theta) //The answer vary due to round off error
+
diff --git a/3751/CH2/EX2.6/Ex2_6.sce b/3751/CH2/EX2.6/Ex2_6.sce new file mode 100644 index 000000000..3c9ad5fcb --- /dev/null +++ b/3751/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,31 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.6
+ clc
+ clear
+//Given Data:-
+ M=13.5; //Mass of plate, kg
+ d=16; //Diameter of the Jet, mm
+ V=20; //Velocity of the Jet, m/s
+ L=300; //Length of Edge of plate, mm
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ d=d/1000; //m
+ L=L/1000; //m
+ W=M*g; //Weight of Plate, N
+ a=(%pi/4)*d^2; //cross sectional area of Jet, m^2
+ //(a)
+ Fx=rho*a*V^2; //Force exerted by Jet normal to plate, N
+ //Taking Moment at 'A',
+ P=Fx*(L/2)/L; //N
+ //(b)
+ theta=asind(rho*a*V^2/W); //Angle of Swing, degrees
+//Results:-
+ printf("(a)Horizontal force applied at Lower edge of plate to keep it vertical, P=%.3f N \n", P) //The answer vary due to round off error
+ printf("(b)Angle of swing, theta=%.2f degrees", theta) //The answer vary due to round off error
+
+
diff --git a/3751/CH2/EX2.7/Ex2_7.sce b/3751/CH2/EX2.7/Ex2_7.sce new file mode 100644 index 000000000..ea35ffc0d --- /dev/null +++ b/3751/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,35 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.7
+
+ clc
+ clear
+
+//Given Data:-
+ W=55.50; //Weight of plate, N
+ V=8; //Velocity of the Jet, m/s
+ d=22; //Diameter of the Jet, mm
+ AG=125; //Distance between centre of gravity of plate from hinge, mm
+ AC=150; //Distance between axis of Jet and hinge, mm
+ theta=35; //Deflection, degrees
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ d=d/1000; //m
+ AC=AC/1000; //m
+ AG=AG/1000; //m
+ a=(%pi/4)*d^2; //cross sectional area of Jet, m^2
+ Fx=rho*a*V^2; //N
+ //Taking moment about hinge point 'A',
+ P=Fx*AC/AG; //N
+ Fn=(W*AG*sind(theta)+P*AG*cosd(theta))/(AC/cosd(theta)); //N
+ V1=sqrt(Fn/(rho*a*cosd(theta))); //Absolute Velocity of Jet, m/s
+ velocity_increase=V1-V; //Velocity Increase of the Jet, m/s
+
+//Results:-
+ printf("(a)Horizontal force applied at centre of gravity to maintain the plate in vertical position, P=%.3f N \n", P) //The answer vary due to round off error
+ printf("(b)Increase in velocity of Jet=%.3f m/s", velocity_increase) //The answer vary due to round off error
+
+
diff --git a/3751/CH2/EX2.8/Ex2_8.sce b/3751/CH2/EX2.8/Ex2_8.sce new file mode 100644 index 000000000..557631668 --- /dev/null +++ b/3751/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,29 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.8
+ clc
+ clear
+
+//Given Data:-
+ d=75; //Diameter of the Jet, mm
+ V=14; //Velocity of the Jet, m/s
+ u=5; //Velocity of plate, m/s
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ d=d/1000; //m
+ a=(%pi/4)*d^2; //cross sectional area of Jet, m^2
+ F=rho*a*(V-u)^2; //N
+ W=F*u; //J/s
+ KE=(1/2)*rho*a*V^3; //N-m/s
+ eta=W/KE*100; //In percentage
+
+//Results:-
+ printf("(a)The Force exerted by the Jet on the plate, F=%.2f N \n", F) //The answer vary due to round off error
+ printf("(b)Work done by the Jet on the plate per second=%.1f N-m/s or J/s \n", W) //The answer vary due to round off error
+ printf("(c)Efficiency of Jet, eta=%.2f percent", eta) //The answer vary due to round off error
+
+
+
diff --git a/3751/CH2/EX2.9/Ex2_9.sce b/3751/CH2/EX2.9/Ex2_9.sce new file mode 100644 index 000000000..5957ec726 --- /dev/null +++ b/3751/CH2/EX2.9/Ex2_9.sce @@ -0,0 +1,30 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 2 - Impact of Jet
+//Example 2.9
+ clc
+ clear
+
+//Given Data:-
+ d=65; //Diameter of the Jet, mm
+ V=20; //Velocity of the Jet, m/s
+ u=8; //Velocity of curved vane, m/s
+ AoD=160; //Angle of Deflection, degrees
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+
+//Computations:-
+ d=d/1000; //m
+ a=(%pi/4)*d^2; //cross-sectional area of Jet, m^2
+ theta=180-AoD; //degrees
+ Fx=rho*a*(V-u)^2*(1+cosd(theta)); //N
+ P=Fx*u/1000; //Power of Jet, KW
+ KE=(1/2)*rho*a*V^3; //Kinetic energy of Jet per second, N-m/s(W)
+ eta=P*1000/KE*100; //In percentage
+
+//Results:-
+ printf("(a)The Force exerted on plate in direction of Jet, Fx=%.2f N \n", Fx) //The answer vary due to round off error
+ printf("(b)Power of Jet=%.3f KW \n", P) //The answer vary due to round off error
+ printf("(c)Efficiency of Jet, eta=%.2f percent", eta)
+
+
diff --git a/3751/CH4/EX4.1/Ex4_1.sce b/3751/CH4/EX4.1/Ex4_1.sce new file mode 100644 index 000000000..bc3c12d25 --- /dev/null +++ b/3751/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,35 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.1
+ clc
+ clear
+
+//Given Data:-
+ P=735.75; //Power Developed, kW
+ H=200; //Head, m
+ N=800; //Speed, rpm
+ eta_O=86/100; //Overall Efficiency
+ d_by_D=1/10; //Ratio of Jet diameter to turbine diameter (d/D)
+ Cv=0.98; //Co-efficienct of velocity
+ Ku=0.45; //Speed ratio
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ Q=P*1000/(rho*g*H*eta_O); //Net discharge, m^3/s
+ //(a)Diameter of Turbine, D
+ D=60*Ku*sqrt(2*g*H)/(%pi*N); //m
+ d=D*d_by_D; //m
+ //(b)The no. of Jets required
+ q=(%pi/4)*d^2*Cv*sqrt(2*g*H); //Discharge of a single Jet, m^3/s
+ n=round(Q/q); //No. of Jets
+ //(c)Diameter of Jet, d
+ d=d_by_D*D; //m
+
+//Results:-
+ printf("(a)Diameter of Turbine, D=%.4f m \n", D) //The answer vary due to round off error
+ printf("(b)The number of Jets required, n=%.f \n", n)
+ printf("(c)Diameter of Jet, d=%.4f m \n", d)
+
diff --git a/3751/CH4/EX4.10/Ex4_10.sce b/3751/CH4/EX4.10/Ex4_10.sce new file mode 100644 index 000000000..19732303c --- /dev/null +++ b/3751/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,51 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.10
+ clc
+ clear
+
+//Given Data:-
+ N=300; //Speed of runner, rpm
+ H=510; //Head, m
+ d=200; //Diameter of the Jet, mm
+ AoD=165; //Angle of Deflection, degrees
+ Vel_per=15; //percentage by which velocity is reduced
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+ Cv=0.98;
+ Ku=0.46;
+
+//Computations:-
+ d=d/1000; //m
+ beta_O=180-AoD; //degrees
+ Vro_by_Vri=1-Vel_per/100; //Vro/Vri
+ Vi=Cv*sqrt(2*g*H); //m/s
+ Vwi=Vi;
+ ui=Ku*sqrt(2*g*H); //m/s
+ uo=ui;
+ u=ui;
+ Vri=Vi-ui; //m/s
+ Vro=Vri*Vro_by_Vri; //m/s
+ Vrwo=Vro*cosd(beta_O); //m/s
+ Vwo=uo-Vrwo; //m/s
+
+ //(i) Water power available at inlet of turbine, P
+ Q=(%pi/4)*d^2*Vi; //m^3.s
+ P=(1/2)*rho*Q*Vi^2/1000; //kW
+ //(ii)Resultant force on the bucket, F
+ F=rho*Q*(Vwi-Vwo)/1000; //kN
+ //(iii)Overall Efficiency, eta_o
+ eta_H=F*u/P; //Hydraulic efficiency
+ //Assume,
+ eta_V=100/100; //Volumetric efficiency is 100%
+ eta_m=98/100 //Mechanical Efficiency is 98%
+
+ eta_O=eta_V*eta_H*eta_m*100; //In percentage
+
+//Results:-
+ printf("(i)Water power available at inlet of turbine=%.2f kW \n", P) //The answer provided in the Textbook is wrong
+ printf("(ii)Resultant force on the bucket, F=%.3f kN \n", F) //The answer vary due to round off error
+ printf("(iii)Overall efficiency, eta_O=%.2f percent", eta_O) //The answer vary due to round off error
+
diff --git a/3751/CH4/EX4.11/Ex4_11.sce b/3751/CH4/EX4.11/Ex4_11.sce new file mode 100644 index 000000000..3cc72e495 --- /dev/null +++ b/3751/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,57 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.11
+
+ clc
+ clear
+
+//Given Data:-
+ N=300; //Speed of runner, rpm
+ H=500; //Head, m
+ d=200; //Diameter of the Jet, mm
+ AoD=165; //Angle of Deflection, degrees
+ Vel_per=15; //percentage by which velocity is reduced
+ Cv=0.98; //Co-efficient of Velocity
+ Ku=0.46; //Speed ratio
+ Loss_per=3; //Percentage of Mechanical losses
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+
+//Computations:-
+ d=d/1000; //m
+ beta_O=180-AoD; //degrees
+ Vro_by_Vri=1-Vel_per/100; //Vro/Vri
+ K=Vro_by_Vri;
+ Vi=Cv*sqrt(2*g*H); //m/s
+ Vwi=Vi;
+ ui=Ku*sqrt(2*g*H); //m/s
+ uo=ui;
+ u=ui;
+ Vri=Vi-ui; //m/s
+ Vro=K*Vri; //m/s
+ Vrwo=Vro*cosd(beta_O); //m/s
+ Vwo=uo-Vrwo; //m/s
+
+ //(a) Water power, WP
+ Q=(%pi/4)*d^2*Vi; //m^3.s
+ WP=rho*Q*g*H/1000; //kW
+
+ //(b)The Force on the bucket in the direction of Jet, F
+ F=rho*Q*(Vwi-Vwo)/1000; //kN
+
+ //(c)Shaft Power, SP
+ Pr=F*u; //Power developed by the Runner, W
+ SP=Pr-Loss_per/100*Pr; //kW
+
+ //(d)Overall Efficiency, eta_o
+ eta_o=SP/WP*100; //In percentage
+
+//Results:-
+ printf("(a) Water power, WP=%.2f kW \n",WP) //The answer provided in the Textbook is wrong
+ printf("(b)The Force on the bucket in the direction of Jet=%.3f kN \n", F) //The answer vary due to round off error
+ printf("(c)Shaft Power, SP=%.3f kW\n",SP) //The answer provided in the Textbook is wrong
+ printf("(d)Overall efficiency, eta_o=%.2f percent", eta_o) //The answer vary due to round off error
+
diff --git a/3751/CH4/EX4.12/Ex4_12.sce b/3751/CH4/EX4.12/Ex4_12.sce new file mode 100644 index 000000000..fb6484d45 --- /dev/null +++ b/3751/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,51 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 4 - Pelton Turbine (Impulse Turbine) +//Example 4.12 + + clc + clear + +//Given Data:- + n=2; //Number of Jets + P=5000; //Shaft Power, HP + N=375; //Speed of Shaft, rpm + Hth=200; //Theoretical Head at Base of Nozzle, m + eta_p=90/100; //Efficiency of Power Transmission + D=1.65; //Diameter of the Runner, m + Vel_per=10; //Percentage by which velocity is decreased + Deflection=165; //Jet Deflection, degrees + eta_o=90/100; //Overall Efficiency + Cv=0.98; + + +//Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + P=P*736; //W + Hact=eta_p*Hth; //Actual Head available at base of Nozzle, m + Vro_by_Vri=1-Vel_per/100; //Vro/Vri + beta_o=180-Deflection; //degrees + + u=%pi*D*N/60; //Velocity of Runner, m/s + ui=u; + uo=u; + Vi=Cv*sqrt(2*g*Hact); //m/s + Vwi=Vi; + Vri=Vi-u; //m/s + Vro=Vri*Vro_by_Vri; //m/s + Vrwo=Vro*cosd(beta_o); //m/s + Vwo=uo-Vrwo; //m/s + + //(a)Efficiency of Runner, eta_H + eta_H=2*(Vwi-Vwo)*u/Vi^2*100; //In Perecentage + + //(b)Diameter of each jet, d + Q=P/(rho*g*Hact*eta_o); //Discharge, m^3/s + d=sqrt(Q/((%pi/4)*n*Vi)); //Diameter of each Jet, m + +//Results:- + printf("(a)Efficiency of the Runner, eta_H=%.2f Percent\n",eta_H) //The answer vary due to round off error + printf("(b)Diameter of each Jet , d=%.3f m\n",d) + diff --git a/3751/CH4/EX4.13/Ex4_13.sce b/3751/CH4/EX4.13/Ex4_13.sce new file mode 100644 index 000000000..ba0e8dad9 --- /dev/null +++ b/3751/CH4/EX4.13/Ex4_13.sce @@ -0,0 +1,45 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 4 - Pelton Turbine (Impulse Turbine) +//Example 4.13 + + clc + clear + +//Given Data:- + H=62; //Effective Head, m + N=225; //Speed of Runner, rpm + P=133.15; //Shaft Power, HP + Ku=0.45; //Speed Ratio + eta_o=86/100; //Overall Efficiency + Cv=0.98; + + +//Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + P=P*736; //W + + Vi=Cv*sqrt(2*g*H); //m/s + u=Ku*sqrt(2*g*H); //m/s + ui=u; + uo=u; + Q=P/(rho*g*H*eta_o); //m^3/s + + d=sqrt(Q/((%pi/4)*Vi))*1000; //Diameter of Jet, mm + D=60*u/(%pi*N); //Diameter of Runner, m + //As per designing range, b=3*d to 4*d + b=3.5*d; //Width of Buckets, mm + //As per designing range, b=0.8*d to 1.2*d + T=1.2*d; //Depth of Buckets, mm + Z=round(0.5*D/(d/1000)+15); //Number of Buckets + +//Results:- + printf(" (a)Diameter of Jet, d=%.2f mm \n",d) //The answer vary due to round off error + printf(" (b) Diameter of Runner, D=%.3f m \n",D) + printf(" (c) Width of Buckets, b=%.2f mm \n",b) //The answer vary due to round off error + printf(" (d) Depth of Buckets, T=%.2f mm \n",T) //The answer vary due to round off error + printf(" (e) Number of Buckets , Z=%.f \n",Z) + + diff --git a/3751/CH4/EX4.14/Ex4_14.sce b/3751/CH4/EX4.14/Ex4_14.sce new file mode 100644 index 000000000..b752a5136 --- /dev/null +++ b/3751/CH4/EX4.14/Ex4_14.sce @@ -0,0 +1,61 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 4 - Pelton Turbine (Impulse Turbine) +//Example 4.14 + + clc + clear + +//Given Data:- + H=452; //Net Head, m + m=12; //Jet Ratio (D/d) + Ku=0.46; //Speed Ratio + AoD=165; //Angle of Jet Deflection, degrees + Cv=0.98; //Co-efficient of Velocity + Loss_f=15; //Percentage of Friction Loss of Buckets + eta_o=86/100; //Overall Efficiency + P_G=10200; //Power developed by Generator, HP + eta_G=95/100; //Generator Efficiency + + +//Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + P_G=P_G*736; //W + Vro_by_Vri=1-Loss_f/100; //Vro/Vri + beta_o=180-AoD; //degrees + + u=Ku*sqrt(2*g*H); //Velocity of Runner, m/s + ui=u; + uo=u; + Vi=Cv*sqrt(2*g*H); //m/s + Vwi=Vi; + Vri=Vi-ui; //m/s + Vro=Vri*Vro_by_Vri; //m/s + Vrwo=Vro*cosd(beta_o); //m/s + Vwo=uo-Vrwo; //m/s + + P=P_G/eta_G; //Shaft Power, W + Q=P/(rho*g*H*eta_o); //Discharge, m^3/s + + //(a) + d=sqrt(Q/((%pi/4)*Vi)); //Diameter of Jet, m + + //(b) + D=m*d; //Diameter of Runner, m + + //(c) + Pr=rho*Q*(Vwi-Vwo)*u/1000; // Power developed by Runner, kW + + //(d) + eta_m=P/(Pr*1000)*100; //Mechanical Efficiency in Percentage + + +//Results:- + printf("(a) Diameter of the Jet, d=%.3f m\n",d) + printf(" (b)Diameter of the Runner, D=%.3f m\n",D) //The answer vary due to round off error + printf(" (c)Power Developed by the Runner=%.3f kW\n",Pr) //The answer provided in the textbook is wrong + printf(" (d)Mechanical Efficiency , eta_m=%.2f Percent\n",eta_m) //The answer vary due to round off error + + diff --git a/3751/CH4/EX4.15/Ex4_15.sce b/3751/CH4/EX4.15/Ex4_15.sce new file mode 100644 index 000000000..2ce45324a --- /dev/null +++ b/3751/CH4/EX4.15/Ex4_15.sce @@ -0,0 +1,35 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 4 - Pelton Turbine (Impulse Turbine) +//Example 4.15 + + clc + clear + +//Given Data:- + H=120; //Head, m + d=74; //Diameter of Jet, mm + Q=200; //Discharge, litres/s + P=202.766; //Shaft Power, kW + P_mr=3.2; //Power lost in mechanical resistance, kW + + +//Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Q=Q/1000; //m^3/s + d=d/1000; //m + P=P*1000; //W + P_mr=P_mr*1000; //W + + Vi=Q/((%pi/4)*d^2); //m/s + P_n=(rho*Q*g*H-rho*Q*Vi^2/2)/1000; //Power lost in Nozzle, kW + P_hr=(rho*Q*g*H-(P+P_n*1000+P_mr))/1000; //Power lost due to hydraulic resistance in Runner, kW + + +//Results:- + printf("(a) Power lost in Nozzle=%.3f kW\n",P_n) //The answer vary due to round off error + printf(" (b)Power lost due to Hydraulic Resistance in Runner =%.2f kW\n",P_hr) //The answer vary due to round off error + + diff --git a/3751/CH4/EX4.16/Ex4_16.sce b/3751/CH4/EX4.16/Ex4_16.sce new file mode 100644 index 000000000..5bcd8666d --- /dev/null +++ b/3751/CH4/EX4.16/Ex4_16.sce @@ -0,0 +1,26 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 4 - Pelton Turbine (Impulse Turbine) +//Example 4.16 + + clc + clear + +//Given Data:- + P=4900; //Shaft Power, kW + P_mr=100; //Power absorbed in mechanical resistance, kW + eta_H=92/100; //Hydraulic Efficiency + P_n=415; //Power lost in Nozzle, kW + + + //Computations:- + P_rd=P+P_mr; //Power Devrloped by Runner, kW + P_rs=P_rd/eta_H; //Power Supplied to Runner, kW + P_an=P_n+P_rs; //Power Available at base of Nozzle, kW + eta_o=P/P_an*100; //Overall Efficiency in Percentage + +//Results:- + printf("(a)Power Available at the Base of Nozzle=%.3f kW\n",P_an) //The answer vary due to round off error + printf("(b)Overall Efficiency, eta_o=%.2f Percent\n",eta_o) + + + diff --git a/3751/CH4/EX4.17/Ex4_17.sce b/3751/CH4/EX4.17/Ex4_17.sce new file mode 100644 index 000000000..6eeaa4c4c --- /dev/null +++ b/3751/CH4/EX4.17/Ex4_17.sce @@ -0,0 +1,41 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.17
+
+ clc
+ clear
+
+//Given Data:-
+ H_G=510; //Gross Head, m
+ h_f=(1/3)*H_G; //Head lost in friction in penstock, m
+ d=170; //Diameter of Jet, mm
+ AoD=165; //Angle of Deflection of Jet, degrees
+ Ku=0.45; //Speed ratio
+ Cv=0.98; //Co-efficient of Velocity
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ H=H_G-h_f; //Effective Head, m
+ Vi=Cv*sqrt(2*g*H); //m/s
+ Vwi=Vi;
+ u=Ku*sqrt(2*g*H); //m/s
+ ui=u;
+ uo=u;
+ Vri=Vi-u; //m/s
+ Vro=Vri;
+ beta_o=180-AoD; //degrees
+ Vrwo=Vro*cosd(beta_o); //m/s
+ Vwo=Vrwo-uo; //m/s
+ Q=(%pi/4)*(d/1000)^2*Vi; //Discharge, m^3/s
+ P=rho*Q*(Vwi+Vwo)*u/1000; //Power developed by runner, kW
+ eta_H=2*(Vwi+Vwo)*u/Vi^2*100; //Hydraulic efficiency, In percentage
+
+//Results:-
+ printf("(a)Power developed by the runner=%.3f kW \n",P) //The answer provided in the Textbook is wrong
+ printf("(b)Hydraulic efficiency, eta_H=%.2f percent", eta_H) //The answer vary due to round off error
+
+
+
diff --git a/3751/CH4/EX4.18/Ex4_18.sce b/3751/CH4/EX4.18/Ex4_18.sce new file mode 100644 index 000000000..420110ebd --- /dev/null +++ b/3751/CH4/EX4.18/Ex4_18.sce @@ -0,0 +1,45 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 4 - Pelton Turbine (Impulse Turbine) +//Example 4.18 + + clc + clear + +//Given Data:- + Ns=15; //Specific Speed + P=1200; //Shaft Power, kW + Ht=500; //Total Head at reservoir, m + Loss_per=5; //Percentage of Head loss in Pipe friction + Cv=0.98; //Co-efficient of Velocity + Ku=0.45; //Speed Ratio + eta_o=85/100; //Overall Efficiency + n=2; //Number of Jets + +//Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + H=Ht-Loss_per/100*Ht; //Effective Head, m + + //(a)Speed of Runner, N + N=Ns*H^(5/4)/sqrt(P/n); //rpm + + //(b)Diameter od each Jet, d + Q=P*1000/(rho*g*H*eta_o); //Net Discharge, m^3/s + q=Q/n; //Net Discharge per Jet, m^3/s + Vi=Cv*sqrt(2*g*H); //m/s + d=sqrt(q/((%pi/4)*Vi)); //m + + //(c)Mean Diameter of Bucket Circle, D + D=Ku*60*sqrt(2*g*H)/(%pi*N); //m + + //(d)Number of Buckets in the Runner, Z + Z=round(0.5*D/d+15); + +//Results:- + printf(" (a)Speed of the Runner, N=%.f rpm\n",N) + printf(" (b)Diameter od each Jet, d =%.3f m\n",d) + printf(" (c)Mean Diameter of Bucket Circle, D =%.3f m\n",D) + printf(" (d)Number of Buckets on the Runner, Z =%.f \n",Z) + diff --git a/3751/CH4/EX4.19/Ex4_19.sce b/3751/CH4/EX4.19/Ex4_19.sce new file mode 100644 index 000000000..3cc34361a --- /dev/null +++ b/3751/CH4/EX4.19/Ex4_19.sce @@ -0,0 +1,43 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 4 - Pelton Turbine (Impulse Turbine) +//Example 4.19 + + clc + clear + +//Given Data:- + Q=2.5; //Total Discharge, m^3/s + Hr=300; //Head from reservoir to base of nozzle, m + n=6; //Total number of Jets + L=1200; //Lenght of Pipe, m + eta_p=92/100; //Efficiency of Power Transmission + eta_o=86/100; //Overall Efficiency + Cv=0.97; //Co-efficient of Velocity + f=0.0025; //Darcy Co-efficient of Friction + + +//Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + h_f=(1-eta_p)*Hr; //m + H=Hr-h_f; //Effective Head, m + Vi=Cv*sqrt(2*g*H); //Velocity of Jet, m/s + + //(a)Shaft Power, P + P=rho*Q*g*H*eta_o/1000; //kW + + //(b)Diameter of the Jet, d + q=Q/n; //Discharge per Jet, m^3/s + d=sqrt(q/((%pi/4)*Vi)); //m + + //(c)Diameter of the Pipe, D_pipe + D_pipe=(64*f*L*Q^2/(h_f*2*g*%pi^2))^(1/5)*1000; //mm + +//Results:- + printf(" (a)Shaft Power, P=%.3f kW\n",P) + printf(" (b)Diameter of the Jet, d=%.4f m\n",d) + printf(" (c)Diameter of the Pipe, D_pipe=%.2f mm\n",D_pipe) //The answer vary due to round off error + + diff --git a/3751/CH4/EX4.2/Ex4_2.sce b/3751/CH4/EX4.2/Ex4_2.sce new file mode 100644 index 000000000..7901582b4 --- /dev/null +++ b/3751/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,38 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.2
+ clc
+ clear
+
+//Given Data:-
+ u=12; //Speed of bucket, m/s
+ ui=u;
+ uo=u;
+ Q=650; //Discharge, liters/s
+ H=40; //Head of water, m
+ AoD=162; //Angle of Deflection, degrees
+ Cv=0.98; //Co-efficient of Velocity
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ Q=Q/1000; //m^3/s
+ beta_O=180-AoD; //Blade angle a outlet, degrees
+ Vi=Cv*sqrt(2*g*H); //Velocity of Jet, m/s
+ Vwi=Vi;
+ Vri=Vi-ui; //m/s
+ Vro=Vri;
+ Vrwo=Vro*cosd(beta_O); //m/s
+ Vwo=Vrwo-uo; //m/s
+ //(a)Power given by water to runner, P
+ P=rho*Q*(Vwi+Vwo)*u/1000; //kW
+ //(b)The hydraulic efficiency, eta_H
+ eta_H=2*(Vwi+Vwo)*u/Vi^2*100; //In percentage
+
+//Results:-
+ printf("(a)The Power given by water to the runner=%.3f kW \n", P) //The answer vary due to round off error
+ printf("(b)The Hydraulic Efficiency of Turbine, eta_H=%.2f percent \n", eta_H) //The answer vary due to round off error
+
+
diff --git a/3751/CH4/EX4.20/Ex4_20.sce b/3751/CH4/EX4.20/Ex4_20.sce new file mode 100644 index 000000000..184934bd5 --- /dev/null +++ b/3751/CH4/EX4.20/Ex4_20.sce @@ -0,0 +1,42 @@ +//Fluid Systems - By - Shiv Kumar +//Chapter 4 - Pelton Turbine (Impulse Turbine) +//Example 4.20 + + clc + clear + +//Given Data:- + D=1.6; //Mean Diameter of Bucket Circle, m + P=3200; //Power Developed, kW + n=2; //Number of Wheels + H=300; //Effective Head, m + N=410; //Speed, rpm + eta_o=0.89; //Overall Efficiency + Cv=0.98; //Co-efficient of Velocity + + +//Data Used:- + rho=1000; //Density of water, kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Q=P*1000/(rho*g*H*eta_o); //Discharge, m^3/s + + //(a)Diameter of the Nozzle, d + Vi=Cv*sqrt(2*g*H); //m/s + d=sqrt(Q/((%pi/4)*Vi))*1000; //mm + + //(b)Speed Ratio, Ku + u=%pi*D*N/60; //m/s + Ku=u/sqrt(2*g*H); + + //(c)Specific Speed, Ns + Ns=N*sqrt(P/n)/(H^(5/4)); // In SI Units + + +//Results:- + printf(" (a)Diameter of the Nozzle, d=%.2f mm\n",d) //The answer vary due to round off error + printf(" (b)Speed Ratio, Ku =%.3f \n",Ku) //The answer vary due to round off error + printf(" (c)Specific Speed, Ns =%.f (SI Units)\n",Ns) + + diff --git a/3751/CH4/EX4.3/Ex4_3.sce b/3751/CH4/EX4.3/Ex4_3.sce new file mode 100644 index 000000000..fc2fc5073 --- /dev/null +++ b/3751/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,44 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.3
+ clc
+ clear
+
+//Given Data:-
+ H=30; //Effective Head, m
