A 5.7 diameter horizontal pipe gradually narrows to 3.6 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 32.5 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?

Feb 1, 2018

The flow rate is $= 2.88 {m}^{3} {s}^{-} 1$

Explanation:

Apply Bernoulli's Principle

${P}_{1} + \frac{1}{2} \rho {v}_{1}^{2} + \rho g {z}_{1} = {P}_{2} + \frac{1}{2} \rho {v}_{2}^{2} + \rho g {z}_{2}$

Since the Pipe is horizontal, ${z}_{1} = {z}_{2}$

So,

${P}_{1} + \frac{1}{2} \rho {v}_{1}^{2} = {P}_{2} + \frac{1}{2} \rho {v}_{2}^{2}$

The flow rate is constant

$Q = {A}_{1} \cdot {v}_{1} = {A}_{2} \cdot {v}_{2}$

Where ${v}_{1}$ and ${v}_{2}$ are the velocities of water in the pipe.

Where ${A}_{1} , {A}_{2}$ are the cross sectional areas of the pipe

${A}_{1} = \pi {d}_{1}^{2} / 4 = \pi \cdot {\left(5.7 \cdot {10}^{-} 2\right)}^{2} / 4$

${A}_{2} = \pi {d}_{2}^{2} / 4 = \pi \cdot {\left(3.6 \cdot {10}^{-} 2\right)}^{2} / 4$

Therefore,

${v}_{1} = {A}_{2} / {A}_{1} \cdot {v}_{2}$

${v}_{1} = \frac{\pi \cdot {\left(3.6 \cdot {10}^{-} 2\right)}^{2} / 4}{\pi \cdot {\left(5.7 \cdot {10}^{-} 2\right)}^{2} / 4} \cdot {v}_{2}$

${v}_{1} = {\left(\frac{3.6}{5.7}\right)}^{2} {v}_{2}$

${v}_{1} = 0.4 {v}_{2}$

The pressures are

${P}_{1} = 32.5 k P a$

${P}_{2} = 24 k P a$

And the density of water is $\rho = 1000 k g {m}^{-} 3$

Therefore,

$32.5 \cdot {10}^{3} + \frac{1}{2} \cdot 1000 \cdot 0.4 {v}_{2} = 24 \cdot {10}^{3} + \frac{1}{2} \cdot 1000 \cdot {v}_{2}$

${v}_{2} \left(\frac{1}{2} \cdot 1000 - \frac{1}{2} \cdot 1000 \cdot 0.4\right) = \left(32.5 - 24\right) \cdot {10}^{3}$

${v}_{2} \cdot \frac{1}{2} \cdot 1000 \cdot 0.6 = 8.5 \cdot 1000$

${v}_{2} = \frac{2 \cdot 8.5}{0.6} = 28.33 m {s}^{-} 1$

Finally,

The flow rate is $Q = {A}_{2} {v}_{2} = \pi \cdot {\left(3.6 \cdot {10}^{-} 2\right)}^{2} / 4 \cdot 28.33$

$= 2.88 {m}^{3} {s}^{-} 1$