TFS practice problem of the week...

[Slay the P.E.]

D?

Yes! 

 

[cornsnicker3]

A steam turbine with an isentropic efficiency of 76% admits steam at 100 psia, 750F. The turbine discharge is at 2 psia. The shaft power developed in kW per lbm/h of steam flowing through the turbine is most nearly:

(A) 0.08

(B) 0.10

(C) 0.27

(D) 262

I can't get anything else other than 0.024 kW. What am I doing wrong?

h1=1404.65 Btu/lbm

Use s1 as 1.8245 Btu/(lbm-R) to find X2 as 94.55%

h2=hf+X2*hfg = 1060.13 Btu/lbm

h2' = h1-eff*(h1-h2) = 1142.8 Btu/lbm

Shaft Power = Delta h = 82.68 Btu/(lbm/hr) = 0.02442 kW/(lbm/hr) 

 

[Slay the P.E.]

Delta h = 82.68 Btu/(lbm/hr) 

This is incorrect. Delta h (using your numbers) is (1404.65 - 1142.8) Btu/lbm = 261.85 Btu/lbm

 

[MikeGlass1969]

Your are using the wrong enthalpies to calculate the shaft power.  You use the h2 real - h2 ideal.  You need h1 - h2real.

 

[cornsnicker3]

This is incorrect. Delta h (using your numbers) is (1404.65 - 1142.8) Btu/lbm = 261.85 Btu/lbm

Way to public humiliate myself...darn

 

[Slay the P.E.]

Way to public humiliate myself...darn

LOL. No worries. Now is the time to make mistakes so you don't make them on test day.

Mike's post above pointed out the real problem -- I couldn't figure out why you were getting 82.68 Btu/lbm -- but he figured out you were using the difference between ideal and real discharge enthalpy to get the shaft power. I hope you know this is wrong.

 

[cornsnicker3]

LOL. No worries. Now is the time to make mistakes so you don't make them on test day.

Mike's post above pointed out the real problem -- I couldn't figure out why you were getting 82.68 Btu/lbm -- but he figured out you were using the difference between ideal and real discharge enthalpy to get the shaft power. I hope you know this is wrong.

Yes, I understand this. I got caught in the frenzy so to speak. The slight panic does not help. 

 

[MikeGlass1969]

Way to public humiliate myself...darn

At least you were smart enough to not use your real name on these boards when you make a mistake.  Unlike myself.

 

[mongolianbbq]

Happy Thursday, TFS peeps.

Air at 10°C and 80 kPa enters the diffuser of a jet engine steadily with a velocity of 200 m/s. The inlet area of the diffuser is 0.4 squared meters. The air leaves the diffuser with a velocity that is very small compared with the inlet velocity.

The mass flow rate of the air (kg/min) is nearest:
(A)  79
(B)  245
(C)  2230
(D)  4728

I also got D. At first glance I was ready to bust out the isentropic flow table since Mach is over .3.

 

[Slay the P.E.]

Hi all. Last one before the big day next Friday. Good luck to all!!!

A gas turbine operates with the products of combustion of methane (CH4), with dry air. The volumetric analysis of the products on a dry basis is CO2, 9.7%; CO, 0.5%; O2, 2.95% and N2, 86.85%. The gases have a constant-pressure specific heat of 0.26 Btu/lbm/°F and are discharged from the turbine at 1 atm and 750°F. Upon leaving the turbine the gases enter a counterflow steam recovery boiler where they are used to generate saturated steam from water at 70°F and 50 psia. The maximum possible ratio of steam mass flow to turbine gas flow ((lbm/h)/(lbm/h)) that can be generated in order to avoid condensation of turbine gas moisture within the boiler is most nearly:

(A) 0.14

(B) 0.28

(C) 0.56

(D) 1.12

 

[Vel2018]

Answer is A 0.1392

 

[Slay the P.E.]

Answer is A 0.1392

That is absolutely correct!

 

[Slay the P.E.]

