a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diameter. Data fo

Question

a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diameter. Data for two tests are available. For a flow rate of 12 ft3/hr the pressure drop is 9.6 psi; for 24 ft3/hr, the pressure drop is 24.1psi. The pump capacity produces a pressure drop of up to 50 psi. Requirements: what is the upper limit on the flow rate

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Tryphena 4 years 2021-07-19T15:19:47+00:00 1 Answers 10 views 0

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  1. Maris
    0
    2021-07-19T15:20:55+00:00
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    Answer:

    The upper limit on the flow rate = 39.46 ft³/hr

    Explanation:

    Using Ergun Equation to calculate the pressure drop across packed bed;

    we have:

    \frac{\delta P}{L}= \frac{150 \mu_oU(1- \epsilon )^2}{d^2p \epsilon^3} + \frac{1.75 \rho U^2(1-\epsilon)}{dp \epsilon^3}

    where;

    L = length of the bed

    \mu = viscosity

    U = superficial velocity

    \epsilon = void fraction

    dp = equivalent spherical diameter of bed material (m)

    \rho = liquid density (kg/m³)

    However, since U ∝ Q and all parameters are constant ; we can write our equation to be :

    ΔP = AQ + BQ²

    where;

    ΔP = pressure drop

    Q = flow rate

    Given that:

    9.6 = A12 + B12²

    Then

    12A + 144B = 9.6       ————–   equation (1)

    24A + 576B = 24.1    —————  equation (2)

    Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So

    288 B = 4.9

           B = 0.017014

    From equation (1)

    12A + 144B  = 9.6

    12A + 144(0.017014) = 9.6

    12 A = 9.6 – 144(0.017014)

    A = \frac{9.6 -144(0.017014}{12}

    A = 0.5958

    Thus;

    ΔP = AQ + BQ²

    Given that ΔP = 50 psi

    Then

    50 = 0.5958 Q + 0.017014 Q²

    Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;

    Q² + 35.02Q – 2938.8 = 0

    Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;

    Q = 39.46 ft³/hr

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