# TFS practice problem of the week...



## Slay the P.E. (Mar 13, 2018)

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


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## Vel2018 (Mar 13, 2018)

Hi,

So this is what I got using Mollier 0.067kW time 3min 46 sec.

Using tables I got 0.077kW time 2min 24 seconds.

I think it made me faster on tables just because I already input the numbers, each step. OR maybe I am just slow in the chart.


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## mongolianbbq (Mar 13, 2018)

a?


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## mongolianbbq (Mar 13, 2018)

Vel2018 said:


> Hi,
> 
> So this is what I got using Mollier 0.067kW time 3min 46 sec.
> 
> ...


I used the chart and got it in about a minute. I think it helps because my super heated steam tables don't have 750 C so i would have to interpolate or estimate and estimating is easier for me with the chart.


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## Vel2018 (Mar 13, 2018)

mongolianbbq said:


> I used the chart and got it in about a minute. I think it helps because my super heated steam tables don't have 750 C so i would have to interpolate or estimate and estimating is easier for me with the chart.


Yes, its just I am bad at the chart, I used the chart in the MERM appendix using only ruler, did not point my pencil or write a line since its not allowed during test, that's what I got from using chart, my estimate was h1=1405 and h2=1105 that't definitely not within 2% from the value calculated using tables. In this case, there are no other options so would choose A.


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## Vel2018 (Mar 13, 2018)

Oh crap I screw it up, was supposed to be about 1060 h2. Probably when I blinked it went up one line lol it should be same value 0.077kW using the chart.


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## mongolianbbq (Mar 13, 2018)

Vel2018 said:


> Oh crap I screw it up, was supposed to be about 1060 h2. Probably when I blinked it went up one line lol it should be same value 0.077kW using the chart.


The merm chart was too small for me. I ended up downloading one online and printing it out and putting it in my resource binder.


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## Slay the P.E. (Mar 13, 2018)

mongolianbbq said:


> The merm chart was too small for me. I ended up downloading one online and printing it out and putting it in my resource binder.


You downloaded this one, right? wink-wink-nod-nod. It's color coded!

https://www.slaythepe.com/free-resources.html


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## Vel2018 (Mar 13, 2018)

Slay the P.E. said:


> You downloaded this one, right? wink-wink-nod-nod. It's color coded!
> 
> https://www.slaythepe.com/free-resources.html


awesome! thanks!


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## MikeGlass1969 (Mar 13, 2018)

I got (B).    h1=1403.4 and h2r=1039.8  h2i=925

Mine is wrong.  your chart reads h2i=1050

Yep.  Should be (A)...   I read the .2 psia curve initially, I think.


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## Slay the P.E. (Mar 16, 2018)

Happy Friday. Here's a fresh practice problem for you.

A military jet flies at a speed of 495 miles/hour at an altitude of 49,000 feet. A commercial airplane is flying at the same Mach number as the military jet but at an altitude of 27,000 feet. A plot showing the variation of temperature with altitude in the atmosphere is provided for your potential use. The speed (ft/s) of the commercial jet is most nearly:

(A)  515

(B)  726

(C)  757

(D)  795


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## MA_PE (Mar 16, 2018)

Mach is speed relative to the speed of sound.  If two planes are flying at the same Mach number then they are flying at the same speed (independent of altitude). 

495 mph converts to 726 ft/s.

Answer is (B)


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## Vel2018 (Mar 16, 2018)

MA_PE said:


> Mach is speed relative to the speed of sound.  If two planes are flying at the same Mach number then they are flying at the same speed (independent of altitude).
> 
> 495 mph converts to 726 ft/s.
> 
> Answer is (B)


I think not, temperature are different just have to equate both v1/(kgcRT1)^1/2=(v2/kgcRT2)^1/2

You get  v2


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## MikeGlass1969 (Mar 16, 2018)

Speed of sound is a function of absolute temperature.  So I came up with 512 ft/sec.

Answer should be (A)


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## Slay the P.E. (Mar 16, 2018)

MikeGlass1969 said:


> Speed of sound is a function of absolute temperature.  So I came up with 512 ft/sec.
> 
> Answer should be (A)


Are you sure your answer is in ft/s?


