To solve the problem, we will use the formula for the generated electromotive force (EMF) in a generator with wave winding, given by:

$E = \frac{N \cdot \phi \cdot P \cdot n}{60 \cdot A}$

Where:

• $E$ is the generated EMF (in volts),
• $N$ is the total number of conductors,
• $\phi$ is the flux per pole (in Weber, Wb),
• $P$ is the number of poles,
• $n$ is the speed (in RPM),
• $A$ is the number of parallel paths in the winding.

For a wave winding, $A = 2$ regardless of the number of poles.

Part 1: Calculate the generated EMF

Given dataTo solve the problem, we will use the formula for the generated electromotive force (EMF) in a generator with wave winding, given by:

$E = \frac{N \cdot \phi \cdot P \cdot n}{60 \cdot A}$

Where:

• $E$ is the generated EMF (in volts),
• $N$ is the total number of conductors,
• $\phi$ is the flux per pole (in Weber, Wb),
• $P$ is the number of poles,
• $n$ is the speed (in RPM),
• $A$ is the number of parallel paths in the winding.

For a wave winding, $A = 2$ regardless of the number of poles.

Part 1: Calculate the generated EMF

Given

• $P = 6$ (number of poles),
• $N = 664$ (number of conductors),
• $\phi = 60 \, \text{mWb} = 60 \times 10^{-3} \, \text{Wb}$,
• $n = 250 \, \text{rpm}$.

Now, let's substitute these values into the formula and calculate the generated EMF:

$E = \frac{664 \times 60 \times 10^{-3} \times 6 \times 250}{60 \times 2}$

Part 2: Find the speed for a given EMF

In this part, we need to find the speed ($n$) to generate an EMF of 550V when the flux per pole is reduced to 58mWb.

We will rearrange the EMF formula to solve for $n$:

$n = \frac{E \times 60 \times A}{N \times \phi \times P}$

Given:

• $E = 550 \, \text{V}$,
• $\phi = 58 \, \text{mWb} = 58 \times 10^{-3} \, \text{Wb}$,
• $P = 6$,
• $N = 664$,
• $A = 2$ (for wave winding).

Now, we substitute these values and calculate the required speed.

I'll solve these steps next.### Part 1: Generated EMF The generated EMF when the flux per pole is $60 \, \text{mWb}$ and the speed is $250 \, \text{rpm}$ is: $E = 83 \, \text{V}$

Part 2: Required Speed for 550V EMF

To generate an EMF of $550 \, \text{V}$ when the flux per pole is reduced to $58 \, \text{mWb}$, the armature must be driven at a speed of approximately: $n = 285.63 \, \text{rpm}$

Let me know if you'd like further details or have any questions!

Here are 5 related questions for further practice:

1. How does the generated EMF change if the speed is doubled while keeping all other parameters constant?
2. What would happen to the EMF if the number of poles is reduced to 4 while keeping the speed constant?
3. How does the EMF change if the flux per pole is halved but the speed is doubled?
4. Calculate the EMF for a generator with lap winding instead of wave winding under the same conditions.
5. How would the required speed change if the number of conductors is increased to 800?

Tip: In wave-wound generators, the number of parallel paths is always 2, regardless of the number of poles, unlike lap windings where the number of parallel paths equals the number of poles.