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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? (b) To what maximum height above the nozzle can this water rise? (The actual height will be significantly smaller due to air resistance.)

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From Bernoulli Equation\textbf{Bernoulli Equation} we know that :

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2P_{1} + \dfrac{1}{2} \rho v_{1}^2 + \rho gh_{1}= P_{2} + \dfrac{1}{2} \rho v_{2}^2 + \rho g h_{2}

When the two point are at the same height : it reduced to

P1+12ρv12=P2+12ρv22P_{1} + \dfrac{1}{2} \rho v_{1}^2 = P_{2} + \dfrac{1}{2} \rho v_{2}^2

Where:

  • P1P_{1} , P2P_{2} are the pressures at points 1, 2 .
  • ρ\rho is the density of the fluid .
  • v1v_{1} ,v2v_{2} are the velocities of the fluids at points 1,2 .

Givens\textbf{Givens} :r2=0.015×mr_{2} = 0.015\times \mathrm{m} , ρ=1000kg/m3\rho = 1000 \mathrm{kg/m^3} , r1=0.045mr_{1} = 0.045 \mathrm{m} , g=9.8m/s2g= 9.8 \mathrm{m/s^2}

Plugging\textbf{Plugging} known information to get :

Q=A1v1v1=QA1=40×103π0.0452=6.29v2=QA2=40×103π0.0152=56.6\begin{align*} Q&= A_{1} v_{1} \\ v_{1} &= \dfrac{Q}{A_{1}} \\ &= \dfrac{40 \times 10^{-3} }{ \pi 0.045^2} \\ &=6.29 \\ v_{2} &= \dfrac{Q}{A_{2}} \\ &= \dfrac{ 40 \times 10^{-3}}{ \pi 0.015^2} \\ &= 56.6 \end{align*}

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