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

Answer 1

Answer:

1581357.68216 Pa

163.21352 m

Explanation:

A = Area = $\pi r^2$

v = Velocity

1 denotes inlet

2 denotes exit

$\rho$ = Density of water = 1000 kg/m³

Flow rate is given by

$Q=A_1v_1\\\Rightarrow v_1=\dfrac{Q}{A_1}\\\Rightarrow v_1=\dfrac{40\times 10^{-3}}{\pi 0.045^2}\\\Rightarrow v_1=6.2876\ m/s$

$Q=A_2v_2\\\Rightarrow v_2=\dfrac{Q}{A_2}\\\Rightarrow v_2=\dfrac{40\times 10^{-3}}{\pi 0.015^2}\\\Rightarrow v_1=56.58842\ m/s$

From the Bernoulli equation we get

$P_1+\dfrac{1}{2}\rho v_1^2=P_2+\dfrac{1}{2}\rho v_2^2\\\Rightarrow P_2-P_1=\dfrac{1}{2}\rho (v_2^2-v_1^2)\\\Rightarrow \Delta P=\dfrac{1}{2}1000 (56.58842^2-6.2876^2)\\\Rightarrow \Delta P=1581357.68216\ Pa$

The pressure drop is 1581357.68216 Pa

From the Bernoulli equation

$P_a+\dfrac{1}{2}\rho v_1^2=P_a+\dfrac{1}{2}\rho v_2+\rho gh\\\Rightarrow h=\dfrac{\dfrac{1}{2}v_1^2}{g}\\\Rightarrow h=\dfrac{\dfrac{1}{2}56.58842^2}{9.81}\\\Rightarrow h=163.21352\ m$

The height is 163.21352 m

Answer 2

Answer:

(a) 1581935 Pa

(b) 161.4 m

Explanation:

diameter, d = 3 cm

r = 1.5 cm

diameter, d = 9 cm

R = 4.5 cm

Volume per second, V = 40 L/s = 0.04 m^3/s

V = a x v1

0.04 = 3.14 x 0.015 x 0.015 x v1

v1 = 56.6 m/s

Now, 0.04 = 3.14 x 0.045 x 0.045 x v2

v2 = 6.3 m/s

(a) By use of Bernoullie's theorem

$P_{1}+\frac{1}{2}dv_{1}^{2}=P_{2}+\frac{1}{2}dv_{2}^{2}$

where, d be the density of water

$P_{1}+\frac{1}{2}\times 1000\times 56.6\times 56.6=P_{2}+\frac{1}{2}\times 1000\times 6.3\times 6.3$

P2 - P1 = 1581935 Pa

(b) Let h be the maximum height

P2 - P1 =  x d x g

1581935 = h x 1000 x 9.8

h = 161.4 m