Cross-posting to the HVAC&R thread as TFS folk should be able to do the last one on that thread:

 

[mongolianbbq]

Hi all. Last one before the big day next Friday. Good luck to all!!!

A gas turbine operates with the products of combustion of methane (CH4), with dry air. The volumetric analysis of the products on a dry basis is CO2, 9.7%; CO, 0.5%; O2, 2.95% and N2, 86.85%. The gases have a constant-pressure specific heat of 0.26 Btu/lbm/°F and are discharged from the turbine at 1 atm and 750°F. Upon leaving the turbine the gases enter a counterflow steam recovery boiler where they are used to generate saturated steam from water at 70°F and 50 psia. The maximum possible ratio of steam mass flow to turbine gas flow ((lbm/h)/(lbm/h)) that can be generated in order to avoid condensation of turbine gas moisture within the boiler is most nearly:

(A) 0.14

(B) 0.28

(C) 0.56

(D) 1.12

Am I approaching this one correctly? I found the dew point of the products and in order to prevent condensation, the temp of the gas in the boiler can't be reduced pass the dew point. So you have (m_gas)(Cp)(750-T_dp) = (m_steam)(h_fg @ 50psi) then solve for the ratio. I got .1733 this way.

 

[Slay the P.E.]

Am I approaching this one correctly? I found the dew point of the products and in order to prevent condensation, the temp of the gas in the boiler can't be reduced pass the dew point. So you have (m_gas)(Cp)(750-T_dp) = (m_steam)(h_fg @ 50psi) then solve for the ratio. I got .1733 this way.

That is the right approach. Nicely done.

However, for the enthalpy change of the water, h_fg(50psi) is not correct. While the water does indeed exit the boiler as saturated vapor, hence h_exit=h_g(50 psi) it enters as a compressed liquid at 70F and 50psi hence h_inlet=h(70F, 50 psi) which is approximated as h_f(70F). Doing this will give you 0.139

 

[ME_VT_PE]

That is the right approach. Nicely done.

However, for the enthalpy change of the water, h_fg(50psi) is not correct. While the water does indeed exit the boiler as saturated vapor, hence h_exit=h_g(50 psi) it enters as a compressed liquid at 70F and 50psi hence h_inlet=h(70F, 50 psi) which is approximated as h_f(70F). Doing this will give you 0.139

how do you find the dew point temperature of methane?

 

[mongolianbbq]

That is the right approach. Nicely done.

However, for the enthalpy change of the water, h_fg(50psi) is not correct. While the water does indeed exit the boiler as saturated vapor, hence h_exit=h_g(50 psi) it enters as a compressed liquid at 70F and 50psi hence h_inlet=h(70F, 50 psi) which is approximated as h_f(70F). Doing this will give you 0.139

So if you are given both temperature and pressure, you should use temperature rather than pressure for finding the enthalpy of a liquid?

 

[mongolianbbq]

how do you find the dew point temperature of methane?

Page 21-14 in the MERM 13th edition should help you out. You basically have to use the flue gas to create a balanced combustion equation and then find the volumetric fraction of the H2O. Equation 21.13 on that page in the merm should take you the rest of the way. I will admit it took me way too long to figure out that I needed to find the dew point in order to solve this problem. In the future, I hope my mind immediately goes to thinking about the dew point when I see a problem asking to prevent condensation of combustion products.

 

[Slay the P.E.]

how do you find the dew point temperature of methane?

Not methane— the products of combustion of methane; for which you’re given the volumetric analysis. Use the given compositions to find the partial pressure of water in the mixture. Use that to find the dew point of the gas mixture.

 

[Slay the P.E.]

So if you are given both temperature and pressure, you should use temperature rather than pressure for finding the enthalpy of a liquid?

If the given conditions (T,p) are such that the thermodynamic state is “compressed liquid” then you could use the compressed liquid table. That table, however, typically only lists moderately high pressures. 50 psi is too low and is not listed (at least in the one in MERM) Therefore, for compressed liquids we use the approximations;

h(T,p) ~ h_f(T)

v(T,p) ~ v_f(T)