Problem 45 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A dc generator is rated at \(16 \mathrm{~kW}, 250 \mathrm{~V}\), and \(64 \mathrm{~A}\) when it rotates at 1000 rpm. The resistance of the armature windings is \(0.40 \Omega\). ( \(a\) ) Calculate the "no-load" voltage at 1000 rpm (when there is no circuit hooked up to the generator). ( \(b\) ) Calculate the full-load voltage (i.e. at 64 A) when the generator is run at 750 rpm. Assume that the magnitude of the magnetic field remains constant.

Step-by-Step Solution

Answer

The no-load voltage at 1000 rpm is 275.6 V. The full-load voltage at 750 rpm is 181.1 V.

Step 1: Understanding Generator Ratings

We're given that the dc generator has specified ratings of 16 kW (kilowatts), 250 V (volts), and 64 A (amperes) while operating at 1000 RPM (rotations per minute). These ratings indicate the generator's maximum power output at the described conditions.

Step 2: Calculating No-Load Voltage

No-load voltage is the generator's output voltage when there is no current (I = 0 A) being drawn from the generator. The internal generated voltage (Eg) can be calculated as follows: Using Ohm's Law and adding the voltage drop across the armature resistance: \[ V = Eg - I \cdot R_a \]Where:- \( V \) is the terminal voltage (250 V)- \( I \) is the current (64 A)- \( R_a \) is the armature resistance (0.40 Ω)Rearranging to find \( Eg \): \[ Eg = V + I \cdot R_a = 250 + 64 \cdot 0.40 = 250 + 25.6 = 275.6 \text{ V} \]At no load, \( I = 0 \): \[ Eg_{\text{no-load}} = 275.6 \text{ V} \]

Step 3: Calculating Full-Load Voltage at Reduced RPM

When the generator is running at 750 RPM and at full load (64 A), the generated voltage will be different due to the changed speed. Using the known relationship that voltage is proportional to speed in a DC generator, Let \( Eg_{1000} \) be the generated voltage at 1000 RPM (275.6 V from Step 2). The generated voltage at 750 RPM is:\[ Eg_{750} = Eg_{1000} \times \left( \frac{750}{1000} \right) = 275.6 \times 0.75 = 206.7 \text{ V} \]Now calculate the terminal voltage at full load (64 A):\[ V_{750} = Eg_{750} - I \cdot R_a = 206.7 - 64 \cdot 0.40 = 206.7 - 25.6 = 181.1 \text{ V} \]

Key Concepts

Armature Resistance

In a DC generator, "armature resistance" refers to the resistance of the winding through which the generated current flows. This resistance plays a crucial role in determining the output voltage of the generator. If the resistance is too high, it causes a significant voltage drop and reduces the efficiency of the generator.

• Analogy: Think of it like the friction in a car engine. More friction means less speed for the same power applied.
• In our example, the armature resistance is given as \(0.40\, \Omega\). This value needs to be accounted for when calculating the terminal voltages at different loads.

Armature resistance can be used to calculate the voltage drop, which is given by the formula:
\[ V_{ ext{drop}} = I \cdot R_a \]
Where \(I\) is the current and \(R_a\) is the armature resistance. This voltage drop affects both no-load and full-load voltages, making it essential for understanding generator behavior.

No-load Voltage

The "no-load voltage" of a DC generator is the voltage when no current is drawn from it. Simply put, it's the maximum potential difference across the generator's terminals with no electrical load attached.

• Pure voltage without any real-world interference.
• Calculated by adding the voltage drop across the resistance to the rated voltage.

In our problem, the no-load voltage at 1000 RPM is calculated by considering the armature resistance and adding the resulting voltage drop to the rated terminal voltage (250 V). As per the step-by-step solution, with the armature resistance factored in, this gives us:
\[ Eg_{\text{no-load}} = 275.6 \text{ V}\]
This showcases the actual potential of the generator under ideal conditions, serving as a crucial reference point for further calculations.

Full-load Voltage

The "full-load voltage" specifies the output voltage of the generator when it is under full load (maximum current). This is important because as current flows through the generator, it produces a voltage drop across the internal resistance, decreasing the terminal voltage.

• The power output is often lower than maximum potential due to real-world conditions.
• A crucial measure to ensure that generators can handle practical applications.

For our exercise at 750 RPM, with full load (64 A), the terminal voltage is computed as:
\[ V_{750} = Eg_{750} - I \cdot R_a\]
Using the values, we find:
\[V_{750} = 206.7 - 25.6 = 181.1 \text{ V}\]
This calculation takes into consideration the fewer RPMs, highlighting the effect of reduced speed on output. The knowledge of full-load voltage equips us to assess generator performance in real operating conditions.

RPM Dependency

The "RPM Dependency" in DC generators reflects how the generated voltage depends on the shaft's rotational speed. When RPMs increase, so does the generated voltage, and vice versa.

• RPM (Rotations per Minute) relates to how fast the generator's rotor spins.
• The faster the rotor spins, the more electromotive force (EMF) is induced.

In our example, when the generator's speed is reduced from 1000 RPM to 750 RPM, the generated voltage drops. This is depicted by using proportional relationships because voltage is directly proportional to rotational speed:
\[ Eg_{750} = Eg_{1000} \times \left( \frac{750}{1000} \right) \]
Where \(Eg_{1000}\) is 275.6 V from the earlier calculation. Thus, we get:
\[ Eg_{750} = 206.7 \text{ V} \]
Understanding RPM dependency is key to predicting the behavior of DC generators under different operating conditions and adjusting to demands.