Question

Water flows through a 4.0-cm-diameter horizontal pipe at a speed of 1.3 m/s. The pipe then narrows down to a diameter of 2.0 cm. Ignoring viscosity, what is the pressure difference between the wide and narrow sections of the pipe?

Answer 1

Answer:

$P_{1}-P_{2}=12675Pa$

Explanation:

From the equation of continuity we know that:

$v_{1}A_{1}=v_{2}A_{2}\\Given \\r_{1}=2.0cm\\r_{2}=1.0cm\\v_{1}=1.3m/s\\Density of water p=1000kg/m^{3}\\Now\\A_{1}=\pi r_{1}^{2} \\A_{2}=\pi r_{2}^{2}\\ Therefore\\v_{2}=v_{1}\frac{A_{1}}{A_{2}}\\v_{2}=v_{1}\frac{\pi r_{1}^{2} }{\pi r_{2}^{2}}\\v_{2}=v_{1}\frac{r_{1}^{2} }{r_{2}^{2}}\\v_{2}=1.3*\frac{2.0^{2} }{1.0^{2} } \\v_{2}=5.2m/s$

From Bernoulli equation we know that:

$P_{1}+1/2pv_{1}^{2}+pgy_{1}=P_{2}+1/2pv_{2}^{2}+pgy_{2}$

Now assuming $y_{1}=y_{2}$

$P_{1}+(1/2)pv_{1}^{2}=P_{2}+(1/2)pv_{2}^{2}\\ P_{1}-P_{2}=1/2pv_{2}^{2} -1/2pv_{1}^{2}\\ P_{1}-P_{2}=1/2p(v_{2}^{2}-v_{1}^{2} )\\ P_{1}-P_{2}=1/2*1000(5.2^{2}-1.3^{2} )\\ P_{1}-P_{2}=12675Pa$