Question

A horizontal pipe 15.0 cm in diameter has a smooth reduction to a pipe 7.50 cm in diameter. If the pressure of the water in the larger pipe is 9.40 104 Pa and the pressure in the smaller pipe is 2.40 104 Pa, at what rate does water flow through the pipes?

Answer 1

Answer:

The rate flow of water through the pipes is 60.34 kg/s.

Explanation:

Given that,

Diameter of pipe d₁= 15.0 cm

Reduce diameter d₂= 7.50 cm

Pressure of large pipe $P_{l}= 9.40\times10^{4}\ Pa$

Pressure of smaller pipe $P_{s}=2.40\times10^{4}\ Pa$

We need to calculate the velocity

Using equation of continuity

$A_{1}v_{1}=A_{2}v_{2}$

$\pi\times r^2\times v_{1}=\pi\times r^2\times v_{2}$

Put the value into the formula

$(7.5\times10^{-2})^2\times v_{1}=(3.75\times10^{-2})^2\times v_{2}$

$v_{1}=\dfrac{(3.75\times10^{-2})^2}{(7.5\times10^{-2})^2}\times v_{2}$

$v_{1}^2=\dfrac{1}{4}\times v_{2}^2$

$v_{1}=\dfrac{1}{2}\times v_{2}$....(I)

We need to calculate the velocity

Using Bernoulli equation

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

Here, $h_{1}=h_{2}$

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

$P_{1}-P_{2}=\dfrac{1}{2}\times\rho (v_{2}^2-v_{1}^2)$

Put the value into the formula

$9.40\times10^{4}-2.40\times10^{4}=\dfrac{1}{2}\times1000\times(v_{2}^2-\dfrac{1}{4}\times v_{2}^2)$

$7.0\times10^{4}=500\times(\dfrac{3}{4}\times v_{2}^2)$

$v_{2}^2=\dfrac{7.0\times10^{4}\times4}{3\times500}$

$v_{2}=\sqrt{\dfrac{7.0\times10^{4}\times4}{500\times3}}$

$v_{2}=13.66\ m/s$

We need to calculate the flow of water through the pipe

Using formula of flow

$flow = Area_{2}\times v_{2}$

Put the value into the formula

$flow=\pi\times(3.75\times10^{-2})^2\times13.66$

$flow=0.06034\ m^3/s$

We need to calculate the flow rate of water

Using formula of flow rate

$M=flow\times\rho_{w}$

Put the value into the formula

$M=0.06034\times1000$

$M=60.34\ kg/s$

Hence, The rate flow of water through the pipes is 60.34 kg/s.