Đề kiểm tra giữa học kỳ II môn Quản lý và sử dụng năng lượng - Năm học 2013-2014

L_A = 1.9000e+006 
Cost_LED_P = 2.1333e+007 
Cost_LED_A = 4.2526e+006 
Cost_LED_All = 4.2526e+006 
____________HPS____________ 
P_HPS = 270 
Cost_HPS_NL_A = 1.7236e+006 
Cost_HPS_P = 1200000 
Cost_HPS_A = 316560 
Cost_MCT_A = 406750 
Cost_HPS_All = 2.4469e+006 
Cau b______________________ 
NL_HPS_TK = 354780 
Cost_HPS_TK = 5.6410e+005 
Cost_TB = 1350000 
SPP_CSCC = 2.3932 
BCR_CSCC = 2.5675
Câu 3: 
3.a) Cost_TH = 81678119.057439 dong = 81.678119 trieu dong 
3.b) Tien tiet kiem 1nam = 13935236.341650 dong = 13.935236 trieu dong 
3.b) SPP = 1.754478 nam 
3.b) BCR = 2.160643 
Chi tiết: 
Cau 3_______________________________ 
TG_TD = 2190 
TG_BT = 5.0057e+003 
TG_CD = 1.5643e+003 
Cost_1kW = 12049380 
Cau a_______________________________ 
S = 823.5294 
beta = 0.8235 
P_TH = 6.7786 
Cost_TH = 8.1678e+007 
Cau b______________________________ 
Cost_kVAr = 120000 
PF_b = 0.9500 
S_b = 736.8421 
beta_b = 0.7368 
P_TH_b = 5.6221 
Cost_TH_b = 6.7743e+007 
Cost_TH_TK = 1.3935e+007 
tan = 0.6197 
tan_b = 0.3287 
C_TB = 203.7422 
Cost_TB = 2.4449e+007 
SPP = 1.7545 
BCR = 2.1606 
Câu 1: 
% Lop QL&SDNL - De thi hoc ky, lop CQ HK2.2014 
% Cau 1 
clc 
clear all 
 % IE1: Dong co tieu chuan (DCTC) 
 % IE2: Dong co hieu suat cao (DCHSC) 
% Bieu gia dien 5/2014 
 % 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 
% 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 
 % Thong so dong co 
Pout=20*0.746 % 20hp=>kWh 
HS_DCTC=0.887 % Hieu suat dong co Tieu Chuan 
HS_DCHSC=0.906 % Hieu suat dong co Hieu Suat 
Cao 
Cost_Dco=5000000 % Chi phi chenh lech giua 2 dong 
co 
 % i=10% - tra bang 
disp('Cau a______________________________') 
 % 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 
Pin_DCTC=Pout/HS_DCTC % Cong suat tieu thu cua 
dong co tieu chuan IE1 
Pin_DCHSC=Pout/HS_DCHSC % Cong suat tieu thu 
cua dong co hieu suat cao IE2 
Pin_TK=Pin_DCTC-Pin_DCHSC % Cong suat tiet kiem 
duoc 
Cost_NL=Pin_TK*Cost_1kW % Chi phi tien dien chenh 
lech giua 2 dong co 
SPP_Dco=Cost_Dco/Cost_NL 
disp('Cau b______________________________') 
ROR1_Dco=Cost_NL/Cost_Dco 
disp('Cau c______________________________') 
 %(P/A, 10%, 10) = 6.1446 <--- tra bang 
BCR_Dco=Cost_NL*6.1446/Cost_Dco 
disp('Cau d____________________________PA1') 
Qd=2/3 
HS_BT=0.94 % Hieu suat bien tan 
Cost_BT=40000000 % Chi phi lap dat bien tan 
Pout_DC_23=Pout*(Qd^3) % Luu luong con 2/3: Q=2/3; 
P~(Q^3) - Cong suat ngo ra cua dong co ty le bac 3 voi 
luu luong. 
Pin_BTHSC_23=Pout_DC_23/HS_DCHSC/HS_BT % 
Cong suat cap vao bien tan khi chay 2/3 toc do 
Pin_BTHSC_TK=Pin_DCHSC-Pin_BTHSC_23 
 % Chi phi nang luong tiet kiem duoc 
Cost_NLHSC_TK= 
Pin_BTHSC_TK*(6*Gia_CN_TD+4*Gia_CN_BT+3*Gia
_CN_CD)*365*(6/7)+... % T2-T7:(TD-BT-CD)=(6-4-3) 
Pin_BTHSC_TK*(6*Gia_CN_TD+7*Gia_CN_BT+0*Gia
_CN_CD)*365*(1/7) % CN:(TD-BT-CD)=(6-7-0) 
SPP_DCHSC_BT=Cost_BT/Cost_NLHSC_TK 
BCR_DCHSC_BT=Cost_NLHSC_TK*3.7908/Cost_BT 
%(P/A, 10%, 5) = 3.7908 <--- tra bang 
disp('Cau d____________________________PA2') 
 % Thong so dong co PA2 
Pout_PA2=7*0.746 % 7hp=>kWh 
HS_DC_PA2=0.86 % Hieu suat dong co 
Cost_Dco_PA2=10000000 % Chi phi dong co PA2 
HS_HT_PA2=0.98 
 % Chi phi nang luong tiet kiem duoc vao BAN NGAY 
Pin_DCHSC_PA2=Pin_DCHSC/HS_HT_PA2 % Hieu 
suat giam do ghep truc 2 dong co 
Pin_PA2_TK_ngay=Pin_DCHSC-Pin_DCHSC_PA2 % 
Chenh lech cong suat truoc va sau khi ghep them dong 
co 7-hp 
Cost_NLPA2_TK_ngay= 
Pin_PA2_TK_ngay*(9*Gia_CN_BT+2*Gia_CN_CD)*36
5*(6/7)+... % T2-T7:(TD-BT-CD)=(0-9-2) 
Pin_PA2_TK_ngay*(11*Gia_CN_BT)*365*(1/7) % 
CN:(TD-BT-CD)=(0-11-0) 
 % Khi dong co chay voi toc do ung voi dong co 7-hp, 
6 cuc (khoang 2/3 toc do dong co 4 cuc) 
Pout_DC_23=Pout*(Qd^3) % Luu luong con 2/3: Q=2/3; 
P~(Q^3) - Cong suat ngo ra cua dong co ty le bac 3 voi 
luu luong. 
Pout_DC7hp_PA2=Pout_DC_23/HS_HT_PA2 % Hieu 
suat giam do ghep truc 2 dong co 
Pin_DC7hp_PA2=Pout_DC7hp_PA2/HS_DC_PA2 % 
Hieu suat dong co nho 7-hp 
 % Chi phi nang luong tiet kiem duoc vao BAN DEM 
Pin_PA2_TK_dem=Pin_DCHSC-Pin_DC7hp_PA2 
Cost_NLPA2_TK_dem= 
Pin_PA2_TK_dem*(6*Gia_CN_TD+4*Gia_CN_BT+3*G
ia_CN_CD)*365*(6/7)+... % T2-T7:(TD-BT-CD)=(6-4-3) 
Pin_PA2_TK_dem*(6*Gia_CN_TD+7*Gia_CN_BT+0*G
ia_CN_CD)*365*(1/7) % CN:(TD-BT-CD)=(6-7-0) 
 % Chi phi nang luong tiet kiem duoc ca ngay dem 
Cost_NLPA2_TK=Cost_NLPA2_TK_ngay+Cost_NLPA
2_TK_dem 
SPP_PA2=Cost_Dco_PA2/Cost_NLPA2_TK 
BCR_PA2=Cost_NLPA2_TK*3.7908/Cost_Dco_PA2 
%(P/A, 10%, 5) = 3.7908 <--- tra bang 
disp('Ket qua ______________________________') 
TEXT = sprintf('1.a) SPP = %f nam = %f thang', 
SPP_Dco, SPP_Dco*12); disp(TEXT) 
TEXT = sprintf('1.b) ROR = %f ', ROR1_Dco); 
disp(TEXT) 
TEXT = sprintf('1.c) BCR = %f', BCR_Dco); disp(TEXT) 
TEXT = sprintf('1.d) Chon dong co hieu suat cao IE2'); 
disp(TEXT) 
TEXT = sprintf('1.d_PA1) SPP_DCHSC_BT = %f = %f 
thang', SPP_DCHSC_BT, SPP_DCHSC_BT*12); 
disp(TEXT) 
TEXT = sprintf('1.d_PA1) BCR_DCHSC_BT = %f', 
BCR_DCHSC_BT); disp(TEXT) 
TEXT = sprintf('1.d_PA2) Cong suat chay dong co nho 
la %f-hp < 7-hp', Pout_DC7hp_PA2/0.746); disp(TEXT) 
TEXT = sprintf('1.d_PA2) SPP_PA2 = %f = %f thang', 
SPP_PA2, SPP_PA2*12); disp(TEXT) 
TEXT = sprintf('1.d_PA2) BCR_PA2 = %f', BCR_PA2); 
disp(TEXT) 
Câu 2: 
% Lop QL&SDNL - De thi hoc ky, lop CQ HK2.2014 
% Cau 2 
clc 
clear all 
% Bieu gia dien 5/2014 
 % Gia dien Cong Cong 
Gia_CSCC=1590 % Gia dien Cong Nghiep gio Thap 
Diem 
 LM=25000; % lm 
 % i=10% - tra bang 
disp('Cau 2_________________________________') 
% Moi ngay chieu sang tu 18:30-5:30 = 11 gio 
% So gio chieu sang mot nam 
 Tnam=11*365; % ~4000 gio 
