Production Engineering Test 1

Explanation:

Given:

x₁ = 8 mm, t₁ = 16 sec, t₂ = 25 sec

\({\rm{Thickness\;of\;casting\;}}\left( x \right) = {C_1}\sqrt t + {C_2}\)

\(8 = {C_1}\sqrt {16} + {C_2}\)

8 = 4C₁ + C₂      ---(1)

Since, at t = 0, x = 0

Thus,

C₂ = 0

8 = 4 C­₁ ⇒ C₁ = 2

So, at t = 25 seconds

\(x = 2\sqrt {25} = 2 \times 5\)

∴ x = 10 mm 

Explanation:

Given:

Edge of casting (a) = 8 mm, Radius of sphere (r) = 6 mm

Calculation:

\({\rm{Freezing\;ratio\;}} = \frac{{{{\left( {\frac{V}{A}} \right)}_{riser}}}}{{{{\left( {\frac{V}{A}} \right)}_{casting}}}}\)

\({\left( {\frac{V}{A}} \right)_{riser}} = \frac{{\frac{4}{3}\pi {r^3}}}{{4\pi {r^2}}} = \frac{r}{3} = \frac{6}{3} = 2\)

\({\left( {\frac{V}{A}} \right)_{casting}} = \frac{{{a^3}}}{{6{a^2}}} = \frac{a}{6} = \frac{8}{6} = \frac{4}{3}\)

\({\rm{Freezing\;ratio}} = \frac{2}{{\left( {\frac{4}{3}} \right)}} = \frac{6}{4}\)

∴ Freezing ratio = 1.5

Concept:

Gating ratio is defined as

Sprue base area : runner area : ingate area

Given ratio is 1 : 2 : 2

⇒ Sprue base area is the minimum area in the gating system.

Choke area is the minimum area of gating system which meters the flow of molten metal into mould cavity.

Choke area (CA) is calculated by

\(CA = \frac{m}{{\rho t\sqrt {2gh} \eta }}\)       ----(1)

In this expression

m = casting mass (kg)

ρ = density of casting (kg/m³)

t = pouring time (s)

h = sprue height (m)

η = Flow efficiency factor

Calculation:

m = 50 kg , t = 15 s, η = 0.92, n = 250 mm = 0.25m,  ρ = 8000 kg/m³

\(CA = \frac{m}{{\rho t\sqrt {2gh} \eta }}\)

\(= \frac{{50}}{{8000 \times 15 \times \sqrt {2 \times 9.81 \times 0.25} \times 0.92}}\)

= 0.0002045 m² = 2.045 cm²

Keypoints:

Concept:

For castings having same length and cross-sectional area, the volume of the casting will be same for both.

The Chvorinov’s rule for solidification time is:

\(t=C{{\left( \frac{V}{A} \right)}^{2}}\)

where C is a constant and V, A are the volume and surface area respectively.

Calculation:

For the rectangular casting, width/thickness = \(\frac{{\rm{b}}}{{\rm{t}}} = 3\)

Since both have the same cross-sectional area:

\(3{{\rm{t}}^2} = \frac{{\rm{\pi }}}{4}{{\rm{d}}^2}{\rm{}}\)

\(3{{\rm{t}}^2} = \frac{{\rm{\pi }}}{4} \times {0.3^2}\)

\(\therefore {\rm{t}} = 0.1534~{\rm{m}}\)

Dimension of rectangular casting: Thickness (t) = 0.1534 mm, Width (b) = 3t = 0.4602 mm, L = 0.5 mm

The surface area of the rectangular casting will be:

A_rect = 2(0.5 × 0.1534 + 0.4602 × 0.1534 + 0.4602 × 0.5) = 0.7548 m²

Surface area of the round casting:

\({{A}_{round}}=2\pi {{r}^{2}}+2\pi rl=2\pi {{\left( 0.15 \right)}^{2}}+2\pi \left( 0.15 \right)\times 0.5=0.6126~{{m}^{2}}\)

Thus, for same volume of the castings, using Chvorinov’s rule:

\(t=C{{\left( \frac{V}{A} \right)}^{2}}\Rightarrow t\propto {{\left( \frac{1}{A} \right)}^{2}}\)

