Question

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?

Answer 1

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, $\dot{V}=40\ L.s^{-1}=0.04\ m^3.s^{-1}$

Now, flow velocity in hose:

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

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

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

Now, flow velocity in nozzle:

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

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

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

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

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

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

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

$\Delta P=1581357.92\ Pa$