Đề kiểm tra giữa học kỳ II môn Quản lý và sử dụng năng lượng - Năm học 2011-2012

phí mỗi lần lắp 
đặt hay thay thế ballast và starter là 300.000đ. Máng chóa đèn 2 triệu đồng, tụ bù 
100.000đ (đã bao gồm chí phí lắp đặt hay thay thế), cả hai có tuổi thọ 100.000 giờ. 
Với loại đèn được lựa chọn trên, để tiết kiệm năng lượng, giải pháp đề xuất là tiết giảm 
bớt 50% công suất đèn từ 22:30 đến 4:30 hàng ngày. Chi phí cho thiết bị điều khiển tiết 
giảm độ sáng là 10.000đ/W, tuổi thọ 20 năm. 
a) Tính tiền tiết kiệm điện năm đầu tiên? (1đ) 
b) Tính SPP và BCR? (1đ) 
c) Tính SPP và BCR nếu hệ thống trên được sử dụng trong công nghiệp? (1đ) 
-------------------------------------------------HẾT------------------------------------------------- 
Bảng tính với lãi suất chiếc khấu i=10% 
Câu 1: 
1.a) SPP = 0.463424 nam = 5.561086 thang 
1.b) ROR = 2.157852 
1.c) BCR = 8.179985 
1.d) SPP_DCHSC_BT = 0.814694 = 9.776331 thang 
1.d) BCR_DCHSC_BT = 4.653034 
1.e) SPP_DCTC_BT = 0.773512 = 9.282143 thang 
1.e) BCR_DCTC_BT = 4.900765 => tot hon 
Chi tiết: 
Pout = 7.4600 
Cau a____________________________________ 
_____Dong co tieu chuan ___________________ 
Cost_1kW = 1.2360e+007 
Pin_DCTC = 8.6343 
Cost_NLTC = 1.0672e+008 
_____Dong co hieu suat cao ___________________ 
Pin_DCHSC = 8.1978 
Cost_NLHSC = 1.0133e+008 
Cost_NL = 5.3946e+006 
SPP_Dco = 0.4634 
Cau b____________________________________ 
ROR1_Dco = 2.1579 
Cau c____________________________________ 
BCR_Dco = 8.1800 
Cau d____________________________________ 
Qd = 0.7000 
HS_BT = 0.9200 
Cost_BT = 25000000 
Pout_DC_70 = 2.5588 
Pin_BTHSC_70 = 3.0564 
Pin_BTHSC_TK = 5.1414 
Cost_NLHSC_TK = 3.0686e+007 
SPP_DCHSC_BT = 0.8147 
BCR_DCHSC_BT = 4.6530 
Cau e____________________________________ 
Pin_BTTC_70 = 3.2191 
Pin_BTTC_TK = 5.4152 
Cost_NLTC_TK = 3.2320e+007 
SPP_DCTC_BT = 0.7735 
BCR_DCTC_BT = 4.9008 
Câu 2: 
2.LED: Cost_LED_(A) = 4087326.333333 
2.MH: Cost_MH_(A) = 1621312.800000 
2.HPS: Cost_HPS_(A) = 1569217.800000 <---chon 
2.a: Cost_TK = 282203.400000 
2.b) SPP_CSCC = 6.024024 nam 
2.b) BCR_CSCC = 1.413275 
2.c) SPP_CSCN = 10.210476 nam 
2.c) BCR_CSCN = 0.833810 
Chi tiết: 
Cost_MC_A = 220400 
Cost_TB_A = 11020 
Cost_BS_A = 2.0840e+005 
____________LED________________________ 
P_LED = 116.6667 
Cost_LED_P = 1.1967e+007 
Cost_LED_NL_A = 7.1012e+005 
Cost_LED_A = 3.8669e+006 
Cost_LED_All = 4.0873e+006 
____________MH_________________________ 
P_MH = 170 
Cost_MH_P = 300000 
Cost_MH_NL_A = 1.0347e+006 
Cost_MH_A = 1.1815e+006 
Cost_MH_All = 1.6213e+006 
____________HPS_________________________ 
P_HPS = 170 
Cost_HPS_P = 300000 
Cost_HPS_NL_A = 1.0347e+006 
Cost_HPS_A = 1.1294e+006 
Cost_HPS_All = 1.5692e+006 
Cau a____________________________________ 
NL_HPS_TK = 186150 
Cost_HPS_TK = 2.8220e+005 
Cau b____________________________________ 
Cost_TB = 1700000 
SPP_CSCC = 6.0240 
BCR_CSCC = 1.4133 
Cau c____________________________________ 
Cost_HPS_TK_CSCN = 1.6650e+005 
SPP_CSCN = 10.2105 
BCR_CSCN = 0.8338 
Câu 1: 
% Lop QL&SDNL - De thi hoc ky, lop CQ HK2.2013 
% Cau 1 
clc 
clear all 
 % Bieu gia dien 5/2013 
Gia_CN_TD=854 % Gia dien Cong Nghiep gio Thap Diem 
Gia_CN_BT=1339 % Gia dien Cong Nghiep gio Binh Thuong 
Gia_CN_CD=2421 % Gia dien Cong Nghiep gio Cao Diem 
 % Van hanh 24gio/ngay, 365ngay/nam 
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=10*0.746 % 10hp=>kWh 
HS_DCTC=0.864 % Hieu suat dong co Tieu Chuan 
HS_DCHSC=0.91 % Hieu suat dong co Hieu Suat Cao 
Cost_Dco=2500000 % Chi phi chenh lech giua 2 dong co 
 % i=10% - tra bang 
disp('Cau a____________________________________') 
disp('_____Dong co tieu chuan ___________________') 
 % Chi phi ten dien mot nam cho moi kW 
Cost_1kW=(6*Gia_CN_TD+13*Gia_CN_BT+5*Gia_CN_CD)*365*(6/7)+... 
