# Difference between revisions of "Hundred dollars a kg"

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− | If we want to solve the energy crisis using space-based solar power, replacement power needs to come on line at a rate of 400-500 GW/year. That replaces about 6 million barrels of oil a day each year over the next 50 years. To make liquid fuels at a dollar a gallon out of electricity requires that the electricity cost be around a penny a kWh. | + | To sum up what's gone before: |

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+ | If we want to solve the energy crisis using space-based solar power, replacement power needs to come on line at a rate of 400-500 GW/year. That replaces about 6 million barrels of oil a day each year over the next 50 years. To make liquid fuels at a dollar a gallon out of electricity requires that the electricity cost be around a penny a kWh. | ||

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+ | ==What can we afford?== | ||

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+ | Penny a kWh power sells for $80/kW/year. Using a simple payoff in ten years, that requires building power sats for $800 a kW or less. The rectenna estimate is $100-200/kW based on inverters costing $60/kW (same as PC power suppies). We can spend $600-700 per kW on parts, lifting them to GEO and assembly. | ||

If you assume 2kg/kW power sats, then we can spend $350 a kg to buy parts, lift them to GEO and install them. At 4kg per kW, (the upper end of the power sat mass range) we can spend $175 per kg for parts and transport. Assume $75 to $250 a kg for parts (perhaps solar cells) and installation. That leaves $100/kg for transport. | If you assume 2kg/kW power sats, then we can spend $350 a kg to buy parts, lift them to GEO and install them. At 4kg per kW, (the upper end of the power sat mass range) we can spend $175 per kg for parts and transport. Assume $75 to $250 a kg for parts (perhaps solar cells) and installation. That leaves $100/kg for transport. |

## Revision as of 02:17, 30 July 2008

To sum up what's gone before:

If we want to solve the energy crisis using space-based solar power, replacement power needs to come on line at a rate of 400-500 GW/year. That replaces about 6 million barrels of oil a day each year over the next 50 years. To make liquid fuels at a dollar a gallon out of electricity requires that the electricity cost be around a penny a kWh.

## What can we afford?

Penny a kWh power sells for $80/kW/year. Using a simple payoff in ten years, that requires building power sats for $800 a kW or less. The rectenna estimate is $100-200/kW based on inverters costing $60/kW (same as PC power suppies). We can spend $600-700 per kW on parts, lifting them to GEO and assembly.

If you assume 2kg/kW power sats, then we can spend $350 a kg to buy parts, lift them to GEO and install them. At 4kg per kW, (the upper end of the power sat mass range) we can spend $175 per kg for parts and transport. Assume $75 to $250 a kg for parts (perhaps solar cells) and installation. That leaves $100/kg for transport.

Four hundred GW of power sats requires 800,000 tons per year lifted to GEO.

Extra terrestrial resources are a better idea, but there isn't time to develop the space industry to tap them. The economic effects (including famines) of the developing energy crisis are so bad that production of SBSP needs to come on line by 2015 to avoid economic collapse. I expect ET resources be exploited by 2025 with this much activity in space.

This works out to lifting 100 tons per hour. The choices examined here are among rockets, lasers or some combination of them.

Rockets are good for high thrust, but the rocket equation is a hard taskmaster when you need delta V that is a multiple of the exhaust velocity.

Lasers can get very high exhaust velocity, but they don't do very good for thrust. Jordin Kare and others have been beating their heads on this problem for decades trying to design lasers to launch from the surface.

Rockets to lift this much cargo would have to launch every hour, each rocket having a lift off mass of 6,000 tons to get 350 tons to LEO and 100 tons of that to GEO. The projected cost is around $500/kg. http://www.ilr.tu-berlin.de/koelle/Neptun/NEP2015.pdf

The energy payback (for making fuel) is about 40 days.

The needed delta V LEO to GEO is about 4.1 km/sec. Using 10% of the LEO payload for reaction mass (35 tons) and a 40 km/sec exhaust velocity provides about 4.2 km/sec delta V.

