The flow rate of liquid metal into the downsprue of a mold = 1 liter/sec. The cross-sectional area at the top of the sprue= 800 mm2, and its length = 175 mm. What area should be used at the base of the sprue to avoid aspiration of the molten metal?

A₂ = 0.00054 m² = 540 mm²

Explanation:

We know that the flow rate of the liquid metal at the top and bottom of th sprue will remain same. Therefore:

Q₁ = Q₂

where,

Q₁ = Flow Rate of Liquid Metal at the top of sprue = 1 liter/s = 0.001 m³/s

Q₂ = Flow Rate of Liquid Metal at the bottom of sprue

0.001 m³/s = A₂V₂

where,

A₂ = Area of the bottom of the sprue = ?

V₂ = Velocity at the bottom

0.001 m³/s = A₂√2gh

where,

g = acceleration due to gravity = 9.8 m/s²

h = length = 175 mm = 0.175 m

Therefore,

0.001 m³/s = A₂√[(2)(9.8 m/s²)(0.175 m)]

A₂ = (0.001 m³/s)/(1.852 m/s)

A₂ = 0.00054 m² = 540 mm²  

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )