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PRODUCT UNGGULAN DIBIDANG ELECTRONICS EE-LAB ( ELECTRICAL ECONOMIZING –LAB ) PENGHEMAT LISTRIK ( 15% ~40 % ) XP-LAB (Xtra Power –LAB ) PENGHEMAT BENSIN UNTUK MOBIL 12~15 VOLT ( 10% ~ 30% ) MASUK BINAAN UKM PELINDO THN 2008 HK 566/17/18/PI.II-08

EE-LAB EE-LAB MEMBERIKAN MANFAAT  DAPAT MEMPERBAIKI COSPHI HINGGA MENDEKATI NILAI IDEAL 1  DAPAT MENGURANGI ARUS START YANG BESAR  DAPAT MEMAKSIMALKAN DAYA LISTRIK TERPASANG DIRUMAH ANDA  TIDAK MELANGGAR HUKUM  DAPAT MENGHILANGKAN NOISE YANG AKAN MERUSAK PERALATAN ELEKTRONIK ANDA  DAPAT MENGURANGI TEMPERATUR JARINGAN LISTRIK DIRUMAH ANDA AKIBAT BEBAN PEMAKAIAN  TIDAK ADA EFEK NEGATIF  DAPAT MENGHEMAT BIAYA LISTRIK BULANAN PADA JENIS METERAN KONVENSIONAL ATAU MODEL PULSA DARI 15% ~40% PADA BEBAN YANG DOMINAN BERBASIS INDUKSI SEPERTI ;AC , POMPA AIR, KIPAS ANGIN, BOR LISTRIK, NEON TRAFO BALAST, KULKAS , LAMPU PIJAR, SOLDER, HAIR DRYER DAN SEBAGAINYA.  SUDAH TERBUKTI HEMAT DAN DIBUAT SEJAK TAHUN 2006  MASA PAKAI 5 TAHUN DAN GARANSI SELAMA 1 TAHUN

CARA PEMASANGAN EE-LAB

DATA HASIL ANALISA KUANTITATIF DAN PENGUKURAN TERHADAP PERALATAN RUMAH TANGGA SETELAH MENGGUNAKAN EE-LAB PERALATAN RUMAH TANGGA ARUS YANG DIBUTUHKAN SEBELUM PASANG EE-LAB/DETIK ARUS YANG DIBUTUHKAN SESUDAH PASANG EE-LAB/DETIK PERSENTASE COST SAVING YANG DIHASILKAN SETELAH MENGGUNAKAN EE-LAB LAMPU PIJAR 5 WATT 0,037 0,023 37,5% LAMPU HEMAT ENERGY 25 WATT 0,053 0,046 13,2% LAMPU LED 4 WATT 0,020 0,018 8,08% SETRIKA 350 WATT 3,977 2,892 27,27% KIPAS ANGIN 25 WATT 0,206 0,120 41,49% SOLDER 20 WATT 0,181 0,096 46,81% KULKAS 90 WATT 0,606 0,328 36,84% MAJICJAR/WARM PST 0,585 0,292 50% NOTEBOOK 90 WATT 0,681 0,454 33,33%

FORMULAS,EQUATIONS AND LAWS   DIRECT CURRENT AMPS= WATTS÷VOLTS I = P ÷ E A = W ÷ V WATTS= VOLTS x AMPS P = E x I W = V x A VOLTS= WATTS ÷ AMPS E = P ÷ I V = W ÷ A HORSEPOWER= (V x A x EFF)÷746 EFFICIENCY= (746 x HP)÷(V x A) AC SINGLE PHASE ~ 1ø WATTS÷(VOLTS x PF) I=P÷(E x PF) A=W÷(V x PF) VOLTS x AMPS x PF P=E x I x PF W=V x A x PF WATTS÷AMPS E=P÷I V=W÷A VOLT-AMPS= VA=E x I VA=V x A (V x A x EFF x PF)÷746 POWERFACTOR= INPUT WATTS÷(V x A) (746 x HP)÷(V x A x PF) AC THREE PHASE ~ 3ø WATTS÷(1.732 x VOLTS x PF) I = P÷(1.732 x E x PF) 1.732 x VOLTS x AMPS x PF P = 1.732 x E x I x PF

Formula from amp to kw DC amps to kilowatts calculation The power P in kilowatts (kW) is equal to the current I in amps (A), times the voltage V in volts (V) divided by 1000: P(kW) = I(A) × V(V) / 1000 AC single phase amps to kilowatts calculation The power P in kilowatts (kW) is equal to the power factor PF times the phase current I in amps (A), times the RMS voltage V in volts (V) divided by 1000: P(kW) = PF × I(A) × V(V) / 1000 AC three phase amps to kilowatts calculation Calculation with line to line voltage The power P in kilowatts (kW) is equal to square root of 3 times the power factor FP times the phase current I in amps (A), times the line to line RMS voltage VL-L in volts (V) divided by 1000: P(kW) = √3 × PF × I(A) × VL-L (V) / 1000 Calculation with line to neutral voltage The power P in kilowatts (kW) is equal to 3 times the power factor FP times the phase current I in amps (A), times the line to neutral RMS voltage VL-N in volts (V) divided by 1000: P(kW) = 3 × PF × I(A) × VL-N (V) / 1000   Amps to kW calculation ►

CONTOH UNTUK SUMBER DC How to convert amps to kilowatts How to convert electric current in amps (A) to electric power in kilowatts (kW). You can calculate kilowatts from amps and volts. You can't convert amps to kilowatts since kilowatts and amps units do not measure the same quantity. DC amps to kilowatts calculation formula The power P in kilowatts is equal to the current I in amps, times the voltage V in volts divided by 1000: P(kW) = I(A) × V(V) / 1000 So kilowatts are equal to amps times volts divided by 1000: kilowatt = amp × volt / 1000 or kW = A × V / 1000 Example What is power consumption in kW when the current is 3A and the voltage supply is 110V? Answer: the power P is equal to current of 3 amps times the voltage of 110 volts, divided by 1000. P = 3A × 110V / 1000 = 0.33kW

CONTOH UNTUK SUMBER AC SINGLE PHASE AC single phase amps to kilowatts calculation formula The real power P in kilowatts is equal to the power factor PF times the phase current I in amps, times the RMS voltage V in volts divided by 1000: P(kW) = PF × I(A) × V(V) / 1000 So kilowatts are equal to power factor times amps times volts divided by 1000: kilowatt = PF × amp × volt / 1000 or kW = PF × A × V / 1000 Example What is power consumption in kW when the power factor is 0.8 and the phase current is 3A and the RMS voltage supply is 110V? Answer: the power P is equal to power factor of 0.8 times current of 3 amps times voltage of 110 volts, divided by 1000. P = 0.8 × 3A × 110V / 1000 = 0.264kW

CONTOH UNTUK SUMBER AC THREE PHASE AC three phase amps to kilowatts calculation formula The real power P in kilowatts is equal to square root of 3 times the power factor PF times the phase current I in amps, times the  line to line RMS voltage VL-L in volts divided by 1000: P(kW) = √3 × PF × I(A) × VL-L(V) / 1000 So kilowatts are equal to square root of 3 times power factor PF times amps times volts divided by 1000: kilowatt = √3 × PF × amp × volt / 1000 or kW = √3 × PF × A × V / 1000 Example What is power consumption in kW when the power factor is 0.8 and the phase current is 3A and the RMS voltage supply is 110V? Answer: the power P is equal to square root of 3 times power factor of 0.8 times current of 3 amps times the voltage of 110 volts, divided by 1000. P = √3 × 0.8 × 3A × 110V / 1000 = 0.457kW

WATT TO KWh How to convert watts to kWh How to convert electric power in watts (W) to energy in kilowatt-hour (kWh). You can calculate kilowatt-hour from watts and hours. You can't convert watt to kilowatt-hour unit since watt and kilowatt-hour unit represent different quantities. Watts to kilowatt-hour calculation formula The energy E in kilowatt-hour (kWh) is equal to the power P in watts (W), times the time period t in hours (hr) divided by 1000: E(kWh) = P(W) × t(hr) / 1000 So kilowatt-hour = watt × hour / 1000 or kWh = W × hr / 1000 Example What is the energy consumption in watt-hour when the power consumption is 5000 watts for time duration of 3 hours? E = 5000W × 3h / 1000 = 15 kWh

Nilai konversi biaya dari kwH ke rupiah CATATAN : HARGA PER KWH DITENTUKAN DARI HARGA JUAL PLN KEPADA PELANGGAN LISTRIK BAIK RUMAHAN MAUPUN INDUSTRI UNTUK JENIS PELANGGAN HARGANYA BERBEDA Perhitungannya adalah Jumlah KWh terpakai X Harga satuan Kwh dari PLN Misal jumlah KWh kita terpakai untuk kebutuhan setrika selama 1 jam misalnya = 8 KWh Maka biaya yg harus dikeluarkan sebesar 8KWh x 1100 = Rp 8.800

XP-LAB XP-LAB MEMBERIKAN MANFAAT MENGURANGI ARUS BOCOR PADA PROSES PENGISIAN ACCU OLEH DINAMO AMPER MENYEBABKAN SOFT SHIFT TRANSMISSION PADA MOBIL MATIC DALAM MENJALANKAN SISTIM ROBOTICS YANG MEMBUTUHKAN KETERSEDIAAN DAYA LISTRIK MESIN LEBIH RESPONSIF DALAM HAL TENAGA KARENA PEMBAKARAN DIDUKUNG TEGANGAN YANG OPTIMAL , DAN UNTUK MOBIL INJECTION AKAN MEMBANTU PEMBUKAAN NEEDLE VALVE INJECTOR LEBIH LEBAR. ( MEMPENGARUHI FORMULA PADA PROGRAM ECU YAITU; Injection signal duration = basic injection duration X injection correction+ voltage correction DAPAT MEMAKSIMALKAN USIA KERJA ACCU LEBIH LAMA DAPAT MENGHILANGKAN SUARA MESIN NGE-LITIK EFEK PENCAHAYAN PADA LAMPU ANDA SEMAKIN TERANG MEMBERIKAN EFEK EFISIENSI KEBUTUHAN BAHAN BAKAR DENGAN RATING 10 ~30 % ( DENGAN AC DAN NON AC ) SUDAH TERBUKTI DIBUAT SEJAK TAHUN 2006 MASA PAKAI 5 TAHUN DAN GARANSI 1 TAHUN 2006 ALAT INI HANYA DAPAT DIGUNAKAN OLEH KENDARAAN NON-TRUCK

CARA PEMASANGAN XP-LAB