Đề kiểm tra giữa học kỳ 1 môn Quản lý và sử dụng năng lượng - Năm học 2014-2015

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Pout_dm = 4*746 % Cong suat 
HS_IE1_dm=0.815 
cos_IE1_dm=0.87 
bta_max_IE1=0.79 
cos_IE1_0=0.12 
Cost_Motor_IE1=8e+6 
HS_IE2_dm=0.846 
cos_IE2_dm=0.84 
bta_max_IE2=0.75 
cos_IE2_0=0.1 
Cost_Motor_IE2=10e+6 
disp('Bai giai ___________________________________') 
Cost_Motor_delta=Cost_Motor_IE2-Cost_Motor_IE1 % Chi phi 
tang them khi mua dong co IE2 
% 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_24_24=TG_TD*Gia_CN_TD+TG_BT*Gia_CN_BT+
TG_CD*Gia_CN_CD % Duoi 6kV, 24/24 gio 
% Van hanh 8gio/ngay, 6 gio binh thuong + 2 gio cao diem, 
nghi Chu nhat 
TG_TD_8=0% Thoi gian Thap Diem trong 1 nam 
TG_BT_8=6*365*(6/7)% Thoi gian Binh Thuong trong 1 nam 
TG_CD_8=2*365*(6/7)% Thoi gian Cao Diem trong 1 nam 
 % Chi phi tien dien mot nam cho moi kW 
Cost_1kW_8_24=TG_TD_8*Gia_CN_TD+TG_BT_8*Gia_CN_
BT+TG_CD_8*Gia_CN_CD % Duoi 6kV, 8/24 gio 
% Cau a 
disp('Cau a: 
24/24_________________________________________') 
Pin_IE1_dm= Pout_dm/HS_IE1_dm % Cong suat dinh muc 
dong co 
Cost_Energy_IE1_1nam_24=Pin_IE1_dm/1000*Cost_1kW_24
_24 % Tinh theo 1000Wh=1kWh, 24/24 gio 
Pin_IE2_dm= Pout_dm/HS_IE2_dm % Cong suat dinh muc 
dong co 
Cost_Energy_IE2_1nam_24=Pin_IE2_dm/1000*Cost_1kW_24
_24 % Tinh theo 1000Wh=1kWh, 24/24 gio 
Cost_Energy_TK_1nam_24=Cost_Energy_IE1_1nam_24-
Cost_Energy_IE2_1nam_24 
Cost_Energy_TK_10nam_24=Cost_Energy_TK_1nam_24*6.1
446 %(P/A, 10%, 10) = 6.1446 <--tra bang 10 nam 
BCR_a_10nam_24=Cost_Energy_TK_10nam_24/Cost_Motor
_delta 
disp('Cau a: 
8/24_________________________________________') 
Cost_Energy_IE1_1nam_8=Pin_IE1_dm/1000*Cost_1kW_8_2
4 % Tinh theo 1000Wh=1kWh, 24/24 gio 
Pin_IE2_dm= Pout_dm/HS_IE2_dm % Cong suat dinh muc 
dong co 
Cost_Energy_IE2_1nam_8=Pin_IE2_dm/1000*Cost_1kW_8_2
4 % Tinh theo 1000Wh=1kWh, 24/24 gio 
Cost_Energy_TK_1nam_8=Cost_Energy_IE1_1nam_8-
Cost_Energy_IE2_1nam_8 
Cost_Energy_TK_10nam_8=Cost_Energy_TK_1nam_8*6.144
6 %(P/A, 10%, 10) = 6.1446 <--tra bang 10 nam 
BCR_a_10nam_8=Cost_Energy_TK_10nam_8/Cost_Motor_d
elta 
% Cau b 
disp('Cau b: 
_________________________________________') 
beta_Load=0.25 % 25% tai 
%beta_Load=1.00 % 100% tai 
Pout_a = beta_Load*Pout_dm % Cong suat 
disp('Cau b: 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 b: 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 b: So sanh tuong quan 
___________________________________') 
Pin_IE1=Pout_a/HS_IE1 
Cost_Energy_IE1_1nam=Pin_IE1/1000*Cost_1kW_24_24 % 
Tinh theo 1000Wh=1kWh, 24/24 gio 
Pin_IE2=Pout_a/HS_IE2 
Cost_Energy_IE2_1nam=Pin_IE2/1000*Cost_1kW_24_24 % 
Tinh theo 1000Wh=1kWh, 24/24 gio 
Cost_Energy_TKb_1nam=Cost_Energy_IE1_1nam-
Cost_Energy_IE2_1nam 
SPP_b=Cost_Motor_delta/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_Motor_delta 
disp('Ket qua 
__________________________________________________
_______') 
TEXT = sprintf('1.a) 24/24: Chi phi tang them khi mua IE2 = %f 
trieu', Cost_Motor_delta/1e+6); disp(TEXT) 
TEXT = sprintf('.'); disp(TEXT) 
TEXT = sprintf('1.a) 24/24: Chi phi nang luong 1 nam cua IE1 
= %f trieu', Cost_Energy_IE1_1nam_24/1e+6); disp(TEXT) 
