Solidification Time MCQ Quiz in తెలుగు - Objective Question with Answer for Solidification Time - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

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, V = volume of the casting, A = surface area of the casting

Calculation:

Given:

D = 400 mm = 0.4 m, ts = 1272 sec

For a sphere:

\(\frac{V}{A} = \frac{{\frac{4}{3}\pi {r^3}}}{{4\pi {r^2}}} = \frac{r}{3} = \frac{D}{6}\)

\({t_s} = K{\left( {\frac{V}{A}} \right)^2} = K{\left( {\frac{D}{6}} \right)^2}\)

\(1272 = K{\left( {\frac{{0.4}}{6}} \right)^2}\)

Explanation:

Riser:

• The riser is a reservoir of molten metal provided in the casting so that hot metal can flow back into the mould cavity when
  there is a reduction in the volume of metal due to solidification.
• A riser must stay in a liquid state at least as long as the casting, and must be able to feed the casting during this time. 
  ∴ Solidification time of riser should be More than the casting.
• In Casting process there are mainly two types of risers are used;
  • Open riser (Top Riser)
  • Blind riser (Side Riser)
• The top riser is placed at the topmost point of the casting so that molten can rise up to the highest point of the casting.
• The top surface of the riser will be open to the atmosphere hence we can visually check filling up of mould cavity in the top riser.
• Blind Riser is completely enclosed in the mould and not exposed to the atmosphere.
• Blind risers maintain heat longer than open risers.

Thus, from the above information, we can conclude that Solidification time of riser should be More than the casting 
So, option 2 is the correct answer.

Chills:

• Chills are metallic objects, which are placed in the mould to increase the cooling rate of castings to provide uniform or desired cooling rate.
• Chills are provided to improve directional solidification.
• The material of the chills should be the same as the casting material.

Concept:

τ ∝ (M)²

\({\rm{Modulus}} = \frac{{Volume}}{{Surface\;Area}}\)

Calculation:

Given:

τ₁ = 3 minutes

\({M_1} = \frac{V}{A} = \frac{{{a^3}}}{{6{a^2}}} = \frac{a}{6}\)

\({M_2} = \frac{{27{a^3}}}{{6\; \times\; {{\left( {3a} \right)}^2}}} = \frac{a}{2}\)

\(\frac{{{\tau _1}}}{{{\tau _2}}} = {\left( {\frac{{{M_1}}}{{{M_2}}}} \right)^2}\)

\(\frac{{{\tau _1}}}{{{\tau _2}}} = {\left( {\frac{{\frac{a}{6}}}{{\frac{a}{2}}}} \right)^2} = \frac{1}{9}\)

∴ τ₂ = 9τ₁

τ₂ = 27 min     

Concept:

Solidification time according to Chvorinov’s rule:

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

where K = solidification factor, Ts = total solidification time, V = volume of the casting and A = surface area of the casting.

Calculation:

Given:

Cylinder diameter = D, Cylinder height = D and Cube side = a.

\({\left( {{t_s}} \right)_{cylinder}} = K{\left[ {\frac{V}{A}} \right]^2} = K{\left[ {\frac{{\frac{\pi }{4}{D^2}H}}{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{\pi } DH + 2.\frac{\pi }{4}{D^2}}}} \right]^2}\)

\(as,\;H = D \Rightarrow \frac{V}{A} = K{\left[ {\frac{{\frac{\pi }{4}{D^3}}}{{\frac{\pi }{2}{D^2} + \pi {D^2}}}} \right]^2} = K{\left[ {\frac{D}{6}} \right]^2}\)

\({\left( {{t_s}} \right)_{cube}} = K{\left[ {\frac{V}{A}} \right]^2} = K{\left[ {\frac{{{a^3}}}{{6{a^2}}}} \right]^2} = K{\left( {\frac{a}{6}} \right)^2}\)

\(\therefore\;\frac{{{{\left( {{t_s}} \right)}_{cylinder}}}}{{{{\left( {{t_s}} \right)}_{cube}}}} = \frac{{K{{\left[ {\frac{D}{6}} \right]}^2}}}{{K{{\left[ {\frac{a}{6}} \right]}^2}}} = {\left[ {\frac{D}{a}} \right]^2}\)

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

Concept:

Solidification Time \(t= k{\left( {\frac{V}{A}} \right)^2}\)

where, V is the volume of casting in m³,

A is surface area of casting in m²

and k is solidification factor in s/m².

