Standard air density or air density @ conditions given

[ksprayberry]

Ok, I'm confused... it's a little nerve racking to be this close to the exam and have a question like this, so I'd really like to have some feedback on this if you can.

I have the Mechanical PE Sample Examination 2nd Edition, Current printing #3. Problem 53 uses standard air density of .075 lbm/ft3 to arrive at the correct solution. My first pass through, I used the specific volume off the psychrometric chart and arrived at a slightly different answer..I may have even used the ideal gas equation to derive at the density and then used that, but nevertheless, the answer I arrived at was a listed answer but was incorrect. Incorrect, but listed as a 1 of 4 choice!

So, I have an older 101 Solved Mechanical Engineering Problems book (Printing 7) and problem 9..Similar problem too, uses the method I used above...DOES NOT USE STANDARD AIR to arrive at the solution. They read 13.84 ft3/lbm off the psych chart and used that to arrive at the proper mass flow.

So which is it? Most every problem I've worked out of the Lindeberg books (Sample exam and Practice) and the NCEES sample exam all seem to use standard air to arrive at the correct answer, I will probably take this approach on the exam, but getting some feedback would be nice, you can see how this would shake your confidence quite a bit.

Thanks for your help

Kelly

 

[Outlaw44]

Does the problem that uses standard air density of 0.075 lbm/ft^3 tell you to use standard or that the conditions are assumed to be standard? Or do they just use it in the solution? I'm not familiar with the sample exams that you listed, but after looking and taking the NCEES practice exams, they specify to use the standard density or give you pressure and temperature to find it using the ideal gas law (assuming it behaves as an ideal gas).

Another question, what are the conditions that drive you to 13.84 ft^3/lbm? That has to be at some air temp higher than 70F (which corresponds to standard density).....looking at 70F on a psych chart, from 0 to 100% RH, it only ranges from 13.35 ft^3/lbm (standard) to just under 13.7 ft^3/lbm.

I guess my thought is, if they tell you to use standard, then obviously use it. But assuming you have everything you need to calculate it or pull it from the psych chart, do that if you need to because the conditions are different.

 

[ksprayberry]

The entering conditions were 80/50%RH. I'll go back and look, but I can't seem to make heads or tails of it. I think I know what's going on though, I'm sure they pull these problems from several sources/authors and as usual, different people go at things different ways. I've been in the manufacturing side of HVAC for 15 years. The 1.08, 4.5 and .68 rough cut estimates have served me well to figure sensible, total, and latent capacities. I don't have a problem with splitting hairs on an exam and I don't have a problem playing by their rules, I'm just trying to figure out what the rules are!

For example:

Problems 50, 53 and 55,59, 60,66 in the Lindeburg sample exam used standard air density, yet the conditions aren't standard. Some of the energy recovery problems used specific volume from the psychrometric chart I can understand that. From all these questions I figured that they assumed you would use standard air on the test like you say, Unless told otherwise...SO?

So, I take the NCEES practice exam this afternoon and they seem to gravitate to using specific volume from the psychrometric chart. Ok, I'm cool with that, but then they seem to have a real mixed bag of problems. They split hairs on some and use the specific volume to calculate the mass flow, then two problems down, they switch to using the rough cut 1.1 x cfm x dT and 4.5 x cfm x dH to calculate their total and sensible loads. Then, on one, they used the specific volume of the leaving air to calculate the capacity, now why would you do that unless you were wanting to err on the higher side since the density is higher.

The problems I see that concern me are 504, 506, 507 (507 I got wrong because I used the density calculated at 77 degrees, I came up with 34-34.5 gph closer to the 36.4 than the correct answer of 31.5...Then you get over to 519, now their using the 1.08..Hey, wait a minute they were using 1.1 a couple problems before this one, what gives? Nothing about this problem seems to be standard..Entering conditions of 60 leaving air of 160, 160 is definitely not standard.

Problem 537 is the fiesta resistance for me. If you look at the solution they offer then standard 4.5 x cfm x dH to come up with the answer then they take the time to say OR....Or, use 13.33 ft3/lbm to solve the problem with. HELLLOOO! where do you think the 4.5 comes from? 13.33 ft3/lbm is the specific volume for standard air. It's sort of like restating the problem a different way. And by the way, It's a energy wheel, so nothing is standard on this one why would they use it anyways, the entering and leaving conditions are well over 68 degrees anyways.

I guess I'll just figure them both ways if I have time. Something tells me I won't.

Any thoughts?

Thanks for letting me vent!

Kelly

 

[MetsFan]

For what it's worth, I've been using the 4.5 and 1.085 on pretty much every problem that doesn't state otherwise and haven't had an issue with it yet. On 507, I used the specific volume read from the psych chart. At 77 degrees/55%rh, the specific volume is right at 13.75ft^3/lbm or a density of .0727lbm/ft^3.

