Problem 2
Question
Consider a velocity vector \(v=\left(x t^{2}-y\right) \vec{i}+\left(x t-y^{2}\right) \vec{j}\). (i) Determine whether this flow is steady (hint: no changes with time). (ii) Determine whether this is an incompressible flow (hint: check if \(\nabla \cdot=0\) ).
Step-by-Step Solution
Verified Answer
The flow is not steady and is compressible.
1Step 1: Identify Components of the Velocity Vector
The velocity vector given is \( v = (x t^2 - y) \vec{i} + (x t - y^2) \vec{j} \). This means the \( x \)-component is \( x t^2 - y \) and the \( y \)-component is \( x t - y^2 \).
2Step 2: Determine if Flow is Steady
A flow is steady if it does not depend on time. We check if the components of the velocity vector are functions of time \( t \). The \( x \)-component \( x t^2 - y \) changes with time, as it includes \( t^2 \). Thus, the flow is not steady.
3Step 3: Check Incompressibility with Divergence
To check if the flow is incompressible, we need to compute the divergence, \( abla \cdot \vec{v} \). The divergence in 2D is: \[ abla \cdot \vec{v} = \frac{\partial}{\partial x}(x t^2 - y) + \frac{\partial}{\partial y}(x t - y^2). \]
4Step 4: Calculate Partial Derivatives
Calculate the partial derivatives: \( \frac{\partial}{\partial x}(x t^2 - y) = t^2 \) and \( \frac{\partial}{\partial y}(x t - y^2) = -2y \).
5Step 5: Evaluate the Divergence
Substitute the partial derivatives into the divergence formula: \[ abla \cdot \vec{v} = t^2 + (-2y) = t^2 - 2y. \] Since \( abla \cdot \vec{v} = t^2 - 2y \) is not zero, the flow is compressible.
Key Concepts
Flow SteadinessIncompressible FlowVelocity Vector Analysis
Flow Steadiness
Understanding whether a flow is steady is crucial in biofluid mechanics. Steady flow implies that the fluid's velocity at any given point does not change over time.
In mathematical terms, if the velocity components of a flow are not functions of time, we classify it as steady. For instance, in our example, the velocity vector is given by \( v = (x t^2 - y) \vec{i} + (x t - y^2) \vec{j} \). Here, the presence of \( t^2 \) in the \( x \)-component clearly indicates that it is a function of time.
Given this dependency, the flow cannot be steady because, as time progresses, the velocity at each point changes.
In mathematical terms, if the velocity components of a flow are not functions of time, we classify it as steady. For instance, in our example, the velocity vector is given by \( v = (x t^2 - y) \vec{i} + (x t - y^2) \vec{j} \). Here, the presence of \( t^2 \) in the \( x \)-component clearly indicates that it is a function of time.
Given this dependency, the flow cannot be steady because, as time progresses, the velocity at each point changes.
- Steady flow: Velocity does not change with time.
- Unsteady flow: Velocity components are functions of time.
Incompressible Flow
Incompressible flow is characterized by the constancy of density in a fluid parcel; in simpler terms, the fluid does not expand or contract significantly. In the context of velocity vectors, a flow is considered incompressible if the divergence of the velocity field \( abla \cdot \vec{v} \) equals zero.
Mathematically, divergence is calculated by taking the partial derivatives of the velocity components.
For our example, the divergence is: \[ abla \cdot \vec{v} = \frac{\partial}{\partial x}(x t^2 - y) + \frac{\partial}{\partial y}(x t - y^2) \].
Therefore, this flow does not maintain a constant volume and is compressible, an important concept in accurately modeling biological flows where incompressibility is often assumed.
Mathematically, divergence is calculated by taking the partial derivatives of the velocity components.
For our example, the divergence is: \[ abla \cdot \vec{v} = \frac{\partial}{\partial x}(x t^2 - y) + \frac{\partial}{\partial y}(x t - y^2) \].
- \( \frac{\partial}{\partial x}(x t^2 - y) = t^2 \)
- \( \frac{\partial}{\partial y}(x t - y^2) = -2y \)
Therefore, this flow does not maintain a constant volume and is compressible, an important concept in accurately modeling biological flows where incompressibility is often assumed.
Velocity Vector Analysis
Examining the velocity vector in biofluid mechanics provides insights into how fluids move. A velocity vector is a representation of the speed and direction of a fluid particle at each point in a flow field.
In our exercise, the vector \( v = (x t^2 - y) \vec{i} + (x t - y^2) \vec{j} \) is made up of \( i \) and \( j \) components corresponding to movement in the \( x \) and \( y \) directions respectively.
By analyzing these components, we determine how fluid particles transit across a plane.
In our exercise, the vector \( v = (x t^2 - y) \vec{i} + (x t - y^2) \vec{j} \) is made up of \( i \) and \( j \) components corresponding to movement in the \( x \) and \( y \) directions respectively.
By analyzing these components, we determine how fluid particles transit across a plane.
- \( (x t^2 - y) \): Influences \( x \)-direction movement; varies with time and position \( y \).
- \( (x t - y^2) \): Governs \( y \)-direction movement; dependent on position \( x \) and \( y \) squared.
Other exercises in this chapter
Problem 1
For a particle with a velocity distribution of \(v_{x}=5 x, v_{y}=-5 y, v_{z}=0\), determine the particle's acceleration vector. Also, determine whether this ve
View solution Problem 3
Given the velocity vector \(\vec{v}=(2 x-y) \vec{i}+(x-2 y) \vec{j}\), determine whether it is irrotational.
View solution Problem 4
The velocity vector for a steady incompressible flow in the \(x y\) plane is given by \(\vec{v}=\) \(\frac{4}{x} \vec{i}+\frac{4 y}{x^{2}} \vec{j}\), where the
View solution Problem 5
Flow in a two-dimensional channel of width \(W\) has a velocity profile defined by $$ v_{x}(y)=k\left[\left(\frac{W}{2}\right)^{2}-y^{2}\right] $$ where \(y=0\)
View solution