Recent content by Rogue

Hi DrDu, thanks for your response. The question comes from research into modes of Naphthalene carry-over in gas systems. I was surprised when I was faced with articles stating Naphthalene to be soluble in Liquid Ammonia.

Hi Please can anyone explain why Naphthalene is soluble in liquid Ammonia when Naphthalene is Non-polar and Ammonia is Polar?

Thanks for the response Chester. Am I right in thinking?: ## \phi = k \iota (C - C \alpha)## Answer = 1.371 x10^-4 ## \phi = k \varrho (P - P \alpha)## In terms of calculating partial pressure - I'll have to re-visit this to refresh my memory along with the relationship between...

Please can someone assist me as this particular question seems to have me completely confounded?

I don't know to be honest. But looking at your previous suggestion, it doesn't seem to take fluid density into account, and I'm really unsure where to take this one.

Apologies for resurrecting an old thread. I am also stuck on part b of this question. I had used one of the formulas above, using 50 as Hp. However, as I move on, it's obvious that this value is incorrect. Please can someone offer a bit of guidance or a nudge in the right direction?

Thanks gents. Realised my mistake, sorted now.

Are you sure? As I follow up ,I can't seem to get it to balance. For leaving the bottom of column 1 I have: 21.6kmol Component C 2.4kmol Component BFor entering column 2 I have : 5.4 kmol component C 42.6kmol component B 108 kmol component ALeaving the bottom of column 2: 5.4 kmol component...

Homework Statement 180kmol/hr of a three component mixture made up of 60 mol % A, 25 mol % B and 15 mol %. Component A is the most volatile ,component C is the least volatile. The mixture is to be passed through a 2 stage distillation process. 80% of component C leaves the bottom of the first...

Definitely a good point. I think the question was worded in such a way to point us down the interpolation route. I had not even considered what you said, suppose that comes with experience/confidence.

This is determined by a previous question where flame temperature needs to be calculated. There will always be some differences in answer to this question as it has to be interpolated from a graph drawn by the student. Accuracy will obviously depend on each student to an extent. Any rounding...

My delta T was 1750 (2050-300), which is different to your original answer.

Yeh, got that. Too many pages with scribbles. My heat load is 101276973 after 5% heat loss. The difference coming from my flame temp difference. Thanks Chester

Thanks Chester. I don't doubt your answer or replies, but I'm not willing to just copy across from something. I need to work through it and gain the understanding where I'm lacking (I have no one to teach me or advise). I'd rather get it wrong on my own merits then blindly repeat something...

Sorry Chester, my mistake. I did have these correct but it would appear I have transferred across onto here incorrectly.