+ AoD=165; //Jet Deflection Angle, degrees
+ Cv=0.98; //Co-efficient of Velocity
+ Ku=0.45; //Speed ratio
+ d=22; //Diameter of Jet, mm
+ //As relative velocity at outlet is 0.98 times relative velocity at inlet,
+ Vro_by_Vri=0.98; // Vro/Vri
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ d=d/1000; //m
+ beta_O=180-AoD; //degrees
+ Vi=Cv*sqrt(2*g*H); //Absolut Velocity of Jet, m/s
+ Vwi=Vi;
+ u=Ku*sqrt(2*g*H); //peripheral velocity of runner, m/s
+ ui=u;
+ uo=u;
+ Vri=Vi-ui; //m/s
+ Vro=Vro_by_Vri*Vri; //m/s
+ Vrwo=Vro*cosd(beta_O); //m/s
+ Vwo=Vrwo-uo; //m/s
+
+ //(a)Power given by water to runner, P
+ Q=(%pi/4)*d^2*Vi; //m^3/s
+ P=rho*Q*(Vwi+Vwo)*u/1000; //kW
+
+ //(b)The hydraulic efficiency, eta_H
+ eta_H=2*(Vwi+Vwo)*u/Vi^2*100; //In percentage
+
+//Results:-
+ printf("(a)The Power given by water to the runner=%.3f kW \n", P) //The answer vary due to round off error
+ printf("(b)The Hydraulic Efficiency, eta_H=%.2f percent \n", eta_H) //The answer vary due to round off error
+
+
diff --git a/3751/CH4/EX4.4/Ex4_4.sce b/3751/CH4/EX4.4/Ex4_4.sce new file mode 100644 index 000000000..73f338fc5 --- /dev/null +++ b/3751/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,39 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.4
+ clc
+ clear
+
+//Given Data:-
+ Cv=0.97;
+ Ku=0.46;
+ K=0.98;
+ m=10.2;
+ beta_o=10; //Bucket angle at exit, degrees
+ eta_m=90.5/100; //Mechanical Efficiency
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ Vi_by_rootH=Cv*sqrt(2*g); //Vi/sqrt(H)
+ Vwi_by_rootH=Vi_by_rootH;
+ ui_by_rootH=Ku*sqrt(2*g); //ui/sqrt(H)
+ Vri_by_rootH=Vi_by_rootH-ui_by_rootH; //Vi/sqrt(H)
+ Vro_by_rootH=K*Vri_by_rootH; //Vro/sqrt(H)
+ Vrwo_by_rootH=Vro_by_rootH*cosd(beta_o); //Vrwo/sqrt(H)
+ Vwo_by_rootH=Vrwo_by_rootH-ui_by_rootH; //Vwo/sqrt(H)
+ Q_by_d2_rootH=(%pi/4)*Vi_by_rootH; //Q/(d^2*sqrt(H))
+ //Pr=Power developed by runner
+ Pr_by_d2_H3_2=rho*Q_by_d2_rootH*(Vwi_by_rootH+Vwo_by_rootH)*ui_by_rootH; //Pr/(d^2*H^(3/2)), P in W
+ //P=Shaft Power
+ P_by_d2_H3_2=eta_m*Pr_by_d2_H3_2/1000; //P/(d^2*H^(3/2)), P in kW
+ N_d_by_rootH=ui_by_rootH*60/(%pi*m); //N*d/sqrt(h), N in rpm
+ Ns=N_d_by_rootH*sqrt(P_by_d2_H3_2); //Specific Speed in SI Units
+
+//Results:-
+ printf("The Specific Speed of the Turbine, Ns=%.f (SI Units)", Ns)
+
+
+
diff --git a/3751/CH4/EX4.5/Ex4_5.sce b/3751/CH4/EX4.5/Ex4_5.sce new file mode 100644 index 000000000..8e86c9580 --- /dev/null +++ b/3751/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,30 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.5
+ clc
+ clear
+
+//Given Data:-
+ n=2; //Number of Jets
+ P=15450; //Shaft Power, kW
+ d=200; //Diameter of each Jet, mm
+ H=400; //Net Head, m
+ Cv=1;
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ P=P*1000; //W
+ d=d/1000; //m
+ Vi=Cv*sqrt(2*g*H); //Absolute Velocity of Jet at Inlet, m/s
+ q=(%pi/4)*d^2*Vi; //Discharge through each Jet, m^3/s
+ Q=n*q; //Net Discharge
+ eta_O=P/(rho*Q*g*H)*100; //Overall Efficiency, in percentage
+
+//Results:-
+ printf("The Overall Efficiency of the Turbine, eta_o=%.2f percent", eta_O) //The answer vary due to round off error
+
+
+
diff --git a/3751/CH4/EX4.6/Ex4_6.sce b/3751/CH4/EX4.6/Ex4_6.sce new file mode 100644 index 000000000..9664981c7 --- /dev/null +++ b/3751/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,57 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.6
+ clc
+ clear
+//Given Data:-
+ N=300; //Speed of runner, rpm
+ H=510; //Head, m
+ d=200; //Diameter of the Jet, mm
+ AoD=165; //Angle of Jet(Deflection inside bucket), degrees
+ Vel_per=15; //Percentage by which velocity is reduced due to friction
+ Loss_per=3; //Percentage of mechanical Losses (of power Supplied)
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+ Cv=0.98; //Co-efficient of Velocity
+ Ku=0.46; //Speed ratio
+
+//Computations:-
+ d=d/1000; //m
+ beta_O=180-AoD; //degrees
+ Vro_by_Vri=1-Vel_per/100; //Vro/Vri
+ Vi=Cv*sqrt(2*g*H); //m/s
+ Vwi=Vi;
+ u=Ku*sqrt(2*g*H); //m/s
+ uo=u;
+ ui=u;
+ Vri=Vi-ui; //m/s
+ Vro=Vri*Vro_by_Vri; //m/s
+ Vrwo=Vro*cosd(beta_O); //m/s
+ Vwo=uo-Vrwo; //m/s
+ Q=(%pi/4)*d^2*Vi; //Discharge, m^3/s
+ //(i) Resultant Force on bucket, F
+ F=rho*Q*(Vwi-Vwo)/1000; //kN
+
+ //Result (i):-
+ printf("(i) Resultant Force on bucket, F=%.3f kN \n", F) //The answer vary due to round off error
+
+ //(ii) Shaft Power, P
+ Pr=F*u; //power developed by runner, kW
+ P=Pr-(Loss_per/100)*Pr; //kW
+
+ //Result (ii)
+ printf("(ii)Shaft Power, P=%.3f kW \n", P) //The answer given in the textbook is wrong (due to round off error in F)
+ //OR
+ eta_m=1-Loss_per/100; //Mechanical Efficiency
+ P=eta_m*Pr; //kW
+
+ //(iii) Overall Efficiency, eta_O
+ eta_O=P*1000/(rho*Q*g*H)*100; //In percentage
+ //Result (iii)
+ printf("(iii)Overall efficiency, eta_O=%.2f percent", eta_O)
+
+
+
+
diff --git a/3751/CH4/EX4.7/Ex4_7.sce b/3751/CH4/EX4.7/Ex4_7.sce new file mode 100644 index 000000000..e9d73eb21 --- /dev/null +++ b/3751/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,39 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.7
+ clc
+ clear
+
+//Given Data:-
+ H_G=500; //Gross Head, m
+ h_f=(1/3)*H_G; //Head lost in friction in penstock, m
+ Q=2; //Discharge, m^3/s
+ AoD=165; //Angle of Deflection of Jet, degrees
+ Ku=0.45; //Speed ratio
+ Cv=1;
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ H=H_G-h_f; //Working Head, m
+ Vi=Cv*sqrt(2*g*H); //m/s
+ Vwi=Vi;
+ u=Ku*sqrt(2*g*H); //m/s
+ ui=u;
+ uo=u;
+ Vri=Vi-u; //m/s
+ Vro=Vri;
+ beta_o=180-AoD; //degrees
+ Vrwo=Vro*cosd(beta_o); //m/s
+ Vwo=Vrwo-uo; //m/s
+ P=rho*Q*(Vwi+Vwo)*u/1000; //power given by water to runner, kW
+ eta_H=2*(Vwi+Vwo)*u/Vi^2*100; //Hydraulic efficiency, In percentage
+
+//Results:-
+ printf("Power given by water to the runner=%.2f kW \n", P) //The answer vary due to round off error
+ printf("Hydraulic efficiency, eta_H=%.2f percent", eta_H) //The answer vary due to round off error
+
+
+
diff --git a/3751/CH4/EX4.8/Ex4_8.sce b/3751/CH4/EX4.8/Ex4_8.sce new file mode 100644 index 000000000..90e1e33af --- /dev/null +++ b/3751/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,49 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.8
+ clc
+ clear
+
+//Given Data:-
+ Cv=0.97; //Co-efficient of Velocity
+ H_l=400; //Head at lake, m
+ d=80; //Diameter of Jet, mm
+ D_pipe=0.6; //Diameter of pipe, m
+ l=4; //Length of pipe, m
+ f_dash=0.032; //Friction factor
+ AoD=165; //Angle of Deflection, degrees
+ beta_o=180-AoD; //degrees
+ // As bucket runs at 0.48 Jet speed
+ u_by_Vi=0.48; //u/Vi
+ Vel_per=15; //percentage by which velocity is reduced
+ eta_m=90/100; //Mechanical Efficiency
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ d=d/1000; //m
+ l=l*1000; //m
+ Vro_by_Vri=1-Vel_per/100; //Vro/Vri
+ //using continuity equation,
+ V_by_Vi=(d/D_pipe)^2; //V/Vi
+ Vi=sqrt((2*g*H_l)/(1/Cv^2+f_dash*l*V_by_Vi^2/D_pipe)); //m/s
+ Vwi=Vi;
+ u=Vi*u_by_Vi; //m/s
+ ui=u;
+ uo=u;
+ Vri=Vi-ui; //m/s
+ Vro=Vri*Vro_by_Vri; //m/s
+ Vrwo=Vro*cosd(beta_o); //m/s
+ //(i) Flow Rate, Q
+ Q=(%pi/4)*d^2*Vi; //m^3/s
+ //(ii) Shaft Power, P
+ Vwo=uo-Vrwo; //m/s
+ Pr=rho*Q*(Vwi-Vwo)*u/1000; //Power developed by the runner, kW
+ P=eta_m*Pr; //kW
+
+//Results:-
+ printf("Flow Rate, Q=%.4f m^3/s \n", Q) //The answer vary due to round off error
+ printf("Shaft power, P=%.2f kW", P) //The answer vary due to round off error
+
diff --git a/3751/CH4/EX4.9/Ex4_9.sce b/3751/CH4/EX4.9/Ex4_9.sce new file mode 100644 index 000000000..5d0fc182b --- /dev/null +++ b/3751/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,32 @@ +//Fluid system - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.9
+ clc
+ clear
+
+//Given Data:-
+ P=3000; //Power developed, kW
+ H=300; //Head, m
+ N=375; //Speed, rpm
+ eta_O=83/100; //Overall efficiency
+ Ku=0.46; //Speed ratio
+ Cv=0.98; //Co-efficient of Velocity
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ //(i) Diameter of Turbine, D
+ D=60*Ku*sqrt(2*g*H)/(%pi*N); //m
+ //(ii) Diameter of Jet, d
+ Q=P*1000/(rho*g*H*eta_O); //m^3/s
+ Vi=Cv*sqrt(2*g*H); //m/s
+ d=(Q/((%pi/4)*Vi))^(1/2); //m
+
+//Results:-
+ printf("(i) Diameter of the Turbine, D=%.2f m \n", D) //The answer vary due to round off error
+ printf("(ii) Diameter of the Jet, d=%.4f m", d) //The answer vary due to round off error
+
+
+
diff --git a/3751/CH5/EX5.1/Ex5_1.sce b/3751/CH5/EX5.1/Ex5_1.sce new file mode 100644 index 000000000..b024b2fc8 --- /dev/null +++ b/3751/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,23 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.1 +//To Find (a)Discharge passing through the Runner and (b) Width of Runner at Outlet + + clc + clear + +//Given Data:- + Do=0.8; //External Diameter of the Runner, m + Di=0.4; //Internal Diameter of the Runner, m + Vfi=1.4; //Velocity of Flow at Inlet, m/s + Vfo=Vfi; //Velocity of Flow at Outlet, m/s + bo=210; //Width of Runner at Inlet, mm + +//Computations:- + Q=%pi*Do*(bo/1000)*Vfi; //Discharge passing through the Runner, m^3/s + bi=Do*bo/Di; //Width of Runner at Outlet, mm + +//Results + printf("(a)Discharge passing through the Runner, Q=%.4f m^3/s\n",Q) //The Answer Vary due to Round off Error + printf("(b)Width of Runner at Outlet, bi=%.f mm",bi) + diff --git a/3751/CH5/EX5.10/Ex5_10.sce b/3751/CH5/EX5.10/Ex5_10.sce new file mode 100644 index 000000000..c991d4b00 --- /dev/null +++ b/3751/CH5/EX5.10/Ex5_10.sce @@ -0,0 +1,44 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.10 +//To Determine (i) The Guide Blade Angle (ii) The Wheel Vane Angle at Inlet (iii) Diameter of Wheel at inlet (iv)Width of Wheel at Inlet + + clc + clear + +//Given Data:- + //Data Required + rho=1000; //Density of water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + + eta_o=75/100; //Overall Efficiency + P=148.25; //Power Produced, kW + H=7.62; // Working Head, m + ui=0.26*sqrt(2*g*H); //Peripheral Velocity at Inlet, m/s + Vfi=0.96*sqrt(2*g*H); //Velocity of Flow at Inlet, m/s + N=150; //Speed, rpm + H_loss=22; //Percentage of Hydraulic Losses in the Turbine (of Available Energy) + //As Discharge is Radial, + alpha_o=90; //Degrees //Vfo=Vo + Vwo=0; + +//Computations:- + Do=ui*60/(%pi*N); //m + Q=P*1000/(rho*g*H*eta_o); //m^3/s + bo=Q/(%pi*Do*Vfi); //m + + //By Energy Balance Equation, + Vwi=(g*H-(H_loss/100)*g*H)/ui; //m/s + + alpha_i=atand(Vfi/Vwi); //degrees + beta_i=atand(Vfi/(Vwi-ui)); //degrees + + +//Results:- + printf(" (i)The Guide Blade Angle , alpha_i=%.2f Degrees\n",alpha_i) + printf(" (ii) The Wheel Vane Angle at Inlet, beta_i =%.2f Degrees\n", beta_i ) //The Answer Vary due to Round off Error + printf(" (iii) Diameter of Wheel at Inlet, Do =%.4f m\n",Do) //The Answer Vary due to Round off Error + printf(" (iv)Width of Wheel at Inlet , bo =%.4f m\n",bo) //The Answer Vary due to Round off Error + + + diff --git a/3751/CH5/EX5.11/Ex5_11.sce b/3751/CH5/EX5.11/Ex5_11.sce new file mode 