19 files changed, 402 insertions, 0 deletions
diff --git a/3733/CH4/EX4.1/Ex4_1.sce b/3733/CH4/EX4.1/Ex4_1.sce
new file mode 100644
index 000000000..1d80bf64c
--- /dev/null
+++ b/3733/CH4/EX4.1/Ex4_1.sce
@@ -0,0 +1,18 @@
+//Example 4_1
+clc;funcprot(0);
+//Given data
+p=25;// kW
+n=480;// rpm
+h=5;// m
+// d_r=D/d
+d_r=10;
+H=40;// m
+
+//Calculation
+N=n*(1/d_r)*(sqrt(H/h));// rpm
+P=p*(d_r)^2*(H/h)^(3/2);// kW
+n_s=(n*sqrt(p))/h^(5/4);
+N_s=(N*sqrt(P))/(H)^(5/4);
+printf('\n N=%0.0f r.p.m\nP=%0.0f kW\nN_s=%0.0f',N,P,N_s);
+printf('\nThe runner is of propeller type');
+// The answer vary due to round off error
diff --git a/3733/CH4/EX4.10/Ex4_10.sce b/3733/CH4/EX4.10/Ex4_10.sce
new file mode 100644
index 000000000..438840511
--- /dev/null
+++ b/3733/CH4/EX4.10/Ex4_10.sce
@@ -0,0 +1,25 @@
+//Example 4_10
+clc;funcprot(0);
+//Given data
+h=36;// m
+p=135;// kW
+q=0.44;// m^3/sec
+H=100;// m
+N=428;// r.p.m
+d_r=4;// d_r=D/d
+
+// Calculation
+w=1000*9.81;// N
+n=N*d_r*sqrt(h/H);// r.p.m
+n_m=(p*1000)/(1000*9.81*q*h);// The efficiency of the model
+n_p=n_m+0.03;//The efficiency of the prototype
+P=p*(n_p/n_m)*(d_r)^2*(H/h)^(3/2);// kW
+printf('\n The power developed by the prototype,P=%0.0f kW',P);
+n_s=(n*sqrt(p))/h^(5/4);
+N_s=(N*sqrt(P))/(H)^(5/4);
+if(N_s~=n_s)
+    printf('\n The runner is of Francis type.');
+else
+    printf('\n Wrong');
+end
+// The answer vary due to round off error
diff --git a/3733/CH4/EX4.11/Ex4_11.sce b/3733/CH4/EX4.11/Ex4_11.sce
new file mode 100644
index 000000000..7b4e02fec
--- /dev/null
+++ b/3733/CH4/EX4.11/Ex4_11.sce
@@ -0,0 +1,30 @@
+//Example 4_11
+clc;funcprot(0);
+//Given data
+Q=70;// m^3/sec
+H=15;// m
+N=200;// r.p.m
+N_s=340;// Specific speed
+n_t=0.90;// The efficiency of the turbine
+rho=1000;// Density in kg/m^3
+g=9.81;// m/s^2
+D=[143 151 158.5 165 172.5];// Diamter of runner in cm
+kW=[66.7 74 82.5 87 92];// (Unit)
+rpm=[53 51 48.5 45.4 42.5];// (Unit)
+
+// Calculation
+//(a)
+printf('\n (a)The type of the runner is Kaplan as the specific speed is 340.');
+//(b)
+w=rho*g;
+P_t=(w*Q*H*n_t)/(1000);// kW
+P=((N_s*H^(5/4))/N)^2;// kW
+T_n=P_t/P;// Number of turbine units required
+printf('\n (b)Number of turbine units required=%0.0f ',T_n);
+//(c)
+P_u=P/(H^(3/2));//The unit Power in kW
+N_u=N/(H^(1/2));//The unit speed in r.p.m.m^-1/2
+// For unit power of 43.35 and unit speed of 51.7,the required diameter can be calculated by interploation from the given data
+D=D(1)+(((rpm(1)-N_u)/(rpm(1)-rpm(2)))*((D(2)-D(1))));// The diameter of the runner in cm
+printf('\n(c)The diameter of the runner=%0.2f cm',D);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH4/EX4.12/Ex4_12.sce b/3733/CH4/EX4.12/Ex4_12.sce
new file mode 100644
index 000000000..bdd6e007e
--- /dev/null
+++ b/3733/CH4/EX4.12/Ex4_12.sce
@@ -0,0 +1,15 @@
+//Example 4_12
+clc;funcprot(0);
+//Given data
+p_a=1.30;// Atmospheric pressure in bar
+p_c=0.5;// bar
+V_c=5;// m/sec
+V_d=2;// m/sec
+h_f=0.2;// m
+g=9.81;// m/s^2
+
+//Calculation
+w=1000*9.81;// N
+h=(((p_a-p_c)*1.03*10^5)/w)-((V_c^2)/(2*g))+(((V_d^2)/(2*g))+h_f);// m
+printf('\n The maximum height of the turbine above tail race,h=%0.3f m',h);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH4/EX4.13/Ex4_13.sce b/3733/CH4/EX4.13/Ex4_13.sce
new file mode 100644
index 000000000..fe5ed0f5e
--- /dev/null
+++ b/3733/CH4/EX4.13/Ex4_13.sce
@@ -0,0 +1,17 @@
+//Example 4_13
+clc;funcprot(0);
+//Given data
+p_a=750;// mm of Hg
+p_v=400;// mm of Hg
+p_d=p_a-p_v;// mm of Hg
+V_c=13; // m/sec
+// Assume Friction loss and exit velocity of water head (V_a^2/(2*g))+h_f=V
+V=1.5;//m
+rho=1000;// kg/m^3
+g=9.81;// m/s^2
+
+//Calculation
+w=rho*g;// N
+h=(((p_a-p_d)*1.03*10^5)/(w*760))-((V_c^2)/(2*g))+V;// m
+printf('\nThe position of the kaplan turbine with respect to tail race,h=%0.2f m',h);
+// The answer vary due to round off error
diff --git a/3733/CH4/EX4.14/Ex4_14.sce b/3733/CH4/EX4.14/Ex4_14.sce
new file mode 100644
index 000000000..89b9ac0f4
--- /dev/null
+++ b/3733/CH4/EX4.14/Ex4_14.sce
@@ -0,0 +1,17 @@
+//Example 4_14
+clc;funcprot(0);
+//Given data
+p_a=755;// mm of Hg
+V_c=8;// m/sec
+V_d=3;// m/sec
+T_a=20;// °C
+g=9.81;// m/s^2
+
+//Calculation
+// The minimum value of p_c corresponds to the saturation pressure of water vapour at 20°C.
+//From steam table,
+p_c=17.6;// mm of Hg
+h_f=0.0;// m
+p_b=760;// mm of Hg
+h=(((p_a-p_c)*13.6)/p_b)-((V_c^2)/(2*g))+(((V_d^2)/(2*g))+h_f);// m
+printf('\n The maximum possible height of the turbine,h=%0.1f meters above tailrace level.',h);
diff --git a/3733/CH4/EX4.15/Ex4_15.sce b/3733/CH4/EX4.15/Ex4_15.sce
new file mode 100644
index 000000000..ff2a587b8
--- /dev/null
+++ b/3733/CH4/EX4.15/Ex4_15.sce
@@ -0,0 +1,21 @@
+//Example 4_15
+clc;funcprot(0);
+//Given data
+N_1=200;// r.p.m
+H_1=25;// m
+Q=9;// m^3/sec
+n_t=0.90;// The turbine efficiency
+g=9.81;// m/s^2
+rho=1000;// kg/m^3
+
+// Calculation
+//(a)
+N_s=(2*%pi*N_1*sqrt(Q))*(sqrt(n_t))/(60*(g*H_1)^(3/4));
+//(b)
+P_1=(rho*g*H_1*Q*n_t)/(1000);// kW
+//(c)
+H_2=15;// m
