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

---
 3831/CH14/EX14.14/Ex14_14.sce | 60 +++++++++++++++++++++++++++++++++++++++++++
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 create mode 100644 3831/CH14/EX14.14/Ex14_14.sce

(limited to '3831/CH14/EX14.14')

diff --git a/3831/CH14/EX14.14/Ex14_14.sce b/3831/CH14/EX14.14/Ex14_14.sce
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+// Example 14_14
+clc;funcprot(0);
+// Given data
+m_ref=0.500;// kg/s
+T_0=25.0;// °C
+n_c=70.0;// The isentropic efficiency of compressor
+// Using Figure 14.36 as the illustration for this example, the properties at the four stations can be found in Tables C.7e, C.7f, and C.8d as
+// Station 1
+// Compressor inlet
+x_1=1.00;// The quality of steam
+T_1=-20.0;// °C
+h_1=235.31;// kJ/kg
+s_1=0.9332;// kJ/kg.K
+p_1=132.99;// kPa
+// Station 2
+// Compressor outlet
+p_2s=800;// kPa
+s_2=s_1;// kJ/kg.K
+h_2s=271.10;// kJ/kg
+T_2s=39.8;// °C
+// Station 3
+// Condenser outlet
+x_3=0.00;// The quality of steam
+p_3=725;// kPa
+h_3=87.46;// kJ/kg
+s_3=0.3257;// kJ/kg.K
+T_3=27.9;// °C
+// Station 4h
+// Expansion valve outlet
+h_4h=h_3;// kJ/kg
+p_4h=160;// kPa
+h_4h=87.46;// kJ/kg
+x_4h=0.280;// The quality of steam
+s_4h=0.3449;// kJ/kg.K
+T_4h=-15.6;// °C
+T_e=-15.6;// °C
+
+// Calculation
+// (a)
+h_2=((h_2s-h_1)/(n_c/100))+h_1;// kJ/kg
+p_2=p_2s;// kPa
+//Interpolation in Table C.7f in Thermodynamic Tables to accompany Modern Engineering Thermodynamics (or through the use of an appropriate computer program) gives the following additional properties at this state:
+s_2=0.9814;// kJ/kg.K
+T_2=54.97;// °C
+Q_condenser=m_ref*(h_3-h_2);// kJ/s
+Q_evaporator=m_ref*(h_1-h_4h);// kJ/s
+Q_compressor=m_ref*(h_2-h_1);// kJ/s
+I_ac=m_ref*(T_0+273.15)*(s_2-s_1);// kW
+I_con=(T_0+273.15)*((m_ref*(s_3-s_2))-(Q_condenser/(T_0+273.15)));// kW
+I_ev=m_ref*(T_0+273.15)*(s_4h-s_3);// kW
+I_e=(T_0+273.15)*((m_ref*(s_1-s_4h))-(Q_evaporator/(T_e+273.15)));// kW
+I_total=I_ac+I_con+I_ev+I_e;// kW
+W_compressor=Q_compressor;// kW
+// (b)
+COP=Q_evaporator/W_compressor;// The system coefficient of performance
+T_L=T_e;// °C
+COP_act=2.85;// The second law efficiency for a refrigeration system
+E_RAC=(abs(1-((T_0+273.15)/(T_e+273.15)))*COP_act)*100;// %
+printf("\n(a)The irreversibility rate of each component in the system are given below: \n   I_adiabatic compressor=%1.2f kW \n   I_condenser=%1.2f kW \n   I_expansion valve=%1.2f kW \n   I_evaporator=%1.2f kW \n   The total irreversibility rate of the system,I_total=%2.0f kW \n(b)The system coefficient of performance,COP=%1.2f \n   The second law efficiency for a refrigeration system,E_RAC=%2.1f percentage",I_ac,I_con,I_ev,I_e,I_total,COP,E_RAC);
+// The answer provided in the text book is wrong(The value of h_2 changed little bit)
-- 
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