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Dmitry [639]
3 years ago
6

What is the pressure drop due to the bernoulli effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

hose while carrying a flow of 40.0 l/s?
Physics
1 answer:
Paul [167]3 years ago
8 0

Answer:

\Delta P=1581357.92\ Pa

Explanation:

Given:

  • diameter of hose pipe, D=0.09\ m
  • diameter of nozzle, d=0.03\ m
  • volume flow rate, \dot{V}=40\ L.s^{-1}=0.04\ m^3.s^{-1}

<u>Now, flow velocity in hose:</u>

v_h=\frac{\dot V}{\pi.D^2\div 4}

v_h=\frac{0.04\times 4}{\pi\times 0.09^2}

v_h=6.2876\ m.s^{-1}

<u>Now, flow velocity in nozzle:</u>

v_n=\frac{\dot V}{\pi.d^2\div 4}

v_n=\frac{0.04\times 4}{\pi\times 0.03^2}

v_n=56.5884\ m.s^{-1}

We know the Bernoulli's equation:

\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}+Z_1=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}+Z_2

when the two points are at same height then the eq. becomes

\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}

\Delta P=\frac{\rho(v_n^2-v_h^2)}{2}

\Delta P=\frac{1000(56.5884^2-6.2876^2)}{2}

\Delta P=1581357.92\ Pa

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