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Slay the P.E.

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Slay the P.E. last won the day on October 7

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About Slay the P.E.

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    Principal in Charge

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    www.slaythepe.com

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  • Location
    Prospect Heights, IL
  • Interests
    Mechanical Engineering Exam Prep

Previous Fields

  • Engineering Field
    CONSULTING
  • License
    PE
  • Discipline
    Mechanical

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  1. Slay the P.E.

    TFS practice problem of the week...

    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.
  2. Slay the P.E.

    TFS practice problem of the week...

    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.
  3. Slay the P.E.

    TFS practice problem of the week...

    This is the correct approach. @MikeGlass1969
  4. Slay the P.E.

    TFS practice problem of the week...

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

    HVAC&R Practice Problem of the week

    Yep. Its badly crafted as is. The factor (3.28/.2) or (.1368/.-0083) is too big and makes the final answer too sensitive to Tdp. I need to change the relative magnitude of the given R-values to improve the problem. I think the problem itself is still pretty cool... just needs better numerical data. We shouldn't get a -7F to 9.3F swing by changing Tdp from 70F to 71F
  6. Slay the P.E.

    HVAC&R Practice Problem of the week

    Thanks. I’m going to have to re-write some of the given data in that problem so the final answer isn’t so sensitive to small variations of Tdp. I’ll make sure that using 70F or 70.2F yields roughly the same answer.
  7. Slay the P.E.

    HVAC&R Practice Problem of the week

    @mcc515 @MikeGlass1969 Thanks for working on the double pane window problem. Here is my solution. I get -4F which is different from what both of you are getting. Let me know if you see anything wrong
  8. Slay the P.E.

    HVAC&R Practice Problem of the week

    Mike, the "extended" Bernoulli equation is actually a statement of conservation of energy. as such, it is essentially this (for steady state): The rate at which energy enters a control volume = The rate at which energy leaves the control volume In this problem a control volume is defined with boundaries just below the water level in the reservoirs. It is at these locations where mass enters/leaves the control volume. The left hand side of the energy balance will have the enthalpy (pressure head, which is zero), kinetic energy (zero), and potential energy (height) entering the control volume at the reservoir in the right side of the drawing. That is all we have on the left side of the equation. (If we had a pump, the pump head would be on this side, because it is energy entering the control volume) The right hand side of the energy balance will have the enthalpy (pressure head, which is zero), kinetic energy (zero), and potential energy (height) leaving the control volume at the reservoir in the left side of the drawing. The right hand side of the equation will also contain the energy that is "lost" (stray heat is an enthalpy, which in hydraulics ends up being pressure loss) so the friction and minor losses are forms of energy leaving the control volume. These terms therefore, belong on the right side of the equation. A much shorter way to say all this is that in the extended Bernoulli equation, the left side of the equation (subscript 1) is the inlet reservoir, and the right side (subscript 2) is the discharge reservoir. Pump head (if non-zero) goes in the left side of the equation. Turbine head (if non-zero) goes in the right side of the equation. Friction and minor losses go in the right side of the equation.
  9. Slay the P.E.

    MD&M practice problem of the week

    It might be a little quicker if you approach it by finding the instantaneous center of rotation for rod BC.
  10. Slay the P.E.

    TFS practice problem of the week...

    Correct.
  11. Slay the P.E.

    MD&M practice problem of the week

    That is correct. The actual answer is 43.5 ft/s
  12. Slay the P.E.

    HVAC&R Practice Problem of the week

    An indoor swimming pool room has a 6 ft × 4 ft double-pane, outside wall window with an effective R-value of 3 (°F ft2 )/(Btu/h). The R-value for the indoor air film is 0.2 (°F ft2 )/(Btu/h), and the convection coefficient for the side exposed to outdoor air is 12 (Btu/h)/(°F ft2 ). The indoor air is maintained at 75°F with a relative humidity of 85%. Under these conditions, the winter outdoor air temperature (°F) below which condensation on the inner face of the window is expected to form is most nearly: (A) -4 (B) 0 (C) 4 (D) 12
  13. Slay the P.E.

    TFS practice problem of the week...

    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
  14. Slay the P.E.

    MD&M practice problem of the week

    In the engine system shown, crank AB is 3 inches long and has a constant clockwise angular velocity of 2000 rpm, and connecting rod BC is 8 inches long. When angle α is 40 degrees, the velocity (ft/s) of piston P is most nearly: (A) 3.7 (B) 21 (C) 41 (D) 44
  15. Slay the P.E.

    MD&M practice problem of the week

    The critical load formula is derived by assuming that buckling will occur at that load, that is when P = Pcrit. Therefore, the factor of safety for buckling is defined as Pcrit/P. Try it this way.
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