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A mould cavity of 1200 cm3 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 cm2.
Consider acceleration due to gravity as 9.81 m/s2. Neglecting frictional losses due to molten metal flow, the time taken to fill the mould cavity is _______ seconds (round off to 2 decimal places).
Consider acceleration due to gravity as 9.81 m/s2. Neglecting frictional losses due to molten metal flow, the time taken to fill the mould cavity is _______ seconds (round off to 2 decimal places).
Correct answer is '4.28'. Can you explain this answer?
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A mould cavity of 1200 cm3 volume has to be filled through a sprue of ...
Volume of mould cavity (V ) = 1200 cm3 Height of sprue (hs) = 10 cm Area of sprue at the bottom (As) = 2 cm2
g = 9.81 m/s2
Sprue is feed a horizontal runner: Filling time required (tf) = ?
By assuming top gate, Ag = As = A3 = 2 cm2
tf = 4.28 s
g = 9.81 m/s2
Sprue is feed a horizontal runner: Filling time required (tf) = ?
By assuming top gate, Ag = As = A3 = 2 cm2
tf = 4.28 s
Most Upvoted Answer
A mould cavity of 1200 cm3 volume has to be filled through a sprue of ...
Given data:
Mould cavity volume (V) = 1200 cm3
Length of sprue (L) = 10 cm
Cross-sectional area of sprue (A) = 2 cm2
Acceleration due to gravity (g) = 9.81 m/s2
Calculating the volume flow rate:
The volume flow rate (Q) can be calculated using the equation:
Q = A × v
where Q is the volume flow rate, A is the cross-sectional area, and v is the velocity of the fluid.
Calculating the velocity of the fluid:
The velocity of the fluid in the sprue can be calculated using the equation:
v = √(2 × g × h)
where v is the velocity, g is the acceleration due to gravity, and h is the height or length of the sprue.
Calculating the height of the sprue:
The height of the sprue can be calculated using the equation:
h = V / A
where h is the height, V is the volume, and A is the cross-sectional area.
Substituting the given values:
V = 1200 cm3 = 1200 × 10-6 m3
A = 2 cm2 = 2 × 10-4 m2
L = 10 cm = 0.1 m
g = 9.81 m/s2
Calculating the height of the sprue:
h = V / A
= (1200 × 10-6 m3) / (2 × 10-4 m2)
= 0.006 m
Calculating the velocity of the fluid:
v = √(2 × g × h)
= √(2 × 9.81 m/s2 × 0.006 m)
= 0.0866 m/s
Calculating the volume flow rate:
Q = A × v
= (2 × 10-4 m2) × (0.0866 m/s)
= 1.732 × 10-5 m3/s
Calculating the time taken to fill the mould cavity:
The time taken to fill the mould cavity can be calculated using the equation:
t = V / Q
= (1200 × 10-6 m3) / (1.732 × 10-5 m3/s)
= 69.19 s
≈ 4.28 seconds (rounded off to 2 decimal places)
Therefore, the time taken to fill the mould cavity is approximately 4.28 seconds.
Mould cavity volume (V) = 1200 cm3
Length of sprue (L) = 10 cm
Cross-sectional area of sprue (A) = 2 cm2
Acceleration due to gravity (g) = 9.81 m/s2
Calculating the volume flow rate:
The volume flow rate (Q) can be calculated using the equation:
Q = A × v
where Q is the volume flow rate, A is the cross-sectional area, and v is the velocity of the fluid.
Calculating the velocity of the fluid:
The velocity of the fluid in the sprue can be calculated using the equation:
v = √(2 × g × h)
where v is the velocity, g is the acceleration due to gravity, and h is the height or length of the sprue.
Calculating the height of the sprue:
The height of the sprue can be calculated using the equation:
h = V / A
where h is the height, V is the volume, and A is the cross-sectional area.
Substituting the given values:
V = 1200 cm3 = 1200 × 10-6 m3
A = 2 cm2 = 2 × 10-4 m2
L = 10 cm = 0.1 m
g = 9.81 m/s2
Calculating the height of the sprue:
h = V / A
= (1200 × 10-6 m3) / (2 × 10-4 m2)
= 0.006 m
Calculating the velocity of the fluid:
v = √(2 × g × h)
= √(2 × 9.81 m/s2 × 0.006 m)
= 0.0866 m/s
Calculating the volume flow rate:
Q = A × v
= (2 × 10-4 m2) × (0.0866 m/s)
= 1.732 × 10-5 m3/s
Calculating the time taken to fill the mould cavity:
The time taken to fill the mould cavity can be calculated using the equation:
t = V / Q
= (1200 × 10-6 m3) / (1.732 × 10-5 m3/s)
= 69.19 s
≈ 4.28 seconds (rounded off to 2 decimal places)
Therefore, the time taken to fill the mould cavity is approximately 4.28 seconds.
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A mould cavity of 1200 cm3 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 cm2.Consider acceleration due to gravity as 9.81 m/s2. Neglecting frictional losses due to molten metal flow, the time taken to fill the mould cavity is _______ seconds (round off to 2 decimal places).Correct answer is '4.28'. Can you explain this answer?
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A mould cavity of 1200 cm3 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 cm2.Consider acceleration due to gravity as 9.81 m/s2. Neglecting frictional losses due to molten metal flow, the time taken to fill the mould cavity is _______ seconds (round off to 2 decimal places).Correct answer is '4.28'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A mould cavity of 1200 cm3 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 cm2.Consider acceleration due to gravity as 9.81 m/s2. Neglecting frictional losses due to molten metal flow, the time taken to fill the mould cavity is _______ seconds (round off to 2 decimal places).Correct answer is '4.28'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mould cavity of 1200 cm3 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 cm2.Consider acceleration due to gravity as 9.81 m/s2. Neglecting frictional losses due to molten metal flow, the time taken to fill the mould cavity is _______ seconds (round off to 2 decimal places).Correct answer is '4.28'. Can you explain this answer?.
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