Nothing Special   »   [go: up one dir, main page]
Scribd Logo
Upload

Welcome to Scribd!

  • 67 views

    Assignment 5 - Pool Fire - s2020027

    Uploaded by

    Sherlyn Lee
    A hydrocarbon liquid is released from confinement at a rate of 0.12 m3/s into a bund wall of 27 m diameter. If the liquid catches fire, the heat flux to a person 55 m away is estimated using two models. The point source model estimates a heat flux of 8.75 kW/m2. The solid plume model accounts for flame obscuration and estimates a lower heat flux of 1.005 kW/m2, showing the importance of accounting for factors like soot in the fire modeling.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd

    Assignment 5 - Pool Fire - s2020027

    Uploaded by

    Sherlyn Lee
    67 views4 pages
    A hydrocarbon liquid is released from confinement at a rate of 0.12 m3/s into a bund wall of 27 m diameter. If the liquid catches fire, the heat flux to a person 55 m away is estimated using two models. The point source model estimates a heat flux of 8.75 kW/m2. The solid plume model accounts for flame obscuration and estimates a lower heat flux of 1.005 kW/m2, showing the importance of accounting for factors like soot in the fire modeling.

    Original Description:

    Original Title

    Assignment 5_ Pool Fire_ s2020027

    Copyright

    © © All Rights Reserved

    Available Formats

    DOCX, PDF, TXT or read online from Scribd

    Did you find this document useful?

    Is this content inappropriate?

    A hydrocarbon liquid is released from confinement at a rate of 0.12 m3/s into a bund wall of 27 m diameter. If the liquid catches fire, the heat flux to a person 55 m away is estimated using two models. The point source model estimates a heat flux of 8.75 kW/m2. The solid plume model accounts for flame obscuration and estimates a lower heat flux of 1.005 kW/m2, showing the importance of accounting for factors like soot in the fire modeling.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    67 views4 pages

    Assignment 5 - Pool Fire - s2020027

    Uploaded by

    Sherlyn Lee
    A hydrocarbon liquid is released from confinement at a rate of 0.12 m3/s into a bund wall of 27 m diameter. If the liquid catches fire, the heat flux to a person 55 m away is estimated using two models. The point source model estimates a heat flux of 8.75 kW/m2. The solid plume model accounts for flame obscuration and estimates a lower heat flux of 1.005 kW/m2, showing the importance of accounting for factors like soot in the fire modeling.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    of 4

    Assignment 5: Pool Fire

    Name : Lee Shei Ching Matric No: S2020027

    A hydrocarbon liquid with a high molecular weight is released from confinement at the
    volumetric rate of 0.12 m3/s. A bund wall of 27 m diameter exist to constrain the spill. If the
    liquid caught fire, estimate the heat flux to a person standing at a distance of 55 m from the
    bund wall. Assume that there is no wind and the relative humidity is 80%. Estimate the
    heat flux using the point source radiation model as well as the solid plume radiation model.
    Given data:
    - Heat of combustion of the liquid: 44,700 kJ/kg
    - Heat of vaporisation at ambient temperature: 320 kJ/kg
    - Liquid boiling point: 375 K = 101.85°C
    - Density of liquid: 760 kg/m3
    - Heat capacity of liquid (constant): 2.8 kJ/kg °C
    - Ambient temperature: 303 K = 29.85°C
    Answer Script
    Heat of vaporisation modified at the liquid boiling point:
    T BP
    o
    Δ H =Δ H v + ∫ C p dT
    Ta

    o
    101.85 C
    kJ kJ kJ kJ
    o
    (
    Δ H = 320 )
    + ∫ 2.8
    kg 29.85 C kg
    dT = 320
    o kg ( ) (
    +2.8 (101.85−29.85)
    kg )
    kJ
    Δ H o=521.6
    kg
    Vertical Burning Rate:

    Δ Hc
    ẏ max =1.27 ×10
    −6

    ΔH
    o
    −6
    =1.27 × 10 ( 44700
    521.6 )

    ẏ max =1.089 ×10−4 m/ s


    Mass Burning Rate:
    kg
    m B =ρ ẏ max=(760 3
    )¿ 1.089 ×10−4 m/s ¿
    m

    m B =0.0828 kg /m2 s

    The maximum, steady state pool diameter:

    V̇ L 0.12m 3 /s
    D max =2
    √ √
    π ẏ
    =2
    π (1.089 ×10−4 m/s)
    =37.457 m

    Since this is larger than the diameter of the diked area, the pool will be constrained by the dike
    with a diameter of 27 m. The area of the pool is:
    2
    π D 2 π ( 27 )
    A= = =572.2 m2
    4 4

    Given that :

    - Density of air = 1.2 kg/m3


    - Gravitational acceleration = 9.81 m/s2
    0.61
    kg
    0.0828

    (( )
    0.61
    H mB m2 s
    D
    =42
    (
    ρa √ gD ) =42
    kg m
    =1. 499
    1.2 3
    m ) √( )
    9.81 2 ( 27 m )
    s

    Thus,
    H
    H= × D=( 1.499 ) ( 27 m )=40.473 m
    D
    Point Source Model

    With the right-


    angled triangle being formed:

    x=√ 20.2372 +68.52=71.427 m


    View Factor:
    1 1
    F p= 2
    = 2
    =1.56 ×10−5 m−2
    4 π x 4 π ( 71.427 )
    Partial pressure of water:
    5328
    80 14.4114− 303 K
    Pw =1013.25 ( )
    100
    e =3395 N /m2 @ 303K

    Transmissivity:
    −0.09
    N
    τ a=2.02 ( Pw X s )−0.09
    [( )
    =2.02 3395 2 ( 71.427 m )
    m ] =0.662

    Thermal flux with conservative value 0.4 for the fraction of the energy converted to
    radiation,

    Er =τ a η m B Δ H c A F p =( 0.662 )( 0.4 ) ( 0.0828 )( 44700 ) (572.56 ) ( 1.56 × 10−5 )


    kJ
    Er =8.75 2
    =8.75 kW /m2
    m s
    Solid Plume Radiation Model
    Estimate of the radiant flux at the source:
    - 𝐸𝑚 is the maximum emissive power of the luminous spots (approximately 140 kW/m2)
    - S is an experimental parameter (0.12m-1)
    - 𝐸𝑠 is the emissive power of smoke (approximately 20 kW/m2)
    kW −(0.12m kW
    ( ) )(27 m)
    −1 −1

    Eav =E m e−SD + E s ( 1−e−SD ) = 140 e +(20 )(1−e−(0.12 m ) ( 27 m)


    )
    m2 2
    m

    Eav =24.7 kW /m2


    H H
    =1.499∴ =2 ( 1.499 )=2.998 ≃ 3
    D R
    The distance from the flame axis to the receiver = 55m+27/2 m = 68.5m

    ¿ 68.5
    ∴ Dimensionless Distance ¿ Flame Axis=Distance¿ Flame Axis = m=5.07
    Pool Radius 13.5
    By referring to Figure 3.33 the Maximum view factor = 0.06

    Atmospheric transmissivity:
    −0.09
    N
    −0.09
    τ a=2.02 ( Pw X s )
    [( )
    =2.02 3395 2 ( 55 m )
    m ] =0.678

    Radiant Flux:
    kW kW
    (
    Er =τ a E av F21=( 0.678 ) 24.7
    m 2)( 0.06 ) =1.005 2
    m

    The result from the solid plume radiation model is smaller than the point source model.
    This is most likely due to consideration of the radiation obscuration by the flame soot, a
    feature not treated directly by the point source model. The differences between the two
    models might be greater at closer distance to the pool fire.

    You might also like