Is the Claim About the Fan's Discharge Rate and Velocity Reasonable?

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SUMMARY

The discussion centers on evaluating the claim that a fan consuming 20W of electric power can discharge air at a rate of 0.25 kg/s with a velocity of 8 m/s. The kinetic energy of the wind is calculated using the formula KE = 0.5 mv², leading to a conclusion that 8J/s is a reasonable output for the fan's power. The participants agree that a 40% efficiency for the fan is plausible, and further exploration of estimating useful power through force and velocity is suggested.

PREREQUISITES
  • Understanding of kinetic energy calculations (KE = 0.5 mv²)
  • Basic principles of power and efficiency in electrical devices
  • Familiarity with fluid dynamics concepts, particularly airflow and discharge rates
  • Knowledge of force and velocity relationships in physics
NEXT STEPS
  • Research the efficiency of various fan models and their performance metrics
  • Learn about fluid dynamics and airflow calculations in HVAC systems
  • Explore the principles of energy conversion in electric motors
  • Investigate methods for measuring airflow rates and velocities in practical applications
USEFUL FOR

Students in physics or engineering, HVAC professionals, and anyone interested in the efficiency and performance of electric fans and airflow systems.

a13x
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Homework Statement


Been at this question for some time an needing a double check to see if things are right.

A fan that consumes 20W of eletric power when operating is claimed to discharge air from a ventilaed room at a rate of 0.25 kg/s at a discharge velocity of 8 m/s. Determine if this claim is reasonable?


Homework Equations


KE of wind = 0.5 mv^2
20W = 20 j/s


The Attempt at a Solution


When i worked it out i calculated it would be 8J. This seems to be posible for a 20W fan to do this work. Is this correct??
 
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You probably mean 8J/s - but yes, 40% efficency sounds reasonable.
 
Sorry, I did mean 8J/s.

Thankyou for the reply
 
Another question, related to the written above -
Is that correct to estimate the "usefull power" of a fan in the following way:

0- Estimating Force applied on fan. we get F.
1 - Estimating the velocity of the wind near the fan. we get Vo.
2 - using this estimation alone (ignoring wind decay & turbulences & whatever happens far from fan) to look at the picture from the wind's frame of reference- where the fan is moving with speed of -Vo, thus calculating P=Vo*F.

It's stage 2 I'm not completely sure of .. is that stage's logic OK?
 

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