# Casting Process

 Question 1
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 1 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 2
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 2 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 3
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 3 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}
 Question 4
The figure shows a pouring arrangement for casting of a metal block. Frictional losses are negligible. The acceleration due to gravity is 9.81 $m/s^2$. The time (in s, round off to two decimal places) to fill up the mold cavity (of size 40 cm x 30 cm x 15 cm) is___
 A 12.55 B 45.28 C 28.92 D 36.28
GATE ME 2019 SET-2   Manufacturing Engineering
Question 4 Explanation:
$\frac{\text { Volume }}{\mathrm{t}}=\mathrm{A}_{\mathrm{c}} \sqrt{2 \mathrm{gh}_{\mathrm{c}}}$
At point (3) is open to atmosphere so (3) is the choke.
$\mathrm{g}=9.81 \mathrm{m} / \mathrm{sec}^{2}=981 \mathrm{cm} / \mathrm{sec}^{2} \\ \frac{40 \times 30 \times 15}{\mathrm{t}}=\frac{\pi}{4} \times 2^{2} \sqrt{2 \times 981 \times 20} \Rightarrow \frac{18000}{\mathrm{t}}=622.32088$
$\Rightarrow \mathrm{t}=28.923 \mathrm{sec}$
 Question 5
The fluidity of molten metal of cast alloys (without any addition of fluxes) increases with increase in
 A viscosity B surface tension C freezing range D degree of superheat
GATE ME 2019 SET-2   Manufacturing Engineering
Question 5 Explanation:
Fluidity increases with increase of degree of super heat.
 Question 6
Match the following sand mold casting defects with their respective causes.
 A P-4, Q-3, R-1, S-2 B P-3, Q-4, R-2, S-1 C P-2, Q-4, R-1, S-3 D P-3, Q-4, R-1, S-2
GATE ME 2019 SET-1   Manufacturing Engineering
Question 6 Explanation:
Blow hole $\Rightarrow$ Poor Permeability
Misrun $\Rightarrow$ Insufficient Fluidity
Hot tearing $\Rightarrow$ Poor Collapsibility
Wash $\Rightarrow$ Mold erosion
 Question 7
In a casting process, a vertical channel through which molten metal flows downward from pouring basin to runner for reaching the mold cavity is called
 A blister B sprue C riser D pin hole
GATE ME 2019 SET-1   Manufacturing Engineering
Question 7 Explanation:
Vertical channel is called sprue and the horizontal channel is called runner
 Question 8
For sand-casting a steel rectangular plate with dimensions 80 mm $\times$ 120 mm $\times$ 20 mm, a cylindrical riser has to be designed. The height of the riser is equal to its diameter. The total solidification time for the casting is 2 minutes. In Chvorinov's law for the estimation of the total solidification time, exponent is to be taken as 2. For a solidification time of 3 minutes in the riser, the diameter (in mm) of the riser is __________ (correct to two decimal places).
 A 51.84 B 41.28 C 85.96 D 74.87
GATE ME 2018 SET-2   Manufacturing Engineering
Question 8 Explanation:
\begin{aligned} \text { casting } \operatorname{size} &=80 \times 120 \times 20 \mathrm{mm} \\ \left(t_{s}\right)_{\text {casting }} &=2 \mathrm{min} \\ \left(t_{s}\right)_{\text {riser }} &=3 \mathrm{min} \\ 2 &=k\left(\frac{80 \times 120 \times 20}{2(80 \times 120+120 \times 20+20 \times 80)}\right)^{2}\\ k &=0.040138 \mathrm{min} / \mathrm{mm}^{2} \\ \text { Riser : } \left(t_{S}\right)_{\text {riser }}&=k\left(\frac{V}{A}\right)^{2} \\ 3 &=0.040138\left(\frac{V}{A}\right)^{2} \\ \left(\frac{V}{A}\right) &=8.64 \\ \frac{d}{6} &=8.64\\ \text{Riser if } h&=d \text{ then }\frac{V}{A}=\frac{d}{6} \\ \frac{V}{A}&=\frac{\frac{\pi}{4} d^{2} h}{2 \frac{\pi}{4} d^{2}+\pi d h}\\ \text{If } h=d, \text{ then }\left(\frac{V}{A}\right)&=\frac{d}{6} \\ h=d&=51.84 \mathrm{mm} \end{aligned}
 Question 9
Match the following products with the suitable manufacturing process
 A $P-4, Q-3, R-1, S-2$ B $P-2, Q-1, R-3, S-4$ C $P-4, Q-1, R-2, S-3$ D $P-1, Q-3, R-4, S-2$
GATE ME 2018 SET-2   Manufacturing Engineering
Question 9 Explanation:
Tooth paste tube is made by impact extrusion process
Centrifugal casting is used to manufacture metallic pipes.
Plastic bottles are produced by blow moulding
Rolling process is used producing threaded bolts.
 Question 10
A sprue in a sand mould has a top diameter of 20mm and height of 200mm. The velocity of the molten metal at the entry of the sprue is 0.5m/s. Assume acceleration due to gravity as $9.8\,m/s^{2}$ and neglect all losses. If the mould is well ventilated, the velocity (upto 3 decimal points accuracy) of the molten metal at the bottom of the sprue is ________ m/s.
 A 1.707 B 2.043 C 2.358 D 2.443
GATE ME 2017 SET-1   Manufacturing Engineering
Question 10 Explanation:
$d^{2}=20\mathrm{mm}$

\begin{aligned} h_{s} &=200 \mathrm{mm} \\ u_{2} &=0.5 \mathrm{m} / \mathrm{s} \\ g &=9.8 \mathrm{m} / \mathrm{s}^{2} \\ u_{3} &=? \\ u_{2} &=\sqrt{2 g h_{c}} \\ 0.5 &=\sqrt{2 \times 9.81 \times h_{c}}\\ \Rightarrow \quad h_{c}&=0.01274 \mathrm{m}\\ h_{t}&=h_{s}+h_{c}=0.200+0.01274\\ &=0.2127 \mathrm{m} \\ u_{3} &=\sqrt{2 g h_{t}}=\sqrt{2 \times 9.81 \times 0.2127} \\ &=2.043 \mathrm{m} / \mathrm{s} \end{aligned}
There are 10 questions to complete.

### 4 thoughts on “Casting Process”

1. Question 11solution is wrong