Consider a unidirectional fluid flow with the velocity field given by

V(𝑥, 𝑦, 𝑧,𝑡) = 𝑢(𝑥,𝑡) 𝑖̂

where 𝑢(0,𝑡) = 1. If the spatially homogeneous density field varies with time 𝑡 as

𝜌(𝑡) = 1 + 0.2𝑒^−𝑡

the value of 𝑢(2, 1) is ______________. (Rounded off to two decimal places) Assume all quantities to be dimensionless.

The correct answer is 1.14.

Step-by-Step Derivation:

We are given a unidirectional fluid flow with the velocity field:

$\overrightarrow{V}(x,y,z,t)=u(x,t)\hat{i}$

with the boundary condition $u(0,t)=1$.

The density field is spatially homogeneous, meaning it varies only with time $t$:

$\rho \left(t\right)=1+0.2e^{-t}$

The continuity equation for a compressible, unsteady flow is given by:

$\frac{\partial \rho }{\partial t}+\nabla ·\left(\rho \overrightarrow{V}\right)=0$

Since the density $\rho$ does not vary in space (spatially homogeneous), its spatial derivatives are zero, i.e., $\frac{\partial \rho }{\partial x}=0$. We can expand the divergence term as follows:

$\nabla ·\left(\rho \overrightarrow{V}\right)=\frac{\partial }{\partial x}\left(\rho u\right)=\rho \frac{\partial u}{\partial x}+u\frac{\partial \rho }{\partial x}=\rho \frac{\partial u}{\partial x}$

Substituting this back into the continuity equation gives:

$\frac{d\rho }{dt}+\rho \frac{\partial u}{\partial x}=0$

Rearranging the equation to solve for the velocity gradient:

$\frac{\partial u}{\partial x}=-\frac{1}{\rho }\frac{d\rho }{dt}$

Integrating this equation with respect to $x$ from $x=0$ to $x$ (since $\rho$ and $\frac{d\rho }{dt}$ are independent of $x$):

$u(x,t)-u(0,t)=-\frac{x}{\rho }\frac{d\rho }{dt}$

Using the boundary condition $u(0,t)=1$, we obtain:

$u(x,t)=1-\frac{x}{\rho }\frac{d\rho }{dt}$

Now, let's find $\frac{d\rho }{dt}$ by differentiating the given density field with respect to time:

$\frac{d\rho }{dt}=\frac{d}{dt}(1+0.2e^{-t})=-0.2e^{-t}$

Substitute $\rho \left(t\right)$ and $\frac{d\rho }{dt}$ into the velocity expression:

$u(x,t)=1-x·\frac{-0.2e^{-t}}{1+0.2e^{-t}}=1+\frac{0.2x{e}^{-t}}{1+0.2e^{-t}}$

To find the value of $u(2,1)$, we substitute $x=2$ and $t=1$:

$u(2,1)=1+\frac{0.2\left(2\right)e^{-1}}{1+0.2e^{-1}}=1+\frac{0.4e^{-1}}{1+0.2e^{-1}}$

Using the value $e^{-1}\approx 0.36788$:

Numerator: $0.4\times 0.36788=0.14715$
Denominator: $1+0.2\times 0.36788=1.07358$

Therefore:

$u(2,1)=1+\frac{0.14715}{1.07358}\approx 1+0.13707=1.13707$

Rounding to two decimal places gives $1.14$.