TFS practice problem of the week...

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Lol! Yea I know what you mean. Just a few more weeks, then hopefully this is all over!

 
Roughly 3 weeks before the big day. I still remember how I am feeling at during this time. hmmmmmm, please practice your speed!

By now, you guys should be finishing the whole 6MS book, NCEES practice book, SlayThePE problem book in 4 hours or less. Once your able to do this, you should be sure passing this test.

Also I advice you guys to look back at the previous problems in this thread. Thanks to @Slay the P.E.!

 
The manufacturing process of thin films on microcircuits uses a perfectly insulated vacuum chamber whose walls are kept at -320°F by a liquid nitrogen bath. An electric resistance heater is embedded inside a 1.5 feet long cylinder of 1½ inch diameter placed inside the vacuum chamber. The surface of the cylinder has an emissivity of 0.25 and is maintained at 80°F by the heater. The nitrogen enters the chamber bath as a saturated liquid and leaves as a saturated vapor. Neglecting any heat transfer from the ends of the cylinder, the required flow rate of nitrogen (pounds-mass per hour) is most nearly (the latent heat of vaporization for N2 is 53.74 Btu per pound): 

(A)   0.05
(B)   0.1
(C)   0.2 
(D)   0.4

Screen Shot 2018-10-05 at 7.13.37 PM.png

 
The manufacturing process of thin films on microcircuits uses a perfectly insulated vacuum chamber whose walls are kept at -320°F by a liquid nitrogen bath. An electric resistance heater is embedded inside a 1.5 feet long cylinder of 1½ inch diameter placed inside the vacuum chamber. The surface of the cylinder has an emissivity of 0.25 and is maintained at 80°F by the heater. The nitrogen enters the chamber bath as a saturated liquid and leaves as a saturated vapor. Neglecting any heat transfer from the ends of the cylinder, the required flow rate of nitrogen (pounds-mass per hour) is most nearly (the latent heat of vaporization for N2 is 53.74 Btu per pound): 

(A)   0.05
(B)   0.1
(C)   0.2 
(D)   0.4

View attachment 11949
I am getting (D) 0.4 lbm/hr.

 
I hope.

The level of activity on this thread (and others) now pales in comparison to what we had for the April exam.  
I was going to say...The april test takers were MUCH more active...Not a good sign for these October test takers....

 
I would like to see the solution for this...   I must be making this more difficult than what it is.
The way I solved it: Qrad=(emissivity)(Stefan-Boltz Constant)(Area)(delta T^4), once you have Q then Q=(m-dot)(hfg). Solve for m-dot.

 
Is there not many TFS exam takers this round?
No idea. But then again, in my circle of mechanical friends I'm the only TFS taker. Literally everyone around me took MDM once and passed (not looking at you @SacMe24). Any person who I heard about anecdotally who took TFS usually had to retake it. I definitely know zero HVAC'ers.

 
This is the correct approach.

@MikeGlass1969
So we were to ignore the convective heat transfer portion of the problem?  The flashing of the Nitrogen should have induced a velocity in the chamber and enhanced the heat transfer. 

Similar to example 37.1 in MERM...  On page 37-4 thru 37-6.

 
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So we were to ignore the convective heat transfer portion of the problem?  The flashing of the Nitrogen should have induced a velocity in the chamber and enhanced the heat transfer. 

Similar to example 37.1 in MERM...  On page 37-4 thru 37-6.
If in that example they had specified the inlet and outlet conditions of the air in the duct, then you wouldn't need to calculate the heat loss with a heat transfer analysis. A simple energy balance for the air in the duct should give you the heat loss.

In our problem, the inlet and outlet conditions of the nitrogen are given, therefore the heat transfer rate into the nitrogen can be calculated with an energy balance on the nitrogen alone.

 
If in that example they had specified the inlet and outlet conditions of the air in the duct, then you wouldn't need to calculate the heat loss with a heat transfer analysis. A simple energy balance for the air in the duct should give you the heat loss.

In our problem, the inlet and outlet conditions of the nitrogen are given, therefore the heat transfer rate into the nitrogen can be calculated with an energy balance on the nitrogen alone.
I just think problem needs to state, neglect the convective heat transfer...    

 
I just think problem needs to state, neglect the convective heat transfer...    
The nitrogen is changing phase because it is being heated by convection het transfer occurring from the chamber walls to the nitrogen. It cannot be neglected. It is just not necessary to calculate it, because you can do an energy balance on the nitrogen.

For example; suppose I ask to calculate the heat transfer rate required for m_dot amount of R-134a to enter an evaporator at x=0.3 and T=-20F and be discharged as saturated vapor at -20F. You could do two things:

a) perform an energy balance and figure out that the heat transfer rate is simply the change in enthalpy for the given end states, or

b) try to do a heat transfer analysis inside the evaporator pipe: find the appropriate Nusselt number correlation, get the convection coefficient inside the pipe and from there get the heat transfer rate.

I think what you're trying to do in our problem is something like approach (b) above.

 
I am looking at the external pipe heat transfer.   With an 80F external wall pipe temperature in a space that is somewhat colder that -320F.  

There should be an increased transfer rate above that of the radiant heat transfer:

(a) natural convection - Assuming no velocity is induced from the expansion of the Nitrogen

(b) quasi forced convection from the expansion of Nitrogen.

Hope that makes sense...

 
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