Question

A mould has a down sprue whose length is 20 cm and the cross-sectional area at the base of the down sprue is 1 cm\(^2\). The down sprue feeds a horizontal runner leading into the mould cavity of volume 1000 cm\(^3\). The time required to fill the mould cavity will be

• A. 4.05 s
• B. 5.05 s
• C. 6.05 s
• D. 7.25 s

Hint:

Using CGS units (cm, $\text{cm}^2$, $\text{cm}^3$, $\text{cm/s}^2$) for this problem avoids potential unit conversion errors.
Remember to use $g = 981\text{ cm/s}^2$ rather than $9.81\text{ m/s}^2$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The goal is to calculate the time required to fill a casting mould cavity when fed by a vertical sprue.

Step 2: Key Formula or Approach:
The velocity \(V\) of the molten metal at the base of a vertical sprue of height \(h\) under gravity is:
\[ V = \sqrt{2gh} \] The volumetric flow rate \(Q\) of metal entering the mould is:
\[ Q = A_{\text{base}} \cdot V \] The total filling time \(t\) for a cavity of volume \(V_{\text{cavity}}\) is:
\[ t = \frac{V_{\text{cavity}}}{Q} \]

Step 3: Detailed Explanation:

• Identify the given parameters in consistent metric units (CGS system):
Sprue height, \(h = 20\text{ cm}\).
Sprue base area, \(A_{\text{base}} = 1\text{ cm}^2\).
Cavity volume, \(V_{\text{cavity}} = 1000\text{ cm}^3\).
Acceleration due to gravity, \(g = 981\text{ cm/s}^2\).

• Calculate the velocity of the molten metal at the base of the sprue:
\[ V = \sqrt{2 \times 981 \times 20} = \sqrt{39240} \approx 198.09\text{ cm/s} \]
• Compute the volumetric flow rate:
\[ Q = 1\text{ cm}^2 \times 198.09\text{ cm/s} = 198.09\text{ cm}^3/\text{s} \]
• Determine the filling time:
\[ t = \frac{V_{\text{cavity}}}{Q} = \frac{1000}{198.09} \approx 5.048\text{ s} \] This value rounds to \(5.05\text{ s}\).

Step 4: Final Answer:
The time required to fill the mould cavity is \(5.05\text{ s}\).