Answer
a) flow is laminar thus paint will exit as a separate streams of Blue and yellow paint.
b) flow will be turbulent hence mixing of paint will occur so we will get green paint at the exit.
Work Step by Step
given
Blue and yellow streams of paint.
$$
\begin{aligned}
& T_{\text {Blue }}=T_{\text {yellow }}=60^{\circ} \mathrm{F} \\
& \rho_{\text {blue }}=P_{\text {yellow }}=1.6 \mathrm{slugs} / \mathrm{ft}^3 \\
& \mu_{\text {blue }}=\mu_{\text {yellow }}=1000 \times \mu_{\mathrm{H}_2 \mathrm{O}} \\
& \left(\mathrm{V}_{\text {avg }}\right)_{\text {entry }}=4 \mathrm{Ft} / \mathrm{sec}
\end{aligned}
$$
Sol
$$
\begin{aligned}
& R_e=\frac{\rho V D}{\mu}=\frac{(1.6)(4)\left(\frac{2}{12}\right)}{1000 \times \mu_{\mathrm{H}_2 \mathrm{O}}}=\frac{(1.6)(4)\left(\frac{2}{12}\right)}{1000\left(2.34 \times 10^{-5}\right)} \quad\left[\therefore \mu_{\mathrm{H}_2 0}=\frac{2.34 \times 10^{-5}}{\frac{\mathrm{Lb} \cdot 5}{\mathrm{Ft}^2}}\right] \\ & R_e=45.6 <2100
{\text { so flow is laminar Thus paint will exit as a separate streams of Blue and yellow paint }} \\
&
\end{aligned}
$$
Now, if the paint were "thinned. So that it is andy 10 times more viscous than water $\left(\mu=10 \mu_{\mathrm{H}_2 \mathrm{O}}\right)$
then
$$
R_e=\frac{P X D}{\mu}=\frac{P Y D}{\left(1000 \times \mu_{H_2 O}\right) \frac{10}{1000}}=\frac{45.6}{(10 / 1000)}=4560
$$
So, $R_e=4560>4000
{\text { so flow will be turbulent hence }}{\text { mixing of paint will occur so we }}$ will get green paint at the exit
