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:

Respuesta :

Answer:

9,000 kg/ms^2

Explanation:

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

It is to be computed by using the following formula

[tex]\Delta P = \frac{1}{2}\times K\times P\times v^2[/tex]

where,

[tex]\Delta P[/tex] = 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

[tex]P = \frac{r}{g}\\\\ = \frac{9,800 N/m^{3}}{9.81 m/s^{2}}\\\\ = 999kg/m^{3}\\\\ = 1000kg/m^{3}[/tex]

Now put all other values to the given formula

So,

[tex]= 2 \times \frac{1}{2} \times 1000 \times 3^2 \\\\ = 9,000 kg/ms^2[/tex]

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