(a) Given a fuel cell efficiency of 75% and a required energy to the wheels of $0.6 \mathrm { MJ } / \mathrm { km }$, calculate the mass of hydrogen fuel required to give a vehicle a range of 300 $\mathrm { km }$.
(b) Calculate the fuel volume if hydrogen is stored in the form of (i) liquid hydrogen, (ii) $\mathrm { CHG }$ at $50 \mathrm { MPa }$, and (iii) $\mathrm { TiH } _ { 2 }$.

Question

Quizplus · Accepted Answer

The Answer is (a) The input energy for the fuel cell is (0.6 MJ/km)/(0.75) = 0.8 MJ/km. Since the energy content of hydrogen in

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

, then the hydrogen requirement is

( 0.8 \mathrm { MJ } / \mathrm { km } ) / ( ( 142 \mathrm { MJ } / \mathrm { kg } ) = 0.0056 \mathrm {~kg} / \mathrm { km }

A range of

300 \mathrm {~km}

would require

( 300 \mathrm {~km} ) imes ( 0.0056 \mathrm {~kg} / \mathrm { km } ) = 1.69 \mathrm {~kg}

(b.i) The density of liquid hydrogen is

71 \mathrm {~kg} / \mathrm { m } ^ { 3 }

, giving a volume of

( 1.69 \mathrm {~kg} ) / \left( 71 \mathrm {~kg} / \mathrm { m } ^ { 3 } ight) = 0.024 \mathrm {~m} ^ { 3 }

(b.ii) For compressed hydrogen gas at

50 \mathrm { MPa }

, Figure

20.3

shows that the density is about

29 \mathrm {~kg} / \mathrm { m } ^ { 3 }

, for a volume of

( 1.69 \mathrm {~kg} ) / \left( 29 \mathrm {~kg} / \mathrm { m } ^ { 3 } ight) = 0.058 \mathrm {~m} ^ { 3 }

(b.iii) From Table

20.3,18 \mathrm {~kg}

of hydrogen requires

0.1 \mathrm {~m} ^ { 3 }

of

\mathrm { TiH } _ { 2 }

. So

1.69 \mathrm {~kg}

hydrogen would require

\left( 0.1 \mathrm {~m} ^ { 3 } ight) imes ( 1.69 \mathrm {~kg} / 18 \mathrm {~kg} ) = 0.009 \mathrm {~m} ^ { 3 }

(A) Given a Fuel Cell Efficiency of 75% and a Required