Calculation:

given V_C = V_S = 1000 cm³ = 1000× 10^-6 m³ and t_c = 4 s.

for cube, V = a³, A = 6a²

\(t_c = 4 = k.\left( {\frac{{{a^3}}}{{6{a^2}}}} \right) = k.{\left( {\frac{a}{6}} \right)^2}\)

vol. of cube = vol of sphere

\(⇒ \frac{4}{3}\pi {r^3} = {a^3}\)

⇒ \(\frac{\pi }{6}{d^3} = {a^3}\)

⇒\(\frac{d}{a} = {\left( {\frac{6}{\pi }} \right)^{\frac{1}{3}}}\)

Solidification time for sphere, \(V = \frac{4}{3}\pi {r^3} = \frac{\pi }{6}{d^3},\;A = 4\pi {r^2} = \pi {d^2}\)

\(t_S=k.{\left( {\frac{{\frac{4}{3}\pi {r^3}}}{{4\pi {r^2}}}} \right)^2} =k. {\left( {\frac{r}{3}} \right)^2}\)\( =k. {\left( {\frac{d}{6}} \right)^2}\)

\(\frac{{{t_S}}}{{{t_C}}} = \frac{{k{{\left( {\frac{d}{6}} \right)}^2}}}{{k{{\left( {\frac{a}{6}} \right)}^2}}} = {\left( {\frac{d}{a}} \right)^2}\)\( = {\left( {\frac{6}{\pi }} \right)^{\frac{2}{3}}}\)

\({t_S} = 1.539 \times 4\)

= 6.157 sec

Concept:

Given that volumes are equal. So, V_cube = V_sphere.

\(Solidification\;Time = {\left( {\frac{V}{A}} \right)^2}\), A = Surface Area

A_cube = 6l²

A_sphere = 4πr²

\(Solidification\;time\;Ratio~(R) = \frac{{\left( {\frac{V}{A}} \right)_{cube}^2}}{{\left( {\frac{V}{A}} \right)_{sphere}^2}}\)

\(\Rightarrow R = {\left( {\frac{{{V_{cube}}}}{{{V_{sphere}}}}} \right)^2} \times{\left( {\frac{{{A_{sphere}}}}{{{A_{cube}}}}} \right)^2}\)

\(R = \frac{{{{\left( {4π {r^2}} \right)}^2}}}{{{{\left( {6{l^2}} \right)}^2}}} = {\left( {\frac{{4π }}{6}} \right)^2}\left( {\frac{{{r^4}}}{{{l^4}}}} \right)\)

Explanation:

Solidification time:

• It is the time required for a casting to solidify after pouring. This time is dependent on the size and shape of the casting.
• There is an empirical relationship known as Chvorinov’s rule –
• \({{\rm{t}}_{\rm{s}}} = {\rm{K }}{\left( {\frac{{\rm{V}}}{{{{\rm{A}}_{\rm{s}}}}}} \right)^2}\)
• where ts = solidification time of casting (min), V = volume of casting (mm3), As = surface area of casting (mm2) and K = solidification factor (min/mm2).
• i.e it indicates that casting with a higher volume-to-surface area ratio will cool and solidify more slowly than one with a lower ratio.

Calculation:

Given:

h = 3d, 

Area of Cylinder, 

\(A=2\frac {\pi}{4}d^2+\pi dh\)

\(A=\frac {\pi}{2}d^2+\pi d(3d)\)

\(A=\frac72 \pi d^2\)

Volume of Cylinder,

\(V=\frac{\pi}{4}d^2h=\frac{\pi}{4}d^2(3d)\)

\(V=\frac{3}{4}\pi d^3\)

Solidification time:

\({{\rm{t}}_{\rm{s}}} = {\rm{K }}{\left( {\frac{{\rm{V}}}{{{{\rm{A}}_{\rm{s}}}}}} \right)^2}\)

\({{\rm{t}}_{\rm{s}}} \propto {\rm{ }}{\left( {\frac{{\rm{V}}}{{{{\rm{A}}_{\rm{s}}}}}} \right)^2}\)

\(t_s \propto ( \frac{\frac34 \pi d^3}{\frac72 \pi d^2} )^2\)

\(t_s \propto d^2\)

Concept:

Solidification time: The time taken for the liquid molten metal into the solid is known as solidification time

Chvorinov's equation for solidification time is given by:

