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Thread: Wind Speed load. API 650. Doubt

  1. #1

    Wind Speed load. API 650. Doubt

    Good morning, fellows

    I am checking the design of some storage tanks in Oman, and I have noticed that the mechanical engenieer has decided to use the ASCE 7, as the API 650 12th edition sets.

    But...the wind speed (design value) that he gets is 82 km/h, and I think that in Oman it is not enough to enusre the stability of the shell against the wind load.

    I remember, that in API 650 11th edition, was clearly that we have to take a normal value for the wind speed of 190 km/h.

    Lookin for more data, I have found this sentences from a web page:

    Para. 5.2.1k. Both ASCE 7-05 and ASCE 7-10 are now referenced as acceptable alternatives for determining the design wind speed. Explicit reference to a 190 km/hr (120 mph) design wind velocity as a “base case” is not included. However, 190 km/hr (120 mph) is still implied when calculating a design wind pressure in this paragraph and later in wind girder calculation requirements.


    I am not saying that the calculation have been made on the wrong way. I am only want to ensure that tanks will last 50 years without structural problemas.

    Anyone can help me?

    Thaks a lot in advance and best regards

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  3. #2

    Re: Wind Speed load. API 650. Doubt

    Your case would be considered under provisions of 5.2.1 Loads. point k Wind, third alternative as in the following API 650 quote:

    Wind (W): The design wind speed (V) shall be either:
    — the 3-sec gust design wind speed determined from ASCE 7-05 multiplied by √I, Figure 6-1; or
    — the 3-sec gust design wind speed determined from ASCE 7-10 for risk category specified by the Purchaser (Figure 26.5-1A, Figure 26.5-1B, or Figure 26.5-1C) multiplied by 0.78; or
    — the 3-sec gust design wind speed specified by the Purchaser, which shall be for a 3-sec gust based on a 2 % annual probability of being exceeded [50-year mean recurrence interval].
    The 3-sec gust wind speed used shall be reported to the Purchaser.

    I think you may study first the figure 82 km/h has the same mean recurrence interval (associated risk of occurrence) and has the same mean time (3-sec).
    In case they used other reference, you would comment/ correct it.
    In case you are the Purchaser, you would ask for supplementary details as above and decide the correct figure
    Last edited by marian_g; 05-27-2015 at 12:25 PM.

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  5. #3

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    Re: Wind Speed load. API 650. Doubt

    In addition; the 190 km/hr is a also a reference factor used in API 650 for wind stiffeners, the correction factor is (V/190)^2.
    Consideration shall also be given to the lifting force for tanks with roofs, so there is a horizontal ans also a vertical UP-LIFT wind force.

  6. #4

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    Re: Wind Speed load. API 650. Doubt

    In addition; the 190 km/hr is a also a reference factor used in API 650 for wind stiffeners, the correction factor is (V/190)^2.
    Consideration shall also be given to the lifting force for tanks with roofs, so there is a horizontal ans also a vertical UP-LIFT wind force.

  7. #5

    Re: Wind Speed load. API 650. Doubt

    Thanks for the answers.

    I have just came back from Oman, where I held some meetings with customer.

    The problem is that they say have an estimate of wind in which, by application of the 3-sec method, according to ASCE 7-05 , is a value of 82 km / h .

    For me , regardless of the value they have calculated ( I asked the calculations to check them ) , should it impose logic and a tank with a calculation of resistance to wind in the desert , less than 144 km / h , seems me recklessness.

    For good practice , I have always gone at speeds exceeding 144 km / h , when calculating the equivalent height wind, I have come to force V / 190 factor to 1, to make sure durability tank windproof .

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  9. #6

    Re: Wind Speed load. API 650. Doubt

    Please see an article found on Internet.
    Please note that the article considers the maximum hourly-mean instead 3-sec gust.


    NOTE. linus mentioned the Durst correlation, so I attached FIGURE C26.5-1 Maximum Speed Averaged over t s to Hourly Mean Speed, from ASCE 7; you may conclude you can take a factor of 1.52 to the hourly-mean speed in order to get 3-sec gust speed. It seems that 25 m/s (90 km/hr) as basic hourly-mean speed wind from article corresponds to a design wind of 137 km/hr as 3-sec gust, so I think your requirement is quite reasonable.
    Last edited by marian_g; 06-08-2015 at 09:08 AM. Reason: added NOTE

  10. #7

    Re: Wind Speed load. API 650. Doubt

    API currently uses 3 sec gust speed . Earlier it was fastest mile. You need to convert between the units using Durst correlation.

  11. #8

    Re: Wind Speed load. API 650. Doubt

    If one can read Note 2 on page 5-84 of API 650 12th Ed Addenda -Sept-2014, Pl read point(d) under Note-2, It is talking about decreasing the unstiffened height H1, based on total pr. 1.72 (basic, = based on 190 kmph + 25 mm H2O vac). It is not talking about increasing the unstiffened height H1, of shell.This means (V/190)^2 is always 1.0 as a minimum.It may be more if site wind velocity is more than 190 kmph. I always interprete it like that.

    KP Sahu

  12. #9

    Re: Wind Speed load. API 650. Doubt

    The actual wording is "When other factors are specified by the Purchaser that are greater than the factors in Items a, b, and c, the total load on the shell shall be modified accordingly, and H1 shall be decreased by the ratio of 1.72 kPa (36 lbf/ft2) to the modified total pressure."

    In my understanding, it refers only to the case when the modified total pressure is greater than 1.72 kPa (36 lbf/ft2) and consequently H1 must be decreased.

    I do not second your logic that the missing wording "increasing" means (V/190)^2 is always 1.0.

  13. Re: Wind Speed load. API 650. Doubt

    I have to recognize that the paragraph of API (mentioned by KP SAHU) is rather confusing. As a detail, the "problem" is that the total pressure has two components: velocity pressure and external pressure (vacuum equivalent)- the last one is not V^2 proportional.


    In my opinion, for a velocity V [in specific units], other factors than considered in 5.9.7.1 Note a- giving a Velocity_pressure [in specific units] plus External_Pressure [in specific units] as external pressure, the result should be amended by factor:
    1.72/[Velocity_pressure +External_Pressure]
    or 36/[[Velocity_pressure +External_Pressure]

    Maybe it appears as a complicated procedure; it may be simplified keeping in mind that the formula
    H=9.47*t*sqrt[(t/D)^3] in SI units
    considers in denominator 1.72 kPa as total pressure.
    When we have the total pressure Velocity_pressure +EP we need to consider Velocity_pressure +EP in denominator; this is equivalent to consider in the first phase 1.72 kPa in denominator and to multiply the results with 1.72/[Velocity_pressure +EP] later.

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