p1ρg+u122g+z1=p2ρg+u222g+z2z1=z2p2=p1+ρ2(u12−u22)\frac{p_1}{\rho g}+\frac{u^2_1}{2 g}+z_1=\frac{p_2}{\rho g}+\frac{u^2_2}{2 g}+z_2\\ z_1=z_2\\ p_2=p_1+\frac{\rho}{2}(u_1^2-u_2^2)\\ρgp1+2gu12+z1=ρgp2+2gu22+z2z1=z2p2=p1+2ρ(u12−u22)
Use the continuity equation to find u2
A1u1=A2u2u2=A1u1A2=d12d22u1=7.8125m/sA_1u_1=A_2u_2\\ u_2= \frac{A_1u_1}{A_2}= \frac{d^2_1}{d^2_2}u_1\\ = 7.8125 m/sA1u1=A2u2u2=A2A1u1=d22d12u1=7.8125m/s
So pressure at section 2
p2=200000−17296.87=182703N/m2p_2= 200000-17296.87\\ =182703 N/m^2p2=200000−17296.87=182703N/m2
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments