Exam 20: Hydrogen

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$$\text { (a) The power required is } 10 ^ { 5 } \times 6 \mathrm {~kW} = 6 \times 10 ^ { 8 } \mathrm {~W} \text {. At } 60 \% \text { efficiency the total }$$ power required $6 \times 10 ^ { 8 } \mathrm {~W} / 0.6 = 10 ^ { 9 } \mathrm {~W}$ . For one month the energy requirement is
$$\left( 10 ^ { 6 } \mathrm {~kW} \right) \times ( 30 \mathrm {~d} / \text { month } ) \times ( 24 \mathrm {~h} / \mathrm { d } ) = 7.2 \times 10 ^ { 8 } \mathrm { kWh }$$
Converting this to Joules gives
$$\left( 7.2 \times 10 ^ { 8 } \mathrm { kWh } \right) \times \left( 3.6 \times 10 ^ { 6 } \mathrm {~J} / \mathrm { kWh } \right) = 2.59 \times 10 ^ { 15 } \mathrm {~J} \text { per month }$$
For an energy content of $142 \mathrm { MJ } / \mathrm { kg }$ , the amount of hydrogen required to provide one month of energy is
$$\left( 2.59 \times 10 ^ { 15 } \right) / \left( 1.42 \times 10 ^ { 8 } \mathrm {~J} / \mathrm { kg } \right) = 1.82 \times 10 ^ { 7 } \mathrm {~kg} \text { hydrogen per month. }$$
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) $\rho = \frac { M P } { R T }$ 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 $1 \mathrm { MPa } = 1.0 \times 10 ^ { 6 } \mathrm {~Pa}$ . Thus equation (20.12) may be written as
$$\begin{aligned}\rho & = \left[ \left( 0.002 \mathrm {~kg} \cdot \mathrm { mol } ^ { - 1 } \right) \times \left( 1.0 \times 10 ^ { 6 } \mathrm {~kg} \cdot \mathrm { m } ^ { - 1 } \cdot \mathrm { s } ^ { - 2 } \right) \right] / \left[ \left( 8.315 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 } \cdot \mathrm { s } ^ { - 2 } \cdot \mathrm { K } ^ { - 1 } \cdot \mathrm { mol } ^ { - 1 } \right) \times ( 293 \mathrm {~K} ) \right] \\& = 0.82 \mathrm {~kg} / \mathrm { m } ^ { 3 }\end{aligned}$$
Thus the volume required for $1.82 \times 10 ^ { 7 } \mathrm {~kg}$ of hydrogen is
$$\left( 1.82 \times 10 ^ { 7 } \mathrm {~kg} \right) / \left( 0.82 \mathrm {~kg} / \mathrm { m } ^ { 3 } \right) = 2.21 \times 10 ^ { 7 } \mathrm {~m} ^ { 3 }$$
(b) In a square array, a $3 \mathrm {~m}$ diameter tank will occupy $3 \times 3 = 9 \mathrm {~m} ^ { 2 }$ 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 $7 \times 5 = 35$ $\left( 2.21 \times 10 ^ { 7 } \mathrm {~m} ^ { 3 } / 35 \mathrm {~m} ^ { 3 } \right) \times \left( 9 \mathrm {~m} ^ { 2 } \right) = 5.68 \times 10 ^ { 6 } \mathrm {~m} ^ { 2 }$ of land area.
This is $5.68 \mathrm {~km} ^ { 2 }$ or $5.68 \%$ of the total land area.

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From Figure 20.3, it is seen that the density at at 35 MPa and 70 MPa is $$24 \mathrm {~kg} / \mathrm { m } ^ { 3 }$$ $$\text { and } 36 \mathrm {~kg} / \mathrm { m } ^ { 3 } \text {, respectively. The increase in density (and energy density) is }$$ $$\left( 36 \mathrm {~kg} / \mathrm { m } ^ { 3 } - 24 \mathrm {~kg} / \mathrm { m } ^ { 3 } \right) / 24 \mathrm {~kg} / \mathrm { m } ^ { 3 } = 0.5 \text { or } 50 \%$$

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$$\text { The atomic mass of } \mathrm { Li } \text { is } 7 \mathrm { u } \text {, so } \mathrm { LiH } \text { has a hydrogen content of }$$ $$100 \times ( 1 \mathrm { u } ) / ( 7 \mathrm { u } + 1 \mathrm { u } ) = 12.5 \%$$
Therefore $1 \mathrm {~kg}$ of $\mathrm { LiH }$ contains
$$( 1 \mathrm {~kg} ) \times ( 0.125 ) = 0.125 \mathrm {~kg} \text { hydrogen }$$
The energy content of hydrogen is $142 \mathrm { MJ } / \mathrm { kg }$ so the energy density of $\mathrm { LiH }$ is
$$( 0.125 \mathrm {~kg} / \mathrm { kg } ) \times ( 142 \mathrm { MJ } / \mathrm { kg } ) = 17.75 \mathrm { MJ } / \mathrm { kg }$$
The energy content of gasoline is $34.8 \mathrm { MJ } / \mathrm { L }$ and its density is about $0.72 \mathrm {~kg} / \mathrm { L }$ so the energy density is
$$( 34.8 \mathrm { MJ } / \mathrm { L } / ( 0.72 \mathrm {~kg} / \mathrm { L } ) = 48.3 \mathrm { MJ } / \mathrm { kg }$$
The relative energy density of $\mathrm { LiH }$ compared to gasoline is
$$100 \times ( 17.75 \mathrm { MJ } / \mathrm { kg } ) / ( 48.3 \mathrm { MJ } / \mathrm { kg } ) = 37 \%$$