(http://en.wikipedia.org/wiki/Rocket_equation )

The power required for 1/20th g (.5 m/sec exp 2) is 1/2 x 350,000 kg x .5m/sec exp 2 x 40,000 m/sec = 3.5 GW

Delta V changes require 4100 m/sec /0.5 m/sec exp 2, or 8200 sec or about 2.7 hours. I.e., 3.5 GW of laser power would raise eight loads a day from LEO to GEO. At 315 t each that is 2500 t/day, somewhat exceeding the 2200 t/day needed to install 400 GW per year.

It also cuts the rocket launches from 24 a day down to a more manageable 7 or 8.

A 3.5-GW space based laser built at GEO with $500/kg parts that massed 10,000 tons would cost $5 billion. It can be expected to more than triple the yearly throughput to GEO, saving 2/3 of 0.8 billion kg x $500/kg or $267 billion a year in transport costs.

Amortized at 10 percent, the additional lift from LEO to GEO would cost $0.5 billion/0.8 billion/kg or $0.63 per kg plus the lift to LEO.

However, there is another way that cuts the liftoff mass by a factor of five.

Injection to geosynchronous transfer orbit is 7.8 k/s (to LEO) plus 2.5 k/s (to GTO), totaling 10.3 k/s. http://en.wikipedia.org/wiki/Delta-v_budget To get to LEO takes 796 s at 1g, 13.2 minutes. At one g 17.5 minutes for GTO, 14 minutes at 1.25 g, plus 2.2 minutes to circularize the orbit (1.6 k/s) at GEO. Assuming half payload and half reaction mass, 0.69 velocity ratio and 11.9 k/s delta v, then the average exhaust velocity needs to be a modest (for lasers) 13.2 k/s.

Because the laser can be cycled close to 4 times an hour, the payload only needs to be 25 tons. Taking the midpoint (the exhaust velocity would be varied for constant thrust) the power required for 1.25 g (12.25 m/sec exp 2) is 1/2 x 37,500 kg x 12.25 m/sec exp 2 x 13,200 m/sec = 6 GW. In the context of building a GW of power sat a day, this is a small piece of hardware.

Of course, the 50-ton payload-plus-reaction-mass has to hang in space long enough to be accelerated. This takes a modest mass ratio rocket.

50 tons payload plus reaction mass. 16.5% 50 tons rocket structure 16.5% 200 tons propellant. 66%

The rocket would lift off on two SSMEs, and if a zeroth turbofan stage was used, it would take eight 50-ton thrust engines, perhaps with afterburners. Simplifying operations, it goes straight up and lands back at the launching site. If the payload is oriented at right angles to vertical, we could avoid wasting time reorienting it for laser acceleration.

This "little" rocket has a mass ratio of two and a delta V of 4 k/sec (less gravity loss and drag). The payload-plus-reaction-mass exits the atmosphere at 2.1 k/sec, goes up to 260 miles, falls to 150 miles before picking up orbital velocity and falls to 55 miles (picking up a little air drag) before losing downward velocity. It enters GTO 1000 seconds after launch. This is conservative because I am not accounting for the curvature of the earth.

It masses 1/20th of a Neptune and (with the aid of the laser) delivers 1/4 as much payload per launch. One part in 12 (rather than one part in 60) is payload. To keep the laser busy and to meet the 100-ton per hour cargo requirement requires a launch every 15 minutes. This rate is common for airlines, but not for spacecraft.

If the laser cost ten billion dollars and was written off at 10% per year, the lift cost from sub orbital to GEO would be $1 B /0.87 billion kg or slightly over a dollar per kg. If cost scales with liftoff mass in this range, then the cost of a kg of payload delivered to GEO will be about $100/kg, thus meeting our penny a kWh and dollar a gallon fuel goals.

Energy payback is 8 days for the rocket fuel and 6 days for the laser. This is about 10% of the theoretical minimum for a space elevator.

This is probably not the optimum design. A shorter "hang time," smaller payloads and higher but shorter acceleration may yield lower cost per kg. (It is possible that laser launch from the ground is the end point of this optimization.) However, the size of the largest part for a power satellite may limit the minimum size of rocket you can use.