100644 index 000000000..75a3035bb --- /dev/null +++ b/3751/CH5/EX5.11/Ex5_11.sce @@ -0,0 +1,46 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.11 +//To Determine The Flow Rate, Guide Vane Angles, Runner Vane Angles and Inner and Outer Diameters of the Runner. + + clc + clear + +//Given Data:- + H=86.4; //Net Head, m + N=650; //Speed, rpm + P=397; //Shaft Power, kW + bo_by_Do=0.1; //Breadth Ratio + Di_by_Do=0.5; //Di/Do + Kf=0.17; //Flow Ratio + eta_H=95/100; //Hydraulic Efficiency + eta_o=85/100; //Overall Efficiency + //As Discharge is Radial and Flow Velocity is Constant, + alpha_o=90; //degrees //Vfi=Vfo=Vo + + +//Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Q=P*1000/(rho*g*H*eta_o); //m^3/s + Vfi=Kf*sqrt(2*g*H); //m/s + Vfo=Vfi; + Do=sqrt(Q/(%pi*bo_by_Do*Vfi)); //m + Di=Do*Di_by_Do; //m + ui=%pi*Do*N/60; //m/s + uo= %pi*Di*N/60; //m/s + Vwi=eta_H*g*H/ui; //m/s + alpha_i=atand(Vfi/Vwi); //degrees + beta_i=atand(Vfi/(Vwi-ui)); //Runner Vane Angle at Inlet, degrees + beta_o=atand(Vfo/uo); //Runner Vane Angle at Outlet, degrees + + +//Results:- + printf(" (i) The Flow Rate, Q=%.3f m^3/s\n",Q ) + printf(" (ii) Guide Vane Angles are: \n alpha_i=%.2f Degrees , alpha_o=%.f Degrees\n",alpha_i,alpha_o) //The Answer Vary due to Round off Error + printf(" (iii) Runner Vane Angles are:- \n beta_i=%.2f Degrees , beta_o =%.2f Degrees \n",beta_i,beta_o ) //The Answer Vary due to Round off Error + printf(" (iv) Inner and Outer Diameters of the Runner are: \n Di=%.2f m , Do=%.2f m \n",Di,Do ) + + diff --git a/3751/CH5/EX5.12/Ex5_12.sce b/3751/CH5/EX5.12/Ex5_12.sce new file mode 100644 index 000000000..6c78bf697 --- /dev/null +++ b/3751/CH5/EX5.12/Ex5_12.sce @@ -0,0 +1,55 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.12 + clc + clear + +//Given Data:- + Dp=5; //Diameter at Penstock, m + P=61000; //Output Power, kW + Q=110; //Discharge, m^3/s + N=160; //Speed, rpm + eta_H=94/100; //Hydraulic Efficiency + Do=4; //Diameter of Runner at Inlet, m + bo=1; //Width of Runner at Inlet, m + Ddi=4.2; //Entry Diameter to Draft Tube, m + V2=2.2; //Velocity in Tail Race, m/s + p_by_rho_g=58; //Static Pressure Head (p/(rho*g)) , m + Z=2.8; //Level of Measurement above Tail Race, m + loss=25; //Percentage of loss in Draft Tube (of Velocity Head at its Entry) + Z1=2.2; //Level of Runner Exit above Tail Race, m + +//Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Vp=4*Q/(%pi*Dp^2); //Velocity in Penstock, m/s + Vo=4*Q/(%pi*Ddi^2); //Velocity at Entry to the Draft Tube, m/s + Hp=p_by_rho_g+Z+Vp^2/(2*g); //Head just before Entry to Runner, m + H=Hp-V2^2/(2*g); //Working Head, m + + //(a)Overall Efficiency + eta_o=P*1000/(rho*Q*g*H)*100; //In Percentage + + // (b) The Direction of Flow relative to the Runner at Inlet + ui=%pi*Do*N/60; //m/s + Vwi=eta_H*g*H/ui; //m/s + Vfi=Q/(%pi*Do*bo); //m/s + beta_i=180-atand(Vfi/(ui-Vwi)); //degrees + + // (c) The Pressure Head at entry to Draft Tube, p1/(rho*g) + //By Applying Bernoulli's Equation with, + Z2=0; + p2_by_rho_g=0; + hf=(loss/100)*Vo^2/(2*g); + + p1_by_rho_g=p2_by_rho_g+(V2^2-Vo^2)/(2*g)+(Z2-Z1)+hf; //m + + +//Results:- + printf("(a)The Overall Efficiency, eta_o=%.2f Percent\n",eta_o) //The Answer Vary due to Round off Error + printf("(b)The Direction of Flow relative to the Runner at Inlet, beta_i=%.2f Degrees \n",beta_i) //The Answer Vary due to Round off Error + printf("(c) The Pressure Head at entry to Draft Tube, p1/(rho*g)=%.2f m (vaccum)\n",abs(p1_by_rho_g)) + + diff --git a/3751/CH5/EX5.13/Ex5_13.sce b/3751/CH5/EX5.13/Ex5_13.sce new file mode 100644 index 000000000..22579ec2c --- /dev/null +++ b/3751/CH5/EX5.13/Ex5_13.sce @@ -0,0 +1,30 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.13 +//To Determine the Blade Angle at Entry and Exit. + + clc + clear + +//Given Data:- + R=0.6; //Degree + ui=15; //Peripheral velocity of Runner at entry, m/s + Vfo=3.2; //m/s + Vfi=Vfo; + Vo=Vfo; + //As the Diameter of rotor at entry is Twice that at exit, + Do_by_Di=2; + + +//Computations:- + Vwi=(1-R)*2*ui; //m/s + beta_i=180-atand(Vfi/(ui-Vwi)); //degrees + uo=ui/Do_by_Di; //Velocity of Runner at Outlet, m/s + beta_o=atand(Vfo/uo); //degrees + + +//Results:- + printf("The Blade Angle at Entry, beta_i=%.2f Degrees\n",beta_i) //The Answer Vary due to Round off Error + printf("The Blade angle at Exit, beta_o=%.3f Degrees \n",beta_o) + + diff --git a/3751/CH5/EX5.14/Ex5_14.sce b/3751/CH5/EX5.14/Ex5_14.sce new file mode 100644 index 000000000..335ef4025 --- /dev/null +++ b/3751/CH5/EX5.14/Ex5_14.sce @@ -0,0 +1,27 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.14 +//To Find the Pressure Head at the Entrance. + + clc + clear + +//Given Data:- + Vo=5.6; //Velocity at Inlet, m/s + Vd=1.4; //Velocity at Outlet, m/s + h_f=0.12; //Friction Losses, m + Ho=5.2; //Vertical Height betwen tail-race and entrance of Draft-tube, m + +//Data Required:- + g=9.81; //Acceleration due to gravity, m/s^2 + +//Calculations:- + //Applying Bernoulli's Equation with, + Z2=0; + pa_by_rho_g=0; + p1_by_rho_g=pa_by_rho_g+(Vd^2-Vo^2)/(2*g)+Z2-Ho+h_f; //m + +//Results:- + printf("The Pressure Head at Entrance, p1/(rho*g)=%.3f m\n",p1_by_rho_g) + + diff --git a/3751/CH5/EX5.15/Ex5_15.sce b/3751/CH5/EX5.15/Ex5_15.sce new file mode 100644 index 000000000..cb4da2776 --- /dev/null +++ b/3751/CH5/EX5.15/Ex5_15.sce @@ -0,0 +1,42 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.15 + clc + clear + +//Given Data:- + Vo=4.5; //Velocity of water at Tube Entrance, m/s + D1=0.4; //Diameter of Tube at Upper End, m + D2=0.65; //Diameter of Tube at Lower End, m + l=4.8; //Length of Tube, m + h_f=0.14; //Head Losses due to Friction, m + h=1; //Length of Tube immersed in Tail-race, m + +//Data Required:- + pa=1.013e5; //Air(Atmospheric) Pressure, Pa + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Calculations:- + A1=(%pi/4)*D1^2; //Cross-sectional Area at Upper End, m^2 + A2=(%pi/4)*D2^2; //Cross-sectional Area at Lower End, m^2 + //Using Continuity Equation, + Vd=A1*Vo/A2; //Velocity of Water at Outlet, m/s + + //(a) Using Bernoulli's Equation, + p1_by_rho_g=pa/(rho*g)+h+(Vd^2-Vo^2)/(2*g)-l+h_f; //Absolute Pressure Head at Inlet, m + + //For Vaccum Pressure Head, + pa_by_rho_g=0; + p1_by_rho_g_v=pa_by_rho_g+h+(Vd^2-Vo^2)/(2*g)-l+h_f; //Vaccum Pressure Head at Inlet, m + + //(b)Efficiency of Draft Tube: + eta_d=(Vo^2-Vd^2-2*h_f*g)*100/Vo^2; //In Percentage + +//Results:- + printf("(a)The Pressure Head at Tube Entrance is\n\t") + printf(" p1/(rho*g)=%.3f m (Absolute)\n\t p1/(rho*g)=%.3f m (Vaccum)\n ",p1_by_rho_g, p1_by_rho_g_v) //The Answer Vary due to Round off Error + printf("(b)Efficiency of Draft Tube, eta_d=%.2f Percent\n",eta_d) //The Answer Vary due to Round off Error + + + diff --git a/3751/CH5/EX5.2/Ex5_2.sce b/3751/CH5/EX5.2/Ex5_2.sce new file mode 100644 index 000000000..afb2f5d9a --- /dev/null +++ b/3751/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,44 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.2 +//To Find (a)Discharge (b) Power Developed and (c)Hydraulic Efficiency + + clc + clear + +//Given Data:- + Do=1.2; //Diameter of Runner at Inlet, m + Ao=0.4; //Area of Flow at Inlet, m^2 + N=500; //Speed of Runner, rpm + H=135; //Head, m + alpha_i=20; //Guide Vane Angle at Inlet, degrees + beta_i=65; //Vane angle at Inlet, degrees + +//Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + + +//Computations:- + ui=%pi*Do*N/60; //m/s + Vri=ui*sind(alpha_i)/sind(beta_i-alpha_i); //m/s + Vfi=Vri*sind(beta_i); //m/s + Vwi=Vfi/tand(alpha_i); //m/s + + //(a)Discharge, Q + Q=Ao*Vfi; //m^3/s + + //(b)Power develpoed by Runner + P=rho*Q*Vwi*ui/1000; //kW + + //(c)Hydraulic Efficiency, eta_H + eta_H=Vwi*ui*100/(g*H); //Percentage + + +//Results:- + +printf(" (a)Discharge, Q=%.3f m^3/s\n ",Q) //The Answer Vary due to Round off Error +printf(" (b)Power develpoed by Runner=%.3f kW\n ",P) //The Answer provided in the Textbook is Wrong +printf(" (c)Hydraulic Efficiency, eta_H =%.2f Percent\n ",eta_H) //The Answer Vary due to Round off Error + + diff --git a/3751/CH5/EX5.3/Ex5_3.sce b/3751/CH5/EX5.3/Ex5_3.sce new file mode 100644 index 000000000..4543b9519 --- /dev/null +++ b/3751/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,77 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.3 +//To Find (a)The Absolute Velocity of Water at Inlet of Runner (b)The Velocity of Whirl at Inlet (c) The Relative Velocity at Inlet + //(d) The Runner Blade Angles (e)Width of Runner at Outlet (f)Weight of Water flowing through the Runner per second + //(g)Head at Inlet of the Turbine (h)Power developed (i) Hydraulic Efficiency of the Turbine + + clc + clear + +//Given Data:- + Do=1; // External Diameter of Runner, m + Di=0.5; //Internal Diameter of Runner, m + N=200; //Speed of Turbine, rpm + bo=225; //Width of Runner at Inlet, mm + + + Vfi=2.15; //Velocity of flow at Inlet, m/s + // As Velocity of Flow is constant through the Runner, + Vfo=Vfi; //Velocity of flow at Outlet, m/s + Vo=Vfo; + alpha_i=11; //Guide Blades Angle at Inlet, degrees + //As Discharge at Outlet of the Turbine is Radial, + alpha_o=90; //Guide Blades angle at Outlet, degrees + Vwo=0; + + +//Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + + +//Computations:- + ui=%pi*Do*N/60; //m/s + + // (a)The Absolute Velocity of Water at Inlet of Runner, + Vi=Vfi/sind(alpha_i); //m/s + + //(b)The Velocity of Whirl at Inlet, + Vwi=Vfi/tand(alpha_i); //m/s + + + // (c) The Relative Velocity at Inlet, + Vri=sqrt(Vfi^2+(Vwi-ui)^2); //m/s + + // (d) The Runner Blade Angles, beta_i, beta_o + beta_i=asind(Vfi/Vri); //Runner Blade Angle at Inlet, degrees + uo=%pi*Di*N/60; //m/s + beta_o=atand(Vfo/uo); //Runner Blade Angle at Outlet, degrees + + // (e)Width of Runner at Outlet , bi + bi=Do*bo/Di; //mm + + // (f)Weight of Water flowing through the Runner per second, W + W=rho*g*%pi*Do*(bo/1000)*Vfi/1000; //kN/s + + //(g)Head at Inlet of the Turbine, H + H=Vwi*ui/g+Vo^2/(2*g); //m + + // (h)Power developed by the Runner, + Q=%pi*Do*(bo/1000)*Vfi; //m^3/s + P=rho*Q*Vwi*ui/1000; //kW + //(i)Hydraulic Efficiency, eta_H + eta_H=Vwi*ui*100/(g*H); //In Percentage + +//Results:- + printf("(a)The Absolute Velocity of Water at Inlet of Runner, Vi=%.3f m/s\n",Vi) //The Answer Vary due to Round off Error + printf(" (b)The Velocity of Whirl at Inlet, Vwi=%.2f m/s\n",Vwi) + printf(" (c) The Relative Velocity at Inlet, Vri=%.2f m/s\n",Vri) + printf(" (d) The Runner Blade Angles are:- \n beta_i =%.2f Degrees and beta_o =%.2f Degrees\n",beta_i,beta_o) //The Answer Vary due to Round off Error + printf(" (e)Width of Runner at Outlet , bi =%.f mm\n",bi) + printf(" (f)Weight of Water flowing through the Runner per second, W =%.2f kN/s\n",W) //The Answer Vary due to Round off Error + printf(" (g)Head at Inlet of the Turbine, H =%.3f m\n",H) //The Answer Vary due to Round off