To determine the flux per pole for the given DC shunt generator, we'll use the following steps and formulas:

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1. Key Data Provided:

• Power output ($P_{\text{out}}$) = $10 \, \text{kW} = 10,000 \, \text{W}$
• Terminal voltage ($V$) = $250 \, \text{V}$
• Speed ($N$) = $1000 \, \text{r.p.m.}$
• Poles ($P$) = $6$
• Armature conductors ($Z$) = $534$
• Lap connection: Number of parallel paths ($A$) = $P = 6$ (since lap winding has $A = P$)
• Full-load armature copper loss ($P_{\text{cu}}$) = $0.64 \, \text{kW} = 640 \, \text{W}$
• Total brush drop = $1 \, \text{V}$
• Neglecting shunt current ($I_{\text{sh}} = 0$).

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2. Calculate Armature Current:

The total output power of the generator is:

$P_{\text{out}} = V \cdot I_{\text{a}}$

Rearranging for $I_{\text{a}}$ (the armature current):

$I_{\text{a}} = \frac{P_{\text{out}}}{V} = \frac{10,000}{250} = 40 \, \text{A}$

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3. Determine the Generated Voltage $E_{\text{g}}$:

The generated voltage $E_{\text{g}}$ is related to the terminal voltage $V$, brush drop, and armature resistance ($R_{\text{a}}$). Using:

$E_{\text{g}} = V + \text{Brush Drop} + I_{\text{a}} R_{\text{a}}$

However, since the armature copper loss is given ($P_{\text{cu}} = I_{\text{a}}^2 R_{\text{a}}$), we can find $R_{\text{a}}$:

$R_{\text{a}} = \frac{P_{\text{cu}}}{I_{\text{a}}^2} = \frac{640}{40^2} = 0.4 \, \Omega$

Now substitute into $E_{\text{g}}$:

$E_{\text{g}} = V + 1 + (40)(0.4) = 250 + 1 + 16 = 267 \, \text{V}$

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4. Use the EMF Equation to Find Flux Per Pole:

The EMF equation for a DC generator is:

$E_{\text{g}} = \frac{N \cdot P \cdot \phi \cdot Z}{60 \cdot A}$

Rearranging for flux per pole ($\phi$):

$\phi = \frac{E_{\text{g}} \cdot 60 \cdot A}{N \cdot P \cdot Z}$

Substitute the known values:

• $E_{\text{g}} = 267 \, \text{V}$,
• $N = 1000 \, \text{r.p.m.}$,
• $P = 6$,
• $Z = 534$,
• $A = 6$ (lap winding).

$\phi = \frac{267 \cdot 60 \cdot 6}{1000 \cdot 6 \cdot 534}$

Simplify:

$\phi = \frac{96120}{3204000} = 0.03 \, \text{Wb}$

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Final Answer:

The flux per pole is:

$\phi = 0.03 \, \text{Wb}$

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Would you like a breakdown of any part of the solution or further related questions answered? Here are some related problems to consider:

1. How would the flux per pole change if the generator speed increased to 1200 r.p.m.?
2. What happens to the generated voltage $E_{\text{g}}$ if the armature resistance is halved?
3. How does the number of poles affect the flux per pole calculation?
4. What would be the effect of shunt current on the calculation?
5. How is the efficiency of the generator calculated in this scenario?

Tip: Always check the generator's operating speed and connection type (lap or wave) when using the EMF equation.