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## Vel2018 (Mar 16, 2018)

Answer should be C


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## MikeGlass1969 (Mar 16, 2018)

Slay the P.E. said:


> Are you sure your answer is in ft/s?


Damn...  I screwed up again.


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## MA_PE (Mar 16, 2018)

Vel2018 said:


> I think not, temperature are different just have to equate both v1/(kgcRT1)^1/2=(v2/kgcRT2)^1/2
> 
> You get  v2


oops.  I forgot that the speed of sound varies with temp/changes in atmosphere.  I'm a structural guy.


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## MikeGlass1969 (Mar 18, 2018)

Slay the P.E. said:


> You downloaded this one, right? wink-wink-nod-nod. It's color coded!
> 
> https://www.slaythepe.com/free-resources.html


Sure!!!  Where was this 2 years ago when I was studying for my exam????   BUT NO.....   Mike had to use the Black and White one and had to blow it up to 11X17 so his eyes wouldn't cross...


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## Slay the P.E. (Mar 18, 2018)

MikeGlass1969 said:


> Sure!!!  Where was this 2 years ago when I was studying for my exam????   BUT NO.....   Mike had to use the Black and White one and had to blow it up to 11X17 so his eyes wouldn't cross...


LOL

well, you know what will really suck? When the exam transitions to closed-book CBT next year. I dunno if the on-screen reference book will have a Mollier diagram, but if it does, just imagine people putting a ruler against the computer monitor... yikes


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## Slay the P.E. (Mar 22, 2018)

Happy Thursday, all. Here is your problem for the week:

A venturi meter is installed in a 2-in diameter, schedule 40 steel pipe (ID = 2.067 in). The fluid in the pipe is ambient temperature No. 4 fuel oil with a specific gravity _SG_=0.85. The meter is provided with a U-tube manometer. From past experience with this meter, it is known that for a manometer fluid height _h_ of 8 inches, the oil flow rate is 400 gallons per minute. The average fluid velocity (ft/min) in the pipe just upstream of the meter when the height _h _is 4 inches is most nearly:

(A) 27

(B) 283

(C) 1,225

(D) 1,623


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## Vel2018 (Mar 22, 2018)

Slay the P.E. said:


> Happy Thursday, all. Here is your problem for the week:
> 
> A venturi meter is installed in a 2-in diameter, schedule 40 steel pipe (ID = 2.067 in). The fluid in the pipe is ambient temperature No. 4 fuel oil with a specific gravity _SG_=0.85. The meter is provided with a U-tube manometer. From past experience with this meter, it is known that for a manometer fluid height _h_ of 8 inches, the oil flow rate is 400 gallons per minute. The average fluid velocity (ft/min) in the pipe just upstream of the meter when the height _h _is 4 inches is most nearly:
> 
> ...


I got D.


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## Slay the P.E. (Mar 22, 2018)

Vel2018 said:


> I got D.


Correct.


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## mongolianbbq (Mar 26, 2018)

This one is confusing me. You aren't given the density of the manometer fluid or the dimensions of the narrow section... are you supposed to use some sort of similarity ratio? Even if you do and find the flow rate for h=4 that way, you dont have A2. Am I going the right direction?


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## Vel2018 (Mar 26, 2018)

mongolianbbq said:


> This one is confusing me. You aren't given the density of the manometer fluid or the dimensions of the narrow section... are you supposed to use some sort of similarity ratio? Even if you do and find the flow rate for h=4 that way, you dont have A2. Am I going the right direction?


As the question said, it used to measure 8in of h at 400GPM. That should tell you something, try again youll get it.


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## mckenz007 (Mar 26, 2018)

mongolianbbq said:


> This one is confusing me. You aren't given the density of the manometer fluid or the dimensions of the narrow section...


Doesn’t say but if you assume the manometer fluid is mercury it works out!


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## Vel2018 (Mar 26, 2018)

mckenz007 said:


> Doesn’t say but if you assume the manometer fluid is mercury it works out!


