Question 11.AE.7: a) The energy of a Pout=5 MW wind-power plant can be used t......
a) The energy of a P_{out}=5 MW wind-power plant can be used to generate hydrogen during 8 hours of operation per day during 365 days per year. The construction cost of the 5 MW plant and the generation of pure water including the electrolysis equipment is $4000/kW installed power capacity. How much hydrogen energy (expressed in MWh) can be obtained during one year taking into account the above-described losses?
b) Calculate the payback period provided one kg-force of hydrogen (weight) can be sold for $4.00. For your payback period calculation you may neglect interest for borrowing the money for the construction cost.
c) How many “equivalent-gasoline gallons” of hydrogen can be produced during one year?
d) Provided a car owner travels 15,000 miles per year with an average mileage of 40 miles/(gallon “equivalent gasoline”), how many car owners can get the “equivalent gasoline gallons of hydrogen” from the 5 MW plant?
a) The hydrogen that can be produced within one year is given by the total energy available from the wind turbine E_{total}=(5 MW)(8 h/day)(365 days/year)=14,600 MWh. The efficiency of electrolysis reduces this energy to E_{useful}=0.8\ E_{total}=11,680 MWh. The energy content of 1 kg-force of hydrogen is E_{H2}=28 kWh/(kg-force), thus the amount of hydrogen produced is H_2=E_{useful}/E_{H2}=417,143 (kg-force). The atomic weight of hydrogen (H_2) is 2 and that of water (H_2O) is 18, therefore the amount of H_2O required for the electrolysis is H_2O=417,143/(2/18)=3.754°10^6 kg-force. The energy loss E_{distillation} due to the production of distilled water is obtained from E_{1kg-forceH2O} = (90°C) (4.186kWs/°C)(1h/3600s)= 0.10465kWh as E_{distillation}= (3.754.10^6 kg-force)(0.10465 kWh/kg-force)=392.9 MWh. The energy available for the electrolysis is now E_{H2}=E_{useful}-E_{distillation}=11,680 MWh-392.9 MWh= 11,287 MWh.
b) The payback period -neglecting interest payments- is based on the earnings per year and the construction cost. For H_2=E_{H2}/(E_{H2\ per\ kg-force})=(11,287.10^6 Wh)/ (28kWh/kg-force)=403,107 kg-force per year the earnings are Earnings/year= (403,107 kg-force/year)($4.00/kg-force)=$1,612,428/year. The construction cost is C_{construction}=($4000)(5,000 kW)=$20,000,000 resulting in a payback period of y=($20,000,000/$1,612,428)=12.4 years.
c) With E_{H2}=11.287106 kWh, the specific weight of gasoline of γ_{gasoline}=0.72 kgforce/liter=2.736 kg-force/gallon, and the energy content of gasoline E_{gasoline}= 12.3 kWh/kg-force the weight of gasoline in kg-force is Weight_{gasoline}=11.287.10^6 kWh/(12.3kWh/kg-force)=917,642 kg-force or the equivalent gallons of gasoline produced per year are Gallons_{gasoline}=917,642 kg-force/(2.736 kg-force/gallon) =335,396 equivalent gallons of gasoline per year.
d) Each car owner uses (15,000 miles/year)/(40 miles/gallon)=375 gallons/year, that is, [335,396 gallons/year]/{375 gallons/(year/owner}=895 owners. In other words the 5 MW wind-power plant produces enough hydrogen to meet the needs of 895 car owners.