Answer:
367.43 mm²
Explanation:
Given:
Flow rate, Q = 0.7 L/s
1000 L = 1 mÂł = 10âš mmÂł
thus,
1 L = 10âś mmÂł
Therefore,
Q = 0.7 Ă 10âś mmÂł/s
Cross-sectional area at the top of the sprue = 750 mm²
Length of the sprue = 185 mm
Now,
Velocity = [tex]\sqrt{2gh}[/tex]
where,  g is the acceleration due to gravity = 9.81 m/s²
h is the height through which flow is taking place = 185 mm = 0.185
thus,
Velocity = [tex]\sqrt{2\times9.81\times0.185}[/tex]
or
velocity = 1.9051 m/s = 1905.1 mm/s
Also,
Q = Area Ă Velocity
thus,
0.7 Ă 10âś = Area Ă 1905.1
or
Area = 367.43 mm²