Question

Problem 2 (a): A mold has a downsprue of length - 6.0 in. The cross-sectional area at the bottom of the sprue is 0.5 in'. The sprue leads into a horizontal runner that feeds the mold cavity, whose volume = 75 in'. Determine (a) the velocity of the molten metal flowing through the base of the downsprue, (b) the volume rate of flow and (c) the time required to fill the mold cavity. Problem 2(b): The total solidification times of two casting shapes are to be compared: (1) a sphere with diameter -1.0 in., (2) a cylinder with diameter and length both = 1.0 in. The same casting alloy is used both cases. (a) determine the relative solidification times for each geometry, (b) Based on the results of part (a), which geometric element would make the best riser? (c) If C - 18.0 min/in in Chvorinov's rule, compute the total solidification time for each casting. Use n-2.

Answer 1

Solution 2 (a) The dota given in the question are aſ follows h = 6 in. A = 0.5 in", V- 75 in* g=32-2143 h V=75in Strue Runner Flowing (@) the Velocity of the Molten metal through the base of the down Spruce is given by varagh v=J2X31'3 #1236 [.lft=12 in] ISec Ang 680094 in Sec V=

(6) the volume of flow rate (0) is given by of Molten metal flow X Area of Q = Velocity Sprue v X 0 = 68.094 x 0.5 in / Sec Sec Ang o = 34.047 in © the time required to fill the mould Carity (t) is given by Volume of Mold Cavity t Flow rate t 75 34.047 2.202 sec ) Ang

Solution 2 (6) The dota given in the question question are as follows H=lin o ( Sphene (Dot in) D= \in" (Cylinder) (a) the relative Soldification time for sphere is given by Tas - CV Where, Tis This is known as Chvorinov's rule Soldification time Cm = mold Constant V = Valume of Casting A- Surface area of area of co costing

for sphere the y the V ratio А ratio is 3 V A TD 6 D 6 Пр2 VS D 6. Put the value of D= 1 in in A bove equal V А 6 V 0.1666 in the relative solidification time for sphere is Its C (x² T7s + Come (0-1666) 2 Tts 0.02775 Cm Ang

for cylinder the V ratio is given by V Пр? Н 4 270² +KDH 4 the diameter and length of Cylinder D=H= 1 in Therefore ? >k 4 27 (12+ 7 V = A 0.25 1.5T 0.1666 in V А. the Melotive Solidification time for Cylinder is Trs - Sn (¥) ² Tts = CM (0.16667 = 0.02775 Cm Are ITTS

(0 the volume of sphere V = RD3 V= (13 7 6 V = 0.52359 in And the Valume of Cylinder V= 20 1²4 V= T (1201) 4 V= 0.78539 in? it is clear see that the volume of molten Metal is less then the cylinder so, the Sphere makes the best riser over Cylinder (d) (n = 18 minlin2 the total Selidification time for each costing TTS = 0.02775 Cm TTS = 0·02775 (18)

T.TS= 0.4995 min And Hence, the total solidification time (TTS) for Sphere and Cylinder is 0.4995 min