Answer:
[tex]V_2=6.17\ m/s[/tex]
[tex]Q=0.49\ m^3/s[/tex]
Explanation:
Given that
[tex]A_1=0.03m^2[/tex]
[tex]A_2=0.08m^2[/tex]
We know that from continuity equation
[tex]Q=A_1V_1=A_2V_2[/tex]
So
[tex]A_1V_1=A_2V_2[/tex]
[tex]0.3V_1=0.08V_2[/tex]
[tex]V_1=2.66V_2[/tex]
Now from energy equation
[tex]P_1+\dfrac{1}{2}\rho V_1^2=P_2+\dfrac{1}{2}\rho V_2^2[/tex]
[tex]P_2-P_1=\dfrac{1}{2}\rho V_1^2-\dfrac{1}{2}\rho V_2^2[/tex]
[tex]140=\dfrac{1}{2}\times 1.2\times 7.11\times V_2^2-\dfrac{1}{2}\times 1.2\times V_2^2[/tex]
[tex]V_2=6.17\ m/s[/tex]
[tex]Q=A_2V_2[/tex]
Q=6.17 x 0.08
[tex]Q=0.49\ m^3/s[/tex]