\({\bf{t}} = {\bf{B}} \times \;{\left[ {\frac{{\bf{V}}}{{\bf{A}}}} \right]^2}\)

where, t = solidification time, B = mould constant, V = volume of the casting, A = surface area of the casting

Calculation:

Given:

\({\bf{t}} = {\bf{B}} \times \;{\left[ {\frac{{\bf{V}}}{{\bf{A}}}} \right]^2}\) ⇒ \({\bf{t}}\; \propto \;{\left[ {\frac{{\bf{V}}}{{\bf{A}}}} \right]^2}\)

\({\left[ {\frac{{\bf{V}}}{{\bf{A}}}} \right]^2} = \;\frac{{\frac{{\bf{\pi }}}{4} \times {{\bf{D}}^2} \times {\bf{h}}}}{{\left[ {\left( {2 \times \frac{{\bf{\pi }}}{4} \times {{\bf{D}}^2}} \right) + \left( {{\bf{\pi }} \times {\bf{D}} \times {\bf{h}}} \right)} \right]}}\)

Initial casting:

Diameter (\(D_i\)) = 1 m, thickness (\(h_i\)) = 50 mm = 0.05 m

\({\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]_{\rm{i}}}^2 = {\rm{\;}}\frac{{\frac{{\rm{\pi }}}{4} \times {1^2} \times 0.05}}{{\left[ {\left( {2 \times \frac{{\rm{\pi }}}{4} \times {1^2}} \right) + \left( {{\rm{\pi }} \times 1 \times 0.05} \right)} \right]}}\)

\({\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]_{\rm{i}}}^2 = 0.0005165\)

Modified casting:

Diameter (\(D_m\)) = 0.5 m, thickness (\(h_m\)) = 25 mm = 0.0025 m

\({\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]_{\rm{m}}}^2 = {\rm{\;}}\frac{{\frac{{\rm{\pi }}}{4} \times {{0.5}^2} \times 0.025}}{{\left[ {\left( {2 \times \frac{{\rm{\pi }}}{4} \times {{0.5}^2}} \right) + \left( {{\rm{\pi }} \times 0.5 \times 0.025} \right)} \right]}}\)

\({\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]_{\rm{m}}}^2 = 0.0001291\)

% reduction of solidification time is:

\(\frac{{{{\rm{t}}_{\rm{i}}} - {{\rm{t}}_{\rm{m}}}}}{{{{\rm{t}}_{\rm{i}}}}} \times 100 = {\rm{\;}}\frac{{{{\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]}_{\rm{i}}}^2 - {\rm{\;}}{{\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]}_{\rm{m}}}^2}}{{{{\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]}_{\rm{i}}}^2}} = {\rm{\;}}\frac{{0.0005165{\rm{\;}} - {\rm{\;}}0.0001291}}{{0.0001565}} \times 100 = 75{\rm{\% }}\)

∴ the % reduction in solidification time is 75%

Concept:

Chvorinov’s law for solidification time.

\(Time = k{\left( {\frac{{Volume}}{{Area}}} \right)^2}\)

\(\left( {\frac{{Volume}}{{Surface\;Area}} = Modulus} \right)\)

For same material 'k' is same.

∴ Time ∝ (Modulus)2

Modulus for cylinder shape \(= \frac{{\frac{{\pi {D^2}}}{4} \times D}}{{\left( {2 \times \frac{{\pi {D^2}}}{4}} \right) \;+ \;(\pi D \times D)}} = \frac{D}{6}\)  (∵ H = D)

Calculation:

Given:

Steel rectangular plate with dimensions = 75 mm × 125 mm × 20 mm, t = 2 minutes

Modulus of steel plate \(= \frac{{75\; \times \;125 \;\times \;20}}{{2\left[ {(75 \;\times \;125) \;+ \;(20 \;\times \;125)\; + \;(75 \;\times \;20)} \right]}} = 7\;mm\) 

Cylinder H = 50 mm, D = 50 mm.

Modulus for cylinder shape \(= \frac{D}{6}=\frac{50}{6}=8.33\;mm\) 

Time ∝ (Modulus)2

\(\Rightarrow \frac{{Time\;of\;cylinder}}{{Time\;of\;plate}} = {\left( {\frac{{Module\;of\;cylinder}}{{Module\;of\;plate}}} \right)^2}\)

\(\Rightarrow \frac{t}{2} = {\left( {\frac{{8.33}}{{{{7}}{{}}}}} \right)^2}\)

∴ t = 2.83 minutes