Weird Steam


#1

So Bernoulli said that v=sqrt(2gh), meaning that velocity of an object is the square root of twice gravitational acceleration times its height. I can buy that, Herr Bernoulli has never lied to me before, and pitot tubes do work. But then one Thomas D. Morton, CEng, FIMarE, Extra First Class Engineer, has written in his book that v=sqrt(2gH), meaning apparently that velocity of steam is the square root of twice gravitational acceleration times the “heat drop” over the nozzle. No further explanation offered. Maybe I can be convinced that heat drop is mathematically like height drop, because potential energy is uncomfortable like that. But why does gravity matter? Wouldn’t a steam turbine work the same way on the moon? Don’t rocket nozzles depend on the same principle? Am I being lied to?


#2

Dunno if this is any help: https://sachinchaturvedi.files.wordpress.com/2012/01/unit-4-flow-through-nozzles.pdf


#3

Yes, actually, thank you. Should be the steady flow energy equation, not Bernoulli.

V2
or, in other words: v=sqrt(2’dh’), where ‘dh’ is delta h or the change in entropy.

I guess you could understand that as a simple typo, if they didn’t double down and define ‘g’ as 9.81 m/s^2.

How does the latest edition of a book that was first published in the 50s have such an nonsensical error?

On the other hand, the similarity between the steady flow energy equation and Bernoulli is striking. I think that Bernoulli is really just a special case of the SFEE, the wrong special case for steam nozzles. So… gravity is a kind of venturi…?


#4

Was the use of “g” meant as an analogy?


#5

No, unfortunately.

The velocity of the steam leaving the nozzle may then be calculated from the following formula: V=sqrt(2gH) where H is the heat drop of the steam as it passes through the nozzle in J/kg, where g is the acceleration due to gravity, i.e. 9.81 m/s^2, and V is the leaving velocity in m/s

That’s all it says about it. This book is far too bare bones to make analogies. But its ok. Now there is this interesting thread to pull on with this gravity-as-venturi idea.


#6

Just out of curiosity, do the units work? Joules is Force x distance…


#7

Entropy is J/K (joules per Kelvin)
Joules is Nm (newton meters)
Newtons is Kg m/s^2 (kilogram meters per second squared)

So Entropy is Kg m^2/s^2 K (kilogram meters squared per second squared per Kelvin

Square root of entropy is sqrt(Kg/K) (m/s), with a root 2 in there somwhere… so it isn’t exactly velocity. I don’t know what sqrt(Kg) and sqrt(K) are. Maybe its temperature and mass weighted velocity. mass weighted velocity is momentum. So momentum divided by temperature, I guess. Unless someone can cleverly cancel the mass and the temperature, in which case we would have pure velocity units. I think they want me to believe that root 2 times root (Kg/K) is unitless… which isn’t obvious, is it?

Temperature is a measure of kinetic energy, and a higher temperature steam does have more energy that we can do work with. And more obviously pressure can be turned into work, too. Brayton cycle seems to describe how pressure and temperature are related to entropy. At the nozzle we drop the pressure, the temperature also drops, and we get velocity from that. We can magically convert the random, not useful temperature energy (internal energy) to nice laminar flow velocity that we can use over the foils (blades) to create roundy-roundy-lift motion.

uh. Need to take more physics… Clearly I missed something important somewhere.


#9

In the original text are you sure it’s not shown as g-sub-c in that position of the equation? In my thermo days g-sub-c had both numerical value and units/dimensions in the English Engineering system of units. It was a constant that related mass,force,distance and time. It appears in many equations (when using that system of units) as 2g-sub-c or the square root of same. This constant 32.174 lbm-ft / lbf-sec^2 lets old timers use lbm and lbf and feet and seconds.

Obviously in the SI system this is not needed because force is defined in terms of the other three quantities unlike English Engineering system where all four are independently and arbitrarily defined.

Is it possible the text you refer to was using the English Engineering system of units and g was anotated in some way to indicate its use as that constant?


#10

For example, here g-sub-c is the constant I mentioned above in a early/mid 70’s thermo book that used lbm and lbf.

That looks like the equation you posted in OP.


#11

It sure does look the same. Mine doesn’t have the c subscript on g, but it could be a typesetting error. except they do explicitly and adjacently define g as the gravitation constant, and say that they are using SI.


#12

I expect “they” was a harried editor who was not entirely up on the subject. I bet if you look in the first edition it will be correct.


#13

Every time I read one of your posts I feel a little dumber haha


#18

This didn’t hold up very well.


#19

Every time I write one, I have the same feeling. Units don’t work?? why like that? :sob:


#20

Apologies if I’m misunderstanding what you are saying and trying to find out (I also don’t have access to the original texts you quote) but let me show you ‘my work’:

Under ideal circumstance, kinetic energy gained equals enthalpy (not entropy) lost during nozzle expansion:
image
In other words:
image
I’d say that your equation in the 2nd post (image) is exactly that if your Hd has units of kJ/kg. The units also work out (since it’s enthalpy units, J, not entropy, J/K).

I have no idea how the gravitational constant came into Mr Morton’s version of this equation but I’m happy to dig if you can provide a bit more context from the text.


#21

There’s no more context than what I quoted in the text, unfortunately. No one else uses gravity when they are doing these problems, so I’m certain that its a flub. I think you might have nailed it with the enthalpy catch, though. That makes a lot more sense: enthalpy is pressure and temperature. Good stuff, sir.


#22

re; working same on moon: terminal velocity equation works equally there doesn’t it? … I think the heat drop statement over nozzel is pretty reliable.


#23

Terminal velocity is different, you have to know the shape and resistance from the fluid that you are falling through.

acceleration constant on the moon is 1.62 m/s^2 (I looked it up), vs. earth (9.81 m/s^2)