Question

Water flows through a valve with inlet and outlet velocities of 3 m/s. If the loss coefficient of the valve is 2.0, and the specific weight of water is 9800 N/m3, the pressure drop across the valve is most nearly:

Answer 1

9,000 kg/ms^2

Step-by-step explanation:

The computation of the pressure fall across the valve is shown below:

It is to be computed by using the following formula

`\Delta P = (1)/(2)* K* P* v^2`

where,

`\Delta P` = Fall in pressure

k = Coefficent loss

P = Loss of density

V = velocity of water

But before reach to the final solution first we have to determine the loss of density which is

`P = (r)/(g)\\\\ = (9,800 N/m^(3))/(9.81 m/s^(2))\\\\ = 999kg/m^(3)\\\\ = 1000kg/m^(3)`

Now put all other values to the given formula

So,

`= 2 * (1)/(2) * 1000 * 3^2 \\\\ = 9,000 kg/ms^2`