Mechanical Engineering Exam > Mechanical Engineering Questions > The height of the down sprue is 175 mm and it...
Start Learning for Free
The height of the down sprue is 175 mm and its CS are at the base is 200 mm2. The CS area of the horizontal runner is also 200 mm2.Assuming no losses, indicate the correct choice for-the time (second) required to fill a mould cavity of volume 106mm3. (Use g = 10 m/s2)
[PI 2002]
- a)2.67
- b)8.45
- c)26.7
- d)84.50
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The height of the down sprue is 175 mm and its CS are at the base is 2...
Most Upvoted Answer
The height of the down sprue is 175 mm and its CS are at the base is 2...
To solve this problem, we can use the formula for volumetric flow rate:
Q = A * v
Where:
Q = volumetric flow rate (mm^3/s)
A = cross-sectional area (mm^2)
v = velocity (mm/s)
We are given the cross-sectional area of the down sprue and the horizontal runner, both of which are 200 mm^2. Therefore, the cross-sectional area for the flow of molten metal into the mould cavity is also 200 mm^2.
Now let's find the velocity of the molten metal. We can use the equation for the volume of a cylinder:
V = A * h
Where:
V = volume (mm^3)
A = cross-sectional area (mm^2)
h = height (mm)
We are given the volume of the mould cavity, which is 106 mm^3. The height of the down sprue is 175 mm. Therefore, we can rearrange the equation to solve for the cross-sectional area of the mould cavity:
A = V / h
A = 106 mm^3 / 175 mm
A = 0.6057 mm^2
Now we can substitute the values into the volumetric flow rate equation:
Q = 0.6057 mm^2 * v
Since the cross-sectional area of the mould cavity is the same as the down sprue and the horizontal runner, the volumetric flow rate is the same for all three sections.
Now we need to find the time required to fill the mould cavity. We can rearrange the equation to solve for time:
t = V / Q
t = 106 mm^3 / (0.6057 mm^2 * v)
Finally, we can substitute the given acceleration due to gravity, g, to find the time in seconds:
t = 106 mm^3 / (0.6057 mm^2 * v * 10 m/s^2)
t = 2.67 s
Therefore, the correct choice is option A, 2.67 seconds.
Q = A * v
Where:
Q = volumetric flow rate (mm^3/s)
A = cross-sectional area (mm^2)
v = velocity (mm/s)
We are given the cross-sectional area of the down sprue and the horizontal runner, both of which are 200 mm^2. Therefore, the cross-sectional area for the flow of molten metal into the mould cavity is also 200 mm^2.
Now let's find the velocity of the molten metal. We can use the equation for the volume of a cylinder:
V = A * h
Where:
V = volume (mm^3)
A = cross-sectional area (mm^2)
h = height (mm)
We are given the volume of the mould cavity, which is 106 mm^3. The height of the down sprue is 175 mm. Therefore, we can rearrange the equation to solve for the cross-sectional area of the mould cavity:
A = V / h
A = 106 mm^3 / 175 mm
A = 0.6057 mm^2
Now we can substitute the values into the volumetric flow rate equation:
Q = 0.6057 mm^2 * v
Since the cross-sectional area of the mould cavity is the same as the down sprue and the horizontal runner, the volumetric flow rate is the same for all three sections.
Now we need to find the time required to fill the mould cavity. We can rearrange the equation to solve for time:
t = V / Q
t = 106 mm^3 / (0.6057 mm^2 * v)
Finally, we can substitute the given acceleration due to gravity, g, to find the time in seconds:
t = 106 mm^3 / (0.6057 mm^2 * v * 10 m/s^2)
t = 2.67 s
Therefore, the correct choice is option A, 2.67 seconds.
Attention Mechanical Engineering Students!
To make sure you are not studying endlessly, EduRev has designed Mechanical Engineering study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Mechanical Engineering.
|
Explore Courses for Mechanical Engineering exam
|
|
Similar Mechanical Engineering Doubts
Top Courses for Mechanical EngineeringView all
The height of the down sprue is 175 mm and its CS are at the base is 200 mm2. The CS area of the horizontal runner is also 200 mm2.Assuming no losses, indicate the correct choice for-the time (second) required to fill a mould cavity of volume 106mm3. (Use g = 10 m/s2)[PI 2002]a)2.67b)8.45c)26.7d)84.50Correct answer is option 'A'. Can you explain this answer?