Probably, but if you don't know the answer yet, its gonna hunt you for a while jajaja

Solution is very short xD


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## MikeGlass1969 (Mar 26, 2018)

See equation 18.41 in MERM....


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## Slay the P.E. (Mar 26, 2018)

MikeGlass1969 said:


> See equation 18.41 in MERM....


@mckenz007 It will work out independently of the density of the manometer fluid.

Look at the equation for flow through a meter. You can see that it simply states that flow rate is proportional to the square root of h. Right? In other words

Q = K  (h)^(1/2)​
where _K_ is some constant (*Side note *and interesting potential exam-type question: What are the units of this constant _K_?)

We know what Q is when h=8... what is it when h=4? Use the above form of the equation to eliminate K and solve for Q when h=4


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## Vel2018 (Mar 26, 2018)

What I did was use the basic equation. 

using v^2/2g=h(rhofluid-rhoOil)/rhoOil there could be a coefficient of flow based on the Q equation of venturi with manometer, but that doesn't matter since everything will turn to a constant.

So there you get the K if you move h=8in on the left side of the equation. Then use that K with an h of 4in to get v2. 

K does not have a unit.


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## Vel2018 (Mar 26, 2018)

Slay the P.E. said:


> @mckenz007 It will work out independently of the density of the manometer fluid.
> 
> Look at the equation for flow through a meter. You can see that it simply states that flow rate is proportional to the square root of h. Right? In other words
> 
> ...


Would that be really a potential question? Since this problem can be solved in many ways. 

If I use the standard Q=CfAo(2gh(rho2-rho1)/rho1)^1/2 I get the unit of K to be ft^5/s^2

But if I do it with velocity I get K without a unit. It's basically the same, one using velocity and one using Q.


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## Slay the P.E. (Mar 26, 2018)

Vel2018 said:


> K does not have a unit.


The units of _K_ has to be something like gpm/(in^(1/2)) for the dimensional homogeneity of the expression Q = K * sqrt(h)

In this case in fact _K_ = [400 gpm / sqrt(8 in)] therefore _K_ = 141.4 gpm/(in^(1/2))


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## Slay the P.E. (Mar 26, 2018)

Vel2018 said:


> Would that be really a potential question? Since this problem can be solved in many ways.
> 
> If I use the standard Q=CfAo(2gh(rho2-rho1)/rho1)^1/2 I get the unit of K to be ft^5/s^2
> 
> But if I do it with velocity I get K without a unit. It's basically the same, one using velocity and one using Q.


Well, but the question I posed was about the constant of proportionality between flow rate and square root of manometer height. By the way you might have a typo because the units would be ft^(5/2)/s

The constant of proportionality between upstream velocity and square root of manometer height has different units, but it is also not dimensionless.


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## Slay the P.E. (Mar 27, 2018)

This discussion about units reminded me of this problem we wrote and briefly considered for including it in our practice exam:

~~~~~~~~~~~~~~~~~~~~~~



In fluid mechanics, the Weber number for a fluid is a dimensionless parameter, defined as:

_We_ =  ( _ρ V_^2_ L_)/_σ_​
where _ρ _is_ _the fluid’s density, _V_ is a velocity, _L_ is a characteristic length, and_ σ_ is the surface tension. Since the Weber number is dimensionless, the units for surface tension in the SI system must be:

(A) kg/(m·s) 

(B) N/m

(C) m/N

(D) s^2/kg

~~~~~~~~~~~~~~~~~~


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## MikeGlass1969 (Mar 27, 2018)

(B)


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## mongolianbbq (Mar 27, 2018)

Slay the P.E. said:


> @mckenz007 It will work out independently of the density of the manometer fluid.
> 
> Look at the equation for flow through a meter. You can see that it simply states that flow rate is proportional to the square root of h. Right? In other words
> 
> ...


This is what I did and I got Q, but I got stuck trying to figure out the area of the throat in order to solve for velocity. Unless I read the question wrong and "just upstream of the meter" means the area before the diameter reduction? I guess the meter starts at the larger diameter since thats where the manometer is introduced? If I do that I get 1673 ft/m.

