NCEES 01 TFS Q504

Ok, I'm truly stumped this time.

The problem is a steam turbine, which is not (to me) difficult to figure out how to work. But I can't get my h3 value to agree with the answer, which makes me question if my methodology is correct.

Based on the S3 value, the problem gives a quality of 1, which suggests you look up h at 50psi and choose hsubg from the table, correct? The value I get in three different tables is 1774.1, 1774.4, and 1774.1 (i.e two tables agree). This is seemingly quite different than their value of 1182.6. Which makes me question if I am doing something wrong? Their look up values for H1 and S2 were only slightly different than mine (which are 1529.1 and 1.66839, the latter being interpolated). Which only further makes me question my value of h3.

Thanks in advance for any help.

Correction: my table values are 1174.4 and 1174.1

You're still in the superheated area of the dome. Look up P = 50 in the superheated steam table and find the h that corresponds to s3. Using ASME steam tables book, it doesn't have s = 1.6698. It has s = 1.6724 which gives an enthalpy of 1184.5, pretty close.

Ok, I see what you did. My table two tables that have 50 psi in the superheated steam region don't go below 300Deg. F, and not seeing a value below the value you use, I didn't think I'd be able to interpolate very well, so I made the assumption I was wrong about what to use. Knowing what to do now, I suppose I could graph the trend and try to determine the temperature to make a guess at the value for h using the same temperature and trendline method.

In retrospect, the saturated steam value isn't much further off than the value you selected. It ended up giving me the right answer in this case anyway. Phew!

I would not waste time graphing and determining a trend to find the enthalpy value. If I were taking the exam, I would do exactly what I did last night, which is find the entropy value closest and use the corresponding enthalpy. The answer choices should not be so close together that a delta of 2 - 3 on one enthalpy will throw you off completely.

Ramnares P.E. said:

I would not waste time graphing and determining a trend to find the enthalpy value. If I were taking the exam, I would do exactly what I did last night, which is find the entropy value closest and use the corresponding enthalpy. The answer choices should not be so close together that a delta of 2 - 3 on one enthalpy will throw you off completely.

Click to expand...

Absolutely agree.

Not to beat this dead horse any further, but I had a separate question related to this problem. I'm not sure where they found their h2 value-- the rest of the problem makes perfect sense. Here are my steps:

1) Assume a throttling, steady-flow system:
> T1 = T2 under ideal gas law (suspected culprit here)
> h1 = h2
>Therefore, p2 = 1000psi, T2 = 1200 degF

2) Look up h value for 1000psi, 1200 degF steam (h=1619.7, s=1.7264)
> Both Appendix 23C and the Mollier Chart (Appendix 23E) seem to be consistent here

The value h=1530.8, s=1.6698 seems more consistent with roughly T=1050 degF

Where am I going wrong here?

Thank you in advance!

I've solved it this way. 

1. Find h1= 1530.8 using 5000 psia & 1200 F

2. Since its throttled (h1 = h2) @1000 psia I find the entropy s = 1.6698 for the enthalphy value 1530.8. 

3. Since Expanded adiabatically to 50 psia I find the quality x = 1.0088. From this quality I find enthalphy h3 = 1182.347

4. Doing rest of the calculaton and find the power required is 89844.99389 kW 

JayhawkerME said:

Not to beat this dead horse any further, but I had a separate question related to this problem. I'm not sure where they found their h2 value-- the rest of the problem makes perfect sense. Here are my steps:

1) Assume a throttling, steady-flow system:
> T1 = T2 under ideal gas law (suspected culprit here)
> h1 = h2
>Therefore, p2 = 1000psi, T2 = 1200 degF

2) Look up h value for 1000psi, 1200 degF steam (h=1619.7, s=1.7264)
> Both Appendix 23C and the Mollier Chart (Appendix 23E) seem to be consistent here

The value h=1530.8, s=1.6698 seems more consistent with roughly T=1050 degF

Where am I going wrong here?

Thank you in advance!

