Đề kiểm tra giữa học kỳ I môn Quản lý và sử dụng năng lượng - Năm học 2014-2015

TD = 2190 
TG_BT = 5.0057e+003 
TG_CD = 1.5643e+003 
Cost_1kW_22kV = 1.2298e+007 
Cost_1kW_6kV = 1.2839e+007 
Cau a___________________________ 
S = 1.1765e+003 
beta = 0.9412 
P_TH = 10.4893 
Cost_TH_22kV = 1.2900e+008 
Cost_TT_22kV = 1.2298e+010 
Cost_tongTT_22kV = 1.2427e+010 
Cost_tongTT_22kV_10nam = 7.6358e+010 
Cost_22kV_10nam = 7.7858e+010 
Cost_TT_6kV = 1.2839e+010 
Cost_6kV_10nam = 7.8891e+010 
Cau a: Cach 2__________________________ 
Cost_TK_1nam = 4.1225e+008 
SPP_MBA = 3.6386 
Cau b__________________________ 
Cost_kVAr = 120000 
PF_b = 0.9500 
S_b = 1.0526e+003 
beta_b = 0.8421 
P_TH_b = 8.6007 
Cost_TH_b = 1.0577e+008 
Cost_TH_TK = 2.3226e+007 
tan = 0.6197 
tan_b = 0.3287 
C_TB = 291.0602 
Cost_TB = 3.4927e+007 
SPP = 1.5038 
BCR = 4.0860 
Câu 1: 
% Lop QL&SDNL - Chinh quy HK1.2014-2015 
% Cau 1 
clc 
clear all 
disp('Gia dien ___________________________________') 
% Bieu gia dien 10/2014 
 % Gia dien Cong Nghiep <6kV 
Gia_CN_TD=890 % Gia dien Cong Nghiep gio Thap Diem 
Gia_CN_BT=1388 % Gia dien Cong Nghiep gio Binh Thuong 
Gia_CN_CD=2520 % Gia dien Cong Nghiep gio Cao Diem 
 % Van hanh 24gio/ngay, 365ngay/nam = 8760 gio. 
TG_TD=6*365% Thoi gian Thap Diem trong 1 nam 
TG_BT=13*365*(6/7)+18*365*1/7% Thoi gian Binh Thuong trong 1 nam 
TG_CD=5*365*(6/7)% Thoi gian Cao Diem trong 1 nam 
 % Chi phi tien dien mot nam cho moi kW 
Cost_1kW=TG_TD*Gia_CN_TD+TG_BT*Gia_CN_BT+TG_CD*Gia_CN_CD % Duoi 6kV 
disp('Thong so ___________________________________') 
Ud_dm=380 
Pout_dm = 2200 % Cong suat 
HS_IE1_dm=0.797 
cos_IE1_dm=0.81 
bta_max_IE1=0.8 
cos_IE1_0=0.1 
Cost_Motor_IE1=7.5e+6 
HS_IE2_dm=0.843 
cos_IE2_dm=0.81 
bta_max_IE2=0.8 
cos_IE2_0=0.1 
Cost_Motor_IE2=9.0e+6 
beta_Load=0.3 % 30% tai 
beta_V=0.60 % 60% dien ap dinh muc 
HS_inverter=0.95 
Cost_Inverter=4e+6 % 4 trieu dong 
% Cau a 
disp('Cau a: _________________________________________') 
Pout_a = beta_Load*Pout_dm % Cong suat 
disp('Cau a: Dong co IE1 ___________________________________') 
Pin_IE1_dm= Pout_dm/HS_IE1_dm % Cong suat dinh muc dong co 
Pth_IE1_dm=(1-HS_IE1_dm)*Pin_IE1_dm 
% P0+Pn=Pth 
% P0=Pn*bta_max^2 
Pn_IE1_dm=Pth_IE1_dm/(1+bta_max_IE1^2) 
P0_IE1_dm=Pth_IE1_dm-Pn_IE1_dm 
PFe_IE1_dm=P0_IE1_dm/2 % PFe=Pc=Ton hao sat bang ton hao co 
Pthco_IE1_dm=P0_IE1_dm/2 % Ton hao sat bang ton hao co 
I0_IE1=P0_IE1_dm/(sqrt(3)*Ud_dm*cos_IE1_0) 
%IFe=PFe/(sqrt(3)*U0) % IFe=Ic, Neu bo qua ton hao co khi chay khong tai! 
IFe_IE1=P0_IE1_dm/(sqrt(3)*Ud_dm) % Neu khong bo qua ton hao co khi chay khong tai! 
Id_IE1_dm=sqrt(I0_IE1^2-IFe_IE1^2) % Id=Im 
I_IE1_dm=Pin_IE1_dm/(sqrt(3)*Ud_dm*cos_IE1_dm) 
Iq_IE1_dm=sqrt(I_IE1_dm^2-Id_IE1_dm^2) 
% Vi dien ap khong doi nen lamda ~ U/f ~ Id khong doi 
Id_IE1=Id_IE1_dm 
% Gia su toc do dong co KDB thay doi khong dang ke, nen Tb ~Pb ~lamda.Iq 
Iq_IE1=Iq_IE1_dm*(Pout_a+Pthco_IE1_dm)/(Pout_dm+Pthco_IE1_dm) 
I_IE1=sqrt(Id_IE1^2+Iq_IE1^2) 
bta_IE1 = I_IE1/I_IE1_dm 
Pn_IE1=bta_IE1^2*Pn_IE1_dm 
P0_IE1=P0_IE1_dm 
HS_IE1=Pout_a/(Pout_a+Pn_IE1+P0_IE1) 
disp('Cau a: Dong co IE2 ___________________________________') 
Pin_IE2_dm= Pout_dm/HS_IE2_dm % Cong suat dinh muc dong co 
Pth_IE2_dm=(1-HS_IE2_dm)*Pin_IE2_dm 
% P0+Pn=Pth 
% P0=Pn*bta_max^2 
Pn_IE2_dm=Pth_IE2_dm/(1+bta_max_IE2^2) 
P0_IE2_dm=Pth_IE2_dm-Pn_IE2_dm 
PFe_IE2_dm=P0_IE2_dm/2 % PFe=Pc=Ton hao sat bang ton hao co 
Pthco_IE2_dm=P0_IE2_dm/2 % Ton hao sat bang ton hao co 
I0_IE2=P0_IE2_dm/(sqrt(3)*Ud_dm*cos_IE2_0) 
%IFe=PFe/(sqrt(3)*U0) % IFe=Ic, Neu bo qua ton hao co khi chay khong tai! 
IFe_IE2=P0_IE2_dm/(sqrt(3)*Ud_dm) % Neu khong bo qua ton hao co khi chay khong tai! 
Id_IE2_dm=sqrt(I0_IE2^2-IFe_IE2^2) % Id=Im 
I_IE2_dm=Pin_IE2_dm/(sqrt(3)*Ud_dm*cos_IE2_dm) 
Iq_IE2_dm=sqrt(I_IE2_dm^2-Id_IE2_dm^2) 
% Vi dien ap khong doi nen lamda ~ U/f ~ Id khong doi 
Id_IE2=Id_IE2_dm 
% Gia su toc do dong co KDB thay doi khong dang ke, nen Tb ~Pb ~lamda.Iq 
Iq_IE2=Iq_IE2_dm*(Pout_a+Pthco_IE2_dm)/(Pout_dm+Pthco_IE2_dm) 
I_IE2=sqrt(Id_IE2^2+Iq_IE2^2) 
bta_IE2 = I_IE2/I_IE2_dm 
Pn_IE2=bta_IE2^2*Pn_IE2_dm 
P0_IE2=P0_IE2_dm 
HS_IE2=Pout_a/(Pout_a+Pn_IE2+P0_IE2) 
disp('Cau a: So sanh tuong quan ___________________________________') 
Pin_IE1=Pout_a/HS_IE1 
Cost_Energy_IE1_1nam=Pin_IE1/1000*Cost_1kW % Tinh theo 1000Wh=1kWh 
Pin_IE2=Pout_a/HS_IE2 
Cost_Energy_IE2_1nam=Pin_IE2/1000*Cost_1kW % Tinh theo 1000Wh=1kWh 
Cost_Motor_delta=Cost_Motor_IE2-Cost_Motor_IE1 % CHi phi tang them khi mua dong co IE2 
Cost_Energy_TKa_1nam=Cost_Energy_IE1_1nam-Cost_Energy_IE2_1nam 
SPP_a=Cost_Motor_delta/Cost_Energy_TKa_1nam 
%(P/A, 10%, 10) = 6.1446 <--- tra bang 10 nam 
Cost_Energy_TKa_10nam=Cost_Energy_TKa_1nam*6.1446 
BCR_a_10nam=Cost_Energy_TKa_10nam/Cost_Motor_delta 
% Cau b 
disp('Cau b: Dong co IE2 ___________________________________') 
Pout_b=Pout_a; % Tai nhu cau a 
Ud_b=beta_V*Ud_dm 
PFe_b = PFe_IE2_dm* (Ud_b/Ud_dm)^2 % Pfe ~ lamda^2 ~ (U/f)^2 
P0_b = PFe_b + Pthco_IE2_dm % Do toc do thay doi khong nhieu nen Ton hao co khong doi 
% lamda ~ U/f ~ Id. 
