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24 September, 21:55

Engine oil (unused) flows at 1.81 x 10^-3 kg/s inside a 1-cm diameter tube that is heated electrically at a rate of 76 W/m. At a particular location where flow and heat transfer are fully developed, the wall temperature is 370K. Determine:

a. The oil mean temperature.

b. The centerline temperature.

c. The axial gradient of the mean temperature.

d. The heat transfer coefficient.

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Answers (1)
  1. 25 September, 00:22
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    (a) Tb = 330.12 K (b) Tc = 304.73 K (c) 19.81 K/m (d) h = 60.65 W/m². K

    Explanation:

    Solution

    Given that:

    The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s

    Diameter of the tube, D = 1cm = 0.01 m

    Electrical heat rate, q = 76 W/m

    Wall Temperature, Ts = 370 K

    Now,

    From the properties table of engine oil we can deduce as follows:

    thermal conductivity, k = 0.139 W/m. K

    Density, ρ = 854 kg/m³

    Specific heat, cp = 2120 J/kg. K

    (a) Thus

    The wall heat flux is given as follows:

    qs = q/πD

    =76/π * 0.01

    = 2419.16 W/m²

    Now

    The oil mean temperature is given as follows:

    Tb = Ts - 11/24 (q. R/k) (R = D/2=0.01/2 = 0.005 m)

    Tb = 370 - 11/24 * (2419.16 * 0.005/0.139)

    Tb = 330.12 K

    (b) The center line temperature is given below:

    Tc = Ts - 3/4 (qs. R/k) = 370 - 3/4 * (2419.16 * 0.005/0.139)

    Tc = 304.73 K

    (c) The flow velocity is given as follows:

    V = m/ρ (πR²)

    Now,

    The The axial gradient of the mean temperature is given below:

    dTb/dx = 2 * qs/ρ * V*cp * R

    =2 * qs/ρ*[m/ρ (πR²) * cp * R

    =2 * qs/[m / (πR) * cp

    dTb/dx = 2 * 2419.16/[1.81 x 10^-3 / (π * 0.005) ] * 2120

    dTb/dx = 19.81 K/m

    (d) The heat transfer coefficient is given below:

    h = 48/11 (k/D)

    =48/11 (0.139/0.01)

    h = 60.65 W/m². K
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