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

---
 172/CH6/EX6.12/ex12.sce | 16 ++++++++++++++++
 1 file changed, 16 insertions(+)
 create mode 100755 172/CH6/EX6.12/ex12.sce

(limited to '172/CH6/EX6.12/ex12.sce')

diff --git a/172/CH6/EX6.12/ex12.sce b/172/CH6/EX6.12/ex12.sce
new file mode 100755
index 000000000..9e1d5283e
--- /dev/null
+++ b/172/CH6/EX6.12/ex12.sce
@@ -0,0 +1,16 @@
+//example 12
+//Calculating mass flow of steam in tank
+clear
+clc
+V1=0.4 //initial volume fo tank in m^3
+v1=0.5243 //initial specific volume in m^3/kg
+h1=3040.4 //initial specific enthalpy in kJ/kg
+u1=2548.9 //initial specific internal energy in kJ/kg
+m1=V1/v1 //initial mass of steam in tank in kg
+V2=0.4 //final volume in m^3
+disp('let x=V*(h1-u2)/v2-m1*(h1-u1).If we assume T2=300C, then v2=0.1823m^3/kg ,u2=2785.2kJ/kg and x=+ve.If we assume T2=350C,then v2=0.2003 m^3/kg, u2=2869.1kJ/kg and x=-ve.Hence,actualt T2 must be between these two assumed values in order that x=0.By interplotation,') 
+T2=342 //final temperature in Celsius
+v2=0.1974 //final specific volume in m^3/kg
+m2=V2/v2 //final mass of the steam in the tank in kg
+m=m2-m1 //mass of steam that flowsinto the tank
+printf(" \n Hence,mass of the steam that flows into the tank is m=%.3f kg. \n",m)
\ No newline at end of file
-- 
cgit