 (6*Gia_CN_TD+18*Gia_CN_BT+0*Gia_CN_CD)*365*(1/7) 
 % Chi phi nang luong cho dong co tieu chuan 
Pin_DCTC=Pout/HS_DCTC 
Cost_NLTC=Pin_DCTC*Cost_1kW 
disp('_____Dong co hieu suat cao ___________________') 
 % Chi phi nang luong cho dong co tieu chuan 
Pin_DCHSC=Pout/HS_DCHSC 
Cost_NLHSC=Pin_DCHSC*Cost_1kW 
Cost_NL=Cost_NLTC-Cost_NLHSC 
SPP_Dco=Cost_Dco/Cost_NL 
disp('Cau b____________________________________') 
ROR1_Dco=Cost_NL/Cost_Dco 
disp('Cau c____________________________________') 
 %(P/A, 10%, 5) = 3.7908 <--- tra bang 
BCR_Dco=Cost_NL*3.7908/Cost_Dco 
disp('Cau d____________________________________') 
Qd=0.7 
HS_BT=0.92 % Hieu suat bien tan 
Cost_BT=25000000 % Chi phi lap dat bien tan 
Pout_DC_70=Pout*(Qd^3) 
 % Luu luong con 70%: Q=0.7; P~(Q)3 - Cong suat ngo ra cua dong co ty le bac 3 voi luu luong. 
Pin_BTHSC_70=Pout_DC_70/HS_DCHSC/HS_BT % Cong suat cap vao bien tan khi chay 70% cong suat 
Pin_BTHSC_TK=Pin_DCHSC-Pin_BTHSC_70 
 % Chi phi nang luong tiet kiem duoc 
Cost_NLHSC_TK=Pin_BTHSC_TK*(6*Gia_CN_TD+5*Gia_CN_BT+2*Gia_CN_CD)*365*(6/7)+... % T2-
T7:(TD-BT-CD)=(6-5-2) 
 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 e____________________________________') 
 % Luu luong con 70%: Q=0.7; P~(Q)3 - Cong suat ngo ra cua dong co ty le bac 3 voi luu luong. 
Pin_BTTC_70=Pout_DC_70/HS_DCTC/HS_BT % Cong suat cap vao bien tan khi chay 70% cong suat 
Pin_BTTC_TK=Pin_DCTC-Pin_BTTC_70 
 % Chi phi nang luong tiet kiem duoc 
Cost_NLTC_TK=Pin_BTTC_TK*(6*Gia_CN_TD+5*Gia_CN_BT+2*Gia_CN_CD)*365*(6/7)+... % T2-T7:(TD-
BT-CD)=(6-5-2) 
 Pin_BTTC_TK*(6*Gia_CN_TD+7*Gia_CN_BT+0*Gia_CN_CD)*365*(1/7) % CN:(TD-BT-CD)=(6-7-
0) 
SPP_DCTC_BT=Cost_BT/Cost_NLTC_TK 
BCR_DCTC_BT=Cost_NLTC_TK*3.7908/Cost_BT %(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) SPP_DCHSC_BT = %f = %f thang', SPP_DCHSC_BT, SPP_DCHSC_BT*12); 
disp(TEXT) 
TEXT = sprintf('1.d) BCR_DCHSC_BT = %f', BCR_DCHSC_BT); disp(TEXT) 
TEXT = sprintf('1.e) SPP_DCTC_BT = %f = %f thang', SPP_DCTC_BT, SPP_DCTC_BT*12); disp(TEXT) 
TEXT = sprintf('1.e) BCR_DCTC_BT = %f => tot hon', BCR_DCTC_BT); disp(TEXT) 
Câu 2: 
% Lop QL&SDNL - De thi hoc ky, lop CQ HK2.2013 
% Cau 2 
clc 
clear all 
 % Bieu gia dien 5/2013 
Gia_CN_TD=854; % Gia dien Cong Nghiep gio Thap Diem 
Gia_CN_BT=1339; % Gia dien Cong Nghiep gio Binh Thuong 
Gia_CN_CD=2421; % Gia dien Cong Nghiep gio Cao Diem 
Gia_CSCC=1516; % Gia dien cho chieu sang cong cong 5/2013 
LM=14000; % 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 
% De so sanh, se tinh chi phi cho nam dau tien! 
