Question

Water is delivered to a 4.0 cm diameter fountain head from a 6.9 cm diameter pipe that is 3.4 m below the fountain head and at a pressure of 94 kPa. The fountain head shoots the water straight up into the air. Neglect all losses.

a.) What is the velocity of water coming out of the fountain head? Hint: Must use Bernoulli AND Continuity

b.) What is the volume flow rate of the water through the fountain head? Hint: Q=vA

Answer 1

Diameter of pipe D₁ = 6.9 cm, D₂ = 4 cm , Height = 3.4 m, P₁ = 94 kPa, P₂ = 101.3 kPa

Part A)

$\pi \left ( \frac{D_{1}}{2} \right )^{2}v_{1} = \pi \left ( \frac{D_{2}}{2} \right )^{2}v_{2}$

$v_{1}= \left (\frac{D_{2}}{D_{1}} \right )^{2}v_{2}$

$v_{1}= \left (\frac{6.9 cm}{4.0 cm} \right )^{2}v_{2} = 2.98v_{2}$

Using Bernoulli's theorem

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

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

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

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

$7.88 v_{2}^{2}=\frac{2(101.3\times 10^{3}Pa - 94 \times 10^{3}Pa)}{1000 kg/m^{3}}+ 2\times 9.8 m/s^{2}\times 3.4 m$

$v_{2}=\sqrt{\frac{81.24 m^{2}/s^{2}}{7.88 }} = 3.21 m/s$

Part B)

$Q = vA = 3.21 m/s \times (\pi \times (2\times 10^{-2}m)^{2})= 4.034\times 10^{-3} m^{3}/s$