Vibrating screen sizing and cost: 24-MMP-B5 Mineral Processing Design and Operations, May 2016, Q3

Question 3: Vibrating screen sizing and cost (3/6)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Screen feed = crusher discharge from Question 2, 507.4 t/h (converted to short tons: $507.4\times1.10231=559.4$ st/h). 15 mm aperture; screen feed is 51.0% passing 15 mm (49.0% oversize) and 18.2% passing 7.5 mm (half-aperture). Ore-characteristics coefficient 1.28; safety factor 20%; two crushers selected in Q2 (1:1 screen ratio). Chart readings (appendix, VSMA-style basic-capacity method): basic capacity at 15 mm aperture ≈ 4.21 short tons/ft²/h; oversize correction factor at 49% oversize ≈ 1.17; half-size correction factor at 18.2% passing half-aperture ≈ 0.62.

Find. Net screen area (with 20% safety factor), screen dimensions, and preliminary current cost.

Approach. Apply the basic-capacity method: required area = feed tonnage ÷ (basic capacity × oversize factor × half-size factor × material factor), inflate by the safety factor, then split across the number of screens set by the 1:1 crusher ratio and select the smallest table dimension that clears the per-screen requirement.

Net area from the basic-capacity method. $$A_{\text{net}}=\frac{\dot m_{\text{screen feed}}}{C_{\text{basic}}\times F_{\text{oversize}}\times F_{\text{half-size}}\times F_{\text{material}}}=\frac{559.4}{4.21\times1.17\times0.62\times1.28}=144.4\ \text{ft}^2$$
Apply the 20% safety factor. $$A_{20\%}=144.4\times1.20=\boxed{173.3\ \text{ft}^2}$$
Split 1:1 with the two crushers. Each of the 2 screens must clear $173.3/2=86.7\ \text{ft}^2$. From the screen dimensions table, the smallest single-deck (top-deck) size clearing this is 6' × 16' (88.0 ft² per unit); 2 units give $2\times88.0=176.0\ \text{ft}^2 \geq 173.3\ \text{ft}^2$.
Preliminary cost. $X=W^2L=6^2\times16=576\ \text{ft}^3$ per screen: $$\text{Cost}_{\text{screen}}=2{,}033\times576^{0.5172}=\$54{,}400/\text{unit}$$ $$\text{Total (2 screens)}=\boxed{\$108{,}900}$$

Final Results — Question 3
Quantity	Value
Net area (before safety factor)	144.4 ft²
Net area (with 20% SF)	173.3 ft²
Selected screen(s)	2 × 6' × 16' single-deck (88.0 ft² each)
Preliminary cost, total	≈ USD 108,900

Check: the basic-capacity, oversize-correction and half-size-correction values are read visually off the appendix charts (short tons/ft²/h vs. aperture; correction factor vs. % oversize; correction factor vs. % feed passing half-aperture) — a ruler-and-eye reading, with roughly ±5% chart-reading tolerance on each factor.

Topic: VSMA/Allis-Chalmers basic-capacity method for vibrating screen sizing.

Key relations: $A=\dot m/(C_{\text{basic}}F_{\text{oversize}}F_{\text{half-size}}F_{\text{material}})$; screen cost $X=W^2L$ (both in ft).

Why this works: The basic capacity chart gives throughput per unit area for a "reference" material at a given aperture; the correction factors scale that reference rate up or down for how coarse the actual feed is (too much oversize blinds the deck), how much near-size material must pass (harder separation, lower effective rate), and how the ore's own density/moisture/shape depart from the reference material.

Common pitfall: Forgetting to apply the safety factor before selecting a screen size, or reading the "% oversize" and "% passing half-aperture" percentages off the wrong distribution (they must come from the screen FEED distribution, not the circuit product).

Source: Wills' Mineral Processing Technology, 8th ed., §8.4 (industrial screens, capacity estimation); Mular & Poulin, CIM Special Volume 47 (1998).

Question 3 of 8: Vibrating screen sizing and cost

Question 3: Vibrating screen sizing and cost (3/6)