Question 501 in the 2011 version of the study materials

[Illi]

This one is driving me nuts; hopefully someone can help. As I understand it, the problem can be summarized as follows:

A wet blanket (6.25% moisture) at 75 degF travelling at a rate of 1fps is to be dried and raised to 400 degF. The blanket has a wet density of 1.6 pcf. The blanket has a heat capacity of 2btu/lb*degF. How much energy is required to evaporate the water and raise the blanket to 400 degF?

If you do some simple maths, you find that you need to evaporate 0.3lb/s of water and heat 4.5lb/s of blanket.

The NCEES solution is in two parts, first evaporate the water and second to heat the blanket to 400 degF:

Part 1:

"Water evaporation heat requirement: 0.3 x (975 - 50) = 277.5 btu/s"

Part 2:

"Blanket drying and curing heat requirement: 4.5 x 2 x (400-75) = 2,925 btu/s"

I have a number of problems with this solution:

1) Part 2 includes "blanket drying"; however the 4.5 lb/s is the mass of blanket ONLY. Water is not included in that mass.

2) In all the reference material I can find, the latent heat of vaporization of water is about 970 btu/lb. I have no idea where the "975" comes from

3) This solution appears to neglect the energy required to raise the temperature of the water from 75degF to 212 degF. My understanding is that to evaporate water you need to 1) provide sufficient energy to raise water to the boiling point and 2) provide the latent heat of vaporization to convert the liquid water (at 212 degF) to water vapor (at 212 degF).

4) The "water evaporation heat requirement" in the solution is LESS than the latent heat of evaporation for water. I'm pretty sure that it should be MORE than the latent heat. I think that the solution should be: massH2O/s x (latentHeat + 1btu/lb*degF x (212 degF - StartTemp)).

Any thoughts would be very much appreciated.

Thanks,

-Jeremy

 

[DavidPE]

I come up with 11.7 million Btu/hr. Here is my solution.

 

[Illi]

David, thanks for the reply. That is exactly how I did the problem, and I got the same answer. However, the NCEES solution is NOT consistent with your (and my) approach. Instead of using Q=m*deltaT+m*hf for the heat requirement to evaporate water, NCEES uses Q=m*(975-50) and I cannot figure out where they get that relationship from.

-Jeremy

 

[DavidPE]

I didn't spend an extensive amount of time looking for where that came from, but I can't find it either. I only know the basics here, so hopefully some else will answer.

I wonder if the answers were structured as they were so that our solution would yield the correct choice?

 

[Illi]

It's probably a mistake to spend as much time as I have ont it: it's just one problem. But it still bugs me because it doesn't make sense to me. I wonder if the difference is something to do with the heat recycling, but I can't seem to get the numbers to make sense.

 

[Dleg]

I went back and looked, and the 2004 edition has the same problem and solution.

On the first part, I solved it the same way you guys did. As you note, we still get the correct answer. I have no idea where the method NCEES used comes from.

On the second part, the NCEES solution makes sense to me - I used the same method.

The basic solution goes as follows:

1. heat to evaporate H2O = mass rate of water in blanket (lb/time) * hfg (Btu/lbm) (I used 970.4 for hfg)

2. heat to raise temp. of blanket to 400 deg F = mass rate of blanket only (minus water - lb/time)*heat capacity of blanket (2.0 Btu/lb-F)*(400-75 degF)

Then just add the two together.

 

[Illi]

I asked the professor I had for thermo; he had no idea where the NCEES solution came from either. I've also tried asking NCEES, but have not gotten a response. That isn't too surprising I suppose, but it is disappointing.