% Mang choa den giong nhau nen co the khong can tinh toan. 
% Chi phi mua va lap dat/thay the mang choa den 
 Cost_MC_P=2000000; 
 Cost_MC_A=Cost_MC_P*0.1102 % (A/P,10%,25)=0.1102<-tra bang 
% Chi phi mua va lap dat/thay the tu bu 
 Cost_TB_P=100000; 
 Cost_TB_A=Cost_TB_P*0.1102 % (A/P,10%,25)=0.1102<-tra bang 
% Chi phi mua va lap dat/thay the ballast, starter 
 Cost_BS_P=360000+130000+300000; 
 Cost_BS_A=Cost_BS_P*0.2638 % (A/P,10%,5)=0.2638<-tra bang 
disp('____________LED________________________') 
 P_LED=LM/120 
 Cost_LED_P=P_LED*100000+300000 % chi phi mua va lap dat/thay the den LED 
 Cost_LED_NL_A=P_LED*Gia_CSCC/1000*Tnam % chi phi nam luong nam dau tien 
 Cost_LED_A=Cost_LED_P*(0.2638)+Cost_LED_NL_A % (A/P,10%,5)=0.2638<-tra bang 
 Cost_LED_All=Cost_LED_A+Cost_MC_A % Chi phi quy doi hang nam 
disp('____________MH_________________________') 
 P_MH=150+20 
 Cost_MH_P=200000+100000 % chi phi mua va lap dat/thay the den MH 
 Cost_MH_NL_A=P_MH*Gia_CSCC/1000*Tnam % chi phi nam luong nam dau tien 
 Cost_MH_A=Cost_MH_P*(0.5762+0.4021)/2+Cost_MH_NL_A 
 % (A/P,10%,2.5)=(0.5762+0.4021)/2<-tra bang 
 Cost_MH_All=Cost_MH_A+Cost_BS_A+Cost_TB_A+Cost_MC_A % Chi phi quy doi hang nam 
disp('____________HPS_________________________') 
 P_HPS=150+20 
 Cost_HPS_P=200000+100000 % chi phi mua va lap dat/thay the den HPS 
 Cost_HPS_NL_A=P_HPS*Gia_CSCC/1000*Tnam % chi phi nam luong nam dau tien 
 Cost_HPS_A=Cost_HPS_P*(0.3155)+Cost_HPS_NL_A % (A/P,10%,4)=0.3155<-tra bang 
 Cost_HPS_All=Cost_HPS_A+Cost_BS_A+Cost_TB_A+Cost_MC_A % Chi phi quy doi hang nam 
disp('Cau a____________________________________') 
%Cost_HPS_NL=P_HPS*Tnam*Gia_CSCC/1000 
NL_HPS_TK=0.5*P_HPS*6*365 % Nang luong tiet kiem torng 1 nam 
Cost_HPS_TK=NL_HPS_TK*Gia_CSCC/1000 
disp('Cau b____________________________________') 
Cost_TB=10000*P_HPS % Chi phi thiet bi dieu khien tiet giam do sang 
SPP_CSCC=Cost_TB/Cost_HPS_TK 
BCR_CSCC=Cost_HPS_TK*8.5136/Cost_TB % (P/A,10%,20)=8.5136<---tra bang 
disp('Cau c____________________________________') 
% Tien dien tiet kiem torng 1 nam: 5.5 gio thap diem, 0.5 gio binh thuong 
Cost_HPS_TK_CSCN=0.5*P_HPS/1000*(Gia_CN_TD*5.5+Gia_CN_BT*0.5)*365 
SPP_CSCN=Cost_TB/Cost_HPS_TK_CSCN 
BCR_CSCN=Cost_HPS_TK_CSCN*8.5136/Cost_TB % (P/A,10%,20)=8.5136<---tra bang 
disp('Ket qua ______________________________') 
TEXT = sprintf('2.LED: Cost_LED_(A) = %f',Cost_LED_All); disp(TEXT) 
TEXT = sprintf('2.MH: Cost_MH_(A) = %f', Cost_MH_All); disp(TEXT) 
TEXT = sprintf('2.HPS: Cost_HPS_(A) = %f <---chon', Cost_HPS_All); disp(TEXT) 
TEXT = sprintf('2.a: Cost_TK = %f ', 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) 
TEXT = sprintf('2.c) SPP_CSCN = %f nam', SPP_CSCN); disp(TEXT) 
TEXT = sprintf('2.c) BCR_CSCN = %f', BCR_CSCN); disp(TEXT)