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SIMPERUMAHAN MAHKOTA PRUG BLOK B7 NO21 TELP: 73455849 PT. IKHSAN SISTEM CIPTA REKAYASA IT. HARDWARE, SOFTWARE SERVICES AND ELECTRONICS SPECILIZE.

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Presentasi berjudul: "SIMPERUMAHAN MAHKOTA PRUG BLOK B7 NO21 TELP: 73455849 PT. IKHSAN SISTEM CIPTA REKAYASA IT. HARDWARE, SOFTWARE SERVICES AND ELECTRONICS SPECILIZE."— Transcript presentasi:

1 SIMPERUMAHAN MAHKOTA PRUG BLOK B7 NO21 TELP: PT. IKHSAN SISTEM CIPTA REKAYASA IT. HARDWARE, SOFTWARE SERVICES AND ELECTRONICS SPECILIZE

2 PT. IKHSAN SISTEM CIPTA REKAYASA IT HARDWARE SOFTWARE SERVICES

3 PT. IKHSAN SISTEM CIPTA REKAYASA ELECTRONICS SPECILIZE AND IT TUTOR  PCB DESIGN BERBASIS CAE  DIGITAL ELECTRONICS DESIGN

4 PT. IKHSAN SISTEM CIPTA REKAYASA PCB FILM

5 PT. IKHSAN SISTEM CIPTA REKAYASA IT TUTORIAL

6 LANGKAH BIJAK

7 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

8 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

9 CARA PEMASANGAN EE-LAB CARA - 1 CARA - 2

10 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 0,0370,02337,5% LAMPU HEMAT ENERGY 25 WATT 0,0530,04613,2% LAMPU LED 4 WATT 0,0200,0188,08% SETRIKA 350 WATT 3,9772,89227,27% KIPAS ANGIN 25 WATT 0,2060,12041,49% SOLDER 20 WATT 0,1810,09646,81% KULKAS 90 WATT 0,6060,32836,84% MAJICJAR/WARM PST 0,5850,29250% NOTEBOOK 90 WATT 0,6810,45433,33%

11 FORMULAS,EQUATIONS AND LAWS DIRECT CURRENT AMPS= WATTS÷VOLTSI = P ÷ EA = W ÷ V WATTS=VOLTS x AMPSP = E x IW = V x A VOLTS=WATTS ÷ AMPSE = P ÷ IV = W ÷ A HORSEPOWER=(V x A x EFF)÷746 EFFICIENCY=(746 x HP)÷(V x A) AC SINGLE PHASE ~ 1ø AMPS= WATTS÷(VOLTS x PF)I=P÷(E x PF)A=W÷(V x PF) WATTS=VOLTS x AMPS x PFP=E x I x PFW=V x A x PF VOLTS=WATTS÷AMPSE=P÷IV=W÷A VOLT-AMPS=VOLTS x AMPSVA=E x IVA=V x A HORSEPOWER=(V x A x EFF x PF)÷746 POWERFACTOR=INPUT WATTS÷(V x A) EFFICIENCY=(746 x HP)÷(V x A x PF) AC THREE PHASE ~ 3ø AMPS= WATTS÷(1.732 x VOLTS x PF)I = P÷(1.732 x E x PF) WATTS=1.732 x VOLTS x AMPS x PFP = x E x I x PF VOLTS=WATTS÷AMPSE=P÷I

12 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:power factor 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 V L-L in volts (V) divided by 1000:power factor P (kW) = √ 3 × PF × I (A) × V L-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 V L-N in volts (V) divided by 1000:power factor P (kW) = 3 × PF × I (A) × V L-N (V) / 1000 Amps to kW calculation ►

13 CONTOH UNTUK SUMBER DC How to convert amps to kilowatts How to convert electric current in amps (A) to electric power in kilowatts (kW).electric currentamps (A)electric powerkilowatts (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.volts 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 P = 3A × 110V / 1000 = 0.33kW

14 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:power factor 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 P = 0.8 × 3A × 110V / 1000 = 0.264kW

15 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 V L-L in volts divided by 1000:power factor P (kW) = √ 3 × PF × I (A) × V L-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 P = √ 3 × 0.8 × 3A × 110V / 1000 = 0.457kW

16 WATT TO KWH How to convert watts to kWh How to convert electric power in watts (W) to energy in kilowatt-hour (kWh).electric powerwatts (W)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

17 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

18 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

19 CARA PEMASANGAN XP-LAB


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