A steel part with surface area of 125 cm² is to be chrome coaled through an electroplating process using chromium acid sulphate as an electrolyte. An increasing current is applied to the part according to the following current time relation :

I = 12 + 0.2t

where, I = current (A) and t = time (minutes). The part is submerged in the plating solution for a duration of 20 minutes for plating purpose. Assuming the cathode efficiency of chromium to be 15% and the plating constant of chromium acid sulphate to be 2.50 × 10^–2 mm³/A·s, the resulting coating thickness on the part surface is _________ μm (round off to one decimal place).

The correct answer is 5.0 (or 5.04, which rounds to 5.0).

Here is a step-by-step breakdown of the electroplating process and calculation to determine the resulting coating thickness on the part surface.

1. Given Parameters:
- Surface area of the steel part, $A=125{\text{cm}}^2$
- Plating constant for chromium acid sulphate, $C=2.50\times {10}^{-2}{\text{mm}}^3\text{/(A·s)}$
- Cathode efficiency, $\eta =15\%=0.15$
- Plating duration, $t=20\text{minutes}$
- Time-varying current equation:

$I(t)=12+0.2t$

where $t$ is in minutes and $I$ is in Amperes (A).

2. Expressing Volume of Plating Material Deposited:
For an infinitesimally small duration $dt$ (in minutes), the corresponding time in seconds is $60\times dt$. The volume of chromium metal deposited $dV$ during this interval (taking into account the cathode efficiency) is given by:

$dV=C\times I(t)\times (60dt)\times \eta$

Substituting the given values into the differential equation:

$dV=(2.50\times {10}^{-2})\times 60\times (12+0.2t)\times 0.15dt$

Simplifying the constant factor:

$2.50\times {10}^{-2}\times 60\times 0.15=1.5\times 0.15=0.225$

Therefore, the differential volume simplifies to:

$dV=0.225\times (12+0.2t)dt$

3. Calculating the Total Deposited Volume by Integration:
We integrate this expression from $t=0$ to $t=20$ minutes to find the total volume $V$:

$V={\int }_0^{20}0.225(12+0.2t)dt$

$V=0.225{[12t+0.1t^2]}_0^{20}$

Evaluating the limits:

$V=0.225\times [(12\times 20)+(0.1\times {20}^2)]$

$V=0.225\times [240+40]$

$V=0.225\times 280=63{\text{mm}}^3$

4. Determining Plating Thickness:
First, convert the surface area from cm² to mm²:

$A=125{\text{cm}}^2=125\times 100{\text{mm}}^2=12,500{\text{mm}}^2$

Plating thickness $T$ is the volume divided by the surface area:

$T=\frac{V}{A}=\frac{63{\text{mm}}^3}{12,500{\text{mm}}^2}=0.00504\text{mm}$

Convert the thickness from millimeters (mm) to micrometers (μm):

$T=0.00504\times 1000\mu \text{m}=5.04\mu \text{m}$

Rounding off to one decimal place, the coating thickness is 5.0 μm.