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The height of the down-sprue is 175 mm and its cross-sectional area at the base is 200 mm2. The cross-sectional area of the horizontal runner is also 200 mm2. Assuming no losses, indicate the correct choice for the time (in seconds) required to fill a mould cavity of volume 106 mm3. (Use g = m/s2).
- a)2.67
- b)8.45
- c)26.72
- d)84.50
Correct answer is option 'B'. Can you explain this answer?
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The height of the down-sprue is 175 mm and its cross-sectional area at...
Given:
- Height of the down-sprue = 175 mm
- Cross-sectional area of the down-sprue = 200 mm^2
- Cross-sectional area of the horizontal runner = 200 mm^2
- Volume of the mold cavity = 10^6 mm^3
To find:
The time required to fill the mold cavity.
Solution:
Step 1: Calculating the volume flow rate
The volume flow rate can be calculated using the equation:
Q = A * V
Where:
Q = Volume flow rate
A = Cross-sectional area
V = Velocity of the fluid
Since the cross-sectional area of the down-sprue and the horizontal runner are the same, the volume flow rate through both of them will be the same.
Let's assume the velocity of fluid through the down-sprue and the horizontal runner is V1.
The volume flow rate through the down-sprue can be calculated as:
Q1 = A1 * V1
The volume flow rate through the horizontal runner can be calculated as:
Q2 = A2 * V1
Since Q1 = Q2, we can equate the two equations:
A1 * V1 = A2 * V1
Since the cross-sectional areas are the same, the velocity of fluid through both the down-sprue and the horizontal runner will be the same.
Step 2: Calculating the time required to fill the mold cavity
The time required to fill the mold cavity can be calculated using the equation:
t = V / Q
Where:
t = Time
V = Volume of the mold cavity
Q = Volume flow rate
Substituting the values:
t = (10^6 mm^3) / Q
Since the volume flow rate is the same through the down-sprue and the horizontal runner, we can substitute Q with A1 * V1:
t = (10^6 mm^3) / (A1 * V1)
Step 3: Finding the velocity of fluid
To find the velocity of the fluid, we need to calculate the velocity head.
The velocity head can be calculated using the equation:
h = (V^2) / (2 * g)
Where:
h = Velocity head
V = Velocity of the fluid
g = Acceleration due to gravity
Since the down-sprue is vertical, the velocity head at the base of the down-sprue will be equal to the height of the down-sprue:
h = 175 mm
Substituting the value of h in the equation and rearranging for V1:
V1 = √(2 * g * h)
Substituting the value of g = 9.8 m/s^2 and h = 175 mm:
V1 = √(2 * 9.8 * 0.175)
V1 ≈ 5.352 m/s
Step 4: Calculating the time
Substituting the values of A1 and V1 in the equation for time:
t = (10^6 mm^3) / (A1 * V1)
Converting the volume and area from mm^3 and mm^2 to m^3 and m^2:
t = (10^-6 m^3) /
- Height of the down-sprue = 175 mm
- Cross-sectional area of the down-sprue = 200 mm^2
- Cross-sectional area of the horizontal runner = 200 mm^2
- Volume of the mold cavity = 10^6 mm^3
To find:
The time required to fill the mold cavity.
Solution:
Step 1: Calculating the volume flow rate
The volume flow rate can be calculated using the equation:
Q = A * V
Where:
Q = Volume flow rate
A = Cross-sectional area
V = Velocity of the fluid
Since the cross-sectional area of the down-sprue and the horizontal runner are the same, the volume flow rate through both of them will be the same.
Let's assume the velocity of fluid through the down-sprue and the horizontal runner is V1.
The volume flow rate through the down-sprue can be calculated as:
Q1 = A1 * V1
The volume flow rate through the horizontal runner can be calculated as:
Q2 = A2 * V1
Since Q1 = Q2, we can equate the two equations:
A1 * V1 = A2 * V1
Since the cross-sectional areas are the same, the velocity of fluid through both the down-sprue and the horizontal runner will be the same.