On 519, I used 1.085 because the entering air temperature of the coil is 62deg. and that is what the coil is seeing.

I wouldn't worry about solving it two ways unless you see that the answers are ambiguous. 537 for example gave me 18.75 using 4.5 and a delta h of 5. I might have been sloppy getting the enthalpy values, but it was close enough to B that I went with it.

 

[ksprayberry]

Thanks for the feedback. I'm sure you're right. I think I'll go back and rework those problems using the 4.5 and the 1,085 and see how they work out. I'm probably getting a touch jumpy over it all. I use those day in and day out and never gave it a thought until I get to studying for the exam. It could also be a touch of my eyesight not being what it used to be. I hate to say it, but I can't read a Mollier diagram for nothing, I need bifocals to do that! I've been working from a copy of a psychrometric chart and not being able to read the enthalpy well, I've been figuring the enthalpy from the formula h = h dry air + WHfg so I can get the enthalpy spot on.

Thanks for the help. I'll go rework them and see how it works out.

thanks

Kelly

 

[MizzouMatt]

I had the same issue on that practice test. I think that this is just a very poorly worded problem. I think that adjusting for the proper density is the correct thing to do. It was just stupid to have two Answers so close together. I asked this same question on the PPI passing zone forum (FYI not at all worth the money, this board has been more helpful) and the adviser said to use the adjusted value on these types of problems and that the solution probably should have been the answer we arrived at. I am guessing that the real test would not be that poorly written to include answers that would be correct under either method unless this was what they were specifically asking about.

So I plan on using the adjusted values for the real test and have confidence in the real testmakers not to have an answer that is in reality "more" correct in the wrong column. That being said I would think that either method would produce the correct answer on a given test question. See NCEES practice test question 511. They do not adjust for density in the problem but if you do you still get the correct answer.

 

[ksprayberry]

I like your answer. I believe that is the way I will approach it. I do hope they space the answers out a bit more than some of these sample problems and will trust that the people designing the test have thought it out. I'm glad you mentioned the Passing Zone thing, I had thought about joining the Exam Cafe thing so I could have access to their instructors, that answers that. I was thinking it would be nice to have access to more problems, but I'm probably down to studying tonight and tomorrow night, so I don't think it will be worth it.

Thanks for your time and thoughts.

Kelly

 

[MetsFan]

ksprayberry, I found this on another forum. Maybe they are taking into account that it is moist air when they use 1.1?

/>http://www.nrel.gov/docs/fy02osti/30152.pdf page 203

There is an apparent minor inconsistency with specific heat of air used for calculating COIL-BF inputs versus SUPPLY-AIR-DT. Both equations utilize the equation:q = m (cp) ΔT, where since the flow rate is given

m = ρ * Q * 60, where Q is volumetric fan air flow rate in cfm.

So that:

q = (ρ (cp) 60) * Q * ΔT.

For developing inputs the term K = "ρ (cp) 60" is used. In general for standard air ρ = 0.075 lb/ft3. For dry air cp = 0.24 Btu/lb°F resulting in K = 1.08 and for moist air w ≈ 0.01 so that cp = 0.244 Btu/lb°F resulting in K = 1.10, where K has units of (Btu*min)/(ft3*F*h). Also note that some references indicate that standard air is dry (e.g. ASHRAE Terminology, Howell et al) while others only specify the density but indicate the possibility that standard air can be moist (e.g. ANSI/ASHRAE 51-1985). For initially calculating SUPPLY-DELTA-T at standard air conditions, the DOE-2.1A Engineers Manual (p.IV.29) uses moist (w=0.01) air (K = 1.10). However, for calculating the COIL-BF input (and data points for COIL-BF-FT), the DOE-2.1A User's Manual, p. IV.247 indicates K = 1.08. Note that an HVAC text published by ASHRAE (Howell et al, p. 3.5) uses K = 1.10 for calculating leaving air conditions using volumetric air flow rates.

 

[ksprayberry]

I think the more I get to digging at it, I think it has more to do with my possible bad eyesight than anything. I can use psycalc or something computerized and hit it using either method. When I go to projecting around on the chart, I skew the lines a little. The answers are so close together, on some of these that you can get it wrong. I hope like MizzouMatt said that the person who wrote the exam thought that out. Wouldn't be an issue if I could use my hp48.

Thanks I'll read that article.

 

[MizzouMatt]

I think the more I get to digging at it, I think it has more to do with my possible bad eyesight than anything. I can use psycalc or something computerized and hit it using either method. When I go to projecting around on the chart, I skew the lines a little. The answers are so close together, on some of these that you can get it wrong. I hope like MizzouMatt said that the person who wrote the exam thought that out. Wouldn't be an issue if I could use my hp48.

Thanks I'll read that article.

It may be worth printing a Psych Chart a bit bigger. I have all of mine at 11x17. see to spread out the lines a bit better