TFS practice problem of the week...

[mongolianbbq]

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!

 

[MikeGlass1969]

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

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

 

[Vel2018]

Might want to cross post this with HVAC&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? 

 

[nathanielnzrn]

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

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?

 

[Vel2018]

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.

 

[nathanielnzrn]

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?

 

[Vel2018]

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

20.4 for H2O 

Pvp 2.49psi

 

[Slay the P.E.]

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&comment=7481226

 

[Slay the P.E.]

Happy Friday all. Brand new practice problem (also posted in the HVAC&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

 

[vitalvi]

Happy Friday all. Brand new practice problem (also posted in the HVAC&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

View attachment 11932

Is the answer D) 98 ft?

 

[ME_VT_PE]

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

 

[McEng PE]

I got answer D (98 ft).

 

[Slay the P.E.]

Is the answer D) 98 ft?

Yes. That is correct.

 

[Slay the P.E.]

I got answer D (98 ft).

Correct!

 

[Slay the P.E.]

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

LOL

why?

 

[Mech_Engineer]

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.

 

[Slay the P.E.]

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.

 

[Mech_Engineer]

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

 

[McEng PE]

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).

 

[Mech_Engineer]

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!