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Re: [Phys-l] Earthquake/Tsunami in Japan



Cal = 1000 X cal
so 7.3E7 kg/day = 7.3E7 / (24X60X60) kg/sec water consumption
leads to 7.3E7/(24X60X60) kg/sec X 540 Cal/kg X 4200 J/Cal
= 1.9GW For reasonable thermodynamic efficiency,
this implies a useful output higher than the 685MW (Pilgrim)
that I had in mind.

Brian W



On 3/20/2011 9:02 AM, brian whatcott wrote:
2E7 (US) gallons of water is 1.6E8 lbs or 7.3E7kg
At 540 Cal/kg that's 1.6E11 joules/day,
a rate of 1.9MW.
Dunno the production rate at these stations, possibly 685MW like Pilgrim??

Brian W

On 3/20/2011 2:03 AM, Hugh Haskell wrote:
At 23:42 -0700 03/19/2011, Bernard Cleyet wrote:
I thought it was a much smaller amount recycled,
as is in an automobile's radiator.

The water drawn into the cooling tower from a
reservoir is sprayed over the pipes to condense
the steam from the low-pressure side of the
turbine, is mostly evaporated. So the water usage
of 20 million gallons a day is evaporated water.
/snip/