Chemical Engineering Sample Exams: For the New Breadth/Depth Exams, First Edition

Solutions to PM Sample Exam 2

Fluids

Control of flow systems:

  1. This is closest to 5:1.

    Answer is (B)

  2. To fill the missing numbers for line and equipment losses, assume the pump curve is flat. Therefore the discharge pressure applicable to second branch is 7.89 kg/cm 2. Then allowable control valve pressure drop [kg/cm 2] in Branch #2 is:

    CV pressure drop = 7.89 - 3.5 - 0 .42 - 1.68 = 2.29 kg/cm 2

    Answer is (C)

  3. Numbers for pressure drops due to friction and equipment are missing. First calculate these losses. For this, use the fact that pressure drop is proportional to square of velocity or flow rate.

    • Frictional loss = 2 0.42 = 0.61 kg/cm 2

    • Equipment loss = 2 1.68 = 2.42 kg/cm 2

    • Therefore CV pressure drop at maximum flow in Branch #2 is:

    • CV pressure drop = 7.89 - 3.5 - 0.61 - 2.42 = 1.36 kg/cm 2

    Answer is (D)

  4. C vc = 1.16(17.4) = 15.5

    If the valve size 1 1/2" is selected , C v = 21

    Then C v/C vc = 21/15.5 = 1.35

    This lies between 1.25 to 2 which is required for good operation of the control valve.

    Answer is (B)

Economics:

  1. Present value, P

    =

    $ 100,000

    Salvage value, L

    =

    $ 10,000

    Useful life, n

    =

    10 years

    Straight line depreciation = = $9000 per year.

    Book value after 5 years = 100000 - 9000(5) = $55,000

    Answer is (A)

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