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

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

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Maciej Morrow

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