Answer:
Change in kinetic energy=0.025\ KJ/Kg[/tex]
Explanation:
Given that
P₁=300 KPa
P₂=275 KPa
Density of water
ρ = 1000 kg/m³
Change in kinetic energy of per unit mass mass given as
Lets take initial velocity V₁ and final velocity V₂
From energy conservation
[tex]P_1+\dfrac{1}{2}\rho V^2_1=P_2+\dfrac{1}{2}\rho V^2_2[/tex]
[tex]300+\dfrac{1}{2}mV^2_1=275+\dfrac{1}{2}mV^2_2[/tex]
[tex]\dfrac{1}{2}\rho V^2_2-\dfrac{1}{2}\rho V^2_1=300-275[/tex]
[tex]\dfrac{1}{2}V^2_2-\dfrac{1}{2} V^2_1=0.025\ KJ/Kg[/tex]