Units of K would be gpm / in^2  (which could be reduced further to in^(5/2)/minute)


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## mongolianbbq (Mar 27, 2018)

I agree with Mike for the weber problem. Even though I looked up surface tension to be force/length in about 30 seconds, I also broke up the problem into units replacing a kg*m/s^2 with N along the way and came up with N/m.


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## Slay the P.E. (Mar 27, 2018)

mongolianbbq said:


> I agree with Mike for the weber problem. Even though I looked up surface tension to be force/length in about 30 seconds, I also broke up the problem into units replacing a kg*m/s^2 with N along the way and came up with N/m.


That’s the flaw in this problem, I think. One can just look up surface tension and notice it is force per unit length. I think for future editions I would say “and σ is a material property” instead of “and σ is the surface tension”


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## Slay the P.E. (Mar 29, 2018)

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


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## mckenz007 (Mar 29, 2018)

D?


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## Slay the P.E. (Mar 29, 2018)

mckenz007 said:


> D?


Yes!


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## cornsnicker3 (Mar 29, 2018)

Slay the P.E. said:


> 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
> 
> ...


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)


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## Slay the P.E. (Mar 29, 2018)

cornsnicker3 said:


> 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


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## MikeGlass1969 (Mar 29, 2018)

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


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## cornsnicker3 (Mar 29, 2018)

Slay the P.E. said:


> 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


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## Slay the P.E. (Mar 29, 2018)

cornsnicker3 said:


> 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.


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## cornsnicker3 (Mar 29, 2018)

Slay the P.E. said:


> 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.


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## MikeGlass1969 (Mar 29, 2018)

cornsnicker3 said:


> 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.


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## mongolianbbq (Mar 31, 2018)

Slay the P.E. said:


> 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.
> 
> ...


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


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## Slay the P.E. (Apr 4, 2018)

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


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## Vel2018 (Apr 4, 2018)

Answer is A 0.1392


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## Slay the P.E. (Apr 4, 2018)

Vel2018 said:


> Answer is A 0.1392


That is absolutely correct!


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## Slay the P.E. (Apr 4, 2018)

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


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## mongolianbbq (Apr 5, 2018)

Slay the P.E. said:


> 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:
> 
> ...


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.


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## Slay the P.E. (Apr 5, 2018)

mongolianbbq said:


> 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


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## ME_VT_PE (Apr 5, 2018)

Slay the P.E. said:


> 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?


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## mongolianbbq (Apr 5, 2018)

Slay the P.E. said:


> 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?


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## mongolianbbq (Apr 5, 2018)

ME_VT said:


> 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.


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## Slay the P.E. (Apr 5, 2018)

ME_VT said:


> 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.


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## Slay the P.E. (Apr 5, 2018)

mongolianbbq said:


> 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)


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## mongolianbbq (Apr 5, 2018)

Slay the P.E. said:


> 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)


Got it. Thanks!


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## MikeGlass1969 (Apr 5, 2018)

Slay the P.E. said:


> 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:
> 
> ...


Might want to cross post this with HVAC&amp;R ...   I saw several of these on my exam in 2016.  I think I had 2 in the morning and 2 in the afternoon.


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## Vel2018 (Apr 5, 2018)

MikeGlass1969 said:


> Might want to cross post this with HVAC&amp;R ...   I saw several of these on my exam in 2016.  I think I had 2 in the morning and 2 in the afternoon.


Holy smokes Mike. And this is difficulty level2?


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## nathanielnzrn (Apr 9, 2018)

Slay the P.E. said:


> 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:
> 
> ...


So for this problem. I need to balance the combustion equation to include H20 so I can find the mole fraction of H20. Then multiply the mole fraction by the gas pressure at 1 atm to find the pressure of water vapor. Then use that water vapor pressure in the steam tables to give use the minimum temperature of the gas exiting the boiler without condensation. Then use the equation (m_gas)(Cp)(750-T_dp) = (m_steam)(h_fg @ 70 F, 50psi)  to find the ration of the mass flow rates?

