# NCEES Thermal and Fluid Sample Questions

### Help Support Engineer Boards:

#### thomas02pe

##### Member
Having a tough time understanding how h=1335.2 for steam on problem 129. It's a mixing question: I'll just type it below for those who don't have it:

Water (100 lb/min, 60 F) and Steam (100 lbm/min, 600 F) both at atmospheric pressure enter an insulated chamber through separate inlets. After thorough mixing, 200 lbm/min of the product is withdrawn at atmospheric pressure. The quality of the steam in the product stream is most nearly:

The solution shows:

hmix = 1/2 (hwater+hsteam)= 1/2 (28.08+1335.2)=681.4 btu/lb

681.64btu/lb=180.1btu/lbm +x(970.4btu/lbm)

x =.516

Any idea how you get 1335.2 btu/lb from 600F/14.7psia steam? Errata sheet doesn't have anything but I'm becoming a little delusional so i might be missing something. Actually I am delusional. I just found it on the mollier diagram. Well I wrote all this so I'll post it anyway.

Anybody else stressing?

#### Anutka2

##### Member
I'm right there with you!

Check out this old post on your same question: http://engineerboards.com/index.php?showtopic=13341

I used Appendix 24.C (Superheated Steam). Find h at P_atm and T=600 deg F. Mollier diagram gives the same result.

Just a few more days left

#### MapuaTech

##### Senior Member
always check if it is above or below saturation so you may know what table you need. use the steam table instead of mollier chart to get accurate enthalpies.

#### Flluterly

##### Member
any idea where is fomular from in MERM? It seems the solution skipping the details. Thank you in advance!

#### Flluterly

##### Member
hmix = 1/2 (hwater+hsteam)=

681.64btu/lb=180.1btu/lbm +x(970.4btu/lbm)

how the formula is derived and how to get the the h value?

#### Unintended Max P.E.

##### Max Collins of Unintended Consequences
The problem indicates uniform mixing, thus one must assume that at the exit the mixture is 50% water and 50% steam.  The values for h come from the saturated steam and superheated steam tables respectively and the formula is derived from basic statistical averages mathematics.  If I have equal parts of two things how much does it weigh?  It weighs the average of the two things.  If I have equal parts of something what is its temperature?  It's the average temperature of the two things.  If I have equal parts of liquid and steam, what is the average entropy?  havg = .5(h1 + h2)

h1= 28.08 (from saturated steam table)

h2 =1335.2 (from superheated steam table: 14.7 psi, 600F)

#### Flluterly

##### Member
Audi Driver, Thank you very much for detail explanation.

#### Flluterly

##### Member
Is  180.1btu/lbm  and 970.4btu/lbm from 330f table?

#### Unintended Max P.E.

##### Max Collins of Unintended Consequences
Is  180.1btu/lbm  and 970.4btu/lbm from 330f table?
I'm not sure what the 330f table is, sorry.  But those values are derived from the Saturated Steam pressure table. According to the steam table I have, @14.696 psi, h(sub f) =180.17 and h(subfg)= 970.3 which is very similar to the values in the NCEES answer (and they differ only because they use a different source than I did, I think they used Kenan&amp;Keyes)

On second review, I believe by 330f, you mean to look up in the superheated steam table at 14.696 psi and over to 330F.  No, that's not the right place to look.

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#### Flluterly

##### Member
Ok. Now it is clear to me. Thanks again

#### cidr1337

##### New member
I am a bit confused on how the solution arrived at h_mixture=.5(h_water+h_steam). While PEmax does a good job explaining the average it doesn't make sense to me given the formula in section 6.1.1. for Open(mixing) Feedwater Heater. This appears to be exactly what this problem is talking about. The formula is (m_1*h_1)+(m_2*h_2)=h_3(m_1+m_2). This yields 757.69Btu/lbm. Additionally there is also a formula that yields the correct answer in section 6.1.3: x = m_vap/(m_liq+m_vap) = 0.5. Why can't we just use this formula? Any help is appreciated.