An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airstream that is

Question

An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airstream that is in parallel flow over the top of the strips. Each strip is 0.2 m wide, and 25 strips are arranged side by side, forming a continuous and smooth surface over which the air flows at 2 m/s. During operation, each strip is maintained at 500°C and the air is at 25°C. What is the rate of convection heat transfer from the first strip? The fifth strip? The tenth strip? All the strips?

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Yến Oanh 3 years 2021-08-10T18:50:52+00:00 1 Answers 417 views 1

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  1. Maris
    0
    2021-08-10T18:52:17+00:00
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    Answer:

    see explanation below

    Explanation:

    Given that,

    T_1 = 500°C

    T_2 = 25°C

    d = 0.2m

    L = 10mm = 0.01m

    U₀ = 2m/s

    Calculate average temperature

    \\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5

    262.5 + 273

    = 535.5K

    From properties of air table A-4 corresponding to T_{avg} = 535.5K \approx 550K

    k = 43.9 × 10⁻³W/m.k

    v = 47.57 × 10⁻⁶ m²/s

    P_r = 0.63

    A)

    Number for the first strips is equal to

    R_e_x = \frac{u_o.L}{v}

    R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4

    Calculating heat transfer coefficient from the first strip

    h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3

    h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2

    The rate of convection heat transfer from the first strip is

    q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W

    The rate of convection heat transfer from the fifth trip is equal to

    q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)

    h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2

    Calculating h_o_-_4

    h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2

    The rate of convection heat transfer from the tenth strip is

    q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)

    h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2

    Calculating

    h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2

    Calculating the rate of convection heat transfer from the tenth strip

    q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W

    The rate of convection heat transfer from 25th strip is equal to

    q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)

    Calculating h_o_-_2_5

    h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2

    Calculating h_o_-_2_4

    h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2

    Calculating the rate of convection heat transfer from the tenth strip

    q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W

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