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 --- 611/CH5/EX5.4/Chap5_Ex4_R1.sce | 24 ++++++++++++++++++++++++ 1 file changed, 24 insertions(+) create mode 100755 611/CH5/EX5.4/Chap5_Ex4_R1.sce (limited to '611/CH5/EX5.4') diff --git a/611/CH5/EX5.4/Chap5_Ex4_R1.sce b/611/CH5/EX5.4/Chap5_Ex4_R1.sce new file mode 100755 index 000000000..789da876f --- /dev/null +++ b/611/CH5/EX5.4/Chap5_Ex4_R1.sce @@ -0,0 +1,24 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India. + +//Chapter-5,Example 4,Page 166 +//Title: Minimum work and maximum possible COP +//================================================================================================================ +clear +clc + +//INPUT +T_L=4.25;//normal boiling point of helium in K +Q_L=0.083;//latent heat of vaporization of helium in kJ/mol +n=1;//amount of liquid helium to be produced in kmol +T_amb=42;//ambient temperature in summer in degree celsius + +//CALCULATION +T_H=T_amb+273.15;//conversion of temperature in K +COP=(T_L)/(T_H-T_L);//calculation of COP of the refrigerator using Eq.(5.20)(no unit) +W=(Q_L)/COP;//calculation of work to be done on the refrigerator unit using Eq.(5.20) in kJ + +//OUTPUT +mprintf("\n The maximum possible COP of the unit=%0.4f \n",COP); +mprintf("\n The minimum amount of work to be done on the refrigerating unit=%f kJ \n",W); + +//===============================================END OF PROGRAM=================================================== -- cgit