Where T_dp is the temperature found using the water vapor pressure of the gas.

Is that right?


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## Vel2018 (Apr 9, 2018)

nathanielnzrn said:


> So for this problem. I need to balance the combustion equation to include H20 so I can find the mole fraction of H20. Then multiply the mole fraction by the gas pressure at 1 atm to find the pressure of water vapor. Then use that water vapor pressure in the steam tables to give use the minimum temperature of the gas exiting the boiler without condensation. Then use the equation (m_gas)(Cp)(750-T_dp) = (m_steam)(h_fg @ 70 F, 50psi)  to find the ration of the mass flow rates?
> 
> Where T_dp is the temperature found using the water vapor pressure of the gas.
> 
> Is that right?


Right except that, hg shall be taken from sat pressure 50psi and hf taken at sat temp at 70F. Not the hfg at sat pressure 50psi.


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## nathanielnzrn (Apr 9, 2018)

Vel2018 said:


> Right except that, hg shall be taken from sat pressure 50psi and hf taken at sat temp at 70F. Not the hfg at sat pressure 50psi.


What values did you get for the number of moles in H2O and water vapor partial pressure?


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## Vel2018 (Apr 9, 2018)

nathanielnzrn said:


> What values did you get for the number of moles in H2O and water vapor partial pressure?


20.4 for H2O 

Pvp 2.49psi


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## Slay the P.E. (Apr 9, 2018)

Vel2018 said:


> Right except that, hg shall be taken from sat pressure 50psi and hf taken at sat temp at 70F. Not the hfg at sat pressure 50psi.


Correct.

@nathanielnzrn see this post: http://engineerboards.com/topic/30175-tfs-practice-problem-of-the-week/?do=findComment&amp;comment=7481226


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## Slay the P.E. (Sep 28, 2018)

Happy Friday all. Brand new practice problem (also posted in the HVAC&amp;R thread):

The flow rate through the pipe system is 5,100 cubic feet per hour and the Darcy friction factor is known to be 0.02 through the entire system. Branch 1 is 3.5” ID, and Branch 2 is 5” ID. The total length of straight pipe in Branch 1 is 85 feet. The length of straight pipe in Branch 2 is also 85 ft. More information is given in the sketch. The difference in elevation (ft) between the water level of the two reservoirs is most nearly:

(A)   35
(B)   70
(C)   85
(D)   98


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## vitalvi (Oct 1, 2018)

Slay the P.E. said:


> Happy Friday all. Brand new practice problem (also posted in the HVAC&amp;R thread):
> 
> The flow rate through the pipe system is 5,100 cubic feet per hour and the Darcy friction factor is known to be 0.02 through the entire system. Branch 1 is 3.5” ID, and Branch 2 is 5” ID. The total length of straight pipe in Branch 1 is 85 feet. The length of straight pipe in Branch 2 is also 85 ft. More information is given in the sketch. The difference in elevation (ft) between the water level of the two reservoirs is most nearly:
> 
> ...


Is the answer D) 98 ft?


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## ME_VT_PE (Oct 1, 2018)

man, you guys are in trouble........


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## McEng PE (Oct 1, 2018)

I got answer D (98 ft).


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## Slay the P.E. (Oct 1, 2018)

vitalvi said:


> Is the answer D) 98 ft?


Yes. That is correct.


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## Slay the P.E. (Oct 1, 2018)

McEng said:


> I got answer D (98 ft).


Correct!


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## Slay the P.E. (Oct 1, 2018)

ME_VT_PE said:


> man, you guys are in trouble........


LOL

why?


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## Mech_Engineer (Oct 2, 2018)

Curious to see what other folks actually got for this latest problem's answer? I came up with 101 ft. and was wondering if I had a rounding error, or if I missed something. Used the energy equation. Found the total head loss in the piping by finding equivalent lengths in each section, then solved for elevation. My state points were the two water surfaces for the energy equation.


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## Slay the P.E. (Oct 2, 2018)

Mech_Engineer said:


> Curious to see what other folks actually got for this latest problem's answer? I came up with 101 ft. and was wondering if I had a rounding error, or if I missed something. Used the energy equation. Found the total head loss in the piping by finding equivalent lengths in each section, then solved for elevation. My state points were the two water surfaces for the energy equation.


That is the right approach. Could be a rounding thing.

If you post your work (photo or scanned PDF) we can look at it and figure it out.


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## Mech_Engineer (Oct 2, 2018)

Slay the P.E. said:


> That is the right approach. Could be a rounding thing.
> 
> If you post your work (photo or scanned PDF) we can look at it and figure it out.


Check out the attached and let me know where I'm going wrong, or if it's just a rounding issue I need to clear up. Thanks!

View attachment Fluids Prob.pdf


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## McEng PE (Oct 2, 2018)

Mech_Engineer said:


> Check out the attached and let me know where I'm going wrong, or if it's just a rounding issue I need to clear up. Thanks!
> 
> View attachment 11942


Look at your denominator for you losses for each branch (62.4 ft/s^2).


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## Mech_Engineer (Oct 2, 2018)

McEng said:


> Look at your denominator for you losses for each branch (62.4 ft/s^2).


Spot on! Thanks! I've been using water density at 62.4 lbm/ft^3 on so many problems lately, that my mind is stuck on that number!


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## McEng PE (Oct 2, 2018)

Lol! Yea I know what you mean. Just a few more weeks, then hopefully this is all over!


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## Slay the P.E. (Oct 2, 2018)

McEng said:


> Look at your denominator for you losses for each branch (62.4 ft/s^2).


Nicely done. Good work catching the error. Thank you.


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## Vel2018 (Oct 4, 2018)

Roughly 3 weeks before the big day. I still remember how I am feeling at during this time. hmmmmmm, please practice your speed!

By now, you guys should be finishing the whole 6MS book, NCEES practice book, SlayThePE problem book in 4 hours or less. Once your able to do this, you should be sure passing this test.

Also I advice you guys to look back at the previous problems in this thread. Thanks to @Slay the P.E.!


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## Slay the P.E. (Oct 5, 2018)

The manufacturing process of thin films on microcircuits uses a perfectly insulated vacuum chamber whose walls are kept at -320°F by a liquid nitrogen bath. An electric resistance heater is embedded inside a 1.5 feet long cylinder of 1½ inch diameter placed inside the vacuum chamber. The surface of the cylinder has an emissivity of 0.25 and is maintained at 80°F by the heater. The nitrogen enters the chamber bath as a saturated liquid and leaves as a saturated vapor. Neglecting any heat transfer from the ends of the cylinder, the required flow rate of nitrogen (pounds-mass per hour) is most nearly (the latent heat of vaporization for N2 is 53.74 Btu per pound): 

(A)   0.05
(B)   0.1
(C)   0.2 
(D)   0.4


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## McEng PE (Oct 6, 2018)

Slay the P.E. said:


> The manufacturing process of thin films on microcircuits uses a perfectly insulated vacuum chamber whose walls are kept at -320°F by a liquid nitrogen bath. An electric resistance heater is embedded inside a 1.5 feet long cylinder of 1½ inch diameter placed inside the vacuum chamber. The surface of the cylinder has an emissivity of 0.25 and is maintained at 80°F by the heater. The nitrogen enters the chamber bath as a saturated liquid and leaves as a saturated vapor. Neglecting any heat transfer from the ends of the cylinder, the required flow rate of nitrogen (pounds-mass per hour) is most nearly (the latent heat of vaporization for N2 is 53.74 Btu per pound):
> 
> (A)   0.05
> (B)   0.1
> ...


I am getting (D) 0.4 lbm/hr.


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## Slay the P.E. (Oct 6, 2018)

McEng said:


> I am getting (D) 0.4 lbm/hr.


Correct.


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## McEng PE (Oct 16, 2018)

Is there not many TFS exam takers this round?


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## ME_VT_PE (Oct 16, 2018)

McEng said:


> Is there not many TFS exam takers this round?


