For Stationary Applications, Mass and Volume Considerations for a Fuel$60 \%$

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ext { (a) The power required is } 10 ^ { 5 } imes 6 \mathrm {~kW} = 6 imes 10 ^ { 8 } \mathrm {~W} ext {. At } 60 \% ext { efficiency the total }

power required

6 imes 10 ^ { 8 } \mathrm {~W} / 0.6 = 10 ^ { 9 } \mathrm {~W}

. For one month the energy requirement is

\left( 10 ^ { 6 } \mathrm {~kW} ight) imes ( 30 \mathrm {~d} / ext { month } ) imes ( 24 \mathrm {~h} / \mathrm { d } ) = 7.2 imes 10 ^ { 8 } \mathrm { kWh }

Converting this to Joules gives

\left( 7.2 imes 10 ^ { 8 } \mathrm { kWh } ight) imes \left( 3.6 imes 10 ^ { 6 } \mathrm {~J} / \mathrm { kWh } ight) = 2.59 imes 10 ^ { 15 } \mathrm {~J} ext { per month }

For an energy content of

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

, the amount of hydrogen required to provide one month of energy is

\left( 2.59 imes 10 ^ { 15 } ight) / \left( 1.42 imes 10 ^ { 8 } \mathrm {~J} / \mathrm { kg } ight) = 1.82 imes 10 ^ { 7 } \mathrm {~kg} ext { hydrogen per month. }

As hydrogen obeys the ideal gas law at

1 \mathrm { MPa }

, the density will follow from the calculation in Example 20.2. We use equation (20.12)

ho = \frac { M P } { R T }

where the molecular mass for hydrogen is

2 \mathrm {~g} / \mathrm { mol } , R = 8.315 \mathrm {~J} \cdot \mathrm { K } ^ { - 1 } \cdot \mathrm { mol } ^ { - 1 } , T = 293 \mathrm {~K}

and

1 \mathrm { MPa } = 1.0 imes 10 ^ { 6 } \mathrm {~Pa}

. Thus equation (20.12) may be written as

\begin{aligned} ho & = \left[ \left( 0.002 \mathrm {~kg} \cdot \mathrm { mol } ^ { - 1 } ight) imes \left( 1.0 imes 10 ^ { 6 } \mathrm {~kg} \cdot \mathrm { m } ^ { - 1 } \cdot \mathrm { s } ^ { - 2 } ight) ight] / \left[ \left( 8.315 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 } \cdot \mathrm { s } ^ { - 2 } \cdot \mathrm { K } ^ { - 1 } \cdot \mathrm { mol } ^ { - 1 } ight) imes ( 293 \mathrm {~K} ) ight] \& = 0.82 \mathrm {~kg} / \mathrm { m } ^ { 3 }\end{aligned}

Thus the volume required for

1.82 imes 10 ^ { 7 } \mathrm {~kg}

of hydrogen is

\left( 1.82 imes 10 ^ { 7 } \mathrm {~kg} ight) / \left( 0.82 \mathrm {~kg} / \mathrm { m } ^ { 3 } ight) = 2.21 imes 10 ^ { 7 } \mathrm {~m} ^ { 3 }

(b) In a square array, a

3 \mathrm {~m}

diameter tank will occupy

3 imes 3 = 9 \mathrm {~m} ^ { 2 }

of land area but will have a cross sectional area of

\pi ( 1.5 \mathrm {~m} ) ^ { 2 } = 7.065 \mathrm {~m} ^ { 2 }

and have a volume of about

7 imes 5 = 35

\left( 2.21 imes 10 ^ { 7 } \mathrm {~m} ^ { 3 } / 35 \mathrm {~m} ^ { 3 } ight) imes \left( 9 \mathrm {~m} ^ { 2 } ight) = 5.68 imes 10 ^ { 6 } \mathrm {~m} ^ { 2 }

of land area.
This is

5.68 \mathrm {~km} ^ { 2 }

or

5.68 \%

of the total land area.

For Stationary Applications, Mass and Volume Considerations for a Fuel $60 \%$