Water at a pressure of 3.00\times10^5 Pa flows through a horizontal pipe at a speed of 1.00 m/s. the pipe narrows to 1/4 its original diameter. Find the following: A. The flow speed in the narrow section B. the pressure in the narrow section

Cheyanne Leigh 2021-04-04 Answered
Water at a pressure of \(\displaystyle{3.00}\times{10}^{{5}}\) Pa flows through a horizontal pipe at a speed of 1.00 m/s. the pipe narrows to 1/4 its original diameter. Find the following:
A. The flow speed in the narrow section
B. the pressure in the narrow section

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Expert Answer

Maciej Morrow
Answered 2021-04-06 Author has 14160 answers

A) this part can be done on the basis of equation of continuity
AV = constant
A is the area of the cross section \(\displaystyle={d}^{{2}}\)
V is the velocity of the water
\(\displaystyle{{d}_{{1}}^{{2}}}\times{V}_{{1}}={{d}_{{2}}^{{2}}}\times{V}_{{2}}\)
\(\displaystyle{d}_{{1}}={d},{d}_{{2}}={\frac{{{d}}}{{{4}}}};{V}_{{1}}={1}\ \frac{{m}}{{s}};{V}_{{2}}=?\)
plug the values and get the answer for velocity at the other end
B) We can do this part by making use of bernoullis theorem
\(\displaystyle{P}+{\frac{{{1}}}{{{2}}}}{\left({{v}_{{1}}^{{2}}}\right)}+{h}{g}={P}_{{2}}+{\frac{{{1}}}{{{2}}}}{\left({{v}_{{2}}^{{2}}}\right)}+{h}{g}\)
remains constant on both sides
therefore, \(\displaystyle{P}_{{1}}+{\frac{{{1}}}{{{2}}}}{\left({{v}_{{1}}^{{2}}}\right)}={P}_{{2}}+{\frac{{{1}}}{{{2}}}}{\left({{v}_{{2}}^{{2}}}\right)}\)
\(\displaystyle{P}_{{1}}={3.00}\times{10}^{{5}}\) Pa; \(\displaystyle{P}_{{2}}=?;{V}_{{1}}={1}\ \frac{{m}}{{s}};{V}_{{2}}=\) (get it from A);

\(\displaystyle ={1000}{k}\frac{{g}}{{m}^{{3}}}\)
plug the vales and get the answer

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