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Notebook1-Visitor
1-Visitor
February 14, 2013
Question

Power required to vapourise Liquid Oxygen.

  • February 14, 2013
  • 1 reply
  • 3523 views

Hello, this is my first post, I hope its in the correct forum!

As the title says I'm interested in the power needed to vapourise LOX(Liquid Oxygen).

Specifically the physics involved in an old UK missile/rocket called Blue Streak;

http://www.spaceuk.org/bstreak/bstreak.htm

I've done some work with Mathcad.

http://i89.photobucket.com/albums/k207/Notebook_04/Lox_tank_31-2_zps7ef40269.jpg

I have a value of 376Kw for this, but whats worrying me is the figure for Gas_Mass_Lox, the gaseous Lox from the heat-exchanger. Its completely different from HE_Lox_Flow. thats the mass going into the heat-exchanger.

I'm not sure any of the work I've done is accurate, just following some text books and using the formulas.

Note: The term "ullage" refers to the empty volume at the top of a tank full of liquid. It comes from the brewery trade I believe...

All help appreciated.

N.

Ther should be a .mcd zip file attached.

Message was edited by: George Douglass editing photobucket image link

1 reply

F
Fred_Kohlhepp23-Emerald I
23-Emerald I
February 14, 2013

A quick web search reveals that the heat of vaporization of oxygen is 2.13 x 10^5 Joules/kg (~51 cal/gm).

This means that vaporizing a gm of liquid O2 requires 51 calories.

N
Notebook1-VisitorAuthor
1-Visitor
February 14, 2013

Thanks for that Fred Kohlhepp.

I did put the latent heat of vaporisation in my working (LHOV = 213.0kJ/kg) in my working. I'm not sure its valid in this project?

I guess I could find out how 51 calories per gram will convert to power in Watts for a certain mass of lox?

My basic problem is my working knowledge of gases and heat, and physics in general is poor.

This was started from curiosity, just to find how much power was needed from the heat exchanger to keep the liquid oxygen tank pressurised during the rocket's flight.

I got a figure, 300KW+ don't know if its within an order of magnitude, seems a bit large? No experience with 1960's rockets!

However. the odd thing from this is the difference in the two mass figures either side of the heat-exchanger. Obviously thats wrong, but is it in the method used, or have I missed something fundamental?

Nice to get a reply, I tried posting on three Physics forums and didn't get any replies at all!

Thanks, George.

F
Fred_Kohlhepp23-Emerald I
23-Emerald I
February 15, 2013

power is cal/sec, so your 300kW converts to a flow rate, not a total mass

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