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

---
 2939/CH9/EX9.1/Ex9_1.sce | 13 +++++++++++++
 2939/CH9/EX9.2/Ex9_2.sce | 16 ++++++++++++++++
 2939/CH9/EX9.3/Ex9_3.sce | 13 +++++++++++++
 2939/CH9/EX9.4/Ex9_4.sce | 17 +++++++++++++++++
 2939/CH9/EX9.5/Ex9_5.sce | 12 ++++++++++++
 2939/CH9/EX9.6/Ex9_6.sce | 12 ++++++++++++
 2939/CH9/EX9.7/Ex9_7.sce | 26 ++++++++++++++++++++++++++
 2939/CH9/EX9.8/Ex9_8.sce | 15 +++++++++++++++
 8 files changed, 124 insertions(+)
 create mode 100755 2939/CH9/EX9.1/Ex9_1.sce
 create mode 100755 2939/CH9/EX9.2/Ex9_2.sce
 create mode 100755 2939/CH9/EX9.3/Ex9_3.sce
 create mode 100755 2939/CH9/EX9.4/Ex9_4.sce
 create mode 100755 2939/CH9/EX9.5/Ex9_5.sce
 create mode 100755 2939/CH9/EX9.6/Ex9_6.sce
 create mode 100755 2939/CH9/EX9.7/Ex9_7.sce
 create mode 100755 2939/CH9/EX9.8/Ex9_8.sce

(limited to '2939/CH9')

diff --git a/2939/CH9/EX9.1/Ex9_1.sce b/2939/CH9/EX9.1/Ex9_1.sce
new file mode 100755
index 000000000..419e6c278
--- /dev/null
+++ b/2939/CH9/EX9.1/Ex9_1.sce
@@ -0,0 +1,13 @@
+
+//Ex9_1
+
+clc;
+
+//Give:
+t=3600;// in seconds
+F=96500;// in columbs
+// Formula: m=0.0373fMit, Faraday's law: m=(itE)/F
+
+//Solution:
+constant=t/F;
+printf("The value of numerical constant is = %f",constant)
diff --git a/2939/CH9/EX9.2/Ex9_2.sce b/2939/CH9/EX9.2/Ex9_2.sce
new file mode 100755
index 000000000..285280b8f
--- /dev/null
+++ b/2939/CH9/EX9.2/Ex9_2.sce
@@ -0,0 +1,16 @@
+
+//Ex9_2
+clc;
+// Given:
+m1=24*10^-6;// g per day
+m2=10^-2;// g per day
+i1=10^-6;// in A
+
+//Formula: i1*m2=m1*i2
+
+//Solution:
+
+i2=(i1*m2)/m1;
+i=i2/10^-3;// in mA
+
+printf("The ion current should be = %f mA",i)
diff --git a/2939/CH9/EX9.3/Ex9_3.sce b/2939/CH9/EX9.3/Ex9_3.sce
new file mode 100755
index 000000000..97a0ed756
--- /dev/null
+++ b/2939/CH9/EX9.3/Ex9_3.sce
@@ -0,0 +1,13 @@
+//Ex9_3
+clc;
+//Given:
+f=1.0014;// seperation factor
+s=4;// series
+p=6;// parallel
+
+// Note: The global yield for s stages in series is(f)^s and each parallel stages simply multiplies the yield of the stage, Hence overall yield with p parallel stages (each with s stages in series) will be Y=p*(f)^s
+
+//Solution:
+Y=p*(f)^s;
+
+printf("The net yield is = %f",Y)
diff --git a/2939/CH9/EX9.4/Ex9_4.sce b/2939/CH9/EX9.4/Ex9_4.sce
new file mode 100755
index 000000000..a1dc20397
--- /dev/null
+++ b/2939/CH9/EX9.4/Ex9_4.sce
@@ -0,0 +1,17 @@
+
+// Ex9_4
+
+clc;
+
+// Given:
+f=1.0014;// seperation factor
+
+//Solution: Part (a)
+A1=3.5/0.72;// total enrichment
+n1=log(A1)/log(f);
+printf("\n The no. of stages will be = %d", n1)
+
+//Solution: Part (b)
+A2=90/0.72;// total enrichment
+n2=log(A2)/log(f);
+printf("\n \n The no. of stages will be = %d", n2)
diff --git a/2939/CH9/EX9.5/Ex9_5.sce b/2939/CH9/EX9.5/Ex9_5.sce
new file mode 100755
index 000000000..ab5f8f5ec
--- /dev/null
+++ b/2939/CH9/EX9.5/Ex9_5.sce
@@ -0,0 +1,12 @@
+//Ex9_5
+
+clc;
+
+// Given:
+f=1.01;// seperation factor
+n=10;// plates
+//Solution: 
+
+A=f^n;
+
+printf("The overall seperation factor is = %f",A)
diff --git a/2939/CH9/EX9.6/Ex9_6.sce b/2939/CH9/EX9.6/Ex9_6.sce
new file mode 100755
index 000000000..ea7e1ddc7
--- /dev/null
+++ b/2939/CH9/EX9.6/Ex9_6.sce
@@ -0,0 +1,12 @@
+//Ex9_6
+
+clc;
+
+// Given:
+f=1.01;// seperation factor
+n=16;// plates
+//Solution: 
+
+A=f^n;
+
+printf("The overall seperation factor is = %f",A)
diff --git a/2939/CH9/EX9.7/Ex9_7.sce b/2939/CH9/EX9.7/Ex9_7.sce
new file mode 100755
index 000000000..72116c301
--- /dev/null
+++ b/2939/CH9/EX9.7/Ex9_7.sce
@@ -0,0 +1,26 @@
+//Ex9_7
+
+clc;
+
+// Given:
+
+k1=3.78;
+k2=2.79;
+t1=298;// in K
+t2=353;// in K
+R=8.314// Gas constant
+
+// Formula: log(k1/k2)=(H/R)*((t2-t1)t1*t2)
+
+// Solution:
+
+H=R*log(k1/k2)/((t2-t1)/(t1*t2));
+
+printf("The enthalpy for the exchange reaction is = %f J",H)
+
+
+
+
+
+
+
diff --git a/2939/CH9/EX9.8/Ex9_8.sce b/2939/CH9/EX9.8/Ex9_8.sce
new file mode 100755
index 000000000..7e795086c
--- /dev/null
+++ b/2939/CH9/EX9.8/Ex9_8.sce
@@ -0,0 +1,15 @@
+
+// Ex9_8
+
+clc;
+
+// Given:
+a1=0.015;
+a2=0.04;
+
+// Solution: Defining the seperation factor f as approximately equal to (a2/a1) where a1, a2 are the relative abundances of the isotope of interest in the initial and final fractions, we have
+
+f=(a2/a1);
+
+printf("The single stage seperation factor is = %f",f)
+ 
-- 
cgit