From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 10 Oct 2017 12:27:19 +0530
Subject: initial commit / add all books

---
 3733/CH24/EX24.13/Ex24_13.sce | 49 +++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 49 insertions(+)
 create mode 100644 3733/CH24/EX24.13/Ex24_13.sce

(limited to '3733/CH24/EX24.13/Ex24_13.sce')

diff --git a/3733/CH24/EX24.13/Ex24_13.sce b/3733/CH24/EX24.13/Ex24_13.sce
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+++ b/3733/CH24/EX24.13/Ex24_13.sce
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+// Example 24_13
+clc;funcprot(0);
+//Given data 
+T_1=27+273;// K
+p_1=1;// bar
+p_2=4;// bar
+n_c=0.80;// Isentropic efficiency of compressor
+n_t=0.85;// Isentropic efficiency of turbine
+e=0.75;// The effectiveness of regenerator
+p_lr=0.1;// Pressure loss in regenerator along air side in bar
+p_lcc=0.05;//  Pressure loss in the combustion chamber in bar
+n_com=0.90;// Combustion efficiency
+n_m=0.90;// Mechanical efficiency
+n_g=0.95;// Generation efficiency
+m_a=25;// kg/sec
+CV=40000;// kJ/kg
+C_pa=1;// kJ/kg.K
+C_pg=1.1;// kJ/kg.K
+r=1.4;// Specific heat ratio
+T_4=700+273;// K
+p_atm=1.03;// bar
+
+// Calculation
+p_i=p_2-(p_lr+p_lcc);// Pressure at the inlet of the turbine in bar
+p_e=p_atm+p_lr;// Pressure at the exit of the turbine in bar
+T_2a=T_1*(p_2/p_1)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_5a=T_4*(p_e/p_i)^((r-1)/r);// K
+T_5=T_4-(n_t*(T_4-T_5a));// K
+// Assume m=(m_a/m_f)
+// m=y(1),T_3=y(2)
+function[X]=airfuelratio(y)
+    X(1)=((y(1)+1)*C_pg*(T_4-y(2)))-(CV*n_com);
+    X(2)=((C_pa*(y(2)-T_2))/(e*C_pg*(T_5-T_2)))-(1+(1/y(1)));
+endfunction
+y=[10 100];
+z=fsolve(y,airfuelratio);
+m=z(1);
+T_3=z(2);// K
+W_c=C_pa*(T_2-T_1);// kJ/kg of air
+W_t=C_pg*(1+(m_a/m))*(T_4-T_5);// kJ/kg of air
+W_a=W_t-W_c;// kJ/kg of air
+W=W_a*n_m*n_g;// Work available per kg of air at the terminals of generator in kJ/kg
+P=(m_a*W)/1000;// Power available at the terminals of generator in kJ/kg
+n_o=((W)/((1/m)*CV))*100;// Over all efficiency
+Fr=m_a*3600*(1/m);// Fuel required per hour in kg/hr
+Sfc=Fr/(P*1000);// Specific fuel consumption in kg/kW.hr
+printf('\nThe over all efficiency of the plant=%0.3f percentage \nSpecific fuel consumption=%0.2f kg/kW.hr',n_o,Sfc);
+// The answers provided in the textbook is wrong
-- 
cgit