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GATE Mechanical Engineering (ME) 2020 Shift 1 Solved Paper

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Question : 55 of 65

Marks: +1, -0

A steel part with surface area of

125\ \mathrm{cm}^2

is to be chrome coated 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

I=

A

t=

time (minutes). The part is submerged in the plating solution for 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 \times 10^{-2}\ \mathrm{mm} / \mathrm{A}\text{-}\mathrm{s}

, the resulting coating thickness on the part surface is ________ µm (round off to one decimal place).

Your Answer:

Solution:

Given

As

=125\ \mathrm{cm}^2

A_s=125 \times 10^{2}\ \mathrm{mm}^2

(\eta)_{\text{cathode}}=0.15

I=12+0.2t

t=20

minutes.

C (Plating Constant)

= B \cdot 0.5 \times 10^{-2}\ \mathrm{mm}^3 / \mathrm{As}

C = A \cdot 0.5\ \mathrm{mm}^3 / (\mathrm{A}\ \mathrm{min})

Since current is changing with time we have to Integrate

T \rightarrow

thickness of coating

\frac{dT}{dt} = \frac{CI}{A_s} h_c

\frac{dT}{dt} = \frac{(1.5)(12+0.2t)(0.15)}{125 \times 100}

dT = \frac{(1.5)(12+0.2t)(0.15) dt}{125 \times 100}

T = \int dT = \int\limits_{0}^{20} \frac{(0.15)(1.5)(12+0.2t) dt}{125 \times 100}

T = \frac{(1.5)(0.15)}{125 \times 100} \left[ 12t + \frac{0.2 t^{2}}{2} \right]_{0}^{20}

T = \frac{(1.5)(0.15)}{125 \times 100} \left[ 240 + 0.1 \times 400 \right]

T = 0.504 \times 10^{-2}\ \mathrm{mm}

T = 5.04\ \mu\mathrm{m}

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