/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 83 Two hoses are connected to the s... [FREE SOLUTION] | 91Ó°ÊÓ

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Two hoses are connected to the same outlet using a Y-connector, as the drawing shows. The hoses \(A\) and \(B\) have the same length, but hose \(B\) has the larger radius. Each is open to the atmosphere at the end where the water exits. Water flows through both hoses as a viscous fluid, and Poiseuille's \(\operatorname{law}\left[Q=\pi R^{4}\left(P_{2}-P_{1}\right) /(8 \eta L)\right]\) applies to each. In this law, \(P_{2}\) is the pressure upstream, \(P_{1}\) is the pressure downstream, and \(Q\) is the volume flow rate. The ratio of the radius of hose \(\mathrm{B}\) to the radius of hose \(\mathrm{A}\) is \(R_{\mathrm{B}} / R_{\mathrm{A}}=1.50 .\) Find the ratio of the speed of the water in hose \(B\) to the speed in hose \(A\).

Short Answer

Expert verified
The speed ratio of water in hose B to hose A is 2.25.

Step by step solution

01

Understanding Poiseuille's Law

Poiseuille's law provides the relationship for the flow rate of a viscous fluid through a pipe. It states that the volume flow rate \( Q \) is proportional to \( R^4 \) where \( R \) is the radius of the hose, inversely proportional to the fluid's viscosity \( \eta \), directly proportional to the pressure difference \( P_2 - P_1 \), and inversely proportional to the length of the hose \( L \). The formula is given by \( Q = \frac{\pi R^4 (P_2 - P_1)}{8 \eta L} \).
02

Flow Rate Comparison of Hoses

Since both hoses are connected to the same source and exit to the atmosphere, the pressure difference \( P_2 - P_1 \) and the lengths are the same for both hoses. Thus, the flow rates are determined primarily by the radii of the hoses. Therefore, \( Q_A = \frac{\pi R_A^4 (P_2 - P_1)}{8 \eta L} \) and \( Q_B = \frac{\pi R_B^4 (P_2 - P_1)}{8 \eta L} \).
03

Calculating Flow Rate Ratio

Divide the flow rate equation for hose \( B \) by that for hose \( A \): \( \frac{Q_B}{Q_A} = \frac{R_B^4}{R_A^4} \). Given \( \frac{R_B}{R_A} = 1.50 \), we get \( \frac{Q_B}{Q_A} = (1.50)^4 \).
04

Flow Speed and Flow Rate Relationship

The flow speed \( v \) in a hose is related to the flow rate via the cross-sectional area \( A \), i.e., \( Q = v \cdot A = v \cdot \pi R^2 \). Solving for \( v \) gives \( v = \frac{Q}{\pi R^2} \).
05

Speed Ratio Calculation

To find the speed ratio \( \frac{v_B}{v_A} \), use \( \frac{v_B}{v_A} = \frac{Q_B / \pi R_B^2}{Q_A / \pi R_A^2} = \frac{Q_B}{Q_A} \cdot \left( \frac{R_A^2}{R_B^2} \right) \). Substitute \( \frac{Q_B}{Q_A} = (1.50)^4 \) and \( \frac{R_A}{R_B} = \frac{1}{1.50} \), yielding \( \frac{v_B}{v_A} = 1.50^4 \cdot \left( \frac{1}{1.50} \right)^2 = 1.50^2 \).
06

Final Speed Ratio

Calculate the final answer: \( 1.50^2 = 2.25 \). Therefore, the speed of the water in hose \( B \) relative to hose \( A \) is \( 2.25 \).