They may not be posting much, but they're lurking here for sure...


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## Slay the P.E. (Oct 16, 2018)

I hope.

The level of activity on this thread (and others) now pales in comparison to what we had for the April exam.


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## ME_VT_PE (Oct 16, 2018)

Slay the P.E. said:


> I hope.
> 
> The level of activity on this thread (and others) now pales in comparison to what we had for the April exam.


I was going to say...The april test takers were MUCH more active...Not a good sign for these October test takers....


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## MikeGlass1969 (Oct 16, 2018)

Slay the P.E. said:


> Correct.


I would like to see the solution for this...   I must be making this more difficult than what it is.


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## McEng PE (Oct 16, 2018)

MikeGlass1969 said:


> I would like to see the solution for this...   I must be making this more difficult than what it is.


The way I solved it: Qrad=(emissivity)(Stefan-Boltz Constant)(Area)(delta T^4), once you have Q then Q=(m-dot)(hfg). Solve for m-dot.


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## Slay the P.E. (Oct 16, 2018)

McEng said:


> The way I solved it: Qrad=(emissivity)(Stefan-Boltz Constant)(Area)(delta T^4), once you have Q then Q=(m-dot)(hfg). Solve for m-dot.


This is the correct approach.

@MikeGlass1969


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## squaretaper LIT AF PE (Oct 16, 2018)

ME_VT_PE said:


> They may not be posting much, but they're lurking here for sure...


It's because I retook the exam so many times they don't need to sign up any more people.


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## squaretaper LIT AF PE (Oct 16, 2018)

McEng said:


> Is there not many TFS exam takers this round?


No idea. But then again, in my circle of mechanical friends I'm the only TFS taker. Literally everyone around me took MDM once and passed (not looking at you @SacMe24). Any person who I heard about anecdotally who took TFS usually had to retake it. I definitely know zero HVAC'ers.


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## MikeGlass1969 (Oct 16, 2018)

Slay the P.E. said:


> This is the correct approach.
> 
> @MikeGlass1969


So we were to ignore the convective heat transfer portion of the problem?  The flashing of the Nitrogen should have induced a velocity in the chamber and enhanced the heat transfer. 

Similar to example 37.1 in MERM...  On page 37-4 thru 37-6.


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## Slay the P.E. (Oct 16, 2018)

MikeGlass1969 said:


> So we were to ignore the convective heat transfer portion of the problem?  The flashing of the Nitrogen should have induced a velocity in the chamber and enhanced the heat transfer.
> 
> Similar to example 37.1 in MERM...  On page 37-4 thru 37-6.


If in that example they had specified the inlet and outlet conditions of the air in the duct, then you wouldn't need to calculate the heat loss with a heat transfer analysis. A simple energy balance for the air in the duct should give you the heat loss.

In our problem, the inlet and outlet conditions of the nitrogen are given, therefore the heat transfer rate into the nitrogen can be calculated with an energy balance on the nitrogen alone.


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## MikeGlass1969 (Oct 16, 2018)

Slay the P.E. said:


> If in that example they had specified the inlet and outlet conditions of the air in the duct, then you wouldn't need to calculate the heat loss with a heat transfer analysis. A simple energy balance for the air in the duct should give you the heat loss.
> 
> In our problem, the inlet and outlet conditions of the nitrogen are given, therefore the heat transfer rate into the nitrogen can be calculated with an energy balance on the nitrogen alone.


I just think problem needs to state, neglect the convective heat transfer...


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## Slay the P.E. (Oct 16, 2018)

MikeGlass1969 said:


> I just think problem needs to state, neglect the convective heat transfer...


The nitrogen is changing phase because it is being heated by convection het transfer occurring from the chamber walls to the nitrogen. It cannot be neglected. It is just not necessary to calculate it, because you can do an energy balance on the nitrogen.

