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1 April, 10:46

Air at 1 atm enters a thin-walled (5-mm diameter) long tube (2 m) at an inlet temperature of 100°C. A constant heat flux is applied to the air from the tube surface. The air mass flow rate is 140 * 10-6 kg/s. If the tube surface temperature at the exit is 160°C, determine the heat rate entering the tube, in W. Evaluate properties at 400 K.

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  1. 1 April, 13:26
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    heat rate = 7.38 W

    Explanation:

    Given dа ta:

    Pressure = 1atm

    diameter (D) = 5mm = 0.005m

    length = 2

    mass flow rate (m) = 140*10^-6 kg/s

    Exit temperature = 160°C,

    At 400K,

    Dynamic viscosity (μ) = 22.87 * 10^-6

    Prandtl number (pr) = 0.688

    Thermal conductivity (k) = 33.65 * 10^-3 W/m-k

    Specific heat (Cp) = 1.013kj/kg. K

    Step 1: Calculating Reynolds number using the formula;

    Re = 4m/πDμ

    = (4*140*10^-6) / (π * 0.005*22.87 * 10^-6)

    = 5.6*10^-4/3.59*10^-7

    = 1559.

    Step 2: Calculating the thermal entry length using the formula

    Le = 0.05*Re*Pr*D

    Substituting, we have

    Le = 0.05 * 1559 * 0.688 * 0.005

    Le = 0.268

    Step 3: Calculate the heat transfer coefficient using the formula;

    Nu = hD/k

    h = Nu*k/D

    Since Le is less than given length, Nusselt number (Nu) for fully developed flow and uniform surface heat flux is 4.36.

    h = 4.36 * 33.65 * 10^-3/0.005

    h = 0.1467/0.005

    h = 29.34 W/m²-k

    Step 4: Calculating the surface area using the formula;

    A = πDl

    =π * 0.005 * 2

    =0.0314 m²

    Step 5: Calculating the temperature Tm

    For energy balance,

    Qc = Qh

    Therefore,

    H*A (Te-Tm) = MCp (Tm - Ti)

    29.34 * 0.0314 (160-Tm) = 140 * 10-6 * 1.013*10^3 (Tm-100)

    0.921 (160-Tm) = 0.14182 (Tm-100)

    147.36 - 0.921Tm = 0.14182Tm - 14.182

    1.06282Tm = 161.542

    Tm = 161.542/1.06282

    Tm = 151.99 K

    Step 6: Calculate the rate of heat transferred using the formula

    Q = H*A (Te-Tm)

    = 29.34 * 0.0314 (160-151.99)

    = 7.38 W

    the Prandtl number using the formula
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