Water flowing out of a horizontal pipe emerges through a nozzle. The radius of the pipe is 1.8 cm, and the radius of the nozzle is 0.53 cm.

Question

Water flowing out of a horizontal pipe emerges through a nozzle. The radius of the pipe is 1.8 cm, and the radius of the nozzle is 0.53 cm. The speed of the water in the pipe is 0.75 m/s. Treat the water as an ideal fluid, and determine the absolute pressure of the water in the pipe.

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King 3 years 2021-08-26T10:59:47+00:00 1 Answers 48 views 0

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  1. Nho
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    2021-08-26T11:01:28+00:00
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    Answer:

    The absolute pressure of the water in the pipe is 1.38 x 10⁵ Pa

    Explanation:

    Given;

    radius of the pipe, r₁ = 1.8 cm = 0.018 m

    radius of the pipe, r₂ = 0.53 cm = 0.0053 m

    speed of water in the pipe, v₁ = 0.75 m/s

    Water absolute pressure can be determined using Bernoulli’s equation;

    P₁ + ¹/₂ρv₁² = P₂ + ¹/₂ρv₂²

    P₁ =  P₂ + ¹/₂ρv₂² – ¹/₂ρv₁²

    P₁ =  P₂ + ¹/₂ρ (v₂² – v₁²)

    where;

    ρ is density of water = 1000 kg/m³

    P₂  is atmospheric pressure = 1.01 x 10⁵ Pa

    From continuity equation; A₁V₁ = A₂V₂

    πr₁²v₁² = πr₂²v₂²

    v_2 = \frac{ r_1^2v_1}{r_2^2}

    P_1 = P_2 + \frac{1}{2} \rho[(\frac{r_1^2v_1}{r_2^2} )^2 - v_1^2]\\\\P_1 = P_2 + \frac{1}{2} \rho[\frac{r_1^4v_1^2}{r_2^4} - v_1^2]\\\\P_1 = P_2 + \frac{1}{2} \rho v_1^2[\frac{r_1^4}{r_2^4} - 1]\\\\P_1 = 1.01*10^5 + \frac{1}{2}* 1000* 0.75^2[\frac{(0.018)^4}{(0.0053)^4} - 1]\\\\P_1 = 1.01*10^5 + \frac{1}{2}* 1000* 0.75^2(132.04)\\\\P_1 = 1.01*10^5 +37136.25 \\\\P_1 = 1.38 *10^5 \ Pa

    Therefore, the absolute pressure of the water in the pipe is 1.38 x 10⁵ Pa

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