Question

A cylindrical riser is to be designed for a sand casting mold. The length of the cylinder is to be 1.5 times its diameter. The casting is a square plate, each side = 12 in and thickness = 1.1 in. If the metal is cast iron, and the mold constant = 13.0 min/in2 in Chvorinov's rule, determine the length of the riser in inches so that it will take 35% longer for the riser to solidify.

Answer 1

Answer:

d = 2.711

H = 4.06 inch

Explanation:

Given Data:

length = 1.5 times of diameter of cylinder

square plate side = 12 inch, thickness = 1.1 inch

mold constant = 13.0 min/in^2

Casting volume

$V= tl^2$

t is thickness and L is length

substitute t = 1.1 inch, L =12 inch

$v = 1.1\times 12^2 = 158.4 inc^3$

casting area

$A = 2L^2 + 4Lt$

$A = 2\times 12^2 + 4\times 12\times 1.1$

A = 340.8 in^2

Ratio between casting volume and area

$\frac{v}{A} = {158.4}{340.8} = 0.464$

Total solidification time TST using chvorinov rule

$TST = Cm [\frac{V}{A}]^2$

Cm is mold constant

$TST = 13\times 0.464^2$[/tex]

TST = 2.798 min

TST for riser

riser $TST = 1.30\times2.798 = 3.63$

Riser volume $v = \frac{\piD^2 H}{4}$

$v = 0.25\pi\times D^2(1.5D)$

$v = 0.375\pi D^3$

Riser area

$A =\frac{2\pi D^2}{4} + \pi DH$

$A = 0.5\pi D^2 +1.5\pi D^2$

$A = 2\pi D^2$

RATIO of riser volume and area

$\frac{v}{A} = \frac{0.375\piD^3}{2.0\pi D^2} = 0.1875 D$

TST for riser

$TST = Cm [\frac{v}{A}]^2[tex]TST= 13 \times (0.1875 D)^2$

$3.36 =13\times (0.1875 D)^2$

d = 2.711

$H = 1.5 \times 2.711 =4.06 inc$

H = 4.06 inch