+N_2=N_1*sqrt(H_2/H_1);// r.p.m
+P_2=P_1*(H_2/H_1)^(3/2);// kW
+printf('\n(a)The specific speed,N_s=%0.3f \n(b)Power generated,P=%0.1f kW \n(c)Speed and Power if the head is reduced to 15m,N_2=%0.1f rpm & P_2=%0.0f kW',N_s,P_1,N_2,P_2);
+printf('\nFrom the range of specific speed it is seen that the turbine to be selected is Francis type');
diff --git a/3733/CH4/EX4.16/Ex4_16.sce b/3733/CH4/EX4.16/Ex4_16.sce
new file mode 100644
index 000000000..e04f3f3ad
--- /dev/null
+++ b/3733/CH4/EX4.16/Ex4_16.sce
@@ -0,0 +1,34 @@
+//Example 4_16
+clc;funcprot(0);
+//Given data
+P=5400;// kW
+N=200;// r.p.m
+D=3;// m
+H=240;// m
+n_t=0.82;
+rho=1000;// kg/m^3
+g=9.81;// m/s^2
+
+//Calculation
+//(a)
+Q=(P*1000)/(rho*g*H*n_t);// m^3/sec
+N_u=(N*D/sqrt(H));// Unit speed
+P_u=(P/(D^2*H^(3/2)));// Unit power
+Q_u=(Q/(D^2*(sqrt(H))));// Unit flow
+N_s=(2*%pi*N*sqrt(Q)*sqrt(n_t))/(60*(g*H)^(3/4));// Specific speed
+printf('\n(a)The flow rate,Q=%0.1f m^3/sec \n   The unit speed,N_u=%0.1f \n   The unit power,P_u=%0.3f \n   The unit flow,Q_u=%0.2f \n   The specific speed,N_s=%0.3f',Q,N_u,P_u,Q_u,N_s);
+//(b)
+// When the head is changed to 160 m,the diameter remains same.
+H=160;// m
+N=(N_u*sqrt(H))/D;// rpm
+P_1=(P_u*D^2*H^(3/2));// kW
+Q=(Q_u*D^2*sqrt(H));// m^3/sec
+printf('\n(b)Speed,N=%0.0f r.p.m\n   Power,P=%0.0f kW\n   The flow rate,Q=%0.2f m^3/sec ',N,P_1,Q);
+//(c)
+H=183;// m
+P=2850;// kW
+D_1=sqrt((P/(P_u*H^(3/2))));// m
+N_1=(N_u*sqrt(H))/(D_1);// r.p.m
+Q=Q_u*D_1^2*sqrt(H);// m^3/sec
+printf('\n(c)Diameter,D_1=%0.2f m\n   Speed,N=%0.0f r.p.m\n   The flow rate,Q=%0.2f m^3/sec ',D_1,N_1,Q);
+// The answer vary due to round off error
diff --git a/3733/CH4/EX4.17/Ex4_17.sce b/3733/CH4/EX4.17/Ex4_17.sce
new file mode 100644
index 000000000..5c683ed67
--- /dev/null
+++ b/3733/CH4/EX4.17/Ex4_17.sce
@@ -0,0 +1,16 @@
+//Example 4_17
+clc;funcprot(0);
+//Given data
+P_1=93;// kW
+H_1=64;// m
+H_2=88;// m
+
+//Calculation
+//(i)For the same size,the speed is proportional to the square root of head and so
+// N=N_2/N_1
+N=sqrt(H_2/H_1);
+// By solving N, it gives the relation N_2=1.173 N_2
+N_i=((N*100)-100);//The speed increases in % 
+//(ii)For the same wheel,power varies as H^3/2 and so
+P_2=P_1*(H_2/H_1)^(3/2);// kW
+printf('\n The speed increases by %0.1f percentage.\n The power developed,P=%0.0f kW',N_i,P_2);
diff --git a/3733/CH4/EX4.18/Ex4_18.sce b/3733/CH4/EX4.18/Ex4_18.sce
new file mode 100644
index 000000000..c2cde5aca
--- /dev/null
+++ b/3733/CH4/EX4.18/Ex4_18.sce
