Question:
Copper ore assaying 10 wt.% Cu is fed to a concentration plant at the rate of 100 tons/h. If the grades of concentrate and tailing are 30 wt.% Cu and 1 wt.% Cu, respectively, the percentage recovery of copper in concentrate is ................... (round off to nearest integer).
Given: 1 ton = 1000 kg
Copper ore assaying 10 wt.% Cu is fed to a concentration plant at the rate of 100 tons/h. If the grades of concentrate and tailing are 30 wt.% Cu and 1 wt.% Cu, respectively, the percentage recovery of copper in concentrate is ................... (round off to nearest integer).
Given: 1 ton = 1000 kg
Given: 1 ton = 1000 kg
Show Hint
The recovery formula \( R = \frac{c(f-t)}{f(c-t)} \times 100% \) is a very useful shortcut for two-product separation problems. It allows you to calculate the recovery directly from the grade assays without needing to know the mass flow rates.
Updated On: Aug 31, 2025
Hide Solution
Verified By Collegedunia
Solution and Explanation
Step 1: Understanding the Concept:
This is a material balance problem in mineral processing. Recovery is the percentage of the valuable metal in the feed that is successfully recovered into the concentrate product. It can be calculated using the grades (metal content percentages) of the feed, concentrate, and tailing streams.
Step 2: Key Formula or Approach:
Let F, C, and T be the mass flow rates of the feed, concentrate, and tailing, respectively. Let f, c, and t be their respective copper grades (as mass fractions).
The recovery (R) is defined as: \[ R = \frac{\text{Mass of Cu in Concentrate}}{\text{Mass of Cu in Feed}} \times 100% = \frac{C \times c}{F \times f} \times 100% \] We can find the ratio \(C/F\) from a copper balance: \(Ff = Cc + Tt\). Since \(F = C+T\), we get the two-product formula: \( \frac{C}{F} = \frac{f-t}{c-t} \). Substituting this into the recovery formula gives a direct equation for recovery based on grades: \[ R = \frac{c(f-t)}{f(c-t)} \times 100% \] Step 3: Detailed Calculation:
Given values:
- Feed grade, \(f = 10% = 0.10\)
- Concentrate grade, \(c = 30% = 0.30\)
- Tailing grade, \(t = 1% = 0.01\)
Substitute these values into the recovery formula: \[ R = \frac{0.30 \times (0.10 - 0.01)}{0.10 \times (0.30 - 0.01)} \times 100% \] \[ R = \frac{0.30 \times 0.09}{0.10 \times 0.29} \times 100% \] \[ R = \frac{0.027}{0.029} \times 100% \] \[ R \approx 0.93103 \times 100% = 93.103% \] Rounding to the nearest integer, the recovery is 93%.
Step 4: Final Answer:
The percentage recovery of copper is 93.
This is a material balance problem in mineral processing. Recovery is the percentage of the valuable metal in the feed that is successfully recovered into the concentrate product. It can be calculated using the grades (metal content percentages) of the feed, concentrate, and tailing streams.
Step 2: Key Formula or Approach:
Let F, C, and T be the mass flow rates of the feed, concentrate, and tailing, respectively. Let f, c, and t be their respective copper grades (as mass fractions).
The recovery (R) is defined as: \[ R = \frac{\text{Mass of Cu in Concentrate}}{\text{Mass of Cu in Feed}} \times 100% = \frac{C \times c}{F \times f} \times 100% \] We can find the ratio \(C/F\) from a copper balance: \(Ff = Cc + Tt\). Since \(F = C+T\), we get the two-product formula: \( \frac{C}{F} = \frac{f-t}{c-t} \). Substituting this into the recovery formula gives a direct equation for recovery based on grades: \[ R = \frac{c(f-t)}{f(c-t)} \times 100% \] Step 3: Detailed Calculation:
Given values:
- Feed grade, \(f = 10% = 0.10\)
- Concentrate grade, \(c = 30% = 0.30\)
- Tailing grade, \(t = 1% = 0.01\)
Substitute these values into the recovery formula: \[ R = \frac{0.30 \times (0.10 - 0.01)}{0.10 \times (0.30 - 0.01)} \times 100% \] \[ R = \frac{0.30 \times 0.09}{0.10 \times 0.29} \times 100% \] \[ R = \frac{0.027}{0.029} \times 100% \] \[ R \approx 0.93103 \times 100% = 93.103% \] Rounding to the nearest integer, the recovery is 93%.
Step 4: Final Answer:
The percentage recovery of copper is 93.
Was this answer helpful?
0
0
Top Questions on Metallurgy
- What do you understand by concentration of an ore? Discuss the different methods of concentration of ores.
- What is slag?
- The process of purification of metal represented by the following equation is called \[ \text{Ti} + \text{I}_2 \xrightarrow{773\,K} \text{TiI}_4 \xrightarrow{1675\,K} \text{Ti} + \text{I}_2 \]
- Which of the following metals is not obtained by electrolysis?
- The process of smelting involves reduction of metal oxide with
View More Questions
Questions Asked in GATE MT exam
- If \( P e^x = Q e^{-x} \) for all real values of \( x \), which one of the following statements is true?
Bird : Nest :: Bee : __________
Select the correct option to complete the analogy.- Two consecutive estimates of the root of a function $f(x)$ obtained using the Newton-Raphson method are $x_i = 8.5$ and $x_{i+1 = 13.5$, and the value of the function at $x_i$ is 15. The numerical value of the first derivative of the function evaluated at $x_i$ is ............ (in integer).}
- GATE MT - 2025
- Calculus
- A cylindrical Al alloy billet of 300 mm diameter is hot extruded to produce a cylindrical rod of 75 mm diameter at a constant true strain rate of 10 s$^{-1$. The flow stress of the alloy at the extrusion temperature is given by $\sigma = 10 \, (\dot{\epsilon})^{0.3}$ MPa. Assume the alloy is perfectly plastic and there is no temperature rise during the extrusion process. The ideal plastic work of deformation per unit volume is ............ × 10$^6$ J m$^{-3}$ (rounded off to one decimal place).}
- GATE MT - 2025
- Metallurgical
- Temperatures at two sides of a 0.4 m thick copper plate are 1000 and 500 °C. Assuming steady state, one-dimensional conductive heat transfer through the wall and ignoring end-effects, the magnitude of the heat flux through the wall is ............... × 10$^5$ W m$^{-2$ (in integer).} Given: Thermal conductivity of copper = 400 W m$^{-1}$ K$^{-1}$
- GATE MT - 2025
- Thermodynamics
View More Questions