Power Sats--PV - Revision history

Hkhenson at 19:22, 15 October 2008

2008-10-15T19:22:44Z

← Older revision		Revision as of 19:22, 15 October 2008
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			==Background numbers==
	Sunlight in space is around 1.3 kW/m^2. Photovoltaic cells of 15% efficiency would allow collecting 200W/m^2. A km^2 would produced 0.2 GW, 5km^2 would generate a GW and 25km^2 would generate 5 GW.		Sunlight in space is around 1.3 kW/m^2. Photovoltaic cells of 15% efficiency would allow collecting 200W/m^2. A km^2 would produced 0.2 GW, 5km^2 would generate a GW and 25km^2 would generate 5 GW.

	Power on the ground is around .8 x .8 x .8 for the transmitter, transmission and rectenna loses. That's close enough to 0.5, so 5GWe would take about 50 km^2 in space or 2 5x5 km surfaces each delivering 5GW to the transmitter.		Power on the ground is around .8 x .8 x .8 for the transmitter, transmission and rectenna loses. That's close enough to 0.5, so 5GWe would take about 50 km^2 in space or 2 5x5 km surfaces each delivering 5GW to the transmitter.

			==Power collection analysis==

	5,000,000,000 watts is 10,000 volts times 500,000 amps or perhaps if someone designs very high voltage klystrons, 100,000 volts and 50,000 amps. Using 10,000 volts, the edge of a 5 km square surface delivers 500,000 amps over 5 km, 100,000 amps per km or 100 amps per meter. If we make the current carrying conductor a thin metal foil 0.1 m wide and let it radiate at roughly room temperature, then we can loose 25 watts per meter of length. (A black surface radiates 250 watts/m^2. I am only counting on one side radiating since the side toward the bulk of the power sat would be at near room temp.)		5,000,000,000 watts is 10,000 volts times 500,000 amps or perhaps if someone designs very high voltage klystrons, 100,000 volts and 50,000 amps. Using 10,000 volts, the edge of a 5 km square surface delivers 500,000 amps over 5 km, 100,000 amps per km or 100 amps per meter. If we make the current carrying conductor a thin metal foil 0.1 m wide and let it radiate at roughly room temperature, then we can loose 25 watts per meter of length. (A black surface radiates 250 watts/m^2. I am only counting on one side radiating since the side toward the bulk of the power sat would be at near room temp.)

	W=I^2R so 25=100^2 R, R = 0.0025 Ω/meter. Aluminum is 26.50 nΩ·m So a nanometer thick layer of Al a meter square would measure about 25 Ω of resistance, one 1/10th meter wide would be 250 Ω. 250/.0025, 100/.001 or 100,000 nm. 10^5 x 10^-9 is 10-4 m or 0.1 mm. In area this would be 500 m x 5 k (the two collectors taper, but together they make a rectangle). This is 1/10 of the area of the power sat. Since the dissipation is 250 W/m^2 and the power sat collection surface is 200 W/m^2, 12.5% would be lost to resistive heating in this design.		W=I^2R so 25=100^2 R, R = 0.0025 Ω/meter. Aluminum is 26.50 nΩ·m So a nanometer thick layer of Al a meter square would measure about 25 Ω of resistance, one 1/10th meter wide would be 250 Ω. 250/.0025, 100/.001 or 100,000 nm. 10^5 x 10^-9 is 10-4 m or 0.1 mm. In area this would be 500 m x 5 k (the two collectors taper, but together they make a rectangle). This is 1/10 of the area of the power sat. Since the dissipation is 250 W/m^2 and the power sat collection surface is 200 W/m^2, 12.5% would be lost to resistive heating in this design. This is excessive.

	The mass would be 500 m x 5000 m x 10^-4 m, 250 cubic meters x 2.7 t/m^3, 675 tons per side or 1350 t. In the context of a 20-40,000 t power sat this is not excessive.		The mass would be 500 m x 5000 m x 10^-4 m, 250 cubic meters x 2.7 t/m^3, 675 tons per side or 1350 t. In the context of a 20-40,000 t power sat this is not excessive.

			==Improvements==

			Going to 100,000 volts and ten amps per meter would cut the power loss by a factor of 100. Or we could cut the conductor mass by ten and still get a 10 fold reduction in loses. Question to be asked: Is there any reason we could not wire ten 10kV Klystrons in series?

Hkhenson at 05:45, 15 October 2008

2008-10-15T05:45:15Z

← Older revision		Revision as of 05:45, 15 October 2008
Line 5:		Line 5:
	5,000,000,000 watts is 10,000 volts times 500,000 amps or perhaps if someone designs very high voltage klystrons, 100,000 volts and 50,000 amps. Using 10,000 volts, the edge of a 5 km square surface delivers 500,000 amps over 5 km, 100,000 amps per km or 100 amps per meter. If we make the current carrying conductor a thin metal foil 0.1 m wide and let it radiate at roughly room temperature, then we can loose 25 watts per meter of length. (A black surface radiates 250 watts/m^2. I am only counting on one side radiating since the side toward the bulk of the power sat would be at near room temp.)		5,000,000,000 watts is 10,000 volts times 500,000 amps or perhaps if someone designs very high voltage klystrons, 100,000 volts and 50,000 amps. Using 10,000 volts, the edge of a 5 km square surface delivers 500,000 amps over 5 km, 100,000 amps per km or 100 amps per meter. If we make the current carrying conductor a thin metal foil 0.1 m wide and let it radiate at roughly room temperature, then we can loose 25 watts per meter of length. (A black surface radiates 250 watts/m^2. I am only counting on one side radiating since the side toward the bulk of the power sat would be at near room temp.)

