Process for deep copper removal through electrodeposition
A production method and copper removal technology, applied in the direction of photography process, photography auxiliary process, process efficiency improvement, etc., can solve the problems of copper metal loss, high energy consumption, low metal recovery rate, etc. Huge losses and the effect of improving metal recovery
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example 1
[0028] Example 1: There are 6 electrowinning cells in one group, and the liquid intake in each group is 1.5m 3 / h, the copper content of the electrolyte is 20g / L (converted to 20kg / m 3 ) The average current efficiency is calculated as 50%, and the volume of electrolyte stored in the last tank is 3m 3 , its copper content is 2g / L (converted to 2 kg / m 3 ) The target value of copper content in the final solution is set to 1 g / L (converted to 1Kg / m 3 ) The current intensity that needs to pass is:
[0029] I=
[0030] = (1.5m 3 / h×20 Kg / m 3 )÷(50%×1.1852g / A·h×6×10 -3 )
[0031] =8437.4A
[0032] The time required to remove the electrolyte in the last tank to 1g / L after the liquid feeding is stopped is:
[0033] t=
[0034] =[3m 3 ×(2-1)Kg / m 3 ]÷(8437.4A×50%×1.1852g / A·h×10 -3 )
[0035] =0.6h
[0036] In this embodiment, the electrolyte is removed through the main liquid inlet pipe into the first electrowinning tank, and then energized and sequentially removed ...
example 2
[0039] Example 2: There are 8 electrowinning cells in one group, and the liquid intake in each group is 2.5m 3 / h, the copper content in the electrolyte is 24g / L (converted to 24kg / m 3 ) The average current efficiency is calculated as 80%, and the volume of electrolyte stored in the last tank is 2.5m 3 , its copper content is 3g / L (converted to 3 kg / m 3 ) The target value of copper content in the final solution is set to 1g / L (converted to 1 Kg / m 3 ) The current intensity that needs to pass is:
[0040] I=
[0041] = (2.5m 3 / h×24 Kg / m 3 )÷(80%×1.1852g / A·h×8×10 -3 )
[0042] =7910A
[0043] The time required to remove the electrolyte in the last tank to 1g / L after the liquid feeding is stopped is:
[0044] t=
[0045] =[2.5m 3 ×(3-1)Kg / m 3 ]÷(7910A×80%×1.1852g / A·h×10 -3 )
[0046] =0.67h
[0047]In this embodiment, the electrolyte is removed through the main liquid inlet pipe into the first electrowinning tank, and then energized and sequentially removed ...
example 3
[0049] Example 3: There are 10 electrowinning cells in one group, and the liquid intake in each group is 2m 3 / h, the copper content in the electrolyte is 22g / L (converted to 22kg / m 3 ) The average current efficiency is calculated as 65%, and the volume of electrolyte stored in the last tank is 3m 3 , its copper content is 5g / L (converted to 5 kg / m 3 ) The target value of copper content in the final solution is set to 1g / L (converted to 1 Kg / m 3 ) The current intensity that needs to pass is:
[0050] I=
[0051] = (2 m 3 / h×22 Kg / m 3 )÷(65%×1.1852g / A·h×10×10 -3 )
[0052] =5711.5A
[0053] The time required to remove the electrolyte in the last tank to 1g / L after the liquid feeding is stopped is:
[0054] t=
[0055] =[3m 3 ×(5-1)Kg / m 3 ]÷(5711A×65%×1.1852g / A·h×10 -3 )
[0056] =2.727h
[0057] In this embodiment, the electrolyte is removed through the main liquid inlet pipe into the first electrowinning tank, and then energized and sequentially removed t...
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