Pure aluminum is poured into a sand mold. The metal level in the pouring basin is 10 in. above the metal level in the mold, and the runner is circular with a 0.4-in. diameter. What are the velocity and rate of the flow of the metal into the mold? Is the flow turbulent or laminar? 10.44 A cylinder with a diameter of 1.5 in. and height of 3 in. solidifies in 3 minutes in a sand casting operation. What is the solidification time if the cylinder height is doubled? What is the time if the diameter is doubled?

Respuesta :

Solution :

Given :

Height at which the metal is poured, h = 10 in

Diameter of the runner , d = 0.4 in

Assume viscosity, μ = 0.004 Pa-s

Now considering Bernoulli's equation to find velocity,

As there is no loss in energy, Δ Pressure energy ≅ 0

So,

[tex]$\frac{v^2_1}{2g}+z_1=\frac{v^2_2}{2g}+z_2$[/tex]

Here 1 and 2 represents top and bottom section of the sprue.

[tex]$\frac{v^2_1}{2g}+z_1=\frac{v^2_2}{2g}+z_2 \ \ \ (v_1=0)$[/tex]

[tex]$v_2=\sqrt{2g \Delta z}$[/tex]

Now substituting [tex]$32.20 \ ft/s^2 = 386.4 \ in/s^2$[/tex] for g and 10 in for Δz in velocity equation,

[tex]$v_2=\sqrt{2 \times 386.4 \times 10}$[/tex]

[tex]$v_2= 87.91 \ in\s = 7.32 \ ft/s$[/tex]

Calculating the area of the basin

[tex]$A=\frac{\pi}{4}d^2$[/tex]

Substitute .04 in  for d in the above equation

[tex]$A =\frac{\pi}{4} \times (0.4)^2$[/tex]

[tex]$A= 0.1256 \ in^2$[/tex]

Calculating the flow rate

Q = 0.1256 x 87.91

[tex]$ Q= 11 .04 \ in^3/s $[/tex]

Hence the viscosity is [tex]$v_2 = 87.91 \ in/s$[/tex] and the flow rate is [tex]$Q=11.04 \ in^3/s$[/tex]

Calculating the Reynolds number of the flow,

[tex]$Re = \frac{\rho v d}{\mu}$[/tex]

[tex]$Re = \frac{0.097544 \times 87.91 \times 0.40}{5.8 \times 10^{-7}}$[/tex]

[tex]$Re=5.9 \times 10^6$[/tex]

Therefore, the flow is turbulent.

Now considering the solidification time,

[tex]$t=c \times \left(\frac{V}{A}\right)^2$[/tex]

[tex]$t=c \times \left(\frac{\frac{\pi}{4}d^2h}{2\left(\frac{\pi}{4}d^2\right)+ \pi dh}\right)^2$[/tex]

[tex]$t=c\left(\frac{dh}{2d+4h}\right)^2$[/tex]

Substituting 1.5 for d and 3 for h and 3 min for t to calculate the value of c is

[tex]$3=c\left(\frac{1.5 \times 3}{2 \times 1.5 + 4 \times 3} \right)^2$[/tex]

c = 33.33

For case when height is double i.e. h = 6 in

[tex]$t_h = 33.33 \times \left(\frac{1.5 \times 6}{2 \times 1.5 + 4 \times 6} \right)^2$[/tex]

[tex]$t_h= 3.70 \ min$[/tex]

For case when the diameter is doubled i.e. 3 in for d and 3 in  for h,

[tex]$t_d = 33.33 \times \left(\frac{3 \times 3}{2 \times 3 + 4 \times 3} \right)^2$[/tex]

[tex]$t_d= 8.3325 \ min$[/tex]