Using Chvorinov’s rule as presented in the text, with n = 2, calculate the dimensions of an effective riser for a casting that is a rectangular plate 2 in. by 4 in. by 6 in. with the dimensions. Assume that the casting and riser are not connected, except through a gate and runner, and that the riser is a cylinder of height/diameter ratio H/D = 1.5. The finished casting is what fraction of the combined weight of the riser and casting?
The correct answer and explanation is :
Alright, let’s dig into this carefully:
First, Chvorinov’s Rule says:
[
t = B\left(\frac{V}{A}\right)^n
]
where:
- ( t ) = solidification time
- ( V ) = volume of the casting or riser
- ( A ) = surface area through which heat is lost
- ( B ) = mold constant (same for casting and riser)
- ( n = 2 ) (as given)
To ensure the riser remains molten longer than the casting, the riser must have a greater solidification time. So we must design the riser so that:
[
\left(\frac{V}{A}\right){\text{riser}} > \left(\frac{V}{A}\right){\text{casting}}
]
Step 1: Find the Casting Volume and Surface Area
The casting is a rectangular plate with dimensions 2 in × 4 in × 6 in.
Volume of casting:
[
V_c = 2 \times 4 \times 6 = 48 \, \text{in}^3
]
Surface area of casting:
There are 6 faces:
- Two faces of ( 4 \times 6 = 24 \, \text{in}^2 ) each
- Two faces of ( 2 \times 6 = 12 \, \text{in}^2 ) each
- Two faces of ( 2 \times 4 = 8 \, \text{in}^2 ) each
Total:
[
A_c = 2(24) + 2(12) + 2(8) = 48 + 24 + 16 = 88 \, \text{in}^2
]
Step 2: Setup Riser Dimensions
The riser is a cylinder with ( H = 1.5D ).
Volume of riser:
[
V_r = \pi \left(\frac{D}{2}\right)^2 H = \pi \left(\frac{D}{2}\right)^2 (1.5D)
]
[
V_r = \pi \times \frac{D^2}{4} \times 1.5D = \frac{3\pi}{8} D^3
]
Surface area of riser (excluding top):
- Side area: ( \pi D H = \pi D (1.5D) = 1.5\pi D^2 )
- Bottom area (connected to runner): ( \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4} )
- Top is free to lose heat, so include it: another ( \frac{\pi D^2}{4} )
Thus, total surface area:
[
A_r = 1.5\pi D^2 + 2\left(\frac{\pi D^2}{4}\right) = 1.5\pi D^2 + 0.5\pi D^2 = 2\pi D^2
]
Step 3: Set up the Solidification Time Ratio
For effective feeding:
[
\left(\frac{V}{A}\right){\text{riser}} \geq \left(\frac{V}{A}\right){\text{casting}}
]
Substituting:
[
\frac{V_r}{A_r} = \frac{\frac{3\pi}{8} D^3}{2\pi D^2} = \frac{3}{16} D
]
Thus:
[
\frac{3}{16} D \geq \frac{48}{88}
]
Simplify:
[
\frac{48}{88} = \frac{12}{22} = \frac{6}{11} \approx 0.545
]
Now solve:
[
\frac{3}{16} D = 0.545
]
Multiply both sides by 16/3:
[
D = 0.545 \times \frac{16}{3}
]
[
D \approx 0.545 \times 5.333
]
[
D \approx 2.91 \, \text{in}
]
Thus:
- ( D \approx 2.91 ) in
- ( H = 1.5D \approx 1.5 \times 2.91 = 4.37 ) in
Step 4: Find Riser Volume
Substitute ( D = 2.91 ):
[
V_r = \frac{3\pi}{8} D^3 = \frac{3\pi}{8} (2.91)^3
]
First, ( (2.91)^3 \approx 24.65 )
Thus:
[
V_r \approx \frac{3\pi}{8} \times 24.65
]
[
V_r \approx \pi \times 9.24
]
[
V_r \approx 29.02 \, \text{in}^3
]
Step 5: Find Weight Fraction
Total volume (casting + riser):
[
V_{\text{total}} = 48 + 29.02 = 77.02 \, \text{in}^3
]
Fraction of finished casting:
[
\text{Fraction} = \frac{48}{77.02} \approx 0.623
]
Thus, the casting is about 62.3% of the combined weight of the casting and riser.
Final Answer:
Fraction = 0.623 or 62.3%
Summary (300+ words):
In foundry operations, it’s critical that the riser remains molten longer than the casting to feed the shrinkage during solidification. Chvorinov’s rule provides a mathematical way to ensure this by comparing the volume-to-surface-area ratios. For this problem, we first found the volume and surface area of the rectangular casting. With a volume of 48 in³ and surface area of 88 in², its V/A ratio determined the minimum needed for the riser.
The riser was specified as a cylinder with a height-to-diameter ratio of 1.5. Using Chvorinov’s formula, we related the riser’s geometry to ensure its V/A ratio exceeded that of the casting. Solving the inequality provided a required diameter of about 2.91 inches and corresponding height of approximately 4.37 inches.
Once the riser size was determined, we calculated its volume (about 29.02 in³) and added it to the casting volume. The casting’s fraction of the total volume was then found: 48 ÷ 77.02 ≈ 0.623. This means about 62.3% of the total metal poured becomes the finished casting, while the remaining 37.7% is consumed by the riser.
This calculation is critical in real-world manufacturing because it affects both material cost and production efficiency. Too large a riser wastes material; too small risks casting defects like shrinkage cavities. Thus, careful riser design ensures quality while minimizing waste.