Question

30. What must be the pressure difference between the two ends of a \( 1.9 \mathrm{~km} \) section of a pipe, \( 29 \mathrm{~cm} \) in diameter, if it is to transfer oil at a rate of \( 4.5 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1} \) ? (For oil: \( \rho=950 \mathrm{~kg} \cdot \mathrm{m}^{-3}, \eta=0.2 \) Pa.s) A) \( 2.76 \times 10^{5} \mathrm{~Pa} \) B) \( 2.07 \times 10^{5} \mathrm{~Pa} \) C) \( 1.04 \times 10^{5} \mathrm{~Pa} \) D) \( 0.98 \times 10^{5} \mathrm{~Pa} \)

          30. What must be the pressure difference between the two ends of a \( 1.9 \mathrm{~km} \) section of a pipe, \( 29 \mathrm{~cm} \) in diameter, if it is to transfer oil at a rate of \( 4.5 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1} \) ? (For oil: \( \rho=950 \mathrm{~kg} \cdot \mathrm{m}^{-3}, \eta=0.2 \) Pa.s)
A) \( 2.76 \times 10^{5} \mathrm{~Pa} \)
B) \( 2.07 \times 10^{5} \mathrm{~Pa} \)
C) \( 1.04 \times 10^{5} \mathrm{~Pa} \)
D) \( 0.98 \times 10^{5} \mathrm{~Pa} \)

30. What must be the pressure difference between the two ends of a 1.9 km section of a pipe, 29 cm in diameter, if it is to transfer oil at a rate of 4.5 m^3·s^-1 ? (For oil: ρ=950 kg·m^-3, η=0.2 Pa.s)
A) 2.76 × 10^5 Pa
B) 2.07 × 10^5 Pa
C) 1.04 × 10^5 Pa
D) 0.98 × 10^5 Pa

Added by David R.

Physics Principles with Applications

Douglas C. Giancoli 6th Edition

Chapter 10

Instant Answer

Solved by Expert Surjit Tewari

09/20/2020

Step 1

Step 1: First, we need to understand that the flow of oil through a pipe can be described by the Hagen-Poiseuille equation, which relates the pressure difference (\( \Delta P \)) to the flow rate (\( Q \)), the viscosity of the fluid (\( \eta \)), the length of the Show more…

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30. What must be the pressure difference between the two ends of a \( 1.9 \mathrm{~km} \) section of a pipe, \( 29 \mathrm{~cm} \) in diameter, if it is to transfer oil at a rate of \( 4.5 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1} \) ? (For oil: \( \rho=950 \mathrm{~kg} \cdot \mathrm{m}^{-3}, \eta=0.2 \) Pa.s) A) \( 2.76 \times 10^{5} \mathrm{~Pa} \) B) \( 2.07 \times 10^{5} \mathrm{~Pa} \) C) \( 1.04 \times 10^{5} \mathrm{~Pa} \) D) \( 0.98 \times 10^{5} \mathrm{~Pa} \)