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LCLS Sector 24 Power Supplies For QE Magnet Splits Paul Bellomo, Antonio de Lira, Dave Mac Nair. LCLS Sector 24 magnet power supplies QE24201 through QE24501 QE24601 QE24701A, B and Q24901A, B. Q24201 through Q24501. Q24201 through Q24501.
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LCLS Sector 24 Power Supplies For QE Magnet Splits Paul Bellomo, Antonio de Lira, Dave Mac Nair
LCLS Sector 24 magnet power supplies • QE24201 through QE24501 • QE24601 • QE24701A, B and Q24901A, B
Q24201 through Q24501 Cable ampacity for new string power cable, 200A needed #350 Cu 90C insulation - 2005 NEC used as reference Conduit Basic ampacity 350A per 310.16 Derating for 113F ambient temperature per 310.16 = 0.87 350A*0.87*=304A Cable Tray Basic ampacity 570A in free air per 310.17 Derating for 113F maximum temperature per 310.17 = 0.87 Derating for use in properly loaded cable tray = 0.65 570A*0.87*0.65=322A 304A > 200A or 300A so 350 AWG is ok
Q24201 through Q24501 • Existing: 510ft (round trip) #350kcmil Al 0.0605 W/1000ft • New: 300ft (one way) #350AWG Cu 0.0382 W/1000ft • The voltage drop change is D E = I* (D R / 1000 ft) * D L ft • E = 200A * ((0.0605 W – 0.0382 W) / 1000 ft) *(255ft – 300 ft)) • E = -0.2 volt which is essentially no change • Material cost estimate • 1/C#350 Cu cable - $5.00/ft *300ft = $1, 500
Q24601 #350kcmil cable good to 304A • Issues • Investigate source of AC power • Investigate existing PPS connection • New Ethernet control • Material cost estimate • $11,800 for 10kW PS, controller, transductions, rack accessories, • 600 ft of #350 Cu cable • $5.00/ft * 600ft = $3,000 • 1/3 rack = $1,000 • Total - $15,800 excluding AC power
Q24601 600ft (round trip) #350kcmil Cu 0.0328 W/1000ft The voltage drop is E = I* ((Cable R / 1000 ft) * L + Magnet R)) E = 200A * ((0.0328 W) / 1000 ft) *600 ft)+0.082 W)) E =20.3V
Q24701A, Q24701B, Q25901A, Q24901B • Issues • Investigate source of AC power • Investigate existing PPS connections • New Ethernet control • Material cost estimate • $23,600 for 2 - 10kW PS, controller, transductions, rack accessories, • 1,200 ft of 4/0 Cu cable • $3.50/ft * 1,200ft = $4,200 • 2/3 rack = $2,000 • Total - $29,800 excluding AC power
Q24701A, Q24701B, Q25901A, Q24901B 625ft (round trip) #350kcmil Cu 0.0328 W/1000ft The voltage drop is E = I* ((Cable R / 1000 ft) * L + Magnet R)) E = 200A * ((0.0328 W) / 1000 ft) *625 ft)+0.082 W*2)) E =36.7V for each of two systems