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

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 2705/CH5/EX5.14/Ex5_14.sce | 54 ++++++++++++++++++++++++++++++++++++++++++++++
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(limited to '2705/CH5/EX5.14')

diff --git a/2705/CH5/EX5.14/Ex5_14.sce b/2705/CH5/EX5.14/Ex5_14.sce
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+clear;
+clc;
+disp('Example 5.14');
+
+//  aim : To determine the
+//  (a)heat transfer
+//  (b)change of internal energy
+//  (c)mass of gas
+
+//   Given values
+V1 = .4;//  initial volume, [m^3]
+P1 = 100;//  initial pressure, [kN/m^2]
+T1 = 273+20;//  temperature, [K]
+P2 = 450;//  final pressure,[kN/m^2]
+cp = 1.0;//  [kJ/kg K]
+Gamma = 1.4; //  heat capacity ratio
+
+//  solution
+
+//  (a)
+//  for the isothermal compression,P*V=constant,so
+V2 = V1*P1/P2;//  [m^3]
+W = P1*V1*log(P1/P2);//  formula of workdone for isothermal process,[kJ]
+
+//  for isothermal process, del_U=0;so
+Q = W;
+mprintf('\n (a) The heat transferred during compression is Q  =  %f kJ\n',Q);
+
+
+//  (b)
+V3 = V1;
+//  for adiabatic expansion
+//  also
+
+P3 = P2*(V2/V3)^Gamma;//  [kN/m^2]
+W = -(P3*V3-P2*V2)/(Gamma-1);//  work done formula for adiabatic process,[kJ]
+//  also, Q=0,so using Q=del_U+W
+del_U = -W;//  [kJ]
+mprintf('\n (b) The change of the internal energy during the expansion is,del_U  =  %f kJ\n',del_U);
+
+//  (c)
+//  for ideal gas
+//  cp-cv=R, and cp/cv=gamma, hence
+R = cp*(1-1/Gamma);//  [kj/kg K]
+
+//  now using ideal gas equation
+m = P1*V1/(R*T1);//  mass of the gas,[kg]
+mprintf('\n (c) The mass of the gas is,m  =  %f  kg\n',m);
+
+// There is calculation mistake in the book
+
+
+//  End
+
-- 
cgit