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Key Concepts

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

Poiseuille's Law
Poiseuille's law is a fundamental principle in fluid dynamics that explains how the flow rate of a viscous fluid, like water, is determined inside a pipe or hose. This law helps us understand how different factors influence the movement of fluid under pressure. In simple terms, Poiseuille's law shows that the flow rate \(Q\) through a pipe is directly related to the fourth power of the pipe's radius \(R\). This means that even small changes in the radius can greatly affect how much fluid flows through. The flow rate is also directly related to the pressure difference \(P_2 - P_1\) between the two ends of the pipe. Lastly, the flow rate is inversely related to both the fluid's viscosity \(\eta\) and the length of the pipe \(L\), meaning thicker fluids or longer pipes decrease the flow. The law is mathematically expressed as:
  • \[ Q = \frac{\pi R^4 (P_2 - P_1)}{8 \, \eta \, L} \]
This equation helps predict how efficiently water will flow through a garden hose, for instance, making it a crucial tool for many engineering fields.
Flow Rate
The flow rate is a measure of how much fluid passes through a section of a pipe or hose in a specific amount of time. It is usually expressed in cubic meters per second or liters per minute and is a crucial aspect in understanding how fluid dynamics operates in real-world applications. In the context of Poiseuille's law, we see that the flow rate depends heavily on the hose's radius.
  • A larger radius means a higher flow rate because more fluid can pass through at once.
  • Conversely, a smaller radius restricts the flow.
Additionally, since the pressure difference and length are constant in our given exercise, the flow rate primarily varies with changes in radius. In practice, flow rate calculations help engineers design systems like plumbing and irrigation that optimize fluid delivery.
Viscous Fluid
Viscous fluid, in simple terms, is a type of fluid that resists motion. Viscosity refers to how "thick" or "sticky" a fluid is. Imagine pouring honey compared with water — honey is much more viscous, meaning it flows slowly because its molecules are more resistant to motion. Viscous fluids are common in everyday life and include not just honey but also things like syrup and oil.
  • In Poiseuille's law, the viscosity \(\eta\) of a fluid plays a critical role in determining the flow rate through a pipe.
  • Higher viscosity will lower the flow rate since the fluid resists being pushed through.
In our example, water is considered a viscous fluid, though its viscosity is relatively low compared to substances like oil. Understanding viscosity is essential, especially in industrial applications, where controlling fluid flow is necessary to ensure optimal performance of systems and machines.

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Most popular questions from this chapter

A hydrometer is a device used to measure the density of a liquid. It is a cylindrical tube weighted at one end, so that it floats with the heavier end downward. The tube is contained inside a large "medicine dropper," into which the liquid is drawn using the squeeze bulb (see the drawing). For use with your car, marks are put on the tube so that the level at which it floats indicates whether the liquid is battery acid (more dense) or antifreeze (less dense). The hydrometer has a weight of \(W=5.88 \times 10^{-2} \mathrm{N}\) and a cross-sectional area of \(A=7.85 \times 10^{-5} \mathrm{m}^{2}\) How far from the bottom of the tube should the mark be put that denotes (a) battery acid \(\left(\rho=1280 \mathrm{kg} / \mathrm{m}^{3}\right)\) and (b) antifreeze \(\left(\rho=1073 \mathrm{kg} / \mathrm{m}^{3}\right) ?\)

The density of ice is \(917 \mathrm{kg} / \mathrm{m}^{3},\) and the density of seawater is 1025 \(\mathrm{kg} / \mathrm{m}^{3} .\) A swimming polar bear climbs onto a piece of floating ice that has a volume of \(5.2 \mathrm{m}^{3} .\) What is the weight of the heaviest bear that the ice can support without sinking completely beneath the water?

Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve that is closed. Both containers are filled initially to the same height of \(1.00 \mathrm{m},\) one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluid level in the left container when equilibrium is reestablished.

A full can of black cherry soda has a mass of \(0.416 \mathrm{kg}\). It contains \(3.54 \times 10^{-4} \mathrm{m}^{3}\) of liquid. Assuming that the soda has the same density as water, find the volume of aluminum used to make the can.

An antifreeze solution is made by mixing ethylene glycol \(\rho=1116\) \(\mathrm{kg} / \mathrm{m}^{3}\) ) with water. Suppose that the specific gravity of such a solution is \(1.0730 .\) Assuming that the total volume of the solution is the sum of its parts, determine the volume percentage of ethylene glycol in the solution.

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