Answer with Explanation:
We are given that
Diameter=0.030 m
Length of sprue=[tex]h_1[/tex]=0.200 m
Metal volume flow rate,Q=0.03[tex]m^3/min[/tex]
Q=[tex]\frac{0.03}{60}=5\times 10^{-4}m^3/s[/tex] because 1 minute=60 seconds
Let 1 for the top and 2 for the bottom
[tex]d_=0.030 m[/tex]
[tex]h_2=0[/tex]
[tex]A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}[/tex]
[tex]A_1=7.065\times 10^{-4} m^2[/tex]
[tex]v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}[/tex]
[tex]v_1=0.708 m/s[/tex]
Pressure at the top and bottom of the sprue is atmospheric
[tex]h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}[/tex]
Substitute the values
[tex]0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}[/tex]
[tex]v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264 [/tex]
[tex]v_2=\sqrt{4.421264}=2.1 m/s[/tex]
[tex]Q=A_2v_2[/tex]
[tex]5\times 10^{-4}=A_2\times 2.1[/tex]
[tex]A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2[/tex]
Reynolds number=[tex]\frac{v_2D\rho}{\eta}[/tex]
[tex]\eta=0.004 N.s/m^2[/tex]
[tex]\rho=2700 kg/m^3[/tex]
Substitute the values then we get
Reynolds number=[tex]\frac{2.1\times 0.03\times 2700}{0.004}[/tex]
Reynolds number=42525
The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.
