Question

The flow rate of liquid metal into the downsprue of a mold = 0.7 L/sec. The cross-sectional area at the top of the sprue = 750 mm^2, and its length = 185 mm. What area should be used at the base of the sprue to avoid aspiration of the molten metal (mm^2)?

Answer 1

367.43 mm²

Step-by-step explanation:

Given:

Flow rate, Q = 0.7 L/s

1000 L = 1 m³ = 10⁹ mm³

thus,

1 L = 10⁶ mm³

Therefore,

Q = 0.7 × 10⁶ mm³/s

Cross-sectional area at the top of the sprue = 750 mm²

Length of the sprue = 185 mm

Now,

Velocity =
`√(2gh)`

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 =
`√(2*9.81*0.185)`

or

velocity = 1.9051 m/s = 1905.1 mm/s

Also,

Q = Area × Velocity

thus,

0.7 × 10⁶ = Area × 1905.1

or

Area = 367.43 mm²