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

---
 2417/CH6/EX6.41/Ex6_41.sce | 23 +++++++++++++++++++++++
 1 file changed, 23 insertions(+)
 create mode 100755 2417/CH6/EX6.41/Ex6_41.sce

(limited to '2417/CH6/EX6.41')

diff --git a/2417/CH6/EX6.41/Ex6_41.sce b/2417/CH6/EX6.41/Ex6_41.sce
new file mode 100755
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+++ b/2417/CH6/EX6.41/Ex6_41.sce
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+clear;
+clc;
+printf("\t\t\tProblem Number 6.41\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.41 (page no. 304) 
+// Solution
+
+//For Methane(CH4,MW=16)
+p=500; //evaluate specific volume at p pressure //Unit:psia
+pc=674; //critical temperature //Unit:psia
+T=50+460; //evaluate specific volume at T temperature //Unit:R
+Tc=343; //critical temperature //Unit:R
+R=1545/16; //gas constant R = 1545/Molecular Weight //ft*lbf/lbm*R
+pr=p/pc; //reduced pressure //unit:psia
+Tr=T/Tc; //reduced temperature //unit:R
+//Reading figure 6.28 at these values gives
+Z=0.93; //compressibility factor
+//Z=(p*v)/(R*T)
+v=Z*((R*T)/(p*144)); //ft^3/lbm //1 ft^2=144 in^2(for conversion of unit) //specific volume
+printf("Using the value of Z=0.93,the specific volume is %f ft^3/lbm\n",v);
+//For ideal gas,
+v=(R*T)/(p*144); //ft^3/lbm //1 ft^2=144 in^2(for conversion of unit) //specific volume
+printf("For the ideal gas,the specific volume is %f ft^3/lbm\n",v);
-- 
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