Casting Process


Question 1
In a metal casting process to manufacture parts, both patterns and moulds provide shape by dictating where the material should or should not go. Which of the option(s) given correctly describe(s) the mould and the pattern?
A
Mould walls indicate boundaries within which the molten part material is allowed, while pattern walls indicate boundaries of regions where mould material is not allowed.
B
Moulds can be used to make patterns.
C
Pattern walls indicate boundaries within which the molten part material is allowed, while mould walls indicate boundaries of regions where mould material is not allowed.
D
Patterns can be used to make moulds.
GATE ME 2023   Manufacturing Engineering
Question 1 Explanation: 
In some moulding processes like investment and full moulding, permanent moulds can be used for preparing the pattern.
Question 2
Consider sand casting of a cube of edge length a. A cylindrical riser is placed at the top of the casting. Assume solidification time, t_s\propto V/A , where V isthe volume and A is the total surface area dissipating heat. If the top of the riser is insulated, which of the following radius/radii of riser is/are acceptable?

MSQ
A
\frac{a}{3}
B
\frac{a}{2}
C
\frac{a}{4}
D
\frac{a}{6}
GATE ME 2022 SET-2   Manufacturing Engineering
Question 2 Explanation: 
Riser should take more time for solidification than casting \left ( \frac{V}{A} \right )_r\geq\left ( \frac{V}{A} \right )_c
For top riser bottom area is not cooling surface and it is given that top cross section is also insulated. So only lateral area is cooling area.
\frac{\frac{\pi}{4}D^2H}{\pi DH}\geq \frac{a^3}{6a^2}\Rightarrow \frac{D}{4}\geq \frac{a}{6}\Rightarrow D\geq \frac{2a}{3}
Therefore, Radius R\geq \frac{a}{3} .
only answer is available \frac{a}{2}.


Question 3
A cast product of a particular material has dimensions 75 mm x 125 mm x 20 mm. The total solidification time for the cast product is found to be 2.0 minutes as calculated using Chvorinov's rule having the index, n = 2. If under the identical casting conditions, the cast product shape is changed to a cylinder having diameter = 50 mm and height = 50 mm, the total solidification time will be __________minutes (round off to two decimal places).
A
1.25
B
2.83
C
2.48
D
3.12
GATE ME 2021 SET-2   Manufacturing Engineering
Question 3 Explanation: 
Casting dimensions =75 \mathrm{~mm} \times 125 \mathrm{~mm} \times 20 \mathrm{~mm}
Total solidification time \left(t_{s}\right)_{\mathrm{cu}}=2.0 \mathrm{~min}.
Cylindrical casting =\mathrm{H}=\mathrm{D}=50 \mathrm{~min}
Total solidification Time \left(t_{s}\right)_{c y}=? Chvorinov's rule,
\begin{aligned} t_{s} &\propto\left(\frac{V}{A}\right)^{n} \quad n=2 \\ \frac{\left(t_{s}\right)_{c u}}{\left(t_{s}\right)_{c y}}&=\frac{\left(\frac{V}{A}\right)_{c u}^{2}}{\left(\frac{V}{A}\right)_{c y}^{2}}\\ \left(\frac{V}{A}\right)_{c y} &=\frac{\frac{\pi}{4} d^{2} h}{2 \frac{\pi}{4} d^{2} h+\pi d h} \\ h &=d \\ \left(\frac{V}{A}\right)_{c y} &=\frac{d}{6} \\ \frac{\left(t_{s}\right)_{c U}}{\left(t_{s}\right)_{c_{y}}} &=\frac{\left[\frac{75 \times 125 \times 20}{2(75 \times 125+125 \times 20+20 \times 75)}\right]^{2}}{\left(\frac{50}{6}\right)^{2}} \\ \frac{2}{\left(t_{s}\right)_{c y}} &=\frac{49.131}{69.44}=0.7075 \\ \left(t_{s}\right)_{c y} &=2.83 \mathrm{~min} . \end{aligned}
Question 4
A true centrifugal casting operation needs to be performed horizontally to make copper tube sections with outer diameter of 250 mm and inner diameter of 230 mm. The value of acceleration due to gravity, g = 10 \; m/s^2. If a G-factor (ratio of centrifugal force to weight) of 60 is used for casting the tube, the rotational speed required is _____rpm
(round off to the nearest integer).
A
124
B
284
C
662
D
847
GATE ME 2021 SET-1   Manufacturing Engineering
Question 4 Explanation: 

\begin{aligned} \text { Given: } D_{0}&=250 \mathrm{~mm} ; \\ D_{i}&=230 \mathrm{~mm} ; \\ g&=10 \mathrm{~m} / \mathrm{s}^{2} ; \\ \text { G-factor }&=60 \\ F_{c} &=m r \omega^{2} \\ F_{g} &=m g \\ \text { G-factor } &=\frac{F_{c}}{F_{g}}=60 \\ \text { G-factor } &=\frac{F_{c}}{F_{g}}=\frac{m r \omega^{2}}{m g}=\frac{D_{0}}{2 g}\left(\frac{2 \pi N}{60}\right)^{2} \\ N &=\sqrt{\frac{2 \times G-\text { Factor } \times g}{D_{0}}} \times\left(\frac{60}{2 \pi}\right) \\ N &=\sqrt{\frac{2 \times 60 \times 10}{0.25}} \times\left(\frac{60}{2 \pi}\right) \\ N &=661.93 \\ N &=662 \mathrm{rpm} \end{aligned}
Question 5
A mould cavity of 1200 cm^3 volume has to be filled through a sprue of 10 cm length feeding a horizontal runner. Cross-sectional area at the base of the sprue is 2 cm^2. Consider acceleration due to gravity as 9.81 m/s^2. Neglecting frictional losses due to molten metal flow, the time taken to fill the mould cavity is _______ seconds (round off to 2 decimal places).
A
10.6
B
2.24
C
4.28
D
15.3
GATE ME 2020 SET-2   Manufacturing Engineering
Question 5 Explanation: 
Volume of mould cavity (V)=1200 \mathrm{cm}^{3}
Height of sprue \left(h_{s}\right)=10 \mathrm{cm}
Area of sprue at the bottom \left(A_{s}\right)=2 \mathrm{cm}^{2}
g=9.81 \mathrm{m} / \mathrm{s}^{2}
Sprue is feed a horizontal runner: Filling time required \left(t_{t}\right)=?


By assuming top gate, A_{g}=A_{s}=A_{3}=2 \mathrm{cm}^{2}
\begin{aligned} \text { Filling time }\left(t_{t}\right) &=\frac{V}{A_{g} v_{g}}=\frac{1200}{2 \sqrt{2 \times 981}}=4.28 \mathrm{s} \\ t_{f} &=4.28 \mathrm{s} \end{aligned}


There are 5 questions to complete.

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