Answer:
The value is [tex]\delta P = 9000 \ Pa[/tex]
Explanation:
From the question we are told that
The inlet and outlet velocity is [tex]v = 3 \ m/s[/tex]
The loss coefficient is [tex]k = 2[/tex]
Th specific weight of water is [tex]w = 9800 \ N/m^3[/tex]
Generally the pressure drop across the valve is mathematically represented as
[tex]\delta P = \frac{1}{2} * k * \rho * v^2[/tex]
Here [tex]\rho[/tex] is the density which is mathematically represented as
[tex]\rho = \frac{w}{g}[/tex]
=> [tex]\rho = \frac{9800}{9.8}[/tex]
=> [tex]\rho = 1000 \ kg/m^3[/tex]
So
[tex]\delta P = \frac{1}{2} * 2* 1000 * 3^2[/tex]
[tex]\delta P = 9000 \ Pa[/tex]