In the year 2004 the USA produced 1787 TWh of electrical energy in conventional thermal plants and 476 TWh in nuclear plants. Assuming 30% efficiency for nuclear plants and 40% for conventional thermal plants, determine the (annual) volume of cooling water required to cool these plants in once-through cooling if the cooling water undergoes a temperature increase of 10°C. (Neglect the heat lost up the chimney in conventional plants, and assume two significant digits in given data.)
η=Useful (electrical) energy/total energy in
1 TWh=10^9 KWh
1 kWh = 3.6*10^6 J
The Attempt at a Solution
I did each plant one separately (of course) and converted the 1787 TWh of the thermal plant to Joules so that I could use it in the specific heat equation. When I converted it to Joules I got 9.5*10^24 Joules. I took that number and plugged it into my specific heat equation to get 2.3*10^20 kg. I plugged that mass into the Volume equation and got a volume of 2.3*10^17 m^3
I did the same steps for the nuclear plant and got 4*10^16 m^3.
Because the book only has one answer, I assumed they added the two up and got 2.7*10^17 m^3. The answer in the book is 2.7*10^11 m^3. So I'm close, but my exponent is off. I'm pretty sure I'm converting it correctly from TWh to Joules.
I also found that when I converted the TWh to MJ instead of Joules, I get the right answer, but I can't have it in MJ because the specific heat equation uses 4186 Jkg-1K-1