Answer to Question #184385 in Mechanics | Relativity for Konoha

Question

What is the pressure drop due to the Bernoulli Effect as water goes into a 4.44-cm-diameter nozzle from a 15.25-cm-diameter fire hose while carrying a flow of 72 L/s? (b) To what maximum height above the nozzle can this water rise? 

Round your answer to 0 decimal places

Answer 1

To be given in question

Radius

R₁=7.625$\times10^{-2}m$

R₂=2.22$\times10^{-2}m$

$Q=72\times10^{-3}m/sec$

To be asked in question

Velocity

v₁=?

v₂=?

Part(a)

We know that

Bernoulli eqution

$P_{1}+\frac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\frac{1}{2}\rho v_{2}^2+\rho gh_{2}$

We know that h=h₁=h₂

$P_{1}+\frac{1}{2}\rho v_{1}^2=P_{2}+\frac{1}{2}\rho v_{2}^2\rightarrow(1)$

Continuity equation

$Q=A_{1}v_{1}=A_{2}v_{2}\rightarrow(2)$

$v_{1}=\frac{Q}{A_{1}}$

Put value

$v_{1}=\frac{72\times10^{-3}}{3.14\times(7.625\times10^{-2})^2}$

$v_{1}=3$ m/sec

$v_{2}=\frac{Q}{A_{2}}$

Put value

$v_{2}=\frac{72\times10^{-3}}{3.14\times(2.22\times10^{-2})^2}$

$v_{2}=46m/sec$

Part (b)

We can re -written as equation (2)

$P_{1}+\frac{1}{2}\rho v_{1}^2=P_{2}+\frac{1}{2}\rho v_{2}^2$

$∆P=\frac{1}{2}\rho v_{2}^2-\frac{1}{2}\rho v_{1}^2$

$∆P=\frac{1}{2}\rho( v_{2}^2-v_{1}^2)\rightarrow(3)$

Put value in eqution (3)

$∆P=1.053\times10^6N/m^2$