A gigawatt (GW) is of course 1000 megawatts (MW), 1 million (10^6) kilowatts (kW), 1 billion (10^9) joules per second (J/s). This is a few times more than a typical large scale power plant puts out.

In BTU units, a GW is about 3.4 Gbtu/hour, and about 1 quad (10^15 BTU) in 33 years.

Here are some calculations regarding a gigawatt:

- How much water vapor carries a GJ? Using enthalpy of evaporation from a program based on the 1967 ASME saturated steam tables, and the GNU units program,
- At 373.15 K (100 °C)
- 1 GJ/(2256.91780000 J/g) = 443.08215 kg

- At 300 K (26.85 °C)
- 1 GJ/(2438.17925840 J/g) = 410.14212 kg

- At 275 K (1.85 °C)
- 1 GJ/(2497.22337280 J/g) = 400.44475 kg

- At 373.15 K (100 °C)

- To carry a GW, over 400 kg of water vapor must flow every second. That would be ok, except that cold water vapor takes a very large volume. Water is not a suitable refrigerant for heat syphons near 0 °C.

- How much propane vapor carries a GJ? Using enthalpy of evaporation from http://www.care-refrigerants.co.uk/,
- At 300 K (26.85 °C) (value for 25 °C)
- liquid: 265.5 kJ/kg vapor: 600.8 kJ/kg
- 1 GJ/((600.8 - 265.5) kJ/kg) = GJ/(335.3 kJ/kg) = 2982.4038 kg

- At 275 K (1.85 °C) (value for 0 °C)
- liquid: 200.0 kJ/kg 528.7 kg/m^3 vapor 574.7 kJ/kg 10.36 kg/m^3
- 1 GJ/(374.7 kJ/kg) = 2668.8017 kg

- At 300 K (26.85 °C) (value for 25 °C)

- To carry a GW, over ? kg of propane must flow every second. The volume does not vary nearly as much as water does over the relevant temperature range.

- How much ideal gas at 300K carries a GJ?
- What diameter pipeline carries a GW of ideal gas at 300K?
- How much should a GW plant cost?

- Coal fired plants cost around $500 to $1000 per kW, or $500 million to $1 billion. A plant requiring no fuel would be competitive at a higher construction cost, depending on the expected useful life of the plant.

- How much gas expanding isothermally absorbs a GW of heat and delivers a GW of work?

- Since the temperature is constant, the heat capacity of the gas is irrelevant. Isothermal expansion is assured if the resistance to expansion is precisely hyperbolic, so that as the gas expands and cools, expansion stops until sufficient heat flows in to overbalance the resistance. The power depends on how fast heat can flow in, which does not appear to depend directly on the amount of gas. With a higher $ \Delta T $, heat flows faster. It is easier to deliver lots of heat into a large volume (given lots of equipment to carry the heat).

- How much liquid piston absorbs a GW of heat from isothermally expanding gas as gravitational potential?

- Smaller amounts of liquid piston raised a larger distance absorb the same amount of energy as larger amounts raised a shorter distance. The power depends also on the flow rate. Smaller amounts would pick up more kinetic energy which would need to be recovered, perhaps by closing a valve as the liquid reaches maximum height and stops moving.

- Assume the liquid flows slowly enough that kinetic energy is negligible.

- This requires larger amounts of liquid rising smaller distances. Suppose by negligible, we mean under 0.1% of the energy appears at kinetic energy. The kinetic energy depends on mass and speed squared. The gravitational potential energy depends on mass and height. The ratio is independent of mass, and negligible kinetic energy thus means $ 500v^2 < gh, v < \sqrt{\frac{gh}{500}} = 0.14\sqrt{h} $, for $ h $ in meters and $ v $ in meters/second. If $ h $ is 100 meters, $ v $ must be less than 1.4 m/s for the kinetic energy to be less than 0.1% of the potential energy.

/usr/bin/units version 1.80 units-1.80-12.src.rpm /usr/share/units.dat Version 1.34 16 June 2002

sqrt(meter gravity / 500) -> 0.14004749 m / s sqrt(100 meter gravity / 500) -> 1.4004749 m / s