@@ -0,0 +1,19 @@
+//Example 4_18
+clc;funcprot(0);
+//Given data
+P=86000;// kW
+N=180;// r.p.m
+H=148;// m
+D=3.4;// m
+Q=66.5;// m^3/sec
+rho=1000;// kg/m^3
+g=9.81;// m/s^2
+
+//Calculation
+N_u=(N*D/sqrt(H));// r.p.m
+P_u=(P/(D^2*H^(3/2)));// kW
+Q_u=(Q/(D^2*(sqrt(H))));// m^3/s
+n_t=((P*1000)/(rho*g*Q*H));// The turbine efficiency
+N_s=((%pi*D*N)/60)*((sqrt(Q*n_t)/(g*H)^(3/4)));// Specific speed
+printf('\nThe unit speed,N_u=%0.0f r.p.m\nThe unit power,P_u=%0.2f kW\nThe unit flow,Q_u=%0.3f m^3/sec\nThe specific speed,N_s=%0.3f',N_u,P_u,Q_u,N_s);
+printf('\nFor this range of specific speed in the SI system,turbine must be francis type');
diff --git a/3733/CH4/EX4.19/Ex4_19.sce b/3733/CH4/EX4.19/Ex4_19.sce
new file mode 100644
index 000000000..302fc5f57
--- /dev/null
+++ b/3733/CH4/EX4.19/Ex4_19.sce
@@ -0,0 +1,21 @@
+//Example 4_19
+clc;funcprot(0);
+//Given data
+P_1=36000;// kW
+P_2=27000;// kW
+N_1=81.8;// r.p.m
+H_1=13;// m
+H_2=11;// m
+D_1=7.82;// m
+
+//Calculation
+//As the specific speeds are the same,using the definition of specific speed in terms of power,
+N_2=((N_1*sqrt(P_1)/(H_1^(5/4)))*((H_2^(5/4))/sqrt(P_2)));// rpm
+// As the unit speeds are same,
+D_2=(D_1*N_1*sqrt(H_2))/(sqrt(H_1)*N_2);// m
+// As the unit flow is same,Q=Q_2/Q_1
+Q=(D_2^2*H_2^(1/2))/((D_1^2*H_1^(1/2)));
+// By solving Q, it gives the relation,Q_2=0.886*Q_1;
+Q_r=(1-Q)*100;
+printf('\n Speed,N_2=%0.1f rpm \n Diameter,D_2=%0.2f m \n There is a reduction in flow by about %0.2f percentage.',N_2,D_2,Q_r);
+// The answer vary due to round off error
diff --git a/3733/CH4/EX4.2/Ex4_2.sce b/3733/CH4/EX4.2/Ex4_2.sce
new file mode 100644
index 000000000..5edddf58c
--- /dev/null
+++ b/3733/CH4/EX4.2/Ex4_2.sce
@@ -0,0 +1,20 @@
+//Example 4_2
+clc;funcprot(0);
+//Given data
+p=4.1;// kW
+n=360;// r.p.m
+h=1.8;// m
+H=6;// m
+// d_r=D/d
+d_r=5
+
+//Calculation
+N=n*(1/d_r)*sqrt(H/h);// r.p.m
+P=p*(d_r)^2*(H/h)^(3/2);// kW
+//Q/q=q_r
+q_r=(d_r)^2*sqrt(H/h);
+n_s=(n*sqrt(p))/h^(5/4);
+N_s=(N*sqrt(P))/(H)^(5/4);
+printf('\nN=%0.0f r.p.m\nP=%0.0f kW\nQ/q=%0.1f',N,P,q_r);
+printf('\nThe runner must be of propeller type');
+// The answer vary due to round off error
diff --git a/3733/CH4/EX4.3/Ex4_3.sce b/3733/CH4/EX4.3/Ex4_3.sce
new file mode 100644
index 000000000..c2a4a2a16
--- /dev/null
+++ b/3733/CH4/EX4.3/Ex4_3.sce
@@ -0,0 +1,20 @@
+//Example 4_3
+clc;funcprot(0);
+//Given data
+P=10000;// kW
+H=12;// m
+N=100;// r.p.m
+// d_r=D/d;
+d_r=10;
+h=8;// m
+n_m=0.8;
+
+//Calculation
+n=N*d_r*sqrt(h/H);// r.p.m
+p=P/((d_r^2*(H/h)^(3/2)));// kW
+w=1000*9.81;// N
+q=(p*1000)/(w*h*n_m);// m^3/sec
+n_s=(n*sqrt(p))/h^(5/4);
+N_s=(N*sqrt(P))/(H)^(5/4);
+printf('\n(a)The running speed of the model,n=%0.0f r.p.m \n(b)B.P,p=%0.1f kW \n   The flow quantity required,q=%0.2f m^3/sec \n   The specific speed of the runner,N_s=%0.0f',n,p,q,N_s );
+// The answer vary due to round off error
diff --git a/3733/CH4/EX4.4/Ex4_4.sce b/3733/CH4/EX4.4/Ex4_4.sce
new file mode 100644
index 000000000..27cabd01c
--- /dev/null
+++ b/3733/CH4/EX4.4/Ex4_4.sce
@@ -0,0 +1,23 @@
+//Example 4_4
+clc;funcprot(0);
+//Given data
+P=40000;//kW
+N=500;//r.p.m
+H=240;// m
+h=30;// m
+SG=1.1;// Specific gravity of water
+q=150;// litres/sec
+q=q*SG;// kg/sec
+n_m=0.88;// The over all efficiency
+
+//Calculation
+w=1000*9.81;// N
+p=(q*w*h*n_m)/(1000*1000);// kW
+//d_r=D/d;
+d_r=sqrt(P/p)*(h/H)^0.75;
+n=N*d_r*sqrt(h/H);// r.p.m
+n_s=(n*sqrt(p))/h^(5/4);
+N_s=(N*sqrt(P))/(H)^(5/4);
+printf('\n(a)The design speed for a turbie,n=%0.0f r.p.m',n);
+printf('\nThe runner is of Francis type');
+// The answer provided in the textbook is wrong
diff --git a/3733/CH4/EX4.5/Ex4_5.sce b/3733/CH4/EX4.5/Ex4_5.sce
new file mode 100644
index 000000000..3155cd3ea
--- /dev/null
+++ b/3733/CH4/EX4.5/Ex4_5.sce
@@ -0,0 +1,21 @@
+//Example 4_5
+clc;funcprot(0);
+//Given data
+P=50000;// kW
+H=225;// m
+N=600;// r.p.m
+h=36;// m
+q=170;// litres/sec
+n_p=0.9;// Over all efficency
+n_m=n_p;
+
+//Calculation
+w=1000*9.81;// N
+Q=(P*1000)/(w*H*n_m);// m^3/s
+// D_r= d/D
+D_r=sqrt(sqrt(h/H)*((Q/q)));
+D=1/D_r;
+p=P*(D_r^2*(h/H)^(3/2));// kW
+n=N*((1/D_r)*(sqrt(h/H)));// r.p.m
+printf('\n(a)The model size is (1/%0.2f)^th of prototype. \n(b)Power developed by the model=%0.1f kW \n(c)Model runner speed=%0.0f r.p.m',D,p,n);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH4/EX4.6/Ex4_6.sce b/3733/CH4/EX4.6/Ex4_6.sce
new file mode 100644
index 000000000..3ed599e89
--- /dev/null
+++ b/3733/CH4/EX4.6/Ex4_6.sce
@@ -0,0 +1,16 @@
+//Example 4_6
+clc;funcprot(0);
+//Given data
+Q=260;//m^3/s
+H=1.7;//m
+n_p=0.825;
+N_s=890;// r.p.m
+N=50;//r.p.m
+
+//Calculation
+w=1000*9.81;// N
+P_t=(Q*w*H*n_p)/(1000);// Total power to be developed in kW
+P=((N_s*H^(5/4))/N)^2;// kW
+n_k=P_t/P;// Number of kaplan turbine required
+printf('Number of Kaplan turbine required=%0.0f\n',n_k);
+// The answer vary due to round off error
diff --git a/3733/CH4/EX4.7/Ex4_7.sce b/3733/CH4/EX4.7/Ex4_7.sce
new file mode 100644
index 000000000..6f8648aa9
--- /dev/null