Id_b = Id_IE2_dm*(Ud_b/Ud_dm) 
% Va Tco ~Pco/w ~ Pco ~lamda.Iq ~Id.Iq 
Iq_b = Iq_IE2*Id_IE2/Id_b 
I_b=sqrt(Id_b^2+Iq_b^2) 
bta_b = I_b/I_IE2_dm 
Pn_b=bta_b^2*Pn_IE2_dm 
HS_b=Pout_b/(Pout_b+Pn_b+P0_b) 
HS_chung=HS_b*HS_inverter 
% Cau bb 
Pin_IE2= Pout_a/HS_IE2 % Cong suat vao cua dong co IE2 o cau a (khong co bien tan) 
Pin_b= Pout_b/HS_chung % Cong suat vao cua dong co o cau b (co dung bien tan) 
P_delta= (Pin_IE2-Pin_b) % Cong suat tiet kiem 
Cost_Energy_TKb_1nam=P_delta/1000*Cost_1kW % Tien dien tiet kiem 1 nam (tinh theo kW) 
SPP_b=Cost_Inverter/Cost_Energy_TKb_1nam 
%(P/A, 10%, 10) = 6.1446 <--- tra bang 10 nam 
Cost_Energy_TKb_10nam=Cost_Energy_TKb_1nam*6.1446 
BCR_b_10nam=Cost_Energy_TKb_10nam/Cost_Inverter 
disp('Ket qua _________________________________________________________') 
TEXT = sprintf('1.a) Hieu suat dong co IE1 = %f', HS_IE1); disp(TEXT) 
TEXT = sprintf('1.a) Hieu suat dong co IE2 = %f', HS_IE2); disp(TEXT) 
TEXT = sprintf('1.a) SPP_a = %f nam', SPP_a); disp(TEXT) 
TEXT = sprintf('1.a) BCR_a_10nam = %f', BCR_a_10nam); disp(TEXT) 
TEXT = sprintf('1.b) Hieu suat dong co IE2 = %f', HS_b); disp(TEXT) 
TEXT = sprintf('1.b) Hieu suat dong co IE2+Inverter = %f', HS_chung); disp(TEXT) 
TEXT = sprintf('1.b) SPP_b = %f nam: =>khong nen dau tu Bien tan', SPP_b); disp(TEXT) 
TEXT = sprintf('1.b) BCR_b_10nam = %f', BCR_b_10nam); disp(TEXT) 
Câu 2: 
% Lop QL&SDNL - De KTGK, lop CQ HK1.2014-2015 
% Cau 2 
clc 
clear all 
% Bieu gia dien 10/2014 
 % Gia dien Cong Nghiep <6kV 
Gia_CN_TD=890 % Gia dien Cong Nghiep gio Thap Diem 
Gia_CN_BT=1388 % Gia dien Cong Nghiep gio Binh Thuong 
Gia_CN_CD=2520 % Gia dien Cong Nghiep gio Cao Diem 
 % Gia dien Cong Nghiep 22kV 
Gia_22kV_TD=845 % Gia dien Cong Nghiep gio Thap Diem 
Gia_22kV_BT=1328 % Gia dien Cong Nghiep gio Binh Thuong 
Gia_22kV_CD=2429 % Gia dien Cong Nghiep gio Cao Diem 
Sdm=1250; % 1250kVA 
Po=1.020; 
Pn=10.690; 
Io=1.5/100; 
Vn=6/100; 
P_TT=1000; % Tai Tieu Thu 1000kW 
PF=0.85; % Tai cam 
Tnam=8760; % gio 
Cost_MBA=1.5e+9 % 1 ty dong, thoi gia hien tai 
 % i=10% - tra bang 
disp('Cau 2___________________________') 
 % Van hanh 24gio/ngay, 365ngay/nam = 8760 gio. 
