"Water at a pressure of3.00×105Pa flows through a..." solved and explained

College
Physics
Advanced Physics

Cheyanne Leigh

Answered question

2021-04-04

Water at a pressure of $3.00 \times 10^{5}$ 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

Answer & Explanation

Maciej Morrow

Skilled2021-04-06Added 98 answers

A) this part can be done on the basis of equation of continuity
AV = constant
A is the area of the cross section $= d^{2}$
V is the velocity of the water
$d_{1}^{2} \times V_{1} = d_{2}^{2} \times V_{2}$
$d_{1} = d, d_{2} = \frac{d}{4}; V_{1} = 1 \frac{m}{s}; V_{2} = ?$
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
$P + \frac{1}{2} (v_{1}^{2}) + h g = P_{2} + \frac{1}{2} (v_{2}^{2}) + h g$
remains constant on both sides
therefore, $P_{1} + \frac{1}{2} (v_{1}^{2}) = P_{2} + \frac{1}{2} (v_{2}^{2})$
$P_{1} = 3.00 \times 10^{5}$ Pa; $P_{2} = ?; V_{1} = 1 \frac{m}{s}; V_{2} =$ (get it from A);

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

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