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A month ago
A 6.8 cm diameter horizontal pipe gradually narrows to 5.2 cm . When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 33.0 kPa and 22.0 kPa , respectively.
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A month ago
We're looking for m^3 / s (volume flow rate per every 1 second).

Use the equation of continuity to relate the volume flow of water at the two locations, and use Bernoulli’s equation to relate the pressure conditions at the two locations. We assume that the two locations are at the same height. Express the pressures as atmospheric pressure plus gauge pressure. We use subscript "1" for the larger diameter, and subscript "2" for the smaller diameter.

\(A_1v_1=\pi r_1^2\times \sqrt{\frac{2\left(P_1-P_2\right)}{\rho \left(\frac{r_1^4}{r_2^4}-1\right)}}\)

Now substitute, make sure you divide each diameter by 2 to get the radius, and convert them to meters:

6.8 / 2 / 100 = 0.034 m
5.2 / 2 / 100 = 0.026 m

33.0 kPa > 33 000 Pa
22.0 kPa > 22 000 Pa

\(A_1v_1=\pi \left(0.034\right)^2\times \sqrt{\frac{2\left(33000-22000\right)}{1000\left(\frac{0.034^4}{0.026^4}-1\right)}}\)

Calculate:

\(A_1v_1=\pi \left(0.034\right)^2\times \sqrt{\frac{2\left(33000-22000\right)}{1000\left(\frac{0.034^4}{0.026^4}-1\right)}}=1.2\cdot 10^{-2}\ \frac{m^3}{\sec }\)
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