Why the mass of steam formed during condensation is equal to 9H2

[Elz]

1. Why the mass of steam formed during condensation is equal to 9H₂

While studying about the lower calorific value of fuel the equation is
L.C.V.= H.C.V. – heat of steam formed during combustion

Now here in the book they have written that,
Let m_s = mass of steam formed in per kg of fuel = 9H₂

Since the amount of heat per kg of steam is the latent heat of vaporization of water
corresponding to standard temperature of 15°C, is 2466 kJ/kg,

∴ L.C.V. = H.C.V.- m_s X 2466 kJ/kg
= H.C.V. - 9H₂ X 2466 kJ/kg …(∵ m_s = 9H₂)

What i don't understand is why this m_s is equal to 9H₂? what is the chemical reaction here? Why this value is considered constant for all fuels? can anyone please explain?

 

[Simon Bridge]

I also found:
http://books.google.co.nz/books?id=...q=9H2 steam for each kilogram of fuel&f=false
The mass of steam produced per kilogram of fuel is $9H_2 + m$ where $m$ is the mass of moisture per kilogram of fuel ... $H_2$ is the mass of hydrogen per kilogram of fuel

The steam comes from the combustion of hydrogen in oxygen to make water.
$H_2$ is roughly 1/9th the mass of water molecules.

 

[Elz]

I also found:
http://books.google.co.nz/books?id=...q=9H2 steam for each kilogram of fuel&f=false
The mass of steam produced per kilogram of fuel is $9H_2 + m$ where $m$ is the mass of moisture per kilogram of fuel ... $H_2$ is the mass of hydrogen per kilogram of fuel

The steam comes from the combustion of hydrogen in oxygen to make water.
$H_2$ is roughly 1/9th the mass of water molecules.

So 1 kg of H₂ is equivalent to 9 kg of water or steam. So if our fuel has 5 kg H₂ then 5X9 kg of steam will be formed. And if we condense it we will get this massX latent heat of vaporization of water=heat of steam formed during condensation.
Did I get it right?

 

[Simon Bridge]

So 1 kg of H2 is equivalent to 9 kg of water or steam. So if our fuel has 5 kg H2 then 5X9 kg of steam will be formed. And if we condense it we will get this massX latent heat of vaporization of water=heat of steam formed during condensation.

That's what I read - yep.

Careful - the ratio is approximate. It assumes all the hydrogen in the fuel gets used to make water - i.e. a very clean burn. The steam is at the boiling point. Stuff like that.
Good enough for a back-of-envelope calculation.

 

[Elz]

Thanks for the help.. :)

 

[Simon Bridge]

No worries.