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28 May, 14:07

A turbine blade 6.3 cm long with cross sectional area 4.6x10-4 m2 and perimeter 0.12 m is made of stainless steel (k=18 w/mk). the temperature of the root is 482oc. the blade is exposed to a hot gas flow at 871oc with a heat transfer coefficient 454 w/m2k. determine the temperature of the blade tip and the rate of heat transfer at the root of the blade. tip could be assumed to be insulated.

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  1. 28 May, 15:37
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    There are two equations to be used for this problem. The firs one is the general formula for heat transfer:

    Q = kAΔT

    where

    Q is the rate of heat transfer

    k is the heat transfer coefficient

    A is the cross-sectional area

    ΔT is the temperature difference

    Substituting the values:

    Q = (454 W/m²·K) (4.6x10⁻⁴ m²) (871°C - 482°C)

    Q = 81.24 W

    Thus, the rate of heat transfer is 81.24 W.

    We use the Q for the second equation. The radial heat transfer would be:

    Q = 2πkL (T₁ - T₂) / ln (r₂/r₁)

    where

    L is the length of the turbine

    r₂ is the distance of tip blade to the center

    r₁ is the distance of root blade to the center

    T₂ is temperature at the tip blade

    T₁ is temperature at the root blade

    Perimeter = 2πr₂

    0.12 m = 2πr₂

    r₂ = 0.019 m

    Cross-sectional area = πr₁²

    4.6x10⁻⁴ m² = πr₁²

    r₁ = 0.012 m

    Substituting to the equation:

    81.24 W = 2π (454 W/m²·K) (6.3 cm * 1m/100 cm) (482°C - T₂) / ln (0.019 m/0.012 m)

    Solving for T₂,

    T₂ = 481.79°C
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