Question Description
The height of the down sprue is 175 mm and its CS are at the base is 200 mm2. The CS area of the horizontal runner is also 200 mm2.Assuming no losses, indicate the correct choice for-the time (second) required to fill a mould cavity of volume 106mm3. (Use g = 10 m/s2)[PI 2002]a)2.67b)8.45c)26.7d)84.50Correct answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The height of the down sprue is 175 mm and its CS are at the base is 200 mm2. The CS area of the horizontal runner is also 200 mm2.Assuming no losses, indicate the correct choice for-the time (second) required to fill a mould cavity of volume 106mm3. (Use g = 10 m/s2)[PI 2002]a)2.67b)8.45c)26.7d)84.50Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The height of the down sprue is 175 mm and its CS are at the base is 200 mm2. The CS area of the horizontal runner is also 200 mm2.Assuming no losses, indicate the correct choice for-the time (second) required to fill a mould cavity of volume 106mm3. (Use g = 10 m/s2)[PI 2002]a)2.67b)8.45c)26.7d)84.50Correct answer is option 'A'. Can you explain this answer?.
The height of the down sprue is 175 mm and its CS are at the base is 200 mm2. The CS area of the horizontal runner is also 200 mm2.Assuming no losses, indicate the correct choice for-the time (second) required to fill a mould cavity of volume 106mm3. (Use g = 10 m/s2)[PI 2002]a)2.67b)8.45c)26.7d)84.50Correct answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The height of the down sprue is 175 mm and its CS are at the base is 200 mm2. The CS area of the horizontal runner is also 200 mm2.Assuming no losses, indicate the correct choice for-the time (second) required to fill a mould cavity of volume 106mm3. (Use g = 10 m/s2)[PI 2002]a)2.67b)8.45c)26.7d)84.50Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The height of the down sprue is 175 mm and its CS are at the base is 200 mm2. The CS area of the horizontal runner is also 200 mm2.Assuming no losses, indicate the correct choice for-the time (second) required to fill a mould cavity of volume 106mm3. (Use g = 10 m/s2)[PI 2002]a)2.67b)8.45c)26.7d)84.50Correct answer is option 'A'. Can you explain this answer?.
Solutions for The height of the down sprue is 175 mm and its CS are at the base is 200 mm2. The CS area of the horizontal runner is also 200 mm2.Assuming no losses, indicate the correct choice for-the time (second) required to fill a mould cavity of volume 106mm3. (Use g = 10 m/s2)[PI 2002]a)2.67b)8.45c)26.7d)84.50Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free.
Here you can find the meaning of The height of the down sprue is 175 mm and its CS are at the base is 200 mm2. The CS area of the horizontal runner is also 200 mm2.Assuming no losses, indicate the correct choice for-the time (second) required to fill a mould cavity of volume 106mm3. (Use g = 10 m/s2)[PI 2002]a)2.67b)8.45c)26.7d)84.50Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The height of the down sprue is 175 mm and its CS are at the base is 200 mm2. The CS area of the horizontal runner is also 200 mm2.Assuming no losses, indicate the correct choice for-the time (second) required to fill a mould cavity of volume 106mm3. (Use g = 10 m/s2)[PI 2002]a)2.67b)8.45c)26.7d)84.50Correct answer is option 'A'. Can you explain this answer?, a detailed solution for The height of the down sprue is 175 mm and its CS are at the base is 200 mm2. The CS area of the horizontal runner is also 200 mm2.Assuming no losses, indicate the correct choice for-the time (second) required to fill a mould cavity of volume 106mm3. (Use g = 10 m/s2)[PI 2002]a)2.67b)8.45c)26.7d)84.50Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of The height of the down sprue is 175 mm and its CS are at the base is 200 mm2. The CS area of the horizontal runner is also 200 mm2.Assuming no losses, indicate the correct choice for-the time (second) required to fill a mould cavity of volume 106mm3. (Use g = 10 m/s2)[PI 2002]a)2.67b)8.45c)26.7d)84.50Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The height of the down sprue is 175 mm and its CS are at the base is 200 mm2. The CS area of the horizontal runner is also 200 mm2.Assuming no losses, indicate the correct choice for-the time (second) required to fill a mould cavity of volume 106mm3. (Use g = 10 m/s2)[PI 2002]a)2.67b)8.45c)26.7d)84.50Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
|
|
Explore Courses for Mechanical Engineering exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup