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

d = 2.711

H = 4.06 inch

Step-by-step 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* 12^2 = 158.4 inc^3`

casting area

`A = 2L^2 + 4Lt`

`A = 2* 12^2 + 4* 12* 1.1`

A = 340.8 in^2

Ratio between casting volume and area

`(v)/(A) = {158.4}{340.8} = 0.464`

Total solidification time TST using chvorinov rule

`TST = Cm [(V)/(A)]^2`

Cm is mold constant

`TST = 13* 0.464^2`[/tex]

TST = 2.798 min

TST for riser

riser
`TST = 1.30*2.798 = 3.63`

Riser volume
`v = (\piD^2 H)/(4)`

`v = 0.25\pi* D^2(1.5D)`

`v = 0.375\pi D^3`

Riser area

`A =(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

`(v)/(A) = (0.375\piD^3)/(2.0\pi D^2) = 0.1875 D`

TST for riser

`TST = Cm [(v)/(A)]^2</p><p>[tex]TST= 13 * (0.1875 D)^2`

`3.36 =13* (0.1875 D)^2`

d = 2.711

`H = 1.5 * 2.711  =4.06 inc`

H = 4.06 inch