an engine pumps water from a river 10m
an below its own level and discharge it through a
nozzle of diameter 10cm with a speed of 50m/
s.find the power required assuming
a.no losses
b.70% efficiency.water weighs 10^3kgm^-3 {g=10ms^-2}
help me
need physics properties
energy/second:
mgh + (1/2) m v^2
where m = mass/second
what is mass/second?
m = 1000 kg/m^3 * pi (.05^2)m^2 * 50 m/s
I need answers please
i need the answer please
Answer
To find the power required by the engine to pump the water, we can use the following formula:
Power = Force × Velocity
First, let's calculate the force exerted by the water on the nozzle.
a) Assuming no losses:
Since there are no losses mentioned, we can assume that all the force exerted by the water on the nozzle is due to the change in momentum. The force can be calculated using the formula:
Force = Mass flow rate × Change in velocity
To calculate the mass flow rate, we need to calculate the cross-sectional area of the nozzle and the velocity of water through the nozzle.
Area = π × (Radius of nozzle)^2
= π × (0.05m)^2
= 0.00785m²
The velocity of water through the nozzle is given as 50m/s.
Mass flow rate = Density × Volume flow rate
= Density × (Area × Velocity)
= (10^3kg/m³) × (0.00785m²) × (50m/s)
= 392.5kg/s
Now, let's calculate the change in velocity. Since the water is being pumped from a river below its own level, we can assume that the initial velocity of the water is zero (as it is not specified).
Change in velocity = Final velocity - Initial velocity
= 50m/s - 0m/s
= 50m/s
Force = (392.5kg/s) × (50m/s)
= 19,625N
Now, let's calculate the power required using the force and velocity values.
Power = Force × Velocity
= 19,625N × 50m/s
= 981,250W
Therefore, the power required, assuming no losses, is 981,250 Watts.
b) Assuming 70% efficiency:
If the engine is assumed to have an efficiency of 70%, the power required would be higher due to energy losses. To find the actual power required, we can use the formula:
Actual Power = Power required / Efficiency
Actual Power = 981,250W / 0.70
= 1,401,786W
Therefore, the power required, considering a 70% efficiency, is 1,401,786 Watts.