Error + printf(" (h)Power developed by the Runner =%.3f kW\n",P) //The Answer Vary due to Round off Error + printf(" (i)Hydraulic Efficiency, eta_H =%.2f Percent\n",eta_H) //The Answer Vary due to Round off Error + + diff --git a/3751/CH5/EX5.4/Ex5_4.sce b/3751/CH5/EX5.4/Ex5_4.sce new file mode 100644 index 000000000..900756ebc --- /dev/null +++ b/3751/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,56 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.4 +//To Find (a)Diameter and Width at Inlet ant Outlet (b)Runner Vane Angles at Inlet and Outlet (c)Guide Blade Angles + + clc + clear + +//Given Data:- + H=70; //Net Head, m + N=700; //Speed, rpm + P=330; //Shaft Power, kW + eta_o=85/100; //Overall Efficiency + eta_H=92/100; //Hydraulic Efficiency + Kf=0.22; //Flow Ratio + bo_by_Do=0.1; //Breadth Ratio + t_per=6; //Percentage of Circumferential Area occupied by the Thickness of Vanes + Kt=1-t_per/100; //Vane Thickness Factor + //As Outer Diameter= 2 times the Inner Diameter, + Do_by_Di=2; //Do/Di + + +//Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Vfi=Kf*sqrt(2*g*H); //m/s + Vfo=Vfi; + Q=P*1000/(rho*g*H*eta_o); //m^3/s + + //(a)Diameter and Width at Inlet ant Outlet, Do,bo, Di and bi. + Do=sqrt(Q/(Kt*%pi*bo_by_Do*Vfi)); //m + Di=Do/Do_by_Di; //m + bo=Do*bo_by_Do*1000; //mm + bi=Do*bo/Di; //mm + + ui=%pi*Do*N/60; //m/s + uo=%pi*Di*N/60; //m/s + Vwi=eta_H*g*H/ui; //m/s + + // (b)Runner Vane Angles at Inlet and Outlet, beta_i,beta_o + beta_i=atand(Vfi/(Vwi-ui)); //Runner Vane Angle at Inlet, degrees + beta_o=atand(Vfo/uo); //Runner Vane Angle at Outlet, degrees + + //(c)Guide Vane Angle, alpha_i + alpha_i=atand(Vfi/Vwi); //degrees + //As flow is radial at outlet, + alpha_o=90; //degrees + +//Results:- + printf(" (a)Diameter and Width at Inlet and Outlet are: \n\t") + printf("Do=%.3f m bo=%.1f mm\n Di=%.3f m bi=%.1f mm\n",Do,bo,Di,bi) //The Answer Vary due to Round off Error + printf(" (b)Runner Vane Angles at Inlet and Outlet are:- \n beta_i=%.2f Degrees , beta_o =%.2f Degrees \n",beta_i,beta_o) //The Answer Vary due to Round off Error + printf(" (c)Guide Vane Angles, \n alpha_i=%.2f Degrees , alpha_o=%.f Degrees\n ",alpha_i,alpha_o) //The Answer Vary due to Round off Error + diff --git a/3751/CH5/EX5.5/Ex5_5.sce b/3751/CH5/EX5.5/Ex5_5.sce new file mode 100644 index 000000000..7e32e8343 --- /dev/null +++ b/3751/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,62 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.5 +//To Determine (a) Guide Blade Angle (b)Runner Vane Angles at Inlet and Outlet (c) Diameter of Runner at Inlet and Outlet (d) Width of Wheel at Inlet. + + clc + clear + +//Given Data:- + H=70; //Net Head, m + N=600; //Speed, rpm + P=367.875; //Shaft Power, kW + eta_o=85/100; //Overall Efficiency + eta_H=95/100; //Hydraulic Efficiency + Kf=0.25; //Flow Ratio + bo_by_Do=0.1; //Breadth Ratio + t_per=10; //Percentage of Circumferential Area occupied by the Thickness of Vanes + Kt=1-t_per/100; //Vane Thickness Factor + //As Outer Diameter= 2 times the Inner Diameter, + Do_by_Di=2; //Do/Di + + +//Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Vfi=Kf*sqrt(2*g*H); //m/s + Vfo=Vfi; + Q=P*1000/(rho*g*H*eta_o); //m^3/s + + Do=sqrt(Q/(Kt*%pi*bo_by_Do*Vfi)); //m + Di=Do/Do_by_Di; //m + bo=Do*bo_by_Do*1000; //mm + bi=Do*bo/Di; //mm + + ui=%pi*Do*N/60; //m/s + uo=%pi*Di*N/60; //m/s + Vwi=eta_H*g*H/ui; //m/s + + //(a)Guide Vane Angle, alpha_i + alpha_i=atand(Vfi/Vwi); //degrees + + // (b)Runner Vane Angles at Inlet and Outlet, beta_i,beta_o + beta_i=atand(Vfi/(Vwi-ui)); //Runner Vane Angle at Inlet, degrees + beta_o=atand(Vfo/uo); //Runner Vane Angle at Outlet, degrees + + + //(c)Diameter of Runner at Inlet ant Outlet, Do and Di . + //Calculated Above + + // (d) Width of Wheel at Inlet, bi. + //Calculated Above + + //Results:- + printf(" (a)Guide Vane Angle, alpha_i=%.2f Degrees\n ",alpha_i) //The Answer Vary due to Round off Error + + printf(" (b)Runner Vane Angles at Inlet and Outlet are:- \n beta_i=%.2f Degrees , beta_o =%.2f Degrees \n",beta_i,beta_o) //The Answer Vary due to Round off Error + + printf(" (c)Diameter of Runner at Inlet and Outlet are: \n Do=%.4f m Di=%.4f m \n",Do,Di) //The Answer(Do) Vary due to Round off Error + printf(" (d) Width of Wheel at Inlet, bo=%.2f mm\n",bo ) //The Answer Vary due to Round off Error + diff --git a/3751/CH5/EX5.6/Ex5_6.sce b/3751/CH5/EX5.6/Ex5_6.sce new file mode 100644 index 000000000..58ede74d6 --- /dev/null +++ b/3751/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,43 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.6 +//To Find (a) Guide Blade Angles (b) Blade Angle at Inlet (c) Power Developed + + clc + clear + +//Given Data:- + Q=300; //Discharge, litres/s + Di=0.36; //Diameter of Runner at Outlet, m + Dp=Di; //Diameter of Outlet Pipe, m + H=36; //Head, m + ui=21; // Velocity of Wheel at Inlet, m/s + +//Data Required:- + rho=1000; //Density of water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Vfo=(Q/1000)/((%pi/4)*Di^2); //m/s + Vo=Vfo; + Vfi=Vfo; + //By Energy Balance Equation, + Vwi=(g*H-Vo^2/2)/ui; //m/s + + // (a) Guide Blade Angles, alpha_i, alpha_o + alpha_i=atand(Vfi/Vwi); //degrees + //As Discharge is Radial, + alpha_o=90; //degrees + + // (b) Blade Angle at Inlet, beta_i + beta_i=180-atand(Vfi/(ui-Vwi)); //degrees + + // (c) Power Developed by Runner + P=rho*(Q/1000)*Vwi*ui/1000; //kW + +//Results:- + printf(" (a) Guide Blade Angles are: \n alpha_i=%.2f Degrees, alpha_o=%.f Degrees\n",alpha_i, alpha_o ) + printf(" (b) Blade Angle at Inlet, beta_i=%.2f Degrees\n",beta_i) + printf(" (c) Power Developed by Runner, P =%.2f kW\n",P) //The Answer Vary due to Round off Error + + diff --git a/3751/CH5/EX5.7/Ex5_7.sce b/3751/CH5/EX5.7/Ex5_7.sce new file mode 100644 index 000000000..bbe086b98 --- /dev/null +++ b/3751/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,59 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.7 +//To Determine (a) The Diameter of Wheel (b) The Quantity of Water Supplied (c) The Guide Blade Angle at Inlet (d) The Runner Vane Angles at Inlet and Exit. + + clc + clear + +//Given Data:- + P=368; //Shaft Power, kW + H=71; //Head, m + N=748; //Speed, rpm + bo_by_Do=0.1; //Breadth Ratio + Kf=0.15; //Flow Ratio + eta_H=95/100; //Hydraulic Efficiency + eta_m=85/100; //Mechanical Efficiency + eta_v=100/100; //Volumetric Efficiency (Assumed to be 100%) + + //As Inner Diameter is Half the Outer Diameter, + Di_by_Do=1/2; //Di/Do + + +//Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + eta_o=eta_H*eta_m*eta_v; //Overall Efficiency + Q=P*1000/(rho*g*H*eta_o); //m^3/s + Vfi=Kf*sqrt(2*g*H); //m/s + Vfo=Vfi; + Do=sqrt(Q/(%pi*bo_by_Do*Vfi)); //m + Di=Do*Di_by_Do; //m + + // (a) The Diameter of Wheel, Do + //Calculated Above + + // (b) The Quantity of Water Supplied, Q + //Calculated Above + + // (c) The Guide Blade Angle at Inlet , alpha_i + + ui=%pi*Do*N/60; //m/s + uo= %pi*Di*N/60; //m/s + Vwi=eta_H*g*H/ui; //m/s + alpha_i=atand(Vfi/Vwi); //degrees + + // (d)Runner Vane Angles at Inlet and Outlet, beta_i, beta_o + beta_i=atand(Vfi/(Vwi-ui)); //Runner Vane Angle at Inlet, degrees + beta_o=atand(Vfo/uo); //Runner Vane Angle at Outlet, degrees + + + //Results:- + printf(" (a) The Diameter of Wheel, Do =%.3f m\n ",Do ) + printf(" (b) The Quantity of Water Supplied, Q=%.4f m^3/s\n",Q ) + printf(" (c) The Guide Blade Angle at Inlet , alpha_i=%.2f Degrees\n",alpha_i ) //The Answer Vary due to Round off Error + printf(" (d)Runner Vane Angles at Inlet and Outlet are:- \n beta_i=%.2f Degrees , beta_o =%.2f Degrees \n",beta_i,beta_o ) //The Answer Vary due to Round off Error + + diff --git a/3751/CH5/EX5.8/Ex5_8.sce b/3751/CH5/EX5.8/Ex5_8.sce new file mode 100644 index 000000000..01d0dc0fa --- /dev/null +++ b/3751/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,4 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.8 + //Theoritical Problem . diff --git a/3751/CH5/EX5.9/Ex5_9.sce b/3751/CH5/EX5.9/Ex5_9.sce new file mode 100644 index 000000000..b9d661320 --- /dev/null +++ b/3751/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,42 @@ +//Fluid Systems - By Shiv Kumar +//Chapter 5- Francis Turbine +//Example 5.9 +//To Find (i) Guide Blade Angle at Inlet (ii) The Wheel Vane Angle at Inlet (iii) Diameter of Wheel at inlet (iv)Width of Wheel at Inlet + + clc + clear + +//Given Data:- + //Data Required + rho=1000; //Density of water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + + eta_o=76/100; //Overall Efficiency + P=150; //Power Produced, kW + H=8; // Working Head, m + ui=0.25*sqrt(2*g*H); //Peripheral Velocity at Inlet, m/s + Vfi=0.95*sqrt(2*g*H); //Velocity of Flow at Inlet, m/s + N=150; //Speed, rpm + H_loss=20; //Percentage of Hydraulic Losses in the Turbine (of Available Energy) + //As Discharge is Radial, + alpha_o=90; //Degrees //Vfo=Vo + Vwo=0; + +//Computations:- + Do=ui*60/(%pi*N); //m + Q=P*1000/(rho*g*H*eta_o); //m^3/s + bo=Q/(%pi*Do*Vfi); //m + + //By Energy Balance Equation, + Vwi=(g*H-(H_loss/100)*g*H)/ui; //m/s + + alpha_i=atand(Vfi/Vwi); //degrees + beta_i=atand(Vfi/(Vwi-ui)); //degrees + + +//Results:- + printf(" (i) Guide Blade Angle at Inlet, alpha_i=%.2f Degrees\n",alpha_i) //The Answer Vary due to Round off Error + printf(" (ii) The Wheel Vane Angle at Inlet, beta_i =%.2f Degrees\n", beta_i ) //The Answer Vary due to Round off Error + printf(" (iii) Diameter of Wheel at Inlet, Do =%.4f m\n",Do) //The Answer Vary due to Round off Error + printf(" (iv)Width of Wheel at Inlet , bo =%.4f m\n",bo) + diff --git a/3751/CH6/EX6.1/Ex6_1.sce b/3751/CH6/EX6.1/Ex6_1.sce new file mode 100644 index 000000000..d43fbab83 --- /dev/null +++ b/3751/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,36 @@ +//Fluid System By Shiv Kumar
+//Chapter 6 - Kaplan and Propeller Turbines
+//Example 6.1
+//To Find (a)Discharge (b)Diameter of Hub And Diameter of Runner (c)Speed
+
+ clc
+ clear
+
+//Given:
+ P=37; //Shaft Power, MW
+ H=22; //Head, m
+ eta_0=92/100; //Overall Efficiency
+ dbyD=1/3; //Ratio of Diameters of Hub and Runner
+ Kf=0.6; //Flow Ratio
+ Ku=2; //Speed Ratio
+//Data Required:
+ rho=1000; //Density of Water, Kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations
+
+ //(a)Discharge, Q
+ Q=P*10^6/(rho*g*H*eta_0); //m^3/s
+
+ //(b)Diameter of Hub(d) And Diameter of Runner(D)
+ d=sqrt(Q/((%pi/4)*Kf*sqrt(2*g*H)*(dbyD^-2-1))); //m
+ D=d/dbyD; //m
+
+ //(c) Speed,N
+ N=Ku*60*sqrt(2*g*H)/(%pi*D); // rpm
+
+//Results
+ printf("(a)Discharge, Q=%.2f m^3/s \n",Q) //The answer vary due to round off error
+ printf(" (b)Diameter of Hub, d=%.2f m \n",d)
+ printf(" Diameter of Runner, D=%.2f m \n",D) //The answer vary due to round off error
+ printf(" (c)Speed, N=%.2f rpm \n",N) //The answer vary due to round off error
diff --git a/3751/CH6/EX6.10/Ex6_10.sce b/3751/CH6/EX6.10/Ex6_10.sce new file mode 100644 index 000000000..958dbe84a --- /dev/null +++ b/3751/CH6/EX6.10/Ex6_10.sce @@ -0,0 +1,44 @@ +//Fluid System By Shiv Kumar
+//Chapter 6 - Kaplan and Propeller Turbines
+//Example 6.10
+//To Determine (i)Hydraulic Efficiency of turbine (ii)Discharge through the turbine (iii)Power Developed by the Runner
+
+clc
+clear
+
+//Given:
+ D=4.5; // Runner Diameter, m
+ N=48; // Speed, rpm
+ Alpha_i=145; //Guide Vane Angle at Inlet, Degrees
+ Beta_o=25; //Runner blade Angle at Outlet
+ A=30; //Flow Area, m^2
+ //As runner blade angle at inlet is radial
+ Beta_i=90 //Degrees
+
+//Data Required:
+ rho=1000; //Density of Water, Kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Calculations
+ u=%pi*D*N/60; //Velocity of Runner,m/s
+ ui=u;
+ uo=u;
+ Vwi=ui;
+ Vfi=ui*tand(180-Alpha_i); //m/s
+ Vfo=Vfi;
+ Vrwo=Vfo/tand(Beta_o); //m/s
+ Vwo=Vrwo-uo; //The answer vary because wrong Value of uo is used to calculate Vwo in the textbook
+ Vo=sqrt(Vfo^2+Vwo^2); //m/s //The answer vary because wrong Value of Vwo is used to calculate Vo in the textbook
+
+//(i)Hydraulic Efficiency, eta_H
+ H= (Vwi-Vwo)*u/g+Vo^2/(2*g); // Head, m //The answer vary because wrong Value of Vo and Vwo is used to calculate H in the textbook
+ eta_H=(Vwi*ui-Vwo*uo)*100/(g*H); //Percent(%)
+
+//(ii) Discharge through the turbine, Q
+ Q=A*Vfi; //m^3/s
+//(iii)Power Developed by the Runner, P
+ P=rho*Q*(Vwi-Vwo)*u/10^6; //MW
+//Results
+ printf("(i)Hydraulic Efficiency, eta_H=%.2f Percent\n",eta_H) //The answer given in the textbook is wrong
+ printf("(ii) Discharge through the turbine, Q =%.1fm^3/s\n",Q) //The answer vary due to round off error
+ printf("(iii)Power Developed by the Runner, P =%.3fMW\n",P) //The answer given in the textbook is wrong
diff --git a/3751/CH6/EX6.2/Ex6_2.sce b/3751/CH6/EX6.2/Ex6_2.sce new file mode 100644 index 000000000..42e18e65f --- /dev/null +++ b/3751/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,32 @@ +//Fluid System By Shiv Kumar
+//Chapter 6 - Kaplan and Propeller Turbines
+//Example 6.2
+//To Find Diameter of Runner, Speed of Runner and Specific Speed of Turbine
+
+ clc
+ clear
+
+//Given:
+ P=8; //Shaft Power, MW
+ H=6; //Head, m
+ Ku=2.09; //Speed Ratio
+ Kf=0.68; //Flow Ratio
+ eta_0=90/100; //Overall Efficiency
+ dbyD=1/3; //Ratio of Diameters of Hub and Runner
+
+//Data Required:
+ rho=1000; //Density of Water, Kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations
+
+ Q=P*10^6/(rho*g*H*eta_0); //Discharge, m^3/s
+ d=sqrt(Q/((%pi/4)*Kf*sqrt(2*g*H)*(dbyD^-2-1))); //Diameter of hub, m
+ D=d/dbyD; //Diameter of runner, m
+ N=Ku*60*sqrt(2*g*H)/(%pi*D); //Speed of Runner, rpm
+ Ns=N*(P*10^3)^(1/2)/(H^(5/4)); //Specific Speed of Turbine, SI Units
+
+//Results
+ printf("Diameter of Runner, D =%.1f m \n",D)
+ printf("Speed of Runner, N = %.2f rpm \n",N) //The answer provided in the textbook is wrong
+ printf("Specific Speed of Turbine, Ns = %.2f (SI Units)",Ns) //The answer provided in the textbook is wrong(Due to error in N)
diff --git a/3751/CH6/EX6.3/Ex6_3.sce b/3751/CH6/EX6.3/Ex6_3.sce new file mode 100644 index 000000000..1efab0a00 --- /dev/null +++ b/3751/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,52 @@ +//Fluid System By Shiv Kumar
+//Chapter 6 - Kaplan and Propeller Turbines
+//Example 6.3
+//To Find (a)Inlet and Outlet blade Angles (b)Mechanical Efficiency (c)Volumetric Efficiency
+
+ clc
+ clear
+
+//Given:
+ D=6; //Outer Diameter of Runner, m
+ d=2; //Inner Diameter of Runner, m
+ P=30; //Shaft Power, MW
+ N=75; //Speed, rpm
+ H=12; //Head, m
+ Q=310 //Discharge through the Runner, m^3/s
+ eta_H=96/100; //Hydraulic Efficiency
+
+//Data Required:
+ rho=1000; //Density of Water, Kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations
+
+ u=%pi*D*N/60; //Velocity of runner, m/s
+ ui=u;
+ uo=u;
+ Vf=Q/((%pi/4)*(D^2-d^2)) // m/s
+ Vfi=Vf;
+ Vfo=Vf;
+ Vwi=eta_H*g*H/ui; // m/s //The Answer Vary Because Value of ui used in book is Wrong
+
+//(a)Inlet and Outlet blade Angles, Beta_i and Beta_o
+
+ Beta_i=180-atand(Vfi/(ui-Vwi)); //Degrees
+ Beta_o=atand(Vfo/uo); //Degrees
+
+//(b)Mechanical Efficiency,eta_m
+
+ eta_M=P*10^6/(rho*Q*Vwi*ui)*100; //percentage(%) //The Answer Vary Because Value of Vwi used in book is Wrong
+
+//(c)Volumetric Efficiency, eta_v
+
+ eta_o=P*10^6/(rho*Q*g*H)*100; //Overall Efficiency, percentage(%)
+ eta_v=eta_o/(eta_M*eta_H); //percentage(%) //The Answer Vary Because Value of eta_m used in book is Wrong
+
+//Results
+
+ printf("(a)Inlet Blade Angle, Beta_i=%.2f degrees and \n",Beta_i) //The answer vary due to round off error
+ printf(" Outlet Blade Angle, Beta_o=%.2f degrees \n",Beta_o) //The answer vary due to round off error
+ printf("(b)Mechanical Efficiency, eta_m=%.2f percent\n",eta_M) //The answer provided in the textbook is wrong
+ printf("(c)Volumetric Efficiency, eta_v=%.2f percent\n ",eta_o) //The answer provided in the textbook is wrong
+
diff --git a/3751/CH6/EX6.4/Ex6_4.sce b/3751/CH6/EX6.4/Ex6_4.sce new file mode 100644 index 000000000..ffbb756b6 --- /dev/null +++ b/3751/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,48 @@ +//Fluid System By Shiv Kumar
+//Chapter 6 - Kaplan and Propeller Turbines
+//Example 6.4
+//To Find (a)Discharge (b)Hydraulic Efficiency (c)Overall Efficiency (d)Specific Speed
+
+ clc
+ clear
+
+//Given:
+ N=30 //Speed, rpm
+ Alpha_i=31; //Inlet Guide Vane Angle, Degrees
+ Beta_i=90; //Inlet Runner Vane Angle, Degrees
+ Beta_o=24; //Outlet Runner Vane Angle, Degrees
+ Dm=4; //Mean Diameter of Runner, m
+ A=31; //Area of Flow, m^2
+ ML=5; //Percent of Mechanical Loss
+//Data Required:
+ rho=1000; //Density of Water, Kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations
+
+ u=%pi*Dm*N/60; //Velocity of runner, m/s
+ ui=u;
+ uo=u;
+ Vwi=ui;
+ Vfi=ui*tand(Alpha_i); //m/s
+ Vf=Vfi;
+ Vfo=Vfi;
+ Vrwo=Vfo/tand(Beta_o); //m/s
+ Vwo=Vrwo-uo;
+ Vo=sqrt(Vfo^2+Vwo^2); //m/s
+//(a)Discharge, Q
+ Q=A*Vfi; //m^3/s
+//(b) Hydraulic Efficiency, eta_H
+ H= (Vwi+Vwo) *u/g+Vo^2/(2*g); // Head, m
+ eta_H=(Vwi*ui+Vwo*uo)*100/(g*H); //Percent(%)
+//(c)Overall Efficiency, eta_o
+ P=rho*Q*(Vwi+Vwo)*u*(1-ML/100); //Shaft Power, Watt(w)
+ eta_o=P/(rho*Q*g*H)*100; //Percent(%)
+//(d)Specific Speed,Ns
+ Ns=N*sqrt(P/1000)/(H^(5/4)); //SI Units
+
+//Results
+ printf("(a)Discharge, Q=%.2f m^3/s\n",Q) //The answer vary due to round off error
+ printf("(b) Hydraulic Efficiency, eta_H =%.2f Percent\n", eta_H) //The answer vary due to round off error
+ printf("(c) Overall Efficiency, eta_o =%.2f Percent\n", eta_o) //The answer vary due to round off error
+ printf("(d)Specific Speed, Ns =%.2f (SI Units)\n", Ns) //The answer vary due to round off error
diff --git a/3751/CH6/EX6.5/Ex6_5.sce b/3751/CH6/EX6.5/Ex6_5.sce new file mode 100644 index 000000000..1d2d317e7 --- /dev/null +++ b/3751/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,45 @@ +//Fluid System By Shiv Kumar
+//Chapter 6 - Kaplan and Propeller Turbines
+//Example 6.5
+//To Find (a)Guide Vane Angle at Inlet (b)Runner Vane Angle at Inlet
+
+clc
+clear
+
+//Given:
+ P=22500 //Shaft Power, KW
+ H=20; //Head, m
+ N=148; //Speed, rpm
+ eta_H=95/100; //Hydraulic Efficiency
+ eta_o=89/100; //Overall Efficiency
+ D=4.5; //Diameter of Runner, m
+ d=2; //Diameter of Hub, m
+ Beta_o=34 //Runner Vane Angle at Outlet, Degrees
+
+//Data Required:
+ rho=1000; //Density of Water, Kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations
+
+ u=%pi*D*N/60; //Velocity of runner, m/s
+ Q=P*10^3/(rho*g*H*eta_o); //Discharge, m^3/s
+ Vfi=Q/((%pi/4)*(D^2-d^2)); // m/s
+ //As Velocity of Flow is Constant
+ ui=u;
+ uo=u;
+ Vfo=Vfi;
+ Vf=Vfo;
+ Vrwo=Vfo/tand(Beta_o); //m/s
+ Vwo=uo-Vrwo;
+ Vo=sqrt(Vfo^2+Vwo^2); //m/s
+ Vwi=(g*H-Vo^2/2)/u+Vwo ; //m/s
+//(a)Guide Vane Angle at Inlet,Alpha_i
+ Alpha_i=atand(Vfi/Vwi); //Degrees
+//(b)Runner Vane Angle at Inlet,Beta_i
+ Beta_i=180-atand(Vfi/(ui-Vwi)); //Degrees
+
+//Results
+ printf("(a)Guide Vane Angle at Inlet, Alpha_i=%.2f Degrees\n",Alpha_i) //The answer vary due to round off error
+ printf("(b)Runner Vane Angle at Inlet, Beta_i =%.f Degrees\n",Beta_i) //The answer provided in the textbook is wrong
+
diff --git a/3751/CH6/EX6.6/Ex6_6.sce b/3751/CH6/EX6.6/Ex6_6.sce new file mode 100644 index 000000000..b89471bc5 --- /dev/null +++ b/3751/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,40 @@ +//Fluid System By Shiv Kumar
+//Chapter 6 - Kaplan and Propeller Turbines
+//Example 6.6
+//To Find (a)Diameter of Runner (b)Speed of Turbine (c)Specific Speed of the turbine
+
+ clc
+ clear
+
+//Given:
+ P=9100; //Shaft Power, KW
+ H=5.6; //Net Available Head, m
+ Ku=2.09; //Speed Ratio
+ Kf=0.68; //Flow Ratio
+ eta_0=86/100; //Overall Efficiency
+ dbyD=1/3; //Ratio of Diameters of Hub and Runner
+
+//Data Required:
+ rho=1000; //Density of Water, Kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations
+
+ Q=P*10^3/(rho*g*H*eta_0); //Discharge, m^3/s
+
+
+ d=sqrt(Q/((%pi/4)*Kf*sqrt(2*g*H)*(dbyD^-2-1))); // Diameter of Hub ,m
+ //(i) Diameter of Runner ,D
+ D=d/dbyD; //m
+
+ //(ii) Speed of Turbine,N
+ N=Ku*60*sqrt(2*g*H)/(%pi*D); // rpm
+//(iii) Specific Speed of Turbine, Ns
+ Ns=N*(P)^(1/2)/(H^(5/4)); // SI Units
+
+
+//Results
+ printf("(i)Diameter of Runner , D=%.2f m\n",D)
+ printf("(ii)Speed of Turbine, N =%.2f rpm\n",N) //The answer vary due to round off error
+ printf("(iii) Specific Speed of Turbine, Ns =%.2f (SI Units)\n",Ns) //The answer provided in the textbook is wrong(Due to error in N)
+
diff --git a/3751/CH6/EX6.7/Ex6_7.sce b/3751/CH6/EX6.7/Ex6_7.sce new file mode 100644 index 000000000..8baafe52a --- /dev/null +++ b/3751/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,43 @@ +//Fluid System By Shiv Kumar
+//Chapter 6 - Kaplan and Propeller Turbines
+//Example 6.7
+//To Find (a)Diameter of Runner (b)Speed (c)Specific Speed
+
+clc
+clear
+
+//Given:
+ H=32; //Head, m
+ P=16000; //Shaft Power, KW
+ D_per=190; //Percentage by which Diameter of Runner(D)is Larger than diameter of Boss(d)
+ eta_0=91/100; //Overall Efficiency
+ Ku=2; //Speed Ratio
+ Kf=0.64; //Flow Ratio
+
+//Data Required:
+ rho=1000; //Density of Water, Kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations
+
+ Vfi=Kf*sqrt(2*g*H); //Velocity of Flow at Inlet, m/s
+ ui= Ku*sqrt(2*g*H); //Velocity of Runner at Inlet, m/s
+
+ Q=P*10^3/(rho*g*H*eta_0); //Discharge, m^3/s
+
+
+ d=sqrt(Q/((%pi/4)*Kf*sqrt(2*g*H)*((D_per/100+1)^2-1))); // Diameter of Hub ,m
+ //(a) Diameter of Runner ,D
+ D=d+(D_per/100)*d; //m
+
+ //(b) Speed,N
+ N=ui*60/(%pi*D); // rpm
+//(iii) Specific Speed of Turbine, Ns
+ Ns=N*P^(1/2)/(H^(5/4)); // SI Units
+
+
+//Results
+ printf("(a)Diameter of Runner , D=%.3f m\n",D)
+ printf(" (b)Speed, N =%.2f rpm\n",N) //The answer vary due to round off error
+ printf(" (c)Specific Speed, Ns =%.2f (SI Units)\n",Ns) //The answer provided in the textbook is wrong.
