Physics, asked by hkspravin, 7 months ago

an oil of viscosity 0.1Ns/m^2 and relative density 0.9 is flowing through s circular pipe of diameter 50mm and length 300mm. the rate of flow through the pipe is 3.5L/s. find the pressure drop in length of 300m and shear stress in pipe wall​

Answers

Answered by Rameshjangid
0

Answer:

$\therefore P_1-P-2=68.43 \mathrm{~N} / \mathrm{cm}^2$

$\tau_0=28.512 \mathrm{~N} / \mathrm{m}^2$

Explanation:

Given: Viscosity, $\mu=0.1 N S / m^2$

Relative density =0.9

$$\begin{aligned}& \rho_o=0.9 \times 1000=900 \mathrm{~kg} / \mathrm{m}^3 \text { (As density of water }=1000 \mathrm{~kg} / \mathrm{m}^3 \text { ) } \\& \mathrm{D}=50 \mathrm{~mm}=0.05 \mathrm{~m} L=300 \mathrm{mQ}=3.5 \mathrm{lit} / \mathrm{s}=0.0035 \mathrm{~m}^3 / \mathrm{s}\end{aligned}$$

Step 1: 1) Pressure Drop $\left(P_1-P_2\right)$

$$\begin{aligned}& =\frac{32 \mu \bar{u} L}{D^2} \\& \therefore \bar{u}=\frac{Q}{A} \\& =\frac{0.0035}{\frac{\pi}{4} D^2} \\& =\frac{0.0035}{\frac{\pi}{4}(0.05)^2}=1.782 \mathrm{~m} / \mathrm{s} \\& R_e=\frac{\rho V D}{\mu}\left(\rho=900 \mathrm{~kg} / \mathrm{m}^3, V=\bar{u}=1.782\right) \\& R_e=\frac{900 \times 1.782 \times 0.05}{0.1}=801.9\end{aligned}$$

As $R_e < 2000$, the flow is laminar,

\begin{aligned}& \therefore P_1-P_2=\frac{32 \times 0.1 \times 1.782 \times 300}{(0.05)^2} \\& =684288 \mathrm{~N} / \mathrm{m}^2=684288 \times 10^{-4} \mathrm{~N} / \mathrm{cm}^2 \\& \therefore P_1-P-2=68.43 \mathrm{~N} / \mathrm{cm}^2\end{aligned}

Step 2: 2) Shear stress at the pipe wall $\left(\tau_0\right)$

$ \therefore \tau=-\frac{\delta P}{\delta x} \cdot \frac{r}{2}(\mathrm{r}=\mathrm{R}) $

$ \therefore \tau=-\frac{\delta P}{\delta x} \cdot \frac{R}{2} $

$ -\frac{\delta P}{\delta x}=\frac{-\left(P_2-P_1\right)}{x_2-x_1} $

$ -\frac{\delta P}{\delta x}=\frac{P_1-P_2}{x_2-x_1} $

$ -\frac{\delta P}{\delta x}=\frac{P_1-P_2}{L}=\frac{684288}{300}=2280.96 \mathrm{~N} / \mathrm{m}^2 $

$ R=\frac{D}{2}=\frac{0.05}{2} $

$ \mathrm{R}=0.025 \mathrm{~m} $

$ \therefore \tau_0=2280.96 \times \frac{0.025}{2} $

$ \tau_0=28.512 \mathrm{~N} / \mathrm{m}^2$

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