Design of Gating System MCQ Quiz in తెలుగు - Objective Question with Answer for Design of Gating System - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Explanation:

Gating ratio:

The term gating ratio is used to describe the relative cross-sectional areas of the components of gating system.

It is defined as the ratio of the sprue area (As) to the total runner area (Ar) to the total gate area (Ag).

i.e. Gating ratio a : b : c = Sprue area : Runner area : Gate area.

Gating ratio is grouped in two classes i.e. pressurised and unpressurised gating system.

Additional Information

Pressurised gating system:

• The proportion of sprue, runner and gate area are so arranged that back-pressure is maintained in the gating system.
• This requires that the total gate area is not greater than the area of the sprue. Eg. 1 : 0.75 : 0.5, 1 : 2 : 1,  2 : 1 : 1.
• A pressurised gating system keeps itself full of metal.
• Air aspiration is minimised.
• Smaller loss of metal and greater yield.
• High metal velocities may cause turbulence at the junctions and corners in mould cavity.
• Method suitable for ferrous material and brass. 

Un-pressurised gating system:

• The unpressurised gating system produces lower metal velocities and permits greater flow rates. 
• Requires careful design to ensure complete filling, and large-sized runners and gates, which reduce the yield and increase the wastage of metal.
• this system is generally adopted for metal such as Aluminium and Magnesium. The ratio used are 1 : 2 : 2, 1 : 3 : 3.

Mould filling time in bottom gating system is given by:

\({t_f} = \frac{{{A_m}}}{{{A_g}}}\sqrt {\frac{2}{g}} \left[ {\sqrt {{h_t}} - \sqrt {{h_t} - {h_m}} } \right]\)

Let height of molten metal column is pouring basin be h_c:

\(\because {V_2} = \sqrt {2g{h_c}} \)

V₂ = 2 m/s

\(200 = \sqrt {2 \times 1000 \times {h_c}}\)

40000 = 2 × 1000 × h_c

h_c = 20 cm

Area of mould (A_m) = π ab

= π × 30 × 10

A_m = 300 π cm²

Area of gate (A_g) = 15 cm²

h_t = 180 + 20

= 200 cm

\({t_f} = \frac{{300\pi }}{{15}}\sqrt {\frac{2}{{1000}}} \left[ {\sqrt {200} - \sqrt {200 - 60} } \right]\)

Explanation

• The design of the gating element is done in such a way that the liquid metal enters into the casting with optimum velocity within the given time without causing turbulence.
• The design of the gating element is such that pure liquid metal enters into the cavity without air aspiration effect.

Components of the gating system:

Concept:

Time of filling is t_f

\({t_f} = \frac{{{V_m}}}{{{A_g} \times {v_g}}}\)

V_m is volume of mold, A_g is area and v_g is velocity

Calculation:

Given d = 10 mm, h_t = 250 mm

\({v_g} = \sqrt {2g{h_t}} \)

\({A_g} = \frac{\pi }{4}{d^2} = \frac{\pi }{4} \times {10^2} = 78.54\;m{m^2}\)

\({V_g} = \sqrt {2g{h_t}} = \sqrt {2 \times 9810 \times 250} = 2214.723\;mm/s\)

V_m = (100)³ mm³

\(\therefore {t_f} = \frac{{{{\left( {100} \right)}^3}}}{{78.54 \times 2214.723}} = 5.75\;sec\)

Points to remember:

i) For top gate \({t_f} = \frac{{{V_m}}}{{{A_g} \times {V_g}}}\)

ii) \({V_g} = \sqrt {2g{h_t}} \)

iii) Filling time \({T_s} = \frac{{Volume\;of\;mould}}{{Flow\;rate}}\)

(i) Concept:

Mould filling time for bottom gating system is given by:

\({{t}_{b}}=\frac{{{A}_{m}}}{{{A}_{g}}}\frac{2}{\sqrt{2g}}\left[ \sqrt{{{h}_{t}}}-\sqrt{{{h}_{t}}-{{h}_{m}}} \right]\)

Where, A_m = cross-sectional area of mould cavity

A_g = cross-sectional area of gate

h_t = total height (height of sprue + height of pouring basin)

h_m = height of mould cavity

(ii) Calculation:

Given data: A_m = 1 × 0.5 = 0.5 m²

A_g = 10 cm² = 10 × 10^-4 m² = 10^-3 m²

h_t = 0.30 m

h_m = 0.25 m

\({{t}_{b}}=\frac{0.5}{{{10}^{-3}}}\frac{2}{\sqrt{2\times 9.81}}\left[ \sqrt{0.30}-\sqrt{0.30-0.25} \right]\)

\(=\frac{0.5\times 2\times {{10}^{3}}}{\sqrt{2\times 9.81}}\left[ \sqrt{0.30}-\sqrt{0.30-0.25} \right]\)

= 73.17298 ≈ 73.17 s

Explanation:

Loam sand:

• Loam sand is a mixture of sand and clay with water to a thin plastic paste.
• Patterns are not used for loam molding and shape is given to mold by sweeps.
• Loam sand consists of 50% silica and 50% clay.

