---
title: "Chvorinov's rule and riser design"
subject: "Manufacturing and Production Processes"
chapter: 4
key: "semester_4/manufacturing_and_production_processes:chvorinov-rule-riser-design"
source: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/chvorinov-rule-riser-design
site: BCT Engineering
---

# Chvorinov's rule and riser design

Manufacturing and Production Processes · Chapter 4: Solidification Process and Powder Metallurgy

- Asked in: year not given, 8 marks
- Video: [Riser Design in Metal casting | Metal Casting Solved Example Problem | Using Chvorinov's rule](https://www.youtube.com/watch?v=c4wP1HkvM3s) — Top Mech Edu
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/chvorinov-rule-riser-design
- Whole chapter: https://www.bctengineer.com/md/semester_4/manufacturing_and_production_processes/chapter-4

## Question

State Chvorinov's rule. A cube-shaped casting of side 80 mm solidifies in 3.2 minutes in a sand mould. Using the same mould material and pouring conditions, (a) find the mould constant, (b) find the solidification time of a rectangular plate casting 200 mm x 100 mm x 30 mm, and (c) design a cylindrical riser with height equal to 1.5 times its diameter so that it solidifies 25 percent longer than the plate. Neglect the area of contact between riser and casting.

## Short answer

Chvorinov's rule: solidification time t = C(V/A)². (a) Mould constant C = 1.8 min/cm². (b) Plate solidification time = 1.93 min. (c) Cylindrical riser with H = 1.5D requires modulus 1.157 cm; solving gives D = 61.7 mm, H = 92.5 mm.

## Answer

**Chvorinov's rule:** the solidification time of a casting is proportional to the square of its volume-to-surface-area ratio (modulus):

$$
t_s = C\left(\frac{V}{A}\right)^{2}
$$

where C is the mould constant (min/cm$^2$), depending on the mould material, metal properties and pouring temperature.

#### (a) Mould constant

For the cube, side $a = 80$ mm $= 8$ cm:

$$
\frac{V}{A} = \frac{a^3}{6a^2} = \frac{a}{6} = \frac{8}{6} = 1.333\ \text{cm}
$$

$$
C = \frac{t_s}{(V/A)^2} = \frac{3.2}{(1.333)^2} = 1.8\ \text{min/cm}^2
$$

#### (b) Solidification time of the plate

$$
V = 200\times100\times30 = 600\,000\ \text{mm}^3
$$

$$
A = 2(200\times100 + 200\times30 + 100\times30) = 2(29\,000) = 58\,000\ \text{mm}^2
$$

$$
\frac{V}{A} = \frac{600\,000}{58\,000} = 10.34\ \text{mm} = 1.034\ \text{cm}
$$

$$
t_{plate} = 1.8\,(1.034)^2 = 1.93\ \text{min}
$$

#### (c) Riser design

Required riser time:

$$
t_r = 1.25\times 1.926 = 2.41\ \text{min}
$$

Required riser modulus:

$$
\left(\frac{V}{A}\right)_r = \sqrt{\frac{2.41}{1.8}} = 1.157\ \text{cm}
$$

For a cylinder with $H = 1.5D$:

$$
V = \frac{\pi}{4}D^2(1.5D) = 0.375\pi D^3,\qquad A = \pi D(1.5D) + 2\cdot\frac{\pi}{4}D^2 = 2\pi D^2
$$

$$
\frac{V}{A} = \frac{0.375\pi D^3}{2\pi D^2} = 0.1875\,D
$$

$$
D = \frac{1.157}{0.1875} = 6.17\ \text{cm},\qquad H = 1.5\times 6.17 = 9.25\ \text{cm}
$$

#### Check of feeding

Riser volume $= 0.375\pi(6.17)^3 = 276.5\ \text{cm}^3$. The casting volume is $600\ \text{cm}^3$. Even with 6 percent solidification shrinkage ($36\ \text{cm}^3$) the riser holds enough liquid, so the riser also satisfies the volume requirement.

**Answer:** $C = 1.8$ min/cm$^2$; plate solidification time $= 1.93$ min; riser diameter $= 61.7$ mm and height $= 92.5$ mm.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Chvorinov's rule states that solidification time is proportional to
   A. V/A
   B. (V/A)^2
   C. A/V
   D. V x A
   Answer: B. t = C (V/A)^2, the square of the volume-to-surface-area ratio. (Key point: Chvorinov's rule)

2. For an 80 mm cube solidifying in 3.2 min, V/A and C are
   A. V/A = 1.333 cm, C = 1.8 min/cm^2
   B. V/A = 8 cm, C = 0.05 min/cm^2
   C. V/A = 1.333 cm, C = 2.4 min/cm^2
   D. V/A = 0.667 cm, C = 7.2 min/cm^2
   Answer: A. V/A = a/6 = 1.333 cm, so C = 3.2/1.777 = 1.8 min/cm^2. (Key point: Mould constant)

3. Why must the riser solidify later than the casting, and how is it specified here?
   A. Riser must freeze first, in 0.75 times plate time
   B. Riser must stay liquid to feed the casting, solidifying 25% longer (1.25 x 1.93 min)
   C. Riser must have equal time
   D. Riser must have a smaller modulus
   Answer: B. The riser must remain liquid while the casting shrinks, so t_r = 1.25 x 1.93 = 2.41 min. (Key point: Riser time requirement)

4. For a cylinder with H = 1.5D, neglecting the contact area, V/A equals
   A. 0.1875 D
   B. 0.375 D
   C. D/4
   D. 1.5 D
   Answer: A. V = 0.375 pi D^3 and A = 2 pi D^2, giving V/A = 0.1875 D (D = 6.17 cm for V/A = 1.157 cm). (Key point: Cylinder riser modulus)

## Related in Manufacturing and Production Processes

- [Sand casting steps and mould parts](https://www.bctengineer.com/question.md?key=semester_4%2Fmanufacturing_and_production_processes%3Asand-casting-process-steps-mould-components) — Manufacturing and Production Processes, chapter 4
- [Solidification, shrinkage and casting defects](https://www.bctengineer.com/question.md?key=semester_4%2Fmanufacturing_and_production_processes%3Asolidification-shrinkage-casting-defects) — Manufacturing and Production Processes, chapter 4

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