+
diff --git a/3751/CH6/EX6.8/Ex6_8.sce b/3751/CH6/EX6.8/Ex6_8.sce new file mode 100644 index 000000000..43cc49b71 --- /dev/null +++ b/3751/CH6/EX6.8/Ex6_8.sce @@ -0,0 +1,37 @@ +//Fluid System By Shiv Kumar
+//Chapter 6 - Kaplan and Propeller Turbines
+//Example 6.8
+//To Find Inlet and outlet Angles of the Runner blades
+
+clc
+clear
+
+//Given:
+ H=25; //Head, m
+ P=23000; //Shaft Power, KW
+ D=5; //External Diameter of Runner, m
+ d=3; //Diameter of Hub, m
+ N=60; //Rotational Speed, rpm
+ eta_H=95/100; //Hydraulic Efficiency
+ eta_0=88/100; //Overall Efficiency
+ Vw=0; //As there is no exit whirl
+
+//Data Required:
+ rho=1000; //Density of Water, Kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations
+ Dm=(D+d)/2; //Mean Diameter of Runner, m
+ ui=%pi*Dm*N/60 //m/s
+ Q=P*10^3/(rho*g*H*eta_0); //Discharge, m^3/s
+ Vfi=Q/((%pi/4)*(D^2-d^2)) // m/s
+ Vwi=eta_H*g*H/ui; //m/s
+ uo=ui;
+ Vfo=Vfi;
+ Beta_i=atand(Vfi/(Vwi-ui)); //Degrees
+ Beta_o=atand(Vfo/uo); //Degrees
+
+//Results
+ printf("At the Mean Radius\n\t")
+ printf("Runner Blade Angle at Inlet, Beta_i=%.2f Degrees\n\t",Beta_i) //The answer vary due to round off error
+ printf("Runner Blade Angle at Outlet, Beta_o=%.2f Degrees\n",Beta_o) //The answer vary due to round off error
diff --git a/3751/CH6/EX6.9/Ex6_9.sce b/3751/CH6/EX6.9/Ex6_9.sce new file mode 100644 index 000000000..874c56af3 --- /dev/null +++ b/3751/CH6/EX6.9/Ex6_9.sce @@ -0,0 +1,50 @@ +//Fluid System By Shiv Kumar
+//Chapter 6 - Kaplan and Propeller Turbines
+//Example 6.9
+//To Determine Runner Vane Angles at the hub and at the Outer Periphery
+
+clc
+clear
+
+//Given:
+ P=22500; //Power Available at Shaft, KW
+ H=20; //Head, m
+ N=150; //Rotational Speed, rpm
+ eta_H=95/100; //Hydraulic Efficiency
+ eta_0=88/100; //Overall Efficiency
+ D=4.5; //Outer Diameter of Runner, m
+ d=2; //Diameter of Hub, m
+ Vw=0; //As there is no exit whirl
+
+//Data Required:
+ rho=1000; //Density of Water, Kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations
+ //Runner Vane Angles at Hub
+ uo=%pi*d*N/60; //m/s
+ ui=uo;
+ Q=P*10^3/(rho*g*H*eta_0); //Discharge, m^3/s
+ Vwi=eta_H*g*H/ui; //m/s
+ Vfi=Q/((%pi/4)*(D^2-d^2)) // m/s
+ Vfo=Vfi;
+ Beta_i=180-atand(Vfi/(ui-Vwi)); //Degrees
+ Beta_o=atand(Vfo/uo); //Degrees
+
+//Result1
+ printf("Runner Vane Angles at the Hub \n\t")
+ printf("Beta_i=%.2f Degrees\n\t",Beta_i) //The answer vary due to round off error
+ printf("Beta_o=%.2f Degrees\n\n",Beta_o) //The answer vary due to round off error
+
+ // Runner Vane Angles at outer periphery
+ uo=%pi*D*N/60; //m/s
+ ui=uo;
+ Vwi=eta_H*g*H/ui; //m/s
+ Beta_i=180-atand(Vfi/(ui-Vwi)); //Degrees
+ Beta_o=atand(Vfo/uo); //Degrees
+
+//Result2
+ printf("Runner Vane Angles at the Outer periphery \n\t")
+ printf("Beta_i=%.2f Degrees\n\t",Beta_i) //The answer vary due to round off error
+ printf("Beta_o=%.2f Degrees\n\n",Beta_o) //The answer vary due to round off error
+
diff --git a/3751/CH7/EX7.1/Ex7_1.sce b/3751/CH7/EX7.1/Ex7_1.sce new file mode 100644 index 000000000..cc766d7c3 --- /dev/null +++ b/3751/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,27 @@ +//Fluid Systems by Shiv Kumar
+//Chapter 7 - Performance of water turbine
+//Example 7.1
+//To Calculate specific speed of turbine and state the type of turbine
+ clc
+ clear
+//Given
+ P=8000; //Power developed, KW
+ H=30; //Head, m
+ N=140; //Speed, rpm
+//Computations
+ Ns=N*P^(1/2)/(H^(5/4)); // specific speed of turbine , in SI UNITS
+
+//Results
+ printf("The Specific speed of Turbine is %.2f (SI Units)\n",Ns)
+ //To Determine the type of turbine
+ if Ns>51 & Ns<=255 then
+ printf(" The type of turbine is Francis")
+ elseif Ns>=8.5 & Ns<=30 then
+ printf("The type of turbine is Pelton Wheel with single jet")
+ elseif Ns>30& Ns<=51 then
+ printf("The type of turbine is Pelton Wheel with multi jet")
+ elseif Ns>255 & Ns<=860 then
+ printf("The type of turbine is Kaplan or Propeller Turbine")
+ end
+
+
diff --git a/3751/CH7/EX7.10/Ex7_10.sce b/3751/CH7/EX7.10/Ex7_10.sce new file mode 100644 index 000000000..283f7c48a --- /dev/null +++ b/3751/CH7/EX7.10/Ex7_10.sce @@ -0,0 +1,29 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 7- Performance of Water Turbine +//Example 7.10 +// To Calculate Speed and Power Developed by the Prototype when working Under a Head of 8 m. + clc + clear + +//Given:- + Lr=1/5; //Scale Ratio + DmbyDp=Lr; + + //For Prototype + Hp=8; //Head, m + + //For Model + Pm=5; //Power, kW + Hm=2; //Head, m + Nm=600; //rpm + +//Computations + Np=Nm*DmbyDp*(Hp/Hm)^(1/2); //rpm + Pp=Pm*(Np/Nm)^3/(DmbyDp^5); //KW + + +//Results + printf("For the Prototype (Working Under a Head of 8 m:\n") + printf(" Speed, Np=%.f rpm\n Power Developed, Pp=%.f kW",Np,Pp) + + diff --git a/3751/CH7/EX7.11/Ex7_11.sce b/3751/CH7/EX7.11/Ex7_11.sce new file mode 100644 index 000000000..4c0f85df2 --- /dev/null +++ b/3751/CH7/EX7.11/Ex7_11.sce @@ -0,0 +1,33 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 7- Performance of Water Turbine +//Example 7.11 +// To Find (a)Power Developed by Model (b)Ratio of Heads and Ratio of Mass Flow Rates between Prototype and Model. + clc + clear + +//Given:- + Pp=12; //Power Developed by Prototype,MW + Lr=1/10; //Scale Ratio + DmbyDp=Lr; + LmbyLp=Lr; + + +//Computations:- + + //(a)Power Developed by the Model + //As Np=Nm and effeciencies of prototype and model are equal + Pm=Pp*10^6*(DmbyDp)^5; //W + + //(b)Ratio of Heads and Ratio of Mass flow Rates + HpbyHm=DmbyDp^(-2); //Dimensionless + QpbyQm=DmbyDp^(-3) + //As m=rho*Q and rho is Constant. So, + m_pbym_m=QpbyQm; + +//Results + + printf("(a)Power Developed by the Model,Pm=%.f W\n",Pm) + printf(" (b)Ratio of Heads, Hp/Hm=%.f\n Ratio of Mass flow rates, m_p/m_m=%.f\n",HpbyHm,m_pbym_m) + + + diff --git a/3751/CH7/EX7.12/Ex7_12.sce b/3751/CH7/EX7.12/Ex7_12.sce new file mode 100644 index 000000000..1583c0430 --- /dev/null +++ b/3751/CH7/EX7.12/Ex7_12.sce @@ -0,0 +1,34 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 7- Performance of Water Turbine +//Example 7.12 +// To Find (a)The Speed ,Discharge and Power required for the Actual Pump (b) The Discharge of the model. + clc + clear + +//Given:- + Lr=5; //Scale Ratio + DpbyDm=Lr; + DmbyDp=1/DpbyDm; + //For Model + Pm=22; //Power Required, kW + Hm=7; //Head, m + Nm=410; //Speed, rpm + eta_m=1; //Assumption that efficiency of the model is 100% + //For Prototype + Hp=35; //Head, m + //Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Np=Nm*DmbyDp*(Hp/Hm)^(1/2); //rpm + Pp=Pm*(Np/Nm)^3*DpbyDm^5; //KW + Qm=Pm*1000*eta_m/(rho*g*Hm); //m^3/s + Qp=Qm*(Np/Nm)^2*DpbyDm^2; //m^3/s + +//Results:- + printf("(a)For the Actual Pump(Prototype):\n Speed, Np=%.2f rpm , \n Discharge, Qp=%.3f m^3/s and \n Power,Pp=%.2fKW\n",Np,Qp,Pp) //The Answer vary due to Round off Error(For Qp), The Answer Provided in the Textbook is Wrong (For Np and Pp). + + printf("(b)The Discharge of the Model, Qm=%.4f m^3/s",Qm) //The Answer vary due to Round off Error + + diff --git a/3751/CH7/EX7.13/Ex7_13.sce b/3751/CH7/EX7.13/Ex7_13.sce new file mode 100644 index 000000000..648937bdb --- /dev/null +++ b/3751/CH7/EX7.13/Ex7_13.sce @@ -0,0 +1,38 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 7- Performance of Water Turbine +//Example 7.13 +// To Determine the maximum flow rate and specific speed for the Turbine and To Find Speed, Power Output and Water Consumption of the Model. + clc + clear + +//Given:- + Lr=10; //Scale Ratio + DpbyDm=Lr; + DmbyDp=1/DpbyDm; + //For Prototype + Pp=1000; //Power , kW + Hp=14; //Head, m + Np=130; //Speed, rpm + eta_o=91/100; //Overall efficiency + //For Model + Hm=6; //Head, m + //Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to gravity, m/s^2 + +//Computations:- + Qp=Pp*1000/(rho*g*Hp *eta_o ); //m^3/s + Ns=Np*Pp^(1/2)/(Hp^(5/4)); //Specific Speed, In SI UNITS + Nm=Np*DpbyDm*(Hm/Hp)^(1/2); //rpm + Qm=Qp*(Nm/Np)*(DmbyDp)^3; //m^3/s + Pm=Pp*(Nm/Np)^3*DmbyDp^5; //KW + +//Results:- + printf("For the Turbine : \n\t") + printf("Maximum Flow Rate, Qp=%.f m^3/s\n\t",Qp) + printf("Specific Speed, Ns=%.2f (SI Units)\n\n",Ns) + printf("For the Model : \n\t") + printf("Speed, Nm=%.2f rpm\n Power Output, Pm=%.2f kW\n Water Consumption, Qm=%.4f m^3/s \n", Nm,Pm,Qm) //The Answer vary due to Round off Error + + + diff --git a/3751/CH7/EX7.14/Ex7_14.sce b/3751/CH7/EX7.14/Ex7_14.sce new file mode 100644 index 000000000..9a4c66d2e --- /dev/null +++ b/3751/CH7/EX7.14/Ex7_14.sce @@ -0,0 +1,36 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 7- Performance of Water Turbine +//Example 7.14 +//To Determine the Size (Scale Ratio) of the Model and To Find the Model Speed and Power. + + clc + clear + +//Given:- + TP=240000; //Total Power Produced, kW + n=4; //No. of Turbines + eta_o=91/100; //Effeciency of each turbine + Np=120; //Speed of each Turbine, rpm + Hp=70; //Head for each Turbine, m + + Qm=0.45; //Discharge for Model, m^3/s + Hm=5; //Head for testing the Model, m + +//Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleratrion due to gravity, m/s^2 + +//Calculations:- + Pp=TP/n; //Power produced from each Turbine, kW + Qp=Pp*1000/(rho*g*Hp*eta_o); //Discharge passing through one Turbine, m^3/s + DmbyDp=(Qm/Qp)^(1/2)*(Hp/Hm)^(1/4); //From Relation of Flow Coefficient + Lr=DmbyDp; //Scale Ratio + Nm=(Np/DmbyDp)*(Hm/Hp)^(1/2); //rpm + Pm=Pp*(Nm/Np)^3*DmbyDp^5; //KW + +//Results + printf("The Scale Ratio is 1:%.2f\n ",1/Lr) + printf("Model Speed, Nm=%.2f rpm\n",Nm) + printf("Model Power, Pm=%.2f kW\n",Pm) //The Answer vary due to Round off Error + + diff --git a/3751/CH7/EX7.15/Ex7_15.sce b/3751/CH7/EX7.15/Ex7_15.sce new file mode 100644 index 000000000..b508479a7 --- /dev/null +++ b/3751/CH7/EX7.15/Ex7_15.sce @@ -0,0 +1,34 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 7- Performance of Water Turbine +//Example 7.15 +//To Determine the rpm of Prototype, Ratio of Power Developed by Model and Prototype and Efficiency of Prototytpe . + + clc + clear + +//Given:- + Lr=1/8; //Scale Ratio + //For Model, + Hm=5; //Head, m + Nm=350; //Speed,rpm + eta_m=78/100; //Effiency of model + //For Prototype, + Hp=100; //Head, m + +//Calculations:- + DpbyDm=1/Lr; // Dp/Dm + //(a) Speed of Prototype, Np + Np=Nm*Lr*(Hp/Hm)^(1/2); //rpm + //(b)Ratio of Power Developed, Pp/Pm + PpbyPm=DpbyDm^5*(Np/Nm)^3; + //(c)Efficiency of Prototype when Scale Effect is Considered + //From Moody's Equation, + eta_p=(1-Lr^0.2*(1-eta_m))*100; //In Percentage + +//Results + +printf(" (a)The Speed of Prototype, Np=%.2f rpm\n",Np) //The Answer vary due to Round off Error +printf(" (b)Ratio of Power Developed, Pp/Pm =%.2f \n",PpbyPm) //The Answer vary due to Round off Error +printf(" (c)Efficiency of Prototype, eta_p =%.2f Percent\n",eta_p) //The Answer vary due to Round off Error + + diff --git a/3751/CH7/EX7.2/Ex7_2.sce b/3751/CH7/EX7.2/Ex7_2.sce new file mode 100644 index 000000000..8f75a0621 --- /dev/null +++ b/3751/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,34 @@ +//Fluid Systems by Shiv Kumar
+//Chapter 7 - Performance of water turbine
+//Example 7.2
+//To Find (a) Specific speed of turbine (b) Power Developed (c) Type of turbine
+ clc
+ clear
+//Given:
+
+ H=28; //Head, m
+ N=225; //Speed, rpm
+ Q=10; //Discharge, cumec=m^3/s
+ eta_o=90/100; //Overall Efficiency
+//Data Required
+ rho=1000 // Density of Water, Kg/m^3
+ g=9.81; // Acceleration due to gravity, m/ s^2
+//Computations
+ P=eta_o*rho*Q*g*H/1000; //Power developed, KW
+ Ns=N*P^(1/2)/(H^(5/4)); // specific speed of turbine , in SI UNITS
+
+//Result1
+
+ printf("(a)The Specific speed of Turbine = %.2f (SI Units)\n",Ns) //The Answer Vary due to Round off Error
+ printf("(b)Power developed =%.2f kW\n",P)
+//To Determine the type of turbine, Result2
+ if Ns>51 & Ns<=255 then
+ printf("(c)The type of turbine is Francis.")
