What are the transformations of energy in a geothermal power plant?

Nuclear power plant produces 4.505 * 10^15 J of energy per year.?

• I know this is a lot, but help would greatly be appreciated. I am so lost. How much power (Watts) does that plant produce if it operates 24/7? (I think that answer is 1.238*10^27 Watts) How much electrical energy does the plant produce if the turbine converts thermal energy to electricity with a 35% efficiency? (I multiplied Watts by 0.38 to get 4.7044*10^26 Watts but not positive that's how you do it) If turbine generator produces current of 1000 Amps, what is the voltage output? If the transmission is only 97% efficient in converting voltage, how much power is lost (Watts)? What would the transmission loss be if you stepped it up to 200,000?

• Answer:

  OK they have stuck you with some really awkward units here. I think that's part of the test. They want to see if you're smart enough to convert to normal sensible units. So... 4.505 x10^15 joules/year divided by 31,536,000 secs/year = 142,852,612 joules/sec = 142,852,612 WATTS. Now apply a "sanity check" to your conversion. Do the numbers actually fit the real world? Here is what we know. The nuclear industry measures reactor heat in megawatts (thermal), or MW(t). They convert at about 33% efficiency give or take --35% is realistic. They measure electricity in megawatts electric, or MW(e). Nuclear plants start at about 10 MWe, become commercially profitable at 200 MWe and the biggest are 1500 MWe. That's actual reality. The industry would say 142.85 Megawatts, thermal. = 142.85 MWt. Now convert to electric. 142,852,162 W(t) x .35 = 49,998,414 W(e). Round it and you have 50 MWe. Now you sanity check. Do these numbers make SENSE? Losing about 2/3 of your energy going from heat to electric - yeah that's typical. 50 MWe, is that a sensible reactor size? If not, go back and check your math. Generator Voltage. Watts = Volts x Amps. 50,000,000 W(e) = X volts * 1000 amps. Easy = X = 50,000 volts. Sanity check it. Is that a reasonable voltage for a generator? http://www.google.com/search?q=nuclear+reactor+main+generator&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a#sclient=psy-ab&hl=en&client=firefox-a&hs=34j&rls=org.mozilla:en-US%3Aofficial&source=hp&q=power+plant+main+generator++voltage&pbx=1&oq=power+plant+main+generator++voltage&aq=f&aqi=&aql=&gs_sm=e&gs_upl=7522l8880l0l9053l12l12l0l0l0l5l167l1330l5.7l12l0&bav=on.2,or.r_gc.r_pw.,cf.osb&fp=7d7a6dcb45a38022&biw=755&bih=878 It seems a little high for me, but close enough for a story problem. If your transmission is 97% efficient it means your transmission loss is 3%. 3% of total watts is what you lost. You can do the math. 50MW x 3% = 1.5 MW. What would transmission loss be if you stepped up the voltage? That's a much more complex problem. I can't believe it's part of the same story problem. Transmission loss is (presumably) resistive loss. E=IR. E (volts of LOSS) = I (current in amps) x R (resistance in ohms). Volts of loss is 3% of 50,000, or 1500. Current is 1000. R we need for an intermediate calculation. Solve. 1500 = 1000 * R R = 1.5 ohms. W=EI. W = 50,000,000. E (volts) they just told us is 200,000. Solve for I and get 250. Transmission loss now would be E=IR E (volts loss) = I (250) * R which we computed before, 1.5. I get 375 volts loss. Now you have your volts loss, you find your wattage loss: W=EI W (watts) = E (volts loss from just above) x I (amps) = 375 * 250 = 93750 watts. Or 0.093 MW. Much better. And that's why they step up voltage for long distance transmission.