From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001 From: prashantsinalkar Date: Tue, 10 Oct 2017 12:27:19 +0530 Subject: initial commit / add all books --- 3751/CH16/EX16.15/Ex16_15.sce | 45 +++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 45 insertions(+) create mode 100644 3751/CH16/EX16.15/Ex16_15.sce (limited to '3751/CH16/EX16.15/Ex16_15.sce') 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) + + -- cgit