+ elseif Ns>=8.5 & Ns<=30 then
+ printf("(c)The type of turbine is Pelton Wheel with single jet.")
+ elseif Ns>30 & Ns<=51 then
+ printf("(c)The type of turbine is Pelton Wheel with multi jet.")
+ elseif Ns>255 & Ns<=860 then
+ printf("(c)The type of turbine is Kaplan or Propeller turbine.")
+
+ end
diff --git a/3751/CH7/EX7.3/Ex7_3.sce b/3751/CH7/EX7.3/Ex7_3.sce new file mode 100644 index 000000000..d044d619e --- /dev/null +++ b/3751/CH7/EX7.3/Ex7_3.sce @@ -0,0 +1,36 @@ +//Fluid Systems by Shiv Kumar
+//Chapter 7 - Performance of water turbine
+//Example 7.3
+//To Find (a) The Discharge required (b) The Diameter of Wheel (c) The Diameter and number of jets required (d)The Specific Speed
+ clc
+ clear
+//Given:
+ P=8200; //Power Developed, kW
+ H=128; // Head , m
+ N=210; // Speed, rpm
+ Cv=0.98; // Co-efficient of Velocity
+ eta_H=89/100; //Hydraulic Efficiency
+ Ku=0.45; // Speed Ratio
+ dbyD=1/10; //Ratio of jet diameter to wheel Diameter
+ eta_m=92/100; //Mechanical Efficiency
+//Data required
+ rho=1000; //Density of Water, Kg/m^3
+ g=9.81; // Acceleration due to gravity, m/s^2
+//Assumptions:
+ eta_v=100/100; // Volumetric efficiency is 100%
+
+//Computations
+ D=Ku*sqrt(2*g*H)*60/(%pi*N); //Wheel Diameter, m
+ d=D*dbyD; // Jet diameter, m
+ eta_o=eta_H*eta_m*eta_v; //Overall Efficiency
+ Q=P*1000/(rho*g*H*eta_o); //Net Discharge, m^3/s
+ q=(%pi/4)*d^2*Cv*sqrt(2*g*H); //Discharge through one jet, m^3/s
+ n=round(Q/q); //Number of jets
+ Ns= N*P^(1/2)/(H^(5/4)); //Specific Speed, SI Units
+
+//Results
+printf("(a) The Discharge required, Q =%.3f m^3/s\n",Q)
+printf("(b) The Diameter of Wheel, D =%.2f m\n",D)
+printf("(c) The Diameter, d=%.3f m and\n number of jets required =%.f \n",d,n)
+printf("(d)The Specific Speed, Ns=%.2f (SI Units)\n",Ns) //The Answer Vary due to Round off Error
+
diff --git a/3751/CH7/EX7.4/Ex7_4.sce b/3751/CH7/EX7.4/Ex7_4.sce new file mode 100644 index 000000000..bbfd9ddbc --- /dev/null +++ b/3751/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,26 @@ +//Fluid Systems by Shiv Kumar
+//Chapter 7 - Performance of water turbine
+//Example 7.4
+//To Find (a) The Diameter of Runner (b) The Diameter of jet
+ clc
+ clear
+//Given:
+ P=3200; //Power Developed, kW
+ H=310; // Effective Head , m
+ eta_o=82/100; //Overall Efficiency
+ Ku=0.46; // Speed Ratio
+ Cv=0.98 // Co-efficient of Velocity
+ Ns=18; //Specific Speed (SI Units)
+//Data required
+ rho=1000; //Density of Water, Kg/m^3
+ g=9.81 // Acceleration due to gravity, m/s^2
+
+//Computations
+ N=Ns*H^(5/4)/sqrt(P); //Speed, rpm
+ D=Ku*sqrt(2*g*H)*60/(%pi*N); //Diameter of runner, m
+ Q=P*1000/(rho*g*H*eta_o); //Discharge, m^3/s
+ d=sqrt(Q/((%pi/4)*Cv*sqrt(2*g*H))); // Diameter of Jet, m
+
+//Results
+ printf("(a) The Diameter of Runner, D =%.2f m\n",D) //The Answer Vary due to Round Off Error
+ printf("(c) The Diameter of Jet, d=%.3f m \n",d)
diff --git a/3751/CH7/EX7.5/Ex7_5.sce b/3751/CH7/EX7.5/Ex7_5.sce new file mode 100644 index 000000000..f2d86c55b --- /dev/null +++ b/3751/CH7/EX7.5/Ex7_5.sce @@ -0,0 +1,47 @@ +//Fluid Systems by Shiv Kumar
+//Chapter 7 - Performance of water turbine
+//Example 7.5
+//To Find (a) Number of Units to be installed (b) Diameter of Wheel (c) Diameter of Jet
+ clc
+ clear
+
+//Given:
+
+ H_G= 310; //Gross Head,m
+ l=2.5; // Length, km
+ h_f=25; // Friction Loses, J/N=m
+ TO=20; //Total Output Power, MW
+ N=660; // Speed, rpm
+ ubyVi=0.46 //Ratio of bucket to jet speed
+ eta_o=88/100; //Overall Efficiency
+ Ns=28; //Specific Speed, SI Units
+ Cv=0.97;
+ Cd=0.94;
+
+
+//Data Required:
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+
+//Computations:
+ H=H_G-h_f; //Effective Head, m
+ P=(Ns*H^(5/4)/N)^2; //Power Output of each Unit, KW
+ //(a) The no. of units to be lnstalled,n
+ n=round(TO*1000/P);
+ //(b)Diameter of Wheel,D
+ Vi=Cv*sqrt(2*g*H); //m/s
+ D=ubyVi*Vi*60/(%pi*N); //m
+
+//(c) Diameter of Jet, d
+ Q=TO*10^6/(rho*g*H*eta_o); //Net Discharge, m^3/s
+ q=Q/n; // Discharge through one unit, m^/s
+ d=sqrt(q/((%pi/4)*Cd*sqrt(2*g*H)))*1000; //mm
+
+
+//Results
+ printf("(a) The no. of units to be Installed=%.f Units\n",n)
+ printf("(b) Diameter of Wheel, D=%.3f m\n",D)
+ printf("(c) Diameter of Jet, d=%.1f mm\n",d) //The Answer Vary due to round off Error
+
+
diff --git a/3751/CH7/EX7.6/Ex7_6.sce b/3751/CH7/EX7.6/Ex7_6.sce new file mode 100644 index 000000000..1599b5c04 --- /dev/null +++ b/3751/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,23 @@ +//Fluid Systems by Shiv Kumar
+//Chapter 7 - Performance of water turbine
+//Example 7.6
+//To Find Speed and Power Developed by the Turbine
+ clc
+ clear
+
+//Given:
+ //Ist Condition
+ P1=8500; //Power Developed, kW
+ N1=120; //Speed, rpm
+ H1=32; //Head, m
+
+ //2nd Condition
+ H2=25; //Head, m
+
+//Computations:
+ P2=P1*(H2/H1)^(3/2); //kW
+ N2=N1*(H2/H1)^(1/2); //rpm
+
+//Results
+ printf("The Speed Developed by the Turbine,N2=%.f rpm\n",N2)
+ printf("The power developed= %.2f kW",P2)
diff --git a/3751/CH7/EX7.7/Ex7_7.sce b/3751/CH7/EX7.7/Ex7_7.sce new file mode 100644 index 000000000..86ba6ebaa --- /dev/null +++ b/3751/CH7/EX7.7/Ex7_7.sce @@ -0,0 +1,39 @@ +//Fluid Systems by Shiv Kumar
+//Chapter 7 - Performance of water turbine
+//Example 7.7
+//To Determine unit power, unit speed and unit discharge and also find speed, discharge and power at condition 2
+ clc
+ clear
+
+//Given:
+ //At Condition 1
+ P1=7200; //Power Developed, KW
+ N1=300; //Speed, rpm
+ H1=350; //Head, m
+ eta_o=85/100; // Overall efficiency
+
+ //At Condition 2
+ H2=300; //Head, m
+
+//Data Used:
+ rho=1000; //Density of Water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^@
+
+
+//Computations:
+ Q1=P1*1000/(rho*g*H1*eta_o); //m^3/s
+ Pu=P1/(H1^(3/2)); //Unit Power, KW
+ Nu=N1/sqrt(H1); //Unit Speed, rpm
+ Qu=Q1/sqrt(H1); //Unit Discharge, m^3/s
+ P2=P1*(H2/H1)^(3/2); //KW
+ N2=N1*(H2/H1)^(1/2); //rpm
+ Q2=Q1*sqrt(H2/H1); //m^3/s
+
+//Results
+ printf("Unit Power, Pu= %.3f kW\n Unit Speed, Nu=%.2f rpm\n Unit Discharge, Qu=%.4f m^/s\n",Pu, Nu, Qu) //The Answer vary due to Round off Error
+
+ printf("At head of 300 m:\n\t")
+ printf("The Speed,N2=%.2f rpm\n\t",N2) //The Answer vary due to Round off Error
+ printf("The power,P2= %.2f kW\n\t",P2)
+ printf("The Discharge, Q2=%.3f m^3/s\n",Q2) //The Answer vary due to Round off Error
+
diff --git a/3751/CH7/EX7.8/Ex7_8.sce b/3751/CH7/EX7.8/Ex7_8.sce new file mode 100644 index 000000000..b4b94bb30 --- /dev/null +++ b/3751/CH7/EX7.8/Ex7_8.sce @@ -0,0 +1,28 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 7- Performance of Water Turbine +//Example 7.8 +// To Find Model Runner Speed and Prototype to Model Scale ratio + clc + clear + +//Given:- + //For Prototype + Pp=30; //Power Developed, MW + Hp=55; //Head, m + Np=100; //Speed, rpm + Pp=Pp*1000; //KW + //For Model + Pm=25 ; //Power Developed, KW + Hm=6; //Head, m + +//Computations:- + Nm=Np*(Hm/Hp)^(5/4)*(Pp/Pm)^(1/2); //rpm + DpbyDm=((Pp/Pm)*(Nm/Np)^3)^(1/5); //A Ratio(Dimensionless) + Lr= DpbyDm; //Scale Ratio + + +//Results + printf("The Model Runner Speed, Nm=%.2f rpm And\n",Nm) + printf("Prototype to Model Scale Ratio,Lr=%.2f",Lr) //The Answer vary due to Round off Error + + diff --git a/3751/CH7/EX7.9/Ex7_9.sce b/3751/CH7/EX7.9/Ex7_9.sce new file mode 100644 index 000000000..e86f4aeb8 --- /dev/null +++ b/3751/CH7/EX7.9/Ex7_9.sce @@ -0,0 +1,34 @@ +//Fluid Systems- By Shiv Kumar +//Chapter 7- Performance of Water Turbine +//Example 7.9 +// To Determine the Performance of the Turbine Under a Head of 20 m + clc + clear + +//Given:- + //Condition 1: + H1=25; //Head, m + N1=200; //Speed, rpm + Q1=9; //Discharge, m^3/s + eta_o=90/100; //Overall Efficiency + + //Condition 2: + H2=20; //Head, m + +//Data Required:- + rho=1000; //Density of Water, Kg/m^3 + g=9.81; //Acceleration due to Gravity, m/s^2 + + +//Calculations:- + P1=rho*Q1*g*H1*eta_o/1000; //KW + N2=N1*sqrt(H2/H1); //rpm + Q2=Q1*sqrt(H2/H1); //m^3/s + P2=P1*(H2/H1)^(3/2); //KW + + +//Results:- + printf("At Condition 2 (Under a Head of 20 m):\n") + printf("\tSpeed, N2=%.2f rpm\n Discharge, Q2=%.2f m^3/s\n Power Developed, P2=%.2f kW",N2,Q2,P2) //The Answer vary due to Round off Error + + |