	W=I^2R so 25=100^2 R, R = 0.0025 Ω/meter. Aluminum is 26.50 nΩ·m So a nanometer thick layer of Al a meter square would measure about 25 Ω of resistance, one 1/10th meter wide would be 250 Ω. 250/.0025, 100/.001 or 100,000 nm. 10^5 x 10^-9 is 10-4 m or 0.1 mm. In area this would be 500 m x 5 k (the two collectors taper, but together they make a rectangle). This is 1/10 of the area of the power sat. Since the dissipation is		W=I^2R so 25=100^2 R, R = 0.0025 Ω/meter. Aluminum is 26.50 nΩ·m So a nanometer thick layer of Al a meter square would measure about 25 Ω of resistance, one 1/10th meter wide would be 250 Ω. 250/.0025, 100/.001 or 100,000 nm. 10^5 x 10^-9 is 10-4 m or 0.1 mm. In area this would be 500 m x 5 k (the two collectors taper, but together they make a rectangle). This is 1/10 of the area of the power sat. Since the dissipation is 250 W/m^2 and the power sat collection surface is 200 W/m^2, 12.5% would be lost to resistive heating in this design.

			The mass would be 500 m x 5000 m x 10^-4 m, 250 cubic meters x 2.7 t/m^3, 675 tons per side or 1350 t. In the context of a 20-40,000 t power sat this is not excessive.

67.170.206.231 at 04:21, 15 October 2008

2008-10-15T04:21:01Z

← Older revision		Revision as of 04:21, 15 October 2008
Line 5:		Line 5:
	5,000,000,000 watts is 10,000 volts times 500,000 amps or perhaps if someone designs very high voltage klystrons, 100,000 volts and 50,000 amps. Using 10,000 volts, the edge of a 5 km square surface delivers 500,000 amps over 5 km, 100,000 amps per km or 100 amps per meter. If we make the current carrying conductor a thin metal foil 0.1 m wide and let it radiate at roughly room temperature, then we can loose 25 watts per meter of length. (A black surface radiates 250 watts/m^2. I am only counting on one side radiating since the side toward the bulk of the power sat would be at near room temp.)		5,000,000,000 watts is 10,000 volts times 500,000 amps or perhaps if someone designs very high voltage klystrons, 100,000 volts and 50,000 amps. Using 10,000 volts, the edge of a 5 km square surface delivers 500,000 amps over 5 km, 100,000 amps per km or 100 amps per meter. If we make the current carrying conductor a thin metal foil 0.1 m wide and let it radiate at roughly room temperature, then we can loose 25 watts per meter of length. (A black surface radiates 250 watts/m^2. I am only counting on one side radiating since the side toward the bulk of the power sat would be at near room temp.)

	W=I^2R so 25=100^2 R, R = 0.0025 ~~ohm~~		W=I^2R so 25=100^2 R, R = 0.0025 Ω/meter. Aluminum is 26.50 nΩ·m So a nanometer thick layer of Al a meter square would measure about 25 Ω of resistance, one 1/10th meter wide would be 250 Ω. 250/.0025, 100/.001 or 100,000 nm. 10^5 x 10^-9 is 10-4 m or 0.1 mm. In area this would be 500 m x 5 k (the two collectors taper, but together they make a rectangle). This is 1/10 of the area of the power sat. Since the dissipation is

67.170.206.231: New page: Sunlight in space is around 1.3 kW/m^2. Photovoltaic cells of 15% efficiency would allow collecting 200W/m^2. A km^2 would produced 0.2 GW, 5km^2 would generate a GW and 25km^2 would gene...

2008-10-14T23:59:39Z

New page: Sunlight in space is around 1.3 kW/m^2. Photovoltaic cells of 15% efficiency would allow collecting 200W/m^2. A km^2 would produced 0.2 GW, 5km^2 would generate a GW and 25km^2 would gene...

New page

Sunlight in space is around 1.3 kW/m^2. Photovoltaic cells of 15% efficiency would allow collecting 200W/m^2. A km^2 would produced 0.2 GW, 5km^2 would generate a GW and 25km^2 would generate 5 GW.

Power on the ground is around .8 x .8 x .8 for the transmitter, transmission and rectenna loses. That's close enough to 0.5, so 5GWe would take about 50 km^2 in space or 2 5x5 km surfaces each delivering 5GW to the transmitter.

5,000,000,000 watts is 10,000 volts times 500,000 amps or perhaps if someone designs very high voltage klystrons, 100,000 volts and 50,000 amps. Using 10,000 volts, the edge of a 5 km square surface delivers 500,000 amps over 5 km, 100,000 amps per km or 100 amps per meter. If we make the current carrying conductor a thin metal foil 0.1 m wide and let it radiate at roughly room temperature, then we can loose 25 watts per meter of length. (A black surface radiates 250 watts/m^2. I am only counting on one side radiating since the side toward the bulk of the power sat would be at near room temp.)

W=I^2R so 25=100^2 R, R = 0.0025 ohm