From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 914/CH7/EX7.13/ex7_13.sce | 24 ++++++++++++++++++++++++
 1 file changed, 24 insertions(+)
 create mode 100755 914/CH7/EX7.13/ex7_13.sce

(limited to '914/CH7/EX7.13')

diff --git a/914/CH7/EX7.13/ex7_13.sce b/914/CH7/EX7.13/ex7_13.sce
new file mode 100755
index 000000000..08775f3e2
--- /dev/null
+++ b/914/CH7/EX7.13/ex7_13.sce
@@ -0,0 +1,24 @@
+clc;
+warning("off");
+printf("\n\n example7.13 - pg303");
+// given
+deltaz=30;  //[ft] - distance between process and the holding tank
+Q=100;  //[gpm] - volumetric flow rate of water
+p1=100;  //[psig]
+p2=0;  //[psig]
+g=32.1;  //[ft/sec] - acceleration due to gravity
+sv=0.0161;  //[ft^3/lb] - specific volume of water
+p=1/sv;  //[lb/ft^3] - density of water
+e=0.77;  // efficiency of centrifugal pump
+deltap=(p1-p2)*(144);  //[lbf/ft^2]
+gc=32.174;
+// using the equation deltap/p+g*(deltaz)+Ws=0;
+Wst=-deltap/p-(g/gc)*(deltaz);
+// using the formula for efficiency e=Ws(theoritical)/Ws(actual)
+// therefore
+Wsa=Wst/e;
+// the calulated shaft work is for a unit mass flow rate of water,therfore for given flow rate multiply it by the flow rate
+w=(Q*p)/(7.48*60);
+Wsactual=Wsa*w;
+power=-Wsactual/(778*0.7070);
+printf("\n\n the required horsepower is %fhp",power);
-- 
cgit