Step 2: Calculating the time required to fill the mold cavity
The time required to fill the mold cavity can be calculated using the equation:
t = V / Q
Where:
t = Time
V = Volume of the mold cavity
Q = Volume flow rate
Substituting the values:
t = (10^6 mm^3) / Q
Since the volume flow rate is the same through the down-sprue and the horizontal runner, we can substitute Q with A1 * V1:
t = (10^6 mm^3) / (A1 * V1)
Step 3: Finding the velocity of fluid
To find the velocity of the fluid, we need to calculate the velocity head.
The velocity head can be calculated using the equation:
h = (V^2) / (2 * g)
Where:
h = Velocity head
V = Velocity of the fluid
g = Acceleration due to gravity
Since the down-sprue is vertical, the velocity head at the base of the down-sprue will be equal to the height of the down-sprue:
h = 175 mm
Substituting the value of h in the equation and rearranging for V1:
V1 = √(2 * g * h)
Substituting the value of g = 9.8 m/s^2 and h = 175 mm:
V1 = √(2 * 9.8 * 0.175)
V1 ≈ 5.352 m/s
Step 4: Calculating the time
Substituting the values of A1 and V1 in the equation for time:
t = (10^6 mm^3) / (A1 * V1)
Converting the volume and area from mm^3 and mm^2 to m^3 and m^2:
t = (10^-6 m^3) /
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The height of the down-sprue is 175 mm and its cross-sectional area at the base is 200 mm2. The cross-sectional area of the horizontal runner is also 200 mm2. Assuming no losses, indicate the correct choice for the time (in seconds) required to fill a mould cavity of volume 106 mm3. (Use g = m/s2).a)2.67 b)8.45 c)26.72 d)84.50Correct answer is option 'B'. Can you explain this answer?
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The height of the down-sprue is 175 mm and its cross-sectional area at the base is 200 mm2. The cross-sectional area of the horizontal runner is also 200 mm2. Assuming no losses, indicate the correct choice for the time (in seconds) required to fill a mould cavity of volume 106 mm3. (Use g = m/s2).a)2.67 b)8.45 c)26.72 d)84.50Correct answer is option 'B'. 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 cross-sectional area at the base is 200 mm2. The cross-sectional area of the horizontal runner is also 200 mm2. Assuming no losses, indicate the correct choice for the time (in seconds) required to fill a mould cavity of volume 106 mm3. (Use g = m/s2).a)2.67 b)8.45 c)26.72 d)84.50Correct answer is option 'B'. 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 cross-sectional area at the base is 200 mm2. The cross-sectional area of the horizontal runner is also 200 mm2. Assuming no losses, indicate the correct choice for the time (in seconds) required to fill a mould cavity of volume 106 mm3. (Use g = m/s2).a)2.67 b)8.45 c)26.72 d)84.50Correct answer is option 'B'. Can you explain this answer?.
The height of the down-sprue is 175 mm and its cross-sectional area at the base is 200 mm2. The cross-sectional area of the horizontal runner is also 200 mm2. Assuming no losses, indicate the correct choice for the time (in seconds) required to fill a mould cavity of volume 106 mm3. (Use g = m/s2).a)2.67 b)8.45 c)26.72 d)84.50Correct answer is option 'B'. 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 cross-sectional area at the base is 200 mm2. The cross-sectional area of the horizontal runner is also 200 mm2. Assuming no losses, indicate the correct choice for the time (in seconds) required to fill a mould cavity of volume 106 mm3. (Use g = m/s2).a)2.67 b)8.45 c)26.72 d)84.50Correct answer is option 'B'. 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 cross-sectional area at the base is 200 mm2. The cross-sectional area of the horizontal runner is also 200 mm2. Assuming no losses, indicate the correct choice for the time (in seconds) required to fill a mould cavity of volume 106 mm3. (Use g = m/s2).a)2.67 b)8.45 c)26.72 d)84.50Correct answer is option 'B'. Can you explain this answer?.
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