Explanation:

Given:

Volume of casting (V) = 0.5 × 0.5 × 0.25

V = 0.0625 m³

H = 25 cm = 0.25 m, Height of runner, h = 30 cm = 0.3 m, Diameter (D) = 4 cm = 0.04 m

Calculation:

For top gating system, filling time

\({t_{top}} = \frac{V}{{{A_g}\sqrt {2gh} }}\)

\({A_g} = \frac{\pi }{4}{D^2} = \frac{\pi }{4} \times {0.04^4} = 0.001256\;{m^2}\)

\({t_{top}} = \frac{{0.0625}}{{\left( {0.001256} \right)\sqrt {2 \times 9.81 \times 0.3} }} = 20.5\;sec\)

Now,

For bottom gating system, filling time

\({t_{bottom}} = \frac{{2A}}{{{A_g}\sqrt {2g} }}\left[ {\sqrt h - \sqrt {h - H} } \right]\)

\({t_{bottom}} = \frac{{2 \times 0.5 \times 0.5}}{{\left( {0.001256} \right)\sqrt {2 \times 9.81} }}\left[ {\sqrt {0.3} - \sqrt {0.3 - 0.25} } \right]\)

Concept:

The buoyancy force that lifts the weight of the casting will be the difference in the weights of the alloy and the core.

One of the hazards during pouring is that the buoyancy of the molten metal will displace the core. Buoyancy results from the weight of molten metal being displaced by the core, according to Archimedes’ principle. The force tending to lift the core is equal to the weight of the displaced liquid less the weight of the core itself.

F_b = W_m – W_c

Where F_b = buoyancy force, W_m = weight of molten metal displaced, W_c = weight of the core

Weights are determined as the volume of the core multiplied by the respective densities of the core material.

Calculation:

Volume of the core:

\(V=\frac{m}{{{\rho }_{sand}}}=\frac{20}{1.6\times {{10}^{3}}}=0.0125~{{m}^{3}}\)

Mass of the displaced Al-Cu alloy:

\(m=V{{\rho }_{alloy}}=0.0125\times 2.81\times 1000=35.125~kg\)

Thus, the buoyance force that tends to lift the casting during pouring process is:

Explanation:

Hot Tears:

They are internal or external ragged discontinuities or crack on the casting surface, caused by rapid contraction occurring immediately after the metal solidified. They may be produced when the casting is poorly designed and abrupt sectional changes take place; no proper fillets and corner radii are provided, and chills are inappropriately placed.

Misrun:

It is caused when the metal is unable to fill the mould cavity completely and thus leaves improperly filled cavities it occurs due to low pouring temperature, lack of fluidity.

Blow Hole:

It is a gas defect that is formed on the surface of the casting is open blows and which formed inside the casting is called as blow holes. They are caused due to the low permeability of the sand mould.

Blisters:

They are caused by gas becoming trapped inside the casting at the time the cavity is filled.

Concept:

Solidification time according to Chvorinov’s rule:

\({T_s} = K{\left( {\frac{V}{A}} \right)^2}\)

where K = Solidification factor, Ts = total solidification time, min; V = volume of the casting, cm3; A = surface area of the casting, cm2

Calculation:

V₁ = (200 × 100 × 70) mm² = 14 × 10⁵ mm³

A₁ = 2 [lb + bh + lh] = 2[(200 × 100) + (100 × 70) + (70 × 200)] = 82 × 10³ mm²

V₂ = (200 × 100 × 35) mm³ = 7 × 10⁵ mm³

A₂ = 2 [(200 × 100) + (100 × 35) + (35 × 200)] = 61 × 10³ mm²

Solidification time t_s

\(\therefore \frac{{{t_{s2}}}}{{{t_{s1}}}} = \frac{{{{\left[ {\frac{{{V_2}}}{{{A_2}}}} \right]}^2}}}{{{{\left[ {\frac{{{V_1}}}{{{A_1}}}} \right]}^2}}} = {\left[ {\frac{{7 \times {{10}^5}}}{{61 \times {{10}^3}}} \times \frac{{82 \times {{10}^3}}}{{14 \times {{10}^5}}}} \right]^2} = 0.452\)

t_s1 = 20 min