Parting sand:

• The sand is employed on the faces of the pattern before molding.
• Parting sand is used without the binder and moisture to keep the green sand not sticking to the pattern and allow the parting surface the cope and drag to separate without clinging.
• Parting sand is clean clay free of silica.

Green sand:

• Green sand is a prepared mixture of silica sand with 18 to 30% clay and moisture content from 6 to 8%.
• Also known as tempered or natural sand.

Dry sand:

• Green sand that has been dried or baked in a suitable oven after making mold and cores is called dry sand
• It possesses more strength, rigidity, and thermal stability.

Explanation:

Casting Pattern Procedure:

• The pattern is a replica of the object to be cast, used to form the cavity in the sand.
• It is a crucial element in the casting process as it determines the shape and size of the final cast product.

Let us understand the role of a pattern in the casting process:

Purpose of a Pattern:

• The primary purpose of a pattern is to create the mould cavity where the molten metal will be poured.
• The pattern is placed in the sand to create an impression, which forms the mould cavity.
• Once the pattern is removed, the cavity retains the shape of the pattern, ready to be filled with molten metal.

Additional Information

There are different types of patterns used in casting, such as:

• Single-piece pattern: Used for simple shapes.
• Split pattern: Used for complex shapes, split into two or more parts.
• Match plate pattern: Both halves of the pattern are mounted on a plate.
• Cope and drag pattern: Similar to split pattern but specifically for cope (top half) and drag (bottom half) parts of the mould.

Concept:

For calculating Time (t) to fill the mould cavity we must know the discharge (Q) through the gate, as the amount of molten metal discharged through gate falls into the mould cavity.

Qgate = Agate × Vgate

where, 

Qgate = Discharge from the gate.

Agate = Area of gate

\({V_{gate}} = Velocity\;at\;the\;gate= \sqrt {2g{h_{ Sprue}}} \)

Qgate × t = Volume of Mould

\(\therefore t =\frac{Volume\;of\;mould\;cavity}{A_g \sqrt{2gh_{sprue}}}\)

Calculation:

Given:

hsprue = 45 cm,  Asprue = 5 cm2, Volume of mould = 3000 cm3, g = 10 m/s2 ⇒ 1000 cm/s2

\(\therefore t =\frac{Volume\;of\;mould\;cavity}{A_g \sqrt{2gh_{sprue}}}\)

\(\therefore t =\dfrac{3000}{5\;\times \sqrt{2\;\times\;1000\;\times\;45}}\)

Concept:

Casting:

Volumetric flow rate is given by 

Q = A × V

Total volume will be 

Volume = Q × T = A × T × \(\sqrt {2gH}\;\)

Including the discharge coefficient, 

Vol_actual = c_d ×  A × T × \(\sqrt {2gH}\;\)      - (1)

Iron Piece:

Density = mass/volume;

⇒ Vol = m/ρ      - (2)

From equations 1 and 2,

 \(A = \frac{m}{{ρ {C_d}T\sqrt {2gh} }}\)

Gating ratio is given as 

Sprue area : Total runner area : Total gate area

Calculation:

Given

Thickness (t) = 12 mm

Coefficient of discharge C_d = 0.9

Pouring time (T) = 12.6 sec

Height of the sprue (h) = 200 mm

Mass (m) = 30 kg

Density \(\left( ρ \right) = 7.9 × {10^{ - 3}} \; kg/cm^3\)

Gating ratio is given as 

Sprue area : Total runner area : Total gate area = 1 : 2 : 2

Choke area = area of the sprue base

\(A = \frac{m}{{ρ {C_d}T√ {2gh} }}\)

\(= \frac{{30}}{{0.9 × 7.9 × {{10}^{ - 3}} × 12.6 × √ {2 × 981 × 20} }} = 1.69\;c{m^2}\)

Area of the runner (A_r) = 2 × 1.69 = 3.38 cm²

Area of gate (A_g) = 2 × 1.69 = 3.38 cm²

Diameter of sprue base

\(A = \frac{\pi }{4}{D^2}\;\)

\(1.69 = \frac{\pi }{4}{D^2}\)

D = 1.47 cm (Sprue diameter at the bottom)

Area of in-gate = 3.38 cm²

\(A = \frac{\pi }{4}{D^2}\;\)

\(3.38 = \frac{\pi }{4}{D^2}\;\)

Din-gate = 2.0744 cm

Concept:

Time taken to fill mould cavity

\(t = \frac{{Volume}}{{Discharge\;Rate}}\)

Discharge rate (Q) = Area of the base of sprue × Velocity at the base of the sprue

Velocity at the base of sprue \(v = \sqrt {2\;g{h_t}}\)

Where h_t is the length of the sprue

Calculation:

Given, V = 1200 cm³, h_t = 10 cm, A = 2 cm²,  g = 9.81 m/s² = 981 cm/s²

Velocity st the base of the sprue \(v = \sqrt {2\;g{h_t}} = \sqrt {2 \times 981 \times 10} \;cm/sec\)

\(t = \frac{V}{Q} = \frac{{1200}}{{2 \times \sqrt {2 \times 981 \times 10} }} = 4.28\;sec\)