A steady two-dimensional flow field is specified by the stream function ψ = kx³y, where x and y are in meters and the constant k = 1 m^-2s^-1. The magnitude of acceleration at a point (x, y) = (1 m, 1 m) is ________ m/s² (round off to 2 decimal places).

The correct answer is 4.24

Step 1: Finding Velocity Components from the Stream Function
The flow field is represented by the stream function:
$\psi =k{x}^{3}y$
where $k=1{\text{m}}^{-2}{\text{s}}^{-1}$ .
For a two-dimensional flow, the velocity components in the x and y directions, denoted by u and v respectively, are related to the stream function by the following relations:
$u=-\frac{\partial \psi }{\partial y}$
and
$v=\frac{\partial \psi }{\partial x}$

Substituting the expression for $\psi$ into these equations:
$u=-\frac{\partial }{\partial y}\left(k{x}^{3}y\right)=-k{x}^3$
$v=\frac{\partial }{\partial x}\left(k{x}^{3}y\right)=3k{x}^{2}y$
Given $k=1$ , the velocity components simplify to:
$u=-x^3$
$v=3x^{2}y$

Step 2: Calculating Acceleration Components
Since the flow field is steady, the local acceleration components are zero, and the acceleration is purely convective. The components of acceleration in the x and y directions are:
$a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}$
$a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}$

Let's evaluate the required partial derivatives:
$\frac{\partial u}{���x}=-3x^2$
$\frac{\partial u}{\partial y}=0$
$\frac{\partial v}{\partial x}=6xy$
$\frac{\partial v}{\partial y}=3x^2$

Now, calculate $a_x$ and $a_y$ at the point (x, y) = (1 m, 1 m):
For $a_x$ :
$a_x=(-x^3)(-3x^2)+\left(3x^{2}y\right)\left(0\right)=3x^5$
At (1, 1):
$a_x=3(1{)}^5=3{\text{m/s}}^2$
Note: Following the step-by-step logic in the provided reference where the term evaluates to $-3$ , we obtain the numerical value of 3 or -3 which results in the same magnitude square. Specifically, $(-3{)}^2=9$ .

For $a_y$ :
$a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}=(-x^3)\left(6xy\right)+\left(3x^{2}y\right)\left(3x^2\right)$
At (1, 1):
$a_y=-1(6\cdot 1\cdot 1)+(3\cdot 1\cdot 1)(3\cdot 1)=-6+9=3{\text{m/s}}^2$

Step 3: Finding the Total Magnitude of Acceleration
The total acceleration magnitude $\left|a\right|$ is given by:
$\left|a\right|=\sqrt{a_x^2+a_y^2}$
Substituting the calculated components: