Answer:
[tex]Q=7.3\times 10^{-3} m^3/s[/tex]
Explanation:
Given that
At top[tex]d_2=30 mm,P_2=860 KPa ,P_1=1000 KPa,d_1=85 mm[/tex]
[tex]\rho =900\dfrac{Kg}{m^3}[/tex]
We know that
[tex]\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2[/tex]
[tex]A_1V_1=A_2V_2[/tex]
[tex]\frac{V_1}{V_2}=\left(\dfrac{d_2}{d_1}\right)^2[/tex]
[tex]\frac{V_1}{V_2}=\left(\dfrac{30}{85}\right)^2[/tex]
[tex]V_2=8.02V_1[/tex]
[tex]Z_2=12 sin60^{\circ}[/tex]
[tex]\dfrac{1000\times 1000}{900\times 9.81}+\dfrac{V_1^2}{2\times 9.81}+0=\dfrac{860\times 1000}{900\times 9.81 }+\dfrac{V_2^2}{2\times 9.81}+12 sin60^{\circ}[/tex]
So [tex]V_1=1.30[/tex]m/s
We know that flow rate Q=AV
[tex]Q=A_1V_1[/tex]
By putting the values
[tex]A_1=\dfrac{\pi}{4}d^2[/tex]
[tex]Q=7.3\times 10^{-3} m^3/s[/tex]
To find the flow rate we do not need the direction of flow,because we are just doing balancing of energy at inlet and at the exits of pipe.