Question
A separately excited dc generator develops $90 \mathrm{~N}-\mathrm{m}$ of torque and $18 \mathrm{~kW}$ of power. Compute the generator speed.
Step 1
We have the torque \( T = 90 \, \text{N-m} \) and the power \( P = 18 \, \text{kW} \). Show more…
Show all steps
Your feedback will help us improve your experience
Paul Gabriel and 73 other educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A motor has a back emf of $110 \mathrm{~V}$ and an armature current of 90 A when running at 1500 rpm. Determine the power and the torque developed within the armature. Power $=($ Armature current $)($ Back emf $)=(90 \mathrm{~A})(110 \mathrm{~V})=9.9 \mathrm{~kW}$ From Chapter 10, power $=\tau \omega$ where $\omega=2 \pi f=2 \pi(1500 \times 1 / 60)$ $\mathrm{rad} / \mathrm{s}$ $$ \text { Torque }=\frac{\text { Power }}{\text { Angular speed }}=\frac{9900 \mathrm{~W}}{(2 \pi \times 25) \mathrm{rad} / \mathrm{s}}=63 \mathrm{~N} \cdot \mathrm{m} $$
A dynamo (generator) delivers $30.0 \mathrm{~A}$ at $120 \mathrm{~V}$ to an external circuit when operating at $1200 \mathrm{rpm} .$ What torque is required to drive the generator at this speed if the total power losses are 400 W?
(II) A dc generator is rated at $16 \mathrm{kW}, 250 \mathrm{V},$ and 64 $\mathrm{A}$ when it rotates at 1000 $\mathrm{rpm} .$ The resistance of the armature windings is 0.40$\Omega .$ (a) Calculate the "no-load" voltage at 1000 $\mathrm{rpm}$ (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.
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD