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

The pressure and speed of the water after the contraction are
`2*10^(5)\ Pa` and 15 m/s.

Step-by-step explanation:

Given that,

Pressure
`P= 3*10^(5)`

Flowing speed = 5.0 m/s

Area = A

Final area
`A'=(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)=(3A*5)/(A)`

`v_(2)=15\ m/s`

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

Using Bernoulli's equation

`P_(1)+(1)/(2)\rho v_(1)^2=P_(2)+(1)/(2)\rho v_(2)^2`

`P_(2)=P_(1)+(1)/(2)\rho (v_(2)^2-v_(2)^2)`

`P_(2)=3*10^(5)+(1)/(2)*1000*(5.0^2-15^2)`

`P_(2)=2*10^(5)\ Pa`

Hence, The pressure and speed of the water after the contraction are
`2*10^(5)\ Pa` and 15 m/s.

Answer 2

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

Step-by-step 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`