+++ b/3733/CH4/EX4.7/Ex4_7.sce
@@ -0,0 +1,26 @@
+//Example 4_7
+clc;funcprot(0);
+//Given data
+Q=400;// m^3/sec
+H=45;// m
+n_t=0.90;// The turbine efficiency
+N=250;// r.p.m
+
+//Calculation
+w=1000*9.81;// N
+P_t=(w*Q*H*n_t)/(1000);// kW
+//(a)
+N_sf=200;// Specific speed
+P=((N_sf*H^(5/4))/N)^2;// kW
+f_n=(P_t/P);// Number of francis turbine required
+// (b)
+N_sk=600;// Specific speed
+P=((N_sk*H^(5/4))/N)^2;// kW
+k_n=(P_t/P);// Number of kaplan turbine required
+printf('\n Number of francis turbines=%0.0f \n Number of kaplan turbine used=%0.0f',f_n,k_n);
+if(f_n>k_n)
+    printf('\n The installation of kaplan turbine is more economical than francis turbine as number of units required is less.');
+else(k_n>f_n)
+    printf('\n The installation of francis turbine is more economical than kaplan turbine as number of units required is less.');
+end
+// The answer provided in the textbook is wrong
diff --git a/3733/CH4/EX4.8/Ex4_8.sce b/3733/CH4/EX4.8/Ex4_8.sce
new file mode 100644
index 000000000..a67bc9df0
--- /dev/null
+++ b/3733/CH4/EX4.8/Ex4_8.sce
@@ -0,0 +1,21 @@
+//Example 4_8
+clc;funcprot(0);
+//Given data
+Q=350;// m^3/sec
+H=30;// m
+n_t=0.88;//// The turbine efficiency
+f=50;// The frequency of generation in cycles/sec
+no_p=24;// Number of poles used
+N_sf=300;// Specific speed
+N_sk=800;// Specific speed
+
+//Calculation
+N=(120*f)/(no_p);// r.p.m
+w=1000*9.81;
+P_t=(w*Q*H*n_t)/(1000);// kW
+P=((N_sf*H^(5/4))/N)^2;// kW
+f_n=(P_t/P);// Number of francis turbine required
+P=((N_sk*H^(5/4))/N)^2;// kW
+k_n=(P_t/P);// Number of kaplan turbine required
+printf('\n Number of francis turbines=%0.0f \n Number of kaplan turbine used=%0.0f',f_n,k_n);
+// The answer vary due to round off error
diff --git a/3733/CH4/EX4.9/Ex4_9.sce b/3733/CH4/EX4.9/Ex4_9.sce
new file mode 100644
index 000000000..821e44c88
--- /dev/null
+++ b/3733/CH4/EX4.9/Ex4_9.sce
@@ -0,0 +1,22 @@
+//Example 4_9
+clc;funcprot(0);
+//Given data
+Q=30;//m^/sec
+H=7.5;// m
+n_t=0.85;
+N=50;///r.p.m
+Sr=0.85;//Speed ratio
+g=9.81;//The acceleration due to gravity in m/s^2
+
+//Calculation
+w=1000*9.81;// N
+P_t=(w*Q*H*n_t)/1000;// kW
+N_s=(N*sqrt(P_t))/(H)^(5/4);//Specific speed
+if(N_s>=174)
+    printf('\n (a)As N_s=340,two turbine units can be used.\n (b)The runner is of Francis type.');
+else
+    printf('\n Wrong');
+end
+D=Sr*60*(sqrt(2*g*H))*(1/(%pi*N));//The diameter of the runner in m
+printf('\n (c)The diameter of the runner,D=%0.2f m',D);
+// The answer vary due to round off error