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/EX24.9/Ex24_9.sce | 33 +++++++++++++++++++++++++++++++++ 1 file changed, 33 insertions(+) create mode 100644 3733/CH24/EX24.9/Ex24_9.sce (limited to '3733/CH24/EX24.9/Ex24_9.sce') diff --git a/3733/CH24/EX24.9/Ex24_9.sce b/3733/CH24/EX24.9/Ex24_9.sce new file mode 100644 index 000000000..c5456cc3f --- /dev/null +++ b/3733/CH24/EX24.9/Ex24_9.sce @@ -0,0 +1,33 @@ +// Example 24_9 +clc;funcprot(0); +//Given data +W=2;// Work done in MW +p_1=1;// bar +p_r=5;// Pressure ratio in bar +p_i=2.5;// bar +T_1=27+273;// K +r=1.4;// Specific heat ratio +CV=40000;// kJ/kg +n_c=85/100;// The isentropic efficiency of the compressor +n_t=85/100;// The isentropic efficiency of the turbine +Q_a=80;// Heat absorbed in kJ/kg of air +m_f=0.01;// kg per kg of air +m_a=1;// kg +r=1.4;// Spcific heat ratio for air and gases +C_p=1;// kJ/kg-k for air and gases +C_pg=C_p; +C_pa=C_p + +//Calculation +T_2a=T_1*(p_r)^((r-1)/r);// K +T_2=((T_2a-T_1)/n_c)+T_1;// K +T_3=T_2+(Q_a/(C_pa*m_a));// K +T_4=((m_f*CV)/((1+m_f)*C_p))+T_3;// K +T_5a=T_4*(1/p_r)^((r-1)/r);// K +T_5=T_4-(n_t*(T_4-T_5a));// K +n_th=(((T_4-T_5)-(T_2-T_1))/(T_4-T_3))*100;// Thermal efficiency in % +Q=(W*10^3)/(n_th/100);//Heat supplied in kJ/sec +F=(Q/CV)*3600;// Fuel required per hour in kg/hr +n_cp=(1-(1/(p_r)^((r-1)/r)))*100;//Efficiency of normal constant pressure cycle +printf('\nThe thermal efficiency of the plant=%0.1f percentage \nEfficiency of normal constant pressure cycle=%0.0f percentage \nFuel consumption per hour=%0.0f kg/hr',n_th,n_cp,F); +// The answer provided in the textbook is wrong -- cgit