An electrochemical machining (ECM) is to be used to cut a through hole into a 12 mm thick aluminum plate. The hole has a rectangular cross-section, 10 mm × 30 mm. The ECM operation will be accomplished in 2 minutes, with efficiency of 90%. Assuming specific removal rate for aluminum as 3.44 × 10^-2 mm³ /(A s), the current (in A) required is __________ (correct to two decimal places).

The correct answer is 968.99.

Step-by-step Explanation:

First, let us identify the given parameters from the problem statement:
Thickness of the plate, $t=12\text{mm}$
Cross-sectional area of the rectangular hole, $A=10\text{mm}\times 30\text{mm}=300{\text{mm}}^2$
Machining time, $T=2\text{minutes}=2\times 60\text{s}=120\text{s}$
Efficiency of the ECM process, $\eta =90\%=0.90$
Specific removal rate for aluminum, $R=3.44\times {10}^{-2}{\text{mm}}^3/(\text{A}·\text{s})$

Step 1: Calculate the total volume of metal to be removed ($V$)
The volume $V$ of the rectangular through hole is the product of its cross-sectional area and the plate thickness:
$V=A\times t=300{\text{mm}}^2\times 12\text{mm}=3600{\text{mm}}^3$

Step 2: Relate volume removal to current and efficiency
In electrochemical machining, the actual volume of metal removed is given by the relation:
$V=\eta \times R\times I\times T$
where $I$ is the required current in Amperes.

Step 3: Calculate the required current ($I$)
Rearranging the equation to solve for current $I$:
$I=\frac{V}{\eta \times R\times T}$
Substituting the values:
$I=\frac{3600}{0.90\times (3.44\times {10}^{-2})\times 120}$
$I=\frac{3600}{108\times 0.0344}$
$I=\frac{3600}{3.7152}\approx 968.9922\text{A}$

Rounding the current to two decimal places, we get:
$I=968.99\text{A}$