Question

Water pressurized to 3 ´ 105 Pa is flowing at 5.0 m/s in a pipe which contracts to 1/3 of its former area. What are the pressure and speed of the water after the contraction

Answer 1

Answer:

=$P' = 2 * 10 ^ 5 Pa$

Explanation:

given data:

Pressure P = 3 * 10^ 5 Pa

speed v = 5 m / s

Area A = A

from the information given in equation final Area is 1/3 of initial area i.e.

A ' = A / 3

we know that density of water = 1000 kg / m^ 3

from continuity equation

Av = A ' v'

so we have

speed v ' = 3*A*v / A

v ' = 3*A*5/ A

v = 15 m / s

from bernoulli's equation we can calculate final pressure

Required pressure P ' = P + ( 1/ 2) \rho [ v^ {2} v'^{ 2}]

= $P ' = P + ( 1/ 2) \rho_{water} [ v^ {2} - v'^{ 2}]$

=$P -  10 ^ {5}$

= $3*10^{5} - 10 ^ {5}$

=$P' = 2 * 10 ^ 5 Pa$

Answer 2

Answer:

The pressure and speed of the water after the contraction are $2\times10^{5}\ Pa$ and 15 m/s.

Explanation:

Given that,

Pressure $P= 3\times10^{5}$

Flowing speed = 5.0 m/s

Area = A

Final area $A'=\dfrac{A}{3}$

We need to calculate the pressure and speed of the water after the contraction

Using equation of continuity

$A_{1}v_{1}=A_{2}v_{2}$

Where, $A_{1}$ = area

$A_{2}$ = final area

$v_{1}$ = speed

Put the value into the formula

$v_{2}=\dfrac{3A\times5}{A}$

$v_{2}=15\ m/s$

We need to calculate the pressure of the water after the contraction

Using Bernoulli's equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2$

$P_{2}=P_{1}+\dfrac{1}{2}\rho (v_{2}^2-v_{2}^2)$

$P_{2}=3\times10^{5}+\dfrac{1}{2}\times1000\times(5.0^2-15^2)$

$P_{2}=2\times10^{5}\ Pa$

Hence, The pressure and speed of the water after the contraction are $2\times10^{5}\ Pa$ and 15 m/s.