An electric fan is used to accelerate the initially quiescent gas up to a veloci
ID: 2993642 • Letter: A
Question
An electric fan is used to accelerate the initially quiescent gas up to a velocity of 20 m/s with a mass flow rate of 9.6 kg/s. The density of the gas is 1.2 kg/m3. The electricity consumption rate is 2.25 kW. The efficiency of the fan in transferring the mechanical work into the kinetic energy of the gas is 87%. Ignoring the change in the potential energy of the gas,
(A) Draw a schematic diagram of this system showing the conversion of the electrical work to
the kinetic energy of the gas.
(B) Determine the volumetric flow rate of the gas.
(C) Determine the specific kinetic energy (J/kg) of the gas entering and exiting the fan.
(D) Determine the rate of kinetic energy (J/s) of the gas exiting the fan.
(E) Determine the power output of the fan.
(F) Determine the efficiency of the electrical to mechanical work conversion by the fan.
.
Explanation / Answer
b)
Volumetric flow rate Q = m / rho = 9.6 / 1.2 = 8 m^3 /s
c)
Specific KE = v^2 /2 = 20^2 /2 = 200 J/kg
d)
Rate of KEof gas exiting the fan = m*specific KE = 9.6*200 = 1920 J/s
e)
Power output = Rate of KE / Efficiency of conversion
= 1920 / 0.87
= 2206.9 W
= 2.2 kW
f)
Efficiency of electrical to mechanical work conversion = Power output / Electricity consumption
= 2.2 / 2.25
= 0.977 or 97.7 %
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.