What must be the pressure difference between the two ends of a 2.3 km section of pipe, 29 cm in diameter, if it is to transport oil (ρ=950kg/m3,η=0.20Pa⋅s) at a rate of 660 cm3/s ?
I'm sorry. We have no tutors here who specialize in Kuwait university as a school subject. If you need a physics tutor, someone may be able to assist you.
Indicate your specific subject in the "School Subject" box, so those with expertise in the area will respond to the question.
To determine the pressure difference between the two ends of the pipe, we can use the Poiseuille's law, which relates the pressure drop to various factors, including viscosity, pipe length, pipe radius, and flow rate.
Poiseuille's law states that the pressure difference (ΔP) is given by:
ΔP = (8ηLQ)/(πr^4),
where:
η = viscosity of the oil (0.20 Pa⋅s),
L = length of the pipe section (2.3 km = 2300 m),
Q = flow rate of the oil (660 cm^3/s = 0.66 L/s = 0.00066 m^3/s),
r = radius of the pipe (29 cm/2 = 14.5 cm = 0.145 m).
Now we can substitute the values into the formula:
ΔP = (8 * 0.20 Pa⋅s * 2300 m * 0.00066 m^3/s) / (π * (0.145 m)^4).
First, let's calculate the denominator:
Denominator = π * (0.145 m)^4
= π * 0.145^4 m^4
≈ 0.0020061 m^4.
Now, let's calculate the pressure difference:
ΔP ≈ (8 * 0.20 Pa⋅s * 2300 m * 0.00066 m^3/s) / 0.0020061 m^4
≈ (3.68 Pa⋅m^2/s) / 0.0020061 m^4
≈ 1831.35 Pa.
Therefore, the pressure difference between the two ends of the 2.3 km section of pipe must be approximately 1831.35 Pa.