TEXT = sprintf('1.a) 24/24: Chi phi nang luong 1 nam cua IE2 
= %f trieu', Cost_Energy_IE2_1nam_24/1e+6); disp(TEXT) 
TEXT = sprintf('1.a) 24/24: BCR_a_10nam = %f ==> chon 
IE2', BCR_a_10nam_24); disp(TEXT) 
TEXT = sprintf('.'); disp(TEXT) 
TEXT = sprintf('1.a) 8/24: Chi phi nang luong 1 nam cua IE1 = 
%f trieu', Cost_Energy_IE1_1nam_8/1e+6); disp(TEXT) 
TEXT = sprintf('1.a) 8/24: Chi phi nang luong 1 nam cua IE2 = 
%f trieu', Cost_Energy_IE2_1nam_8/1e+6); disp(TEXT) 
TEXT = sprintf('1.a) 8/24: BCR_a_10nam = %f ==> co the 
chon IE1 hoac IE2 tuy theo nha dau tu', BCR_a_10nam_8); 
disp(TEXT) 
TEXT = sprintf('.'); disp(TEXT) 
TEXT = sprintf('1.b) Hieu suat dong co IE1 o 1-hp = %f', 
HS_IE1); disp(TEXT) 
TEXT = sprintf('1.b) Hieu suat dong co IE2 o 1-hp = %f', 
HS_IE2); disp(TEXT) 
TEXT = sprintf('1.b) SPP_b = %f nam', SPP_b); disp(TEXT) 
TEXT = sprintf('1.b) BCR_b_10nam = %f ==> chon IE2', 
BCR_b_10nam); disp(TEXT) 
Câu 2: 
% De thi HK Lop QL&SDNL - Chinh quy HK1.2014-2015 
% Cau 2 
clc 
clear all 
% Bieu gia dien 
 % Gia dien Cong Nghiep <6kV 
Gia_CN_TD=897 % Gia dien Cong Nghiep gio Thap Diem 
Gia_CN_BT=1406 % Gia dien Cong Nghiep gio Binh Thuong 
Gia_CN_CD=2542 % Gia dien Cong Nghiep gio Cao Diem 
 % Gia dien Cong Nghiep 22kV 
Gia_22kV_TD=822 % Gia dien Cong Nghiep gio Thap Diem 
Gia_22kV_BT=1305 % Gia dien Cong Nghiep gio Binh 
Thuong 
Gia_22kV_CD=2376 % Gia dien Cong Nghiep gio Cao Diem 
Sdm=2000; % 2000kVA 
Po=1.500; 
Pn=17.100; 
Io=1.0/100; 
Vn=6/100; 
P_TT=1500; % Tai Tieu Thu 1000kW 
PF=0.9; % Tai cam 
Tnam=24*365; % 8760 gio 
Cost_MBA=2.0e+9 % 2 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_B
T+TG_CD*Gia_22kV_CD % 6kV-22kV 
Cost_1kW_6kV=TG_TD*Gia_CN_TD+TG_BT*Gia_CN_BT+T
G_CD*Gia_CN_CD % Duoi 6kV 
disp('Cau a___________________________') 
Cost_TT_6kV=P_TT*Cost_1kW_6kV % Chi phi cho nang 
luong tren tai trong 1 nam neu mua dien duoi 6kV 
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_MBA_1nam=Cost_MBA*0.1627% Chi phi lap dat MBA 
cho 1 nam %(A/P, 10%, 10) = 0.1627 <--- tra bang 10 nam 
Cost_tong_22kV=Cost_TT_22kV+Cost_TH_22kV+Cost_MBA
_1nam % Chi phi Tong trong 1 nam neu mua dien 22kV, bao 
gom ca MBA 
disp('Cau a: Cach 2__________________________') 
Cost_TK_1nam=P_TT*(Cost_1kW_6kV-Cost_1kW_22kV)-
Cost_TH_22kV % Tinh cho 1 nam 
SPP_MBA=Cost_MBA/Cost_TK_1nam % SPP 
disp('Cau b__________________________') 
Cost_kVAr=125000 % 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 0.9 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_6kV (1 nam) = %f ty dong', 
Cost_TT_6kV/1e+9); disp(TEXT) 
TEXT = sprintf('2.a) Cost_22kV (1 nam, bao gom chi phi lap 
dat MBA) = %f ty dong ==>chon 22kV', 
Cost_tong_22kV/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 ==> nen 
chon mua dien 22kV', 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) C = %f kVAr', C_TB); disp(TEXT) 
TEXT = sprintf('2.a) Cost_Tu. dien = %f trieu dong', 
Cost_TB/1e+6); disp(TEXT) 
TEXT = sprintf('2.b) SPP = %f nam', SPP); disp(TEXT) 
TEXT = sprintf('2.b) BCR = %f ==> Nen dau tu neu mua dien 
trung the (22kV), truoc MBA', BCR); disp(TEXT) 
Câu 3: 
?? 
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