A fan is to accelerate quiescent air to a velocity of 12.5 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup plied to the fan. Take the density of air to be 1.18 kg/m3.
1 answer:
Answer:
= 829.69 Watt
≅ 830 Watt
Explanation:
Given that,
Velocity of air flow = 12.5m/s
Rate of flow of air = 9m³/s
Density of air = 1.18kg/m³
power by kinetic energy = 1/2(mv²)
mass = density × volume
m = 1.18 × 9
= 10.62 kg/s
power = 1/2 mV²
= 1/2 (10.62 × 12.5²)
= 829.69 Watt
≅ 830 Watt
Flow rate
u
=
9
m
3
/
s
Velocity of the air
V
=
8
m/s
Density of the air
ρ
=
1.18
kg
/
m
3
You might be interested in
10,000 units of momentum.
Answer:
F4.0
Explanation:
To obtain a shutter speed of 1/1000 s to avoid any blur motion the f-number should be changed to F4.0 because the light intensity goes up by a factor of 2 when the f-number is decreased by the square root of 2.
Answer:
ω = 4.07 rad/s
Explanation:
By conservation of the energy:
W = ΔK
where
Solving for ω:
Answer:
the hypotenuse = 13.78 cm
Ф = 27.44°
θ = 62.56°
explanation:
Yes bc its going from one to another solid-liquid-gas
