initial commit / add all books

18 files changed, 545 insertions, 0 deletions

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
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+//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
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+//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  
+
+