The speed(ω) - torque(T) characteristics of a fan type load is given byans.T ∝ ωT ∝ 1/ωT = constantT ∝ ω2 Previous Marked for Review Next
Question
The speed(ω) - torque(T) characteristics of a fan type load is given byans.T ∝ ωT ∝ 1/ωT = constantT ∝ ω2 Previous Marked for Review Next
Solution
It seems like you've provided some relationships between torque (T) and speed (ω) for a fan type load, but you haven't asked a specific question. Here's what I can tell you based on the information you've given:
-
T ∝ ω: This means that torque is directly proportional to speed. If the speed increases, the torque also increases, and vice versa.
-
T ∝ 1/ω: This means that torque is inversely proportional to speed. If the speed increases, the torque decreases, and vice versa.
-
T = constant: This means that the torque is constant, regardless of the speed.
-
T ∝ ω2: This means that torque is proportional to the square of the speed. If the speed increases, the torque increases by the square of that increase.
Please provide more context or a specific question for a more detailed answer.
Similar Questions
When a ceiling fan rotating with a angular speed of 2.60 rad/s is turned off, a frictional torque of 0.221 N*m slows it to a stop in 5.80 s. What is the moment of inertia of the fan?
Variable Fan Speed Control
A ventilation fan with a moment of inertia of 0.034 kg·m2 has a net torque of 0.14 N·m applied to it. What angular acceleration does it experience?Select one:a.0.31 rad/s2b.4.1 rad/s2c.3.2 rad/s2d.5.3 rad/s2
A fan blade, initially at rest, rotates with a constant acceleration of 0.025 rad/s2. What is its angular speed at the instant it goes through an angular displacement of 10 rad?Select one:a.0.11 rad/sb.0.025 rad/sc.0.71 rad/sd.0.46 rad/s
A fan blade, initially at rest, rotates with a constant acceleration of 0.025 rad/s2. What is its angular speed at the instant it goes through an angular displacement of 10 rad?Select one:a.0.46 rad/sb.0.11 rad/sc.0.71 rad/sd.0.025 rad/s
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.