Solution of Assignment 5

Transcription

1 Solution of Assignment 5 1. Chills are used in moulds to (a) Achieve directional solidification (b) Reduce possibility of blow holes (c) Increase the freezing time (d) Smoothen the metal for reducing spatter 2. The directional solidification in casting can be improved by using (a) Chills and chaplets (b) Chills and padding (c) Chaplets and padding (d) Chills, chaplets and padding 3. For a constant volume, the shape of riser that has minimum surface area exposed for heat transfer is (a) Cylindrical (b) Cubical (c) Rectangular (d) Hexagonal 4. Riser is designed so as to (a) Freeze after the casting freezes (b) Freeze before the casting freezes (c) Freeze at the same time as the casting (e) Minimize the time of pouring 5. Chills are made by those metal having (a) Higher melting point than that of cast metal (b) Lower melting point than that of cast metal (c) Equal melting point as that of cast metal (d) None of these 6. Chaplets are placed between moulds in order to (a) Promote directional solidification (b) Help alloying the metal (c) Facilitate easy removal of core from casting (d) Prevent core movement due to buoyancy 7. While cooling, a cubical casting of side 40 mm undergoes 3%, 4% and 5% volume shrinkage during the liquid state, phase transition and solid state respectively. The volume of metal compensated from the riser is (a) 2%

2 (b) 7% (c) 8% (d) 9% 8. According to Naval Research method, shape factor of casting is quantified as (a) (L+W)/t (b) (L+2W)/t (c) (W+2L)/L (d) (L+t)/W Where L= length of casting, W= width of casting and t= thickness of casting 9. Freezing ratio is defined as (a) Modulus of riser/ Modulus of casting (b) Modulus of casting/ Modulus of riser (c) Modulus of riser + Modulus of casting (d) Modulus of riser Modulus of casting 10. Modulus of casting is evaluated as (a) Volume of casting/ surface area of casting (b) Surface area of casting/ Volume of casting (c) Surface area of riser/ Volume of casting (d) Volume of casting/ Surface area of riser 11. A cast steel slab of dimension cm is poured horizontally using a side riser. The riser is cylindrical in shape with diameter and height, both equal to D. The freezing ratio of the mould is (a) 8D/75 (b) 4D/75 (c) 75/8D (d) 75/4D Solution: Freezing ratio (FR) = (Ac / Vc) / (Ar / Vr) (1)

3 Ac = Area of cast = 2 [(30 20) +(20 5) +(5 30)] cm 2 Vc = Volume of cast = cm 3 Ar = Area of riser = [2 (πd 2 /4)] +(πd 2 ), Here L =D Where L = Height of cylinder D = Diameter of cylinder Vr = Volume of riser = (πd 3 /4) though L = D Putting all values in equation (1), we get that FR = (8D/ 75) 12. A mould having dimensions 100 mm 90 mm 20 mm is filled with molten metal through a gate with height h and cross-sectional area A, the mould filling time is t1. The height is now quadrupled and the cross-sectional area is halved. The corresponding filling time is t2. The ratio t2/t1 is (a) 3 (b) 1 (c) 4 (d) 5 Solution: T1 = (V) / (A 2gh) Now, for T2 we have h2 = 4 h, A2 = A/2 T2 = (V) / [(A/2) 2g 4h] So T2 / T1 = The shape factor for a casting in the form of an annular cylinder of outside diameter 30 cm, inside diameter 20 cm and height 30 cm (correction factor k = 1) will be (a) (b) 6.28 (c) 9.42 (d) Solution: Shape factor = (Length + Width)/ Thickness Length = Height of annular cylinder = 30 cm Thickness = (Do Di)/ 2 = 5 cm Do = outer diameter of annular cylinder = 30cm Di = inner diameter of annular cylinder 20 cm

4 Width = π Dav = 25π cm Where Dav = average diameter = (Do + Di)/ 2 So, shape factor = In a sand casting operation, the total liquid head is maintained constant such that it is equal to the mould height. The time taken to fill the mould with a top gate is ta. If the same mould is filled with a bottom gate, then the time taken is tb. Ignore the time required to fill the runner and frictional effects. Assume atmospheric pressure at the top molten metal surfaces. The relation between ta and tb is (a) tb= 5 ta (b) tb = 2tA (c) tb = 3tA (d) tb = 4 ta Solution: Time taken to fill the mould with top gate is ta ta = (AH) / (Ag 2gh m ) (1) Where A= Area of mould H = Height of mould Ag = Area of gate hm = total liquid head Given that, hm = H, so equation (1) become: ta = (A h m )/ (Ag 2g) (2) Time taken to fill the mould with bottom gate tb = [(2A)/(Ag 2g)] ( h m - h m H) though hm = H tb = [(2A)/(Ag 2g)] h m (3) From equation (2) and (3) tb / ta =2 15. A casting of size 100 mm 100 mm 50 mm is required. Assume volume shrinkage of casting as 2.6%. If the height of the riser is 80 mm and the riser volume desired is 4 times the shrinkage in casting, the appropriate riser diameter in mm will be (a) (b) (c) 28.76

5 (d) Solution: Shrinkage volume of casting = (2.6/100) ( ) = mm 3 Volume of riser = Vr = = mm 3 Height of riser = h = 80 mm Vr = (π /4) d 2 h (1) From equation (1) diameter of riser, d = mm But we know that solidification of riser should be more than or equal to that of casting. (V/A) riser (V/A) casting (V/A) casting = ( ) / {2 [( ) +(100 50) +(50 100)]} (2) (V/A) riser = [(πd 2 /4) 80] / {[2 (πd 2 /4)] +(80πd)} (3) Equating equation (2) and (3) we get that, d Hence taking the limiting case d = mm