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Chapter 1: Problem 19

In his study of the circular hydraulic jump formed by a faucet flowing into a sink, Watson [53] proposed a parameter combining volume flow rate \(Q,\) density \(\rho,\) and viscosity \(\mu\) of the fluid, and depth \(h\) of the water in the sink. He claims that his grouping is dimensionless, with \(Q\) in the numerator. Can you verify this?

Short Answer

Expert verified
Yes, the parameter \( \frac{Q \cdot \rho}{\mu h} \) is dimensionless.

Step by step solution

01

Identify the dimensions of each parameter

First, we determine the dimensions of each variable: - Volume flow rate \(Q\) has dimensions of \([L^3 T^{-1}]\).- Density \(\rho\) has dimensions of \([M L^{-3}]\).- Viscosity \(\mu\) has dimensions of \([M L^{-1} T^{-1}]\).- Depth \(h\) has dimensions of \([L]\).
02

Construct the dimensionless group

Watson claims a dimensionless parameter that involves these variables. To form a dimensionless group, we follow the principle that the dimensions should cancel each other out. Assume the dimensionless group is structured as \(\frac{Q}{\rho^a \mu^b h^c}\) and solve for the exponents \(a, b, \) and \(c\) so that the result is dimensionless.
03

Solve for the exponents

Set up the equation for dimensions:\[ [L^3 T^{-1}] = [M L^{-3}]^a [M L^{-1} T^{-1}]^b [L]^c \]Simplify this to:\[ [L^3 T^{-1}] = [M^{a+b} L^{-3a-b+c} T^{-b}] \]Now, equate the powers of \(M, L,\) and \(T\) from both sides:- For \(M: a + b = 0\)- For \(L: -3a - b + c = 3\)- For \(T: -b = -1\)
04

Solve the exponent equations

- From \(-b = -1\), we get \(b = 1\).- Substitute \(b = 1\) into \(a + b = 0\), gives \(a = -1\).- Substitute \(a = -1\) and \(b = 1\) into \(-3a - b + c = 3\): \[ -3(-1) - 1 + c = 3 \] \[ 3 - 1 + c = 3 \] \[ c = 1 \]So, \(a = -1\), \(b = 1\), and \(c = 1\).
05

Verify the dimensionless group

With the found exponents, the dimensionless group can be expressed as:\[ \frac{Q}{\rho^{-1} \mu^1 h^1} = \frac{Q}{\frac{1}{\rho} \mu h} = \frac{Q \cdot \rho}{\mu h} \]This resulting group is indeed dimensionless since the dimensions cancel out, verifying Watson's claim.

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Key Concepts

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

Hydraulic Jump
A hydraulic jump is an intriguing phenomenon in fluid mechanics. It occurs when a high-velocity liquid stream shifts to a lower velocity flow, suddenly causing an increase in the fluid depth. This is often visible as a circular step-up in the water level, like what might occur if water flows rapidly from a faucet into a shallow sink.
Understanding hydraulic jumps provides insights into energy dissipation and turbulent behavior in fluids. In practical terms, this phenomenon is crucial in civil and hydraulic engineering, influencing the design of structures like spillways and sewer systems where chaotic water flow needs to be controlled.
  • Key features of hydraulic jumps:
    • Occurs in open channel flow when supercritical flow turns subcritical.
    • Visible as a sudden rise in the water surface.
    • Converts kinetic energy into potential energy, leading to energy loss.
Description of hydraulic jumps in fluid systems can also enhance our understanding of shock waves in compressible fluids or atmospheric phenomena.
Fluid Mechanics
Fluid mechanics is the study of how fluids鈥攍iquids and gases鈥攂ehave and interact with their environment. This field underpins many natural phenomena and engineering applications, from weather patterns to airplane design.
The core principles of fluid mechanics involve understanding fluid motion, the forces acting upon fluids, and the properties that dictate their behavior. These principles help engineers optimize systems where fluid motion is key, like hydraulics, pipelines, and energy systems.
  • Foundational concepts in fluid mechanics include:
    • Conservation laws: mass, momentum, and energy.
    • Equations of fluid motion, such as the Navier-Stokes equations.
    • Flow types: laminar and turbulent flow.
This comprehensive study of fluid behavior is crucial for solving problems involving fluid flow in both natural and man-made systems.
Dimensional Analysis
Dimensional analysis is a powerful method used in engineering and physics to simplify complex problems by reducing units and dimensions into manageable, dimensionless groups. It ensures that equations and models are consistent and physically meaningful.
By analyzing the dimensions of physical quantities, dimensional analysis helps in deriving relationships between different variables in a system without detailed knowledge of the system model. This method relies on the Buckingham Pi theorem, which states that one can simplify a relationship by reducing parameters to dimensionless terms.
  • Benefits of dimensional analysis include:
    • Identifying key variables affecting a system.
    • Simplifying mathematical models.
    • Validating experimental and theoretical results.
In the hydraulic jump example, dimensional analysis verifies that combining various fluid parameters into a single dimensionless group can predict fluid behavior.
Fluid Properties
Understanding fluid properties is crucial in predicting how fluids behave under different conditions. These properties include density, viscosity, surface tension, and compressibility, all of which impact how fluids flow and interact with their surroundings.
Each property provides insight into the fluid's character and its potential interactions in environments ranging from the tiny confines of microfluidics to the expanses of ocean currents.
  • Key fluid properties:
    • Density \((\rho)\): Mass per unit volume, affecting buoyancy and pressure.
    • Viscosity \(\mu\): Resistance to flow, influencing how easily a fluid deforms.
    • Surface tension: The cohesive force at a fluid's surface, affecting droplet formation.
These properties are foundational in modeling and engineering designs where flow dynamics need precise control.

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Most popular questions from this chapter

A simple viscometer measures the time \(t\) for a solid sphere to fall a distance \(L\) through a test fluid of density \(\rho .\) The fluid viscosity \(\mu\) is then given by \\[\mu \approx \frac{W_{\mathrm{net}} t}{3 \pi D L} \quad \text { if } \quad t \geq \frac{2 \rho D L}{\mu}\\] where \(D\) is the sphere diameter and \(W_{\text {net }}\) is the sphere net weight in the fluid. (a) Prove that both of these formulas are dimensionally homogeneous. ( \(b\) ) Suppose that a \(2.5 \mathrm{mm}\) diameter aluminum sphere (density \(2700 \mathrm{kg} / \mathrm{m}^{3}\) ) falls in an oil of density \(875 \mathrm{kg} / \mathrm{m}^{3} .\) If the time to fall \(50 \mathrm{cm}\) is \(32 \mathrm{s}\) estimate the oil viscosity and verify that the inequality is valid.

The Stokes-Oseen formula [33] for drag force \(F\) on a sphere of diameter \(D\) in a fluid stream of low velocity \(V\), density \(\rho,\) and viscosity \(\mu\) is \\[F=3 \pi \mu D V+\frac{9 \pi}{16} \rho V^{2} D^{2}\\] Is this formula dimensionally homogeneous?

Books on porous media and atomization claim that the viscosity \(\mu\) and surface tension \(Y\) of a fluid can be combined with a characteristic velocity \(U\) to form an important dimensionless parameter. (a) Verify that this is so. (b) Evaluate this parameter for water at \(20^{\circ} \mathrm{C}\) and a velocity of \(3.5 \mathrm{cm} / \mathrm{s} .\) Note: You get extra credit if you know the name of this parameter.

An amazing number of commercial and laboratory devices have been developed to measure fluid viscosity, as described in Refs. 29 and \(49 .\) Consider a concentric shaft, fixed axially and rotated inside the sleeve. Let the inner and outer cylinders have radii \(r_{i}\) and \(r_{o},\) respectively, with total sleeve length \(L\). Let the rotational rate be \(\Omega(\mathrm{rad} / \mathrm{s})\) and the applied torque be \(M .\) Using these parameters, derive a theoretical relation for the viscosity \(\mu\) of the fluid between the cylinders.

A solid cylindrical needle of diameter \(d,\) length \(L,\) and density \(\rho_{n}\) may float in liquid of surface tension \(Y\). Neglect buoyancy and assume a contact angle of \(0^{\circ} .\) Derive a formula for the maximum diameter \(d_{\max }\) able to float in the liquid. Calculate \(d_{\max }\) for a steel needle \((\mathrm{SG}=7.84)\) in water at \(20^{\circ} \mathrm{C}\).
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