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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:
- P = ρ × Q × g × h
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:
Answer:
1) The pressure difference is 4.207 kilopascals.
2) 2.5 pounds per square inch equals 5.093 inches of mercury and 5.768 feet of water.
Explanation:
1) We can calculate the gas pressure difference from the U-tube manometer by using the following hydrostatic formula:
(Eq. 1)
Where:
- Relative density, dimensionless.
- Density of water, measured in kilograms per cubic meter.
- Gravitational acceleration, measured in meters per square second.
- Height difference in the U-tube manometer, measured in meters.
- Gas pressure difference, measured in kilopascals.
If we know that ,
,
and
, then the pressure difference is:
The pressure difference is 4.207 kilopascals.
2) From Physics we remember that a pound per square unit equals 2.036 inches of mercury and 2.307 feet of water and we must multiply the given pressure by corresponding conversion unit: ()
2.5 pounds per square inch equals 5.093 inches of mercury and 5.768 feet of water.