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?
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. | |
Moulds can be used to make patterns. | |
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. | |
Patterns can be used to make moulds. |
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
MSQ
\frac{a}{3} | |
\frac{a}{2} | |
\frac{a}{4} | |
\frac{a}{6} |
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}.
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).
1.25 | |
2.83 | |
2.48 | |
3.12 |
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}
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).
(round off to the nearest integer).
124 | |
284 | |
662 | |
847 |
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).
10.6 | |
2.24 | |
4.28 | |
15.3 |
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}
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.
Question 11solution is wrong
Thank you Rahul for your suggestions.
Sir,I think question no 3 not put height of the sprue plz Correct other wise thank you for your effort❤️
Sir,plz check the answer of the question no 12 because cavity of the sphere neglected
This is very useful for every gate student, thanks
Solution for 38 question is wrong ..it should be 0.97
Paper is insufficient
In question no. 17. (Neglect heat loss of cylinder top and bottom) this line is missing.Please add this it.
Dear Swapnil Sonawane,
This statement is not mentioned in official GATE 2015 Paper.
Q16, Answer given in Solution is Pouring basin or choke ?
Answer choke is correct. We have updated the answer description of this question
In the centrifugal casting question the r is the mean radius. Not D/2
Q4