For example; suppose I ask to calculate the heat transfer rate required for m_dot amount of R-134a to enter an evaporator at x=0.3 and T=-20F and be discharged as saturated vapor at -20F. You could do two things:

a) perform an energy balance and figure out that the heat transfer rate is simply the change in enthalpy for the given end states, or

b) try to do a heat transfer analysis inside the evaporator pipe: find the appropriate Nusselt number correlation, get the convection coefficient inside the pipe and from there get the heat transfer rate.

I think what you're trying to do in our problem is something like approach (b) above.


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## MikeGlass1969 (Oct 16, 2018)

I am looking at the external pipe heat transfer.   With an 80F external wall pipe temperature in a space that is somewhat colder that -320F.  

There should be an increased transfer rate above that of the radiant heat transfer:

(a) natural convection - Assuming no velocity is induced from the expansion of the Nitrogen

(b) quasi forced convection from the expansion of Nitrogen.

Hope that makes sense...


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## McEng PE (Oct 17, 2018)

squaretaper PE said:


> No idea. But then again, in my circle of mechanical friends I'm the only TFS taker. Literally everyone around me took MDM once and passed (not looking at you @SacMe24). Any person who I heard about anecdotally who took TFS usually had to retake it. I definitely know zero HVAC'ers.


I have only worked with and have met mechanical PE's that took the TFS exam, and the consensus is that everyone they know that took the exam had to take it multiply times. Same, never met an HVAC exam taker. 

This is my second go at the TFS exam, I was pretty close the first time (60%) but the HVAC and power cycles killed it for me. I feel much more prepared this time though, and covered in depth power cycles and HVAC this time lol.


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## Slay the P.E. (Oct 17, 2018)

MikeGlass1969 said:


> I am looking at the external pipe heat transfer.   With an 80F external wall pipe temperature in a space that is somewhat colder that -320F.
> 
> There should be an increased transfer rate above that of the radiant heat transfer:
> 
> ...


No.

the nitrogen is flowing through an external jacket around the chamber. Look at the figure. One surface of the jacket is at -320F the other is adjacent to insulation.

the 80F is the surface temperature of a heat source (because it has a heating element inside) inside the vacuum chamber. There is no nitrogen in the heater. There is no media between the heater and the jacket.

heat is transferred by radiation between the heater surface and the nitrogen-cooled jacket. This heat must equal the latent heat of vaporization for nitrogen.


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## MikeGlass1969 (Oct 17, 2018)

Slay the P.E. said:


> No.
> 
> the nitrogen is flowing through an external jacket around the chamber. Look at the figure. One surface of the jacket is at -320F the other is adjacent to insulation.
> 
> ...


OHHH!!!!!  ...   Now I see it.


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## squaretaper LIT AF PE (Oct 17, 2018)

McEng said:


> I have only worked with and have met mechanical PE's that took the TFS exam, and the consensus is that everyone they know that took the exam had to take it multiply times. Same, never met an HVAC exam taker.
> 
> This is my second go at the TFS exam, I was pretty close the first time (60%) but the HVAC and power cycles killed it for me. I feel much more prepared this time though, and covered in depth power cycles and HVAC this time lol.


You got this @McEng!! Remember to go to Mollier first. I don't bother with the steam tables unless I have to.


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## McEng PE (Oct 17, 2018)

squaretaper PE said:


> You got this @McEng!! Remember to go to Mollier first. I don't bother with the steam tables unless I have to.


Thanks @squaretaper PE! Yes!! Mollier is much faster route.


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## Slay the P.E. (Oct 17, 2018)

McEng said:


> Thanks @squaretaper PE! Yes!! Mollier is much faster route.


We also advocate for this approach. The very first post in this thread is a good example showing how much time can be saved.

Of course, Mollier diagrams are useful for superheated steam and high quality water-steam mixtures, that is, the typical steam turbine problem.


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## Vel2018 (Oct 25, 2018)

Tomorrow is your big day. Today, go walk in the park for some fresh air, run 10 miles and give your calculators a break. Go eat your favorite meal, relax, drink a glass of wine, sleep early and pray. 

Good luck everyone!


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## Slay the P.E. (Feb 15, 2019)

Don’t know if people are aware of this thread. We won’t be adding more problems to it, but all problems here are still very relevant.


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