Effective Molecular Weight and Universal Gas Constant related to propellant c*

[dvdsnyd]

Hi all,

I have been working through Richard Nakka's theory section.

My question is mostly about units.

The meat of my questions arise from the following page:
https://www.nakka-rocketry.net/techs.html

Nakka calculates the effective molecular weight by dividing the number of gas moles into the system mass, simply -



This makes perfect sense to me up to this point. The units are literally grams/mole

What I don't understand is how the molecular weight is all of a sudden shown with units of kg/kmol with the same value as above, see below -



Plugging the numbers into the equation yields the correct value, but I don't understand how the effective molecular weight, M goes from having units of grams/mole to kg/kmol, but the value of 41.98 doesn't change. What am I missing here? Is changing the units some way being multiplied by 1/1 or something equivalent? The same thing is shown in the other ideal performance calculations for other sugar types, so I don't believe it is a mistake, just trying to figure out how/why it was done.

Thanks for your help,

Dave

 

[dave carver]

I know it says Propulsion but that's a question for Research. About all you can get here is what load fits what case. To get into the Research section you have to be Tripoli 2nd level and have prior approval.

You might be able to get the answer but I'd bet you'd get it in a PM if at all.

 

[BLKKROW]

It seems they just multiplied each by a Kilo.

I don't know if I am able to answer this.

 

[bandman444]

You have the right idea with multiplying by 1/1. Really in your example it is by 1000/1000.

1000grams equals 1kilogram
And
1000mol equals 1kilo mol

A kilo means 1000. Hence the above.

Does that clear it up?

 

[4kids49]

Hi all,

I have been working through Richard Nakka's theory section.

My question is mostly about units.

The meat of my questions arise from the following page:
https://www.nakka-rocketry.net/techs.html

Nakka calculates the effective molecular weight by dividing the number of gas moles into the system mass, simply -

View attachment 168592

This makes perfect sense to me up to this point. The units are literally grams/mole

What I don't understand is how the molecular weight is all of a sudden shown with units of kg/kmol with the same value as above, see below -

View attachment 168591

Plugging the numbers into the equation yields the correct value, but I don't understand how the effective molecular weight, M goes from having units of grams/mole to kg/kmol, but the value of 41.98 doesn't change. What am I missing here? Is changing the units some way being multiplied by 1/1 or something equivalent? The same thing is shown in the other ideal performance calculations for other sugar types, so I don't believe it is a mistake, just trying to figure out how/why it was done.

Thanks for your help,

Dave

I haven't followed the whole theoretical argument, but yes g/mole is the same as kg/kmole. The R value is in J/kmole-K, so maybe the MW was converted to same units as R units. Kmole should cancel in the resulting equation if i jotted it down correctly.

 

[SpaceManMat]

Plugging the numbers into the equation yields the correct value, but I don't understand how the effective molecular weight, M goes from having units of grams/mole to kg/kmol, but the value of 41.98 doesn't change. What am I missing here? Is changing the units some way being multiplied by 1/1 or something equivalent? The same thing is shown in the other ideal performance calculations for other sugar types, so I don't believe it is a mistake, just trying to figure out how/why it was done.

Dave if you were brought up using the metric system this would not surprise you.

The maths is quite simply 41.98/1 = (41.98 * 1000)/(1 * 1000)

 

[dvdsnyd]

I know it says Propulsion but that's a question for Research. About all you can get here is what load fits what case. To get into the Research section you have to be Tripoli 2nd level and have prior approval.

You might be able to get the answer but I'd bet you'd get it in a PM if at all.

Thanks for your response,
I am a member of the research section on this forum. I did not post my questions in the research section because I did not ask anything about propellant formulations, ingredients, or processes. My question was simply about clarification with some unit conversions applied to propulsion.
Dave


Thanks for all the other replies as well! I knew it was something simple that I wasn't seeing..

Dave if you were brought up using the metric system this would not surprise you.

The maths is quite simply 41.98/1 = (41.98 * 1000)/(1 * 1000)

Thanks!
I feel a little dopey. I got caught up in unit conversion once again. I will expand on your post, just to make it clear to all:

[41.98 grams/mol] * [1 kg/1000 grams] * [1000 mol/1 kmol] = 41.98 kg/kmol

When written out it is a little easier to see, but everything either cancels out or reduces down to 1 leaving, 41.98 kg/kmol

Thanks again for helping me clear this up!

Dave