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Chapter 15: Problem 102

A horizontal pipe carries oil whose coefficient of viscosity is \(0.00012 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\). The diameter of the pipe is \(5.2 \mathrm{cm},\) and its length is \(55 \mathrm{m}\) (a) What pressure difference is required between the ends of this pipe if the oil is to flow with an average speed of \(1.2 \mathrm{m} / \mathrm{s}\) ? (b) What is the volume flow rate in this case?

Short Answer

Expert verified
(a) 7608.3 N/m虏 (b) 0.0025484 m鲁/s

Step by step solution

01

Convert Units

First, convert the diameter of the pipe from centimeters to meters.\[\text{Diameter} = 5.2\, \text{cm} = 0.052\, \text{m}\]The radius will be half of the diameter:\[\text{Radius} = \frac{0.052}{2} = 0.026\, \text{m}\]
02

Use Poiseuille鈥檚 Law

Poiseuille鈥檚 law relates the pressure difference \( \Delta P \) to the flow properties in a pipe:\[\Delta P = \frac{8 \eta L v_{avg}}{\pi r^4}\]where:\( \eta = 0.00012\, \mathrm{N} \cdot \mathrm{s}/\mathrm{m}^2 \) is the viscosity,\( L = 55\, \mathrm{m} \) is the length of the pipe,\( v_{avg} = 1.2\, \mathrm{m/s} \) is the average flow speed, and\( r = 0.026\, \mathrm{m} \) is the radius.
03

Calculate Pressure Difference

Substitute the known values into Poiseuille鈥檚 law to find \( \Delta P \):\[\Delta P = \frac{8 \times 0.00012 \times 55 \times 1.2}{\pi \times (0.026)^4}\]Calculate the expression:\[\Delta P \approx 7608.3\, \text{N/m}^2\]
04

Calculate Volume Flow Rate

The volume flow rate \( Q \) can be found using the formula:\[Q = \text{Cross-sectional area} \times \text{Average velocity}\]The cross-sectional area of the pipe is:\[A = \pi r^2 = \pi (0.026)^2\]Calculate \( A \):\[A \approx 0.0021237\, \mathrm{m}^2\]Thus, the volume flow rate:\[Q = 0.0021237 \times 1.2 \approx 0.0025484\, \mathrm{m}^3/\mathrm{s}\]
05

Final Result

The pressure difference required is approximately 7608.3 N/m虏, and the volume flow rate is approximately 0.0025484 m鲁/s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Poiseuille's Law
In fluid dynamics, Poiseuille's Law describes how fluid flows through a cylindrical pipe. It connects several important factors affecting the flow, including the fluid's viscosity and the pipe鈥檚 dimensions. This law is expressed mathematically as: \[\Delta P = \frac{8 \eta L v_{avg}}{\pi r^4}\]where \( \Delta P \) is the pressure difference needed across the pipe length \( L \) to maintain a flow with average speed \( v_{avg} \). The coefficient of viscosity is represented by \( \eta \), and \( r \) is the pipe's radius.
  • Having a thicker fluid or a longer pipe increases the pressure needed.
  • Conversely, a larger radius significantly reduces the required pressure difference.
This formula is essential for calculating the pressure needed to maintain a steady oil flow in a pipe.
Viscosity
Viscosity is the measure of a fluid's resistance to flow. Imagine pouring honey and water; honey flows much slower, which means it has higher viscosity. In scientific terms, it's the internal friction inside the fluid.
  • High viscosity fluids require more force to move through a pipe.
  • In our example, the oil's viscosity is given as 0.00012 N路s/m虏.
This parameter is crucial in Poiseuille's Law, as a higher viscosity means a larger pressure difference is necessary to achieve the same flow rate.
Pressure Difference
The pressure difference (\( \Delta P \)) across the ends of a pipe is a driving force for fluid flow. It's like the push that keeps things moving. Because of fluid dynamics principles, you need to maintain a certain pressure to balance the viscous resistance as the fluid flows.
  • In Poiseuille's Law, the pressure difference is determined using factors like viscosity, length, and radius.
  • For a fixed flow rate in a smaller pipe, the required pressure is much higher.
In the original problem, we calculated the pressure difference to be approximately 7608.3 N/m虏 to keep oil flowing steadily.
Volume Flow Rate
Volume flow rate, denoted by \( Q \), is the amount of fluid passing through a section of the pipe per unit time. It shows how "fast" a fluid fills up a space. If you're filling a pool, a higher flow rate means quicker filling. Mathematically, it is determined by: \[Q = A \times v_{avg}\]where \( A \) is the cross-sectional area of the pipe and \( v_{avg} \) is the average velocity of the fluid.
  • The larger the cross-sectional area, the greater the volume flow rate.
  • In our case, we calculated the flow rate to be approximately 0.0025484 m鲁/s.
Understanding flow rate is crucial as it affects how quickly fluids can be transported through pipes.

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Most popular questions from this chapter

Two drinking glasses, 1 and 2 , are filled with water to the same depth. Glass 1 has twice the diameter of glass \(2 .\) (a) Is the weight of the water in glass 1 greater than, less than, or equal to the weight of the water in glass \(2 ?\) (b) Is the pressure at the bottom of glass 1 greater than, less than, or equal to the pressure at the bottom of glass \(2 ?\)

A hot-air balloon plus cargo has a mass of \(1890 \mathrm{kg}\) and a volume of \(11,430 \mathrm{m}^{3} .\) The balloon is floating at a constant height of \(6.25 \mathrm{m}\) above the ground. What is the density of the hot air in the balloon?

A hollow cubical box, \(0.29 \mathrm{m}\) on a side, with walls of negligible thickness floats with \(35 \%\) of its volume submerged. What mass of water can be added to the box before it sinks?

Water flows through a pipe with a speed of \(2.1 \mathrm{m} / \mathrm{s}\). Find the flow rate in \(\mathrm{kg} / \mathrm{s}\) if the diameter of the pipe is \(3.8 \mathrm{cm} .\)

Predict/Explain A person floats in a boat in a small backyard swimming pool. Inside the boat with the person are some bricks. (a) If the person drops the bricks overboard to the bottom of the pool, does the water level in the pool increase, decrease, or stay the same? (b) Choose the best explanation from among the following: When the bricks sink they displace less water than when they were floating in the boat; hence, the water level decreases. II. The same mass (boat \(+\) bricks \(+\) person) is in the pool in either case, and therefore the water level remains the same. III. The bricks displace more water when they sink to the bottom than they did when they were above the water in the boat; therefore the water level increases.
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