# A 6cm diameter horizontal pipe gradually narrows to 4cm. Whenwater flows throught this pipe at a certain rate, the guagepressure in these two sections is 32.0kPa and 24kPa respectively.What is the volume rate of flow?

Reggie 2020-12-29 Answered
A 6cm diameter horizontal pipe gradually narrows to 4cm. Whenwater flows throught this pipe at a certain rate, the guagepressure in these two sections is 32.0kPa and 24kPa respectively.What is the volume rate of flow?
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## Expert Answer

Tasneem Almond
Answered 2020-12-30 Author has 91 answers

diameters d = 6 cm
d' = 4 cm
radii $r=\frac{d}{2}=3cm=0.03m$

Gauge pressures P ' = 32 kPa $=32×{10}^{3}Pa$
P = 24 kPa $=24×{10}^{3}Pa$
from bernoulli's equation $\left(\frac{1}{2}\right)D\left[{v}^{\prime 2}-{v}^{2}\right]={P}^{\prime }-P$
$\left(\frac{1}{2}\right)D\left[{v}^{\prime 2}-{\left({A}^{\prime }v\frac{{}^{\prime }}{A}\right)}^{2}\right]={P}^{\prime }-P$
$\left(\frac{1}{2}\right)D\left[{v}^{\prime 2}-\left({r}^{\prime 2}v\frac{{}^{\prime }}{{r}^{2}}\right]={P}^{\prime }-P$ since Area $A=\pi ×{r}^{2}$
where D = density of water $=1000k\frac{g}{{m}^{3}}$
$500\left[{v}^{\prime 2}-0.444{v}^{\prime 2}\right]=8×{10}^{3}Pa$
$0.5555{v}^{\prime 2}=16$
v ' = 5.366 m / s
volume flow rate = A ' v ' $=\mathrm{\Pi }×{r}^{\prime 2}×{v}^{\prime }=6.74×{10}^{-3}{m}^{4}=6.74lit$

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