Answer:246.57 m/s
Explanation:
Given data
[tex]inlet pressure\left ( P_1\right )=101.3KPa[/tex]
[tex]outlet pressure\left ( P_2\right )=300 MPa[/tex]
[tex]Outlet Temperature\left ( T_0\right )=40^{\circ}C[/tex]
[tex]Specific Work into the pump=0.5 KJ/kg[/tex]
Applying Steady flow energy equation
[tex]h_1+\frac{C_1^2}{2000}\frac{gz_1}{1000}+Q=h_2+\frac{C_2^2}{2000}\frac{gz_2}{1000}[/tex]+W
[tex]C_1=0[/tex]
[tex]Z_1&Z_2[/tex] is zero
Q=0
[tex]h_1-h_2-W=\frac{C_2^2}{2000}[/tex]
[tex]\frac{C_2^2}{2000}=\int \nu dP+W[/tex]
[tex]\frac{C_2^2}{2000}[/tex]=[tex]\frac{P_1-P_2}{1000}-\left ( -0.5\right )[/tex]
[tex]\frac{C_2^2}{2000}[/tex]=30.39
[tex]C_2^2=60,797.36[/tex]
[tex]C_2=246.57m/s[/tex]