diff options
Diffstat (limited to '3751/CH12')
| -rw-r--r-- | 3751/CH12/EX12.1/Ex12_1.sce | 46 | ||||
| -rw-r--r-- | 3751/CH12/EX12.10/Ex12_10.sce | 50 | ||||
| -rw-r--r-- | 3751/CH12/EX12.11/Ex12_11.sce | 43 | ||||
| -rw-r--r-- | 3751/CH12/EX12.12/Ex12_12.sce | 57 | ||||
| -rw-r--r-- | 3751/CH12/EX12.13/Ex12_13.sce | 30 | ||||
| -rw-r--r-- | 3751/CH12/EX12.2/Ex12_2.sce | 54 | ||||
| -rw-r--r-- | 3751/CH12/EX12.3/Ex12_3.sce | 59 | ||||
| -rw-r--r-- | 3751/CH12/EX12.4/Ex12_4.sce | 42 | ||||
| -rw-r--r-- | 3751/CH12/EX12.5/Ex12_5.sce | 62 | ||||
| -rw-r--r-- | 3751/CH12/EX12.6/Ex12_6.sce | 81 | ||||
| -rw-r--r-- | 3751/CH12/EX12.7/Ex12_7.sce | 67 | ||||
| -rw-r--r-- | 3751/CH12/EX12.7/Ex12_7_Indicator_Diagram.jpg | bin | 0 -> 201063 bytes | |||
| -rw-r--r-- | 3751/CH12/EX12.8/Ex12_8.sce | 52 | ||||
| -rw-r--r-- | 3751/CH12/EX12.9/Ex12_9.sce | 71 |
14 files changed, 714 insertions, 0 deletions
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 + + + + |