disp('Cau a_________________________________') 
% De so sanh, se tinh chi phi hang nam: 
disp('____________LED______________________') 
 P_LED=LM/84 % 84 lm/W 
 Cost_LED_NL_A=P_LED*Gia_CSCC/1000*Tnam % 
chi phi nam luong nam dau tien 
 Cost_LED_P=P_LED*70000+500000 % chi phi mua 
va lap dat/thay the den LED 
Cost_LED_A=Cost_LED_P*(0.11028)+Cost_LED_NL_
A % (A/P,10%,25)=0.1102<-tra bang 25 nam 
 Cost_LED_All=Cost_LED_A % Chi phi quy doi hang 
nam 
 disp('____________HPS_______________________') 
 P_HPS=250+20 
 Cost_HPS_NL_A=P_HPS*Gia_CSCC/1000*Tnam % 
chi phi nam luong nam dau tien 
 Cost_HPS_P=1200000 % chi phi mua va lap 
dat/thay the den HPS 
 Cost_HPS_A=Cost_HPS_P*(0.2638) % 
(A/P,10%,5)=0.2638<-tra bang 5 nam 
 Cost_MCT_P=2500000; % Chi phi mua va lap 
dat/thay the mang choa den+tu bu 
 Cost_MCT_A=Cost_MCT_P*0.1627 % 
(A/P,10%,10)=0.1627<-tra bang 10 nam 
Cost_HPS_All=Cost_HPS_A+Cost_MCT_A+Cost_HPS
_NL_A % Chi phi quy doi hang nam 
disp('Cau b__________________________________') 
%Cost_HPS_NL=P_HPS*Tnam*Gia_CSCC/1000 
NL_HPS_TK=0.6*P_HPS*6*365 % Nang luong tiet 
kiem trong 1 nam 
Cost_HPS_TK=NL_HPS_TK*Gia_CSCC/1000 
Cost_TB=5000*P_HPS % Chi phi thiet bi dieu khien tiet 
giam do sang 
SPP_CSCC=Cost_TB/Cost_HPS_TK 
BCR_CSCC=Cost_HPS_TK*6.1446/Cost_TB % 
(P/A,10%,10)=6.1446<---tra bang 
disp('Ket qua ______________________________') 
TEXT = sprintf('2.a) LED: Cost_LED_(A) = 
%f',Cost_LED_All); disp(TEXT) 
TEXT = sprintf('2.a) HPS: Cost_HPS_(A) = %f <---chon 
HPS vi chi phi thap hon', Cost_HPS_All); disp(TEXT) 
TEXT = sprintf('2.b) Cost_TK = %f dong', 
Cost_HPS_TK); disp(TEXT) 
TEXT = sprintf('2.b) SPP_CSCC = %f nam', 
SPP_CSCC); disp(TEXT) 
TEXT = sprintf('2.b) BCR_CSCC = %f', BCR_CSCC); 
disp(TEXT) 
Câu 3: 
% Lop QL&SDNL - De thi hoc ky, lop CQ HK2.2014 
% Cau 3 
clc 
clear all 
% Bieu gia dien 5/2014 
 % 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=1000; % 100kVA 
Po=0.980; 
Pn=8.550; 
Io=1/100; 
Vn=6/100; 
P=700; % 700kW 
PF=0.85; % Tai cam 
Tnam=8760; % gio 
 % i=10% - tra bang 
disp('Cau 3______________________________') 
 % 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_22kV_TD+TG_BT*Gia_22kV_
BT+TG_CD*Gia_22kV_CD 
disp('Cau a___________________________') 
S=P/PF 
beta=S/Sdm 
P_TH=Po+Pn*beta^2 
Cost_TH=P_TH*Cost_1kW % Chi phi cho nang luong 
ton hao tren MBA trong 1 nam 
disp('Cau b__________________________') 
Cost_kVAr=120000 % d/kVAr 
PF_b=0.95 
S_b=P/PF_b 
beta_b=S_b/Sdm 
P_TH_b=Po+Pn*beta_b^2 
Cost_TH_b=P_TH_b*Cost_1kW % Chi phi cho nang 
luong ton hao tren MBA trong 1 nam 
Cost_TH_TK=Cost_TH-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*(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*3.7908/Cost_TB %(P/A, 10%, 5) = 
3.7908 <--- tra bang 5 nam 
disp('Ket qua _______________________') 
TEXT = sprintf('3.a) Cost_TH = %f dong = %f trieu 
dong', Cost_TH, Cost_TH/1000000); disp(TEXT) 
TEXT = sprintf('3.b) Tien tiet kiem 1nam = %f dong = 
%f trieu dong', Cost_TH_TK, Cost_TH_TK/1000000); 
disp(TEXT) 
TEXT = sprintf('3.b) SPP = %f nam', SPP); disp(TEXT) 
TEXT = sprintf('3.b) BCR = %f', BCR); disp(TEXT)