Click to expand...

No, you need to use the initial conditions for the determination of h1, as Phenomenon083 did.  The steam has a constant h value during the throttling process, but (and here is the key to understanding why your approach is wrong) not a constant temperature and the process is not reversible.  It's going to be important for you to understand underlying principles like this, or you're going to make invalid assumptions, like the one you posted.

Phenomenon083 said:

I've solved it this way. 

1. Find h1= 1530.8 using 5000 psia & 1200 F

2. Since its throttled (h1 = h2) @1000 psia I find the entropy s = 1.6698 for the enthalphy value 1530.8. 

3. Since Expanded adiabatically to 50 psia I find the quality x = 1.0088. From this quality I find enthalphy h3 = 1182.347

4. Doing rest of the calculaton and find the power required is 89844.99389 kW 

Click to expand...

since you seem to have the same exact table values, what source are you and the NCEES using?  It doesn't seem to be ASME.

This was a good post. I was thrown off for a second and followed a similar path as Jay (I wasn't thinking ideal gas law though). I just used 1200*F and 1000PSIA to find entropy as they were given in a manner that made you want to use them (without even thinking about it... even knowing I had just found the enthalpy...). Then thinking about it after getting the wrong answer, I saw that my 2 points to find entropy should have been the H1=H2 enthalpy and the pressure at State 2 (and not the Pressure and T1 Temperature combo). Looked up the enthalpy at 1000 PSIA and got the correct entropy. Worked the rest of the problem out from there. These are usually the small details that kill me on some of the sample questions. Thanks for beating a dead horse Jay, it helped me get another problem in.

Audi driver said:

since you seem to have the same exact table values, what source are you and the NCEES using?  It doesn't seem to be ASME

Click to expand...

I was using the steam tables by keenan and keyes. I didn't bother to look at the NCEES 01 solution procedure. As you've mentioned on a different post and I agree with you that compare to NCEES 2008, NCEES 01 doesn't do a good job about explaining the solution procedure. Also I've noticed that the thermodynamic values for the power cycles on the NCEES solution procedure are very similar to the ones in the steam table by keenan and keyes.

Thanks to everyone on their input. A few takeaways here:

* This problem is an example of where the MERM falls short on the steam tables. The steam charts don't cover the superheated region required for this problem. I've augmented my reference materials accordingly.

* For those using the 2013 MERM, Eqn 24.156 can be a trap (T1 = T2 under Ideal Gas for Throttling, Stead-Flow Systems). Stick with the isenthalpic relationship. Also a good reminder of the inherent limits of the ideal gas laws.

* Call a PE if you encounter 5000psi, 1200 degF applications.

Thanks again, and good luck to my fellow test takers.

JayhawkerME, I wasn't aware about this but I just checked and found it now. MERM Page 23-16, the ideal gas definition says that ideal gas temperature is much higher than critical temperature which is true in this case but the ideal gas pressure has to be very low, in this case the pressure is 5000 psi but the critical pressure is 218.2 atm for water vapor (MERM page 23-8). So this substance is in vapor form and cannot be treated as an ideal gas. Hopefully my reasoning is right

JayhawkerME said:

Thanks to everyone on their input. A few takeaways here:

* This problem is an example of where the MERM falls short on the steam tables. The steam charts don't cover the superheated region required for this problem. I've augmented my reference materials accordingly.

* For those using the 2013 MERM, Eqn 24.156 can be a trap (T1 = T2 under Ideal Gas for Throttling, Stead-Flow Systems). Stick with the isenthalpic relationship. Also a good reminder of the inherent limits of the ideal gas laws.

* Call a PE if you encounter 5000psi, 1200 degF applications.

Thanks again, and good luck to my fellow test takers.

Click to expand...

It's not a "trap" per se, it's just that you have no given information in this problem that suggests it's a steady flow system or that the steam is behaving like an ideal gas.  See  also "4. Throttling Processes" on 24-3 in the MERM, and look at the bullet at the top of column 2 on 24-2.