Consider the possibility of extracting energy from 1 m3 of water. One
approach would be lift the water to some elevation and then to generate electricity hydroelectrically (i.e., pumped hydroelectric storage). The second approach would be to produce hydrogen from the water by electrolysis and then generate electricity from the hydrogen in a proton exchange membrane fuel cell. Using typical efficiencies as given in Chapters 18 and 20, how high would the cubic meter of water have to be lifted to provide the same total electrical energy output as the fuel cell?

Question

Quizplus · Accepted Answer

The Answer is The gravitational potential energy is

E = \eta m g h

where

\eta

is the efficiency of the turbine/generator, typically about

0.85

. So solving for

h

for a

1 \mathrm {~m} ^ { 3 } = 1000 \mathrm {~kg}

parcel of water gives

h = E / ( \eta m g ) = \left( 1.2 imes 10 ^ { - 4 } \mathrm {~s} ^ { 2 } \mathrm {~kg} ^ { - 1 } \mathrm {~m} ^ { - 1 } ight) E

The energy available from a proton exchange membrane fuel cell is obtained from the energy content of the hydrogen obtained in

1000 \mathrm {~kg}

of water. The molecular mass of water is

18 \mathrm {~g} / \mathrm { mol }

and the molecular mass of

\mathrm { H } _ { 2 }

is

2 \mathrm {~g} / \mathrm { mol }

, so water contains (by mass)

100 imes ( 2 \mathrm {~g} / \mathrm { mol } ) / 18 \mathrm {~g} / \mathrm { mol } ) = 11.1 \% ext { hydrogen }

So

1000 \mathrm {~kg}

of water contains

( 100 \mathrm {~kg} ) imes ( 0.111 ) = 111 \mathrm {~kg}

hydrogen. The energy content of hydrogen is

142 \mathrm { MJ } / \mathrm { kg }

so the total energy content of the hydrogen in

1000 \mathrm {~kg}

water is

( 111 \mathrm {~kg} ) imes ( 142 \mathrm { MJ } / \mathrm { kg } ) = 1.58 imes 10 ^ { 10 } \mathrm {~J}

We assume that electrolysis is about

70 \%

efficient and the efficiency pf a PEM fuel cell is about

50 \%

for an overall efficiency of

( 0.7 ) imes ( 0.5 ) = 0.35

or

35 \%

. So the net energy available from the fuel cell is

( 0.35 ) imes \left( 1.58 imes 10 ^ { 10 } \mathrm {~J} ight) = 5.53 imes 10 ^ { 9 } \mathrm {~J}

Therefore the height required for hydroelectric generation is

h = \left( 1.2 imes 10 ^ { - 4 } \mathrm {~s} ^ { 2 } \mathrm {~kg} ^ { - 1 } \mathrm {~m} ^ { - 1 } ight) imes \left( 5.53 imes 10 ^ { 9 } \mathrm {~J} ight) = 6.64 imes 10 ^ { 5 } \mathrm {~m}

Not reasonable.

Consider the Possibility of Extracting Energy from 1 M3 of Water