Question

A 200 mm long down sprue has an area of cross-section of 650 $m{m}^2$ where the pouring basin meets the down sprue (i.e. at the beginning of the down sprue). A constant head of molten metal is maintained by the pouring basin. The molten metal flow rate is 6.5 × ${10}^{5}m{m}^3/s$. Considering the end of down sprue to be open to atmosphere and acceleration due to gravity of ${10}^{4}mm/s^2,$ the area of the down sprue in $m{m}^2$ at its end (avoiding aspiration effect) should be

• A. $650.0$
• B. $190.0$
• C. $290.7$
• D. $350.0$

Answer 1

The correct option is C $290.7$



$\text{Given data:}$

$\text{Metal flow rate}=6.5\times {10}^{5}m{m}^3/s$

$\therefore A_1{v}_1=6.5\times {10}^5$

$\therefore v_1=\frac{6.5\times {10}^5}{650}=1000mm/s$

$\text{Applying Bernoulli's theorem at points A and B, we get}$

$h+\frac{v_1^2}{2g}=\frac{v_2^2}{2g}[P_A-P_B=P_{atm}]$

$200+\frac{{\left(1000\right)}^2}{2\times {10}^4}=\frac{v_2^2}{2\times {10}^4}$

$\therefore v_2=2236.068mm/s$

$\text{Now, area of down sprue at its end,}$

$A_2=\frac{Q}{v_2}=\frac{6.5\times {10}^5}{2236.068}$

$=290.7m{m}^2$