The maximum volume flow rate of kerosene is 8.3 L/s
<h3>What is the maximum volume flow rate?</h3>
In fluid dynamics, the maximum volume flow rate (Q) is the volume or amount of fluid flowing via a required cross-sectional area per unit time.
In fluid mechanics, using the following relation, we can determine the maximum volume flow rate of kerosene.
- Power = mass flow rate(m) × specific work(w) --- (1)
- Specific work = acceleration due to gravity (g) × head (h) ---- (2)
- Mass flow rate (m) = density (ρ) × volume flow rate (Q) --- (3)
By combining the three equations together, we have:
The power gained through the fluid pump to be:
Making Q the subject, we have:
where:
- P = 2 kW = 2000 W
- ρ = 0.820 kg/L
- g = 9.8 m/s
- h = 30 m
Q = 0.008296 m³/s
Q ≅ 8.3 L/s
Learn more about the maximum volume flow rate here:
brainly.com/question/19091787
Answer:
1.The velocity of fluid
2.Fluid properties.
3.Projected area of object(geometry of the object).
Explanation:
Drag force:
Drag force is a frictional force which offered by fluid when a object is moving in it.Drag force try to oppose the motion of object when object is moving in a medium.
Drag force given as
So we can say that drag force depends on following properties
1.The velocity of fluid
2.Fluid properties.
3.Projected area of object(geometry of the object).
The answer is b, I hope this helps you
Answer:
1709.07 ft^3/s
Explanation:
Annual peak streamflow = Log10(Q [ft^3/s] )
mean = 1.835
standard deviation = 0.65
Probability of levee been overtopped in the next 15 years = 1/5
<u>Determine the design flow ins ft^3/s </u>
P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2
∴ T = 67.72 years
Q₁₅ = 1 - 0.2 = 0.8
Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )
K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )
= 2.1504
back to equation 1
Zt = 1.835 + ( 2.1504 * 0.65 ) = 3.23276
hence:
Log₁₀ ( Qt(ft^3/s) ) = Zt = 3.23276
hence ; Qt = 10^3.23276
= 1709.07 ft^3/s
