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

---
 479/CH3/EX3.7/Example_3_7.sce | 27 +++++++++++++++++++++++++++
 479/CH3/EX3.7/Example_3_7.txt |  3 +++
 2 files changed, 30 insertions(+)
 create mode 100755 479/CH3/EX3.7/Example_3_7.sce
 create mode 100755 479/CH3/EX3.7/Example_3_7.txt

(limited to '479/CH3/EX3.7')

diff --git a/479/CH3/EX3.7/Example_3_7.sce b/479/CH3/EX3.7/Example_3_7.sce
new file mode 100755
index 000000000..81f9e705d
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+++ b/479/CH3/EX3.7/Example_3_7.sce
@@ -0,0 +1,27 @@
+//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.7
+clear;
+clc;
+
+//Given
+m = 5000;//Amount of steam recived per hour in Kg
+H1 = 666;//Specific enthalpy when steam entered in the turbine in Kcal/Kg
+H2 = 540;//Specific enthalpy when steam left the turbine in Kcal/Kg
+u1 = 3000/60;//velocity at which steam entered in m/sec
+u2 = 600/60;//velocity at which steam left in m/sec
+Z1 = 5;//height at which steam entered in m
+Z2 = 1;//height at which steam left in m
+Q = -4000;//heat lost in Kcal
+g = 9.81;
+
+//To calculate the horsepuwer output of the turbine
+delH = H2-H1;//change in enthalpy in Kcal
+delKE = ((u2^2)-(u1^2)/(2*g))/(9.8065*427);//change in kinetic energy in Kcal; 1kgf = 9.8065 N
+delPE = ((Z2-Z1)*g)/(9.8065*427);//change in potential energy in Kcal
+W = -(m*(delH+delKE+delPE))+Q;//work delivered in Kcal/hr
+W1 = W*(427/(3600*75));//work delivered by turbine in hp
+mprintf('Work delivered by turbine is %f hp',W1);
+//end
\ No newline at end of file
diff --git a/479/CH3/EX3.7/Example_3_7.txt b/479/CH3/EX3.7/Example_3_7.txt
new file mode 100755
index 000000000..62da72829
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+++ b/479/CH3/EX3.7/Example_3_7.txt
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+For Example 3.7,
+the answer given in the book is 999 hp but i am getting 990.13329 hp.
+There might be some calculation mistake in the book.
\ No newline at end of file
-- 
cgit