In a direct current arc welding process, the power source has an open circuit voltage of 100 V and short circuit current of 1000 A. Assume a linear relationship between voltage and current. The arc voltage (V) varies with the arc length (l) as V = 10 + 5l, where V is in volts and l is in mm. The maximum available arc power during the process is _________ kVA (in integer).

The correct answer is 25.

Step 1: Understand the Given Parameters
In a direct current (DC) arc welding process, we are given the characteristics of the power source:
- Open circuit voltage, V_o = 100 V
- Short circuit current, I_s = 1000 A
We are also given the relationship between the arc voltage (V) and the arc length (l, in mm):
V = 10 + 5l

Step 2: Establish the Power Source Characteristic Equation
Assuming a linear relationship between the operating voltage (V) and current (I) for the power source, we can write the linear power source characteristic equation as:

$V=V_o-\left(\frac{V_o}{I_s}\right)I$

Substituting the given values of V_o and I_s:

$V=100-\left(\frac{100}{1000}\right)I$

Simplifying this relation:

$V=100-\frac{I}{10}$

Step 3: Relate Arc Current and Arc Length for a Stable Arc
For a stable welding arc, the voltage supplied by the power source must equal the arc voltage. Therefore, we equate the two voltage equations:

$10+5l=100-\frac{I}{10}$

Rearranging this equation to solve for the current I as a function of the arc length l:

$\frac{I}{10}=100-10-5l$

$\frac{I}{10}=90-5l$

Multiplying both sides by 10 gives:

$I=900-50l$

Step 4: Formulate the Arc Power Equation
The electrical power (P) generated in the arc is the product of the arc voltage (V) and the arc current (I):

$P=V\times I$

Substituting the expressions for V and I in terms of the arc length l:

$P=(10+5l)(900-50l)$

Expanding the product:

$P=9000-500l+4500l-250l^2$

$P=9000+4000l-250l^2$

Step 5: Maximize the Arc Power
To find the maximum available arc power, we differentiate the power equation with respect to l and set it to zero:

$\frac{dP}{dl}=4000-500l=0$

Solving for the optimal arc length l:

$500l=4000\Rightarrow l=8\text{mm}$

Step 6: Calculate the Maximum Arc Power
Substitute l = 8 mm back into the voltage and current equations:
- Arc voltage at maximum power:

$V_{\text{max}}=10+5\left(8\right)=10+40=50\text{V}$

- Arc current at maximum power:

$I_{\text{max}}=900-50\left(8\right)=900-400=500\text{A}$

Now, calculate the maximum power:

$P_{\text{max}}=V_{\text{max}}\times I_{\text{max}}=50\text{V}\times 500\text{A}=25000\text{W}$

Converting the power into kVA (or kW, since power factor is assumed to be 1 in a DC circuit):

$P_{\text{max}}=\frac{25000}{1000}=25\text{kVA}$