Respuesta :
Answer:
Value of pressure at upper portion is 38.123kPa
Explanation:
We shall use Bernoull's equation to solve
According to Bernoulli's equation we have
[tex]\frac{P}{\gamma }+\frac{8Q^{2}}{\pi ^{2}D^{4}g}+z=constant\\\\\frac{P_{1}}{\gamma }+\frac{8Q^{2}}{\pi ^{2}D_{1}^{4}g}+z_{1}=\frac{P_{2}}{\gamma }+\frac{8Q^{2}}{\pi ^{2}D_{2}^{4}g}+z_{2}[/tex]
Taking [tex]P_{1}[/tex] pressure at upper end
[tex]P_{2}[/tex] pressure at lower end
Applying values we get
[tex]\frac{50\times 10^{3}}{\gamma _{w}\times 0.85}+\frac{8\times (0.1)^{2}}{\pi ^{2}\times (0.25)^{4}\times 9.81}+z_{1}=\frac{P_{2}}{\gamma _{w}\times 0.85}+\frac{8\times (0.1)^{2}}{\pi ^{2}\times (0.15)^{4}\times 9.81}+z_{2}\\\\\therefore \frac{P_{2}}{\gamma _{w}\times 0.85}=\frac{50\times 10^{3}}{\gamma _{w}\times 0.85}+\frac{8\times (0.1)^{2}}{\pi ^{2}\times (0.25)^{4}\times 9.81}+z_{1}-\frac{8\times (0.1)^{2}}{\pi ^{2}\times (0.15)^{4}\times 9.81}-z_{2}[/tex]
Solving for [tex]P_{2}[/tex] we get
[tex]\frac{P_{2}}{0.85\times \gamma _{w}}=4.575+(z_{1}-z_{2})\\\\\therefore P_{2}=0.85\times 9810\times (4.575+0.002)\\\\\therefore P_{2}=38.123kPa[/tex]