TG_TD=6*365% Thoi gian Thap Diem trong 1 nam 
TG_BT=13*365*(6/7)+18*365*1/7% Thoi gian Binh Thuong trong 1 nam 
TG_CD=5*365*(6/7)% Thoi gian Cao Diem trong 1 nam 
 % Chi phi tien dien mot nam cho moi kW 
Cost_1kW_22kV=TG_TD*Gia_22kV_TD+TG_BT*Gia_22kV_BT+TG_CD*Gia_22kV_CD % 6kV-22kV 
Cost_1kW_6kV=TG_TD*Gia_CN_TD+TG_BT*Gia_CN_BT+TG_CD*Gia_CN_CD % Duoi 6kV 
disp('Cau a___________________________') 
S=P_TT/PF 
beta=S/Sdm 
P_TH=Po+Pn*beta^2 
Cost_TH_22kV=P_TH*Cost_1kW_22kV % Chi phi cho nang luong ton hao tren MBA trong 1 nam 
Cost_TT_22kV=P_TT*Cost_1kW_22kV % Chi phi cho nang luong tren tai trong 1 nam neu mua dien 22kV 
Cost_tongTT_22kV=Cost_TT_22kV+Cost_TH_22kV % Chi phi cho nang luong Tong trong 1 nam neu mua dien 22kV 
%(P/A, 10%, 10) = 6.1446 <--- tra bang 10 nam 
Cost_tongTT_22kV_10nam=6.1446*Cost_tongTT_22kV % Chi phi cho nang luong Tong trong 10 nam neu mua dien 22kV 
Cost_22kV_10nam=Cost_tongTT_22kV_10nam+Cost_MBA % Chi phi cho nang luong Tong + chi phi MBA trong 10 nam neu mua dien 
22kV 
Cost_TT_6kV=P_TT*Cost_1kW_6kV % Chi phi cho nang luong tren tai trong 1 nam neu mua dien duoi 6kV 
Cost_6kV_10nam=6.1446*Cost_TT_6kV % Chi phi cho nang luong tren tai trong 10 nam neu mua dien 6kV 
disp('Cau a: Cach 2__________________________') 
Cost_TK_1nam=P_TT*(Cost_1kW_6kV-Cost_1kW_22kV)-Cost_TH_22kV 
SPP_MBA=Cost_MBA/Cost_TK_1nam 
disp('Cau b__________________________') 
Cost_kVAr=120000 % d/kVAr 
PF_b=0.95 
S_b=P_TT/PF_b 
beta_b=S_b/Sdm 
P_TH_b=Po+Pn*beta_b^2 
Cost_TH_b=P_TH_b*Cost_1kW_22kV % Chi phi cho nang luong ton hao tren MBA trong 1 nam 
Cost_TH_TK=Cost_TH_22kV-Cost_TH_b % Chi phi cho nang luong tiet kiem trong 1 nam 
% Tinh chi dung luong tu bu de nang PF tu 0.85 len 0.95 
tan=sqrt(1-PF^2)/PF 
tan_b=sqrt(1-PF_b^2)/PF_b 
C_TB=P_TT*(tan-tan_b) % kVAr 
Cost_TB=C_TB*Cost_kVAr % Chi phi lap dat tu bu 
SPP=Cost_TB/Cost_TH_TK 
BCR=Cost_TH_TK*6.1446/Cost_TB %(P/A, 10%, 10) = 6.1446 <--- tra bang 10 nam 
disp('Ket qua _______________________') 
TEXT = sprintf('2.a) Cost_22kV (10 nam, bao gom chi phi lap dat MBA) = %f dong = %f ty dong=>chon', Cost_22kV_10nam, 
Cost_22kV_10nam/1e+9); disp(TEXT) 
TEXT = sprintf('2.a) Cost_6kV (10 nam) = %f dong = %f ty dong', Cost_6kV_10nam, Cost_6kV_10nam/1e+9); disp(TEXT) 
TEXT = sprintf('2.a_Cach 2) Cost_TK_1nam = %f trieu dong', Cost_TK_1nam/1e+6); disp(TEXT) 
TEXT = sprintf('2.a_Cach 2) SPP_MBA = %f nam < 10 nam', SPP_MBA); disp(TEXT) 
TEXT = sprintf('2.b) Tien tiet kiem 1nam = %f dong = %f trieu dong', Cost_TH_TK, Cost_TH_TK/1000000); disp(TEXT) 
TEXT = sprintf('2.b) SPP = %f nam', SPP); disp(TEXT) 
TEXT = sprintf('2.b) BCR = %f', BCR); disp(TEXT)