Question

A two-dimensional reducing bend has a linear velocity profile at section 1 . the flow is uniform at sections 2 and 3. the fluid is incompressible and the flow is steady. find the maximum velocity, v1max at section 1.

Answer 1

Answer:

The answer is "$\bold{3.8\ \frac{m}{s}}$"

Explanation:

$0 = \int_{cs} \overrightarrow{V} \overrightarrow{dn} \\\\= \int_{h_1} \overrightarrow{V_1} \overrightarrow{dA_1} + V_2A_2 +V_3A_3V_3A_3 \\\\ = -\int_{h_1} V_1dA_1 + V_2 dA_2 \\\\\to \int^{h_1}_{0} V_1 max \frac{y}{h_1} w dy -V_2 wh_2 \ \ \ \ \ \ \ \_{where} \ \ \ v_1=\frac{y}{h_1}$

$\to V_3A_3= V_1 max [\frac{y^2}{2h_1}]^{h_1}_{0} -V_2wh_2 \\\\= \frac{V_1 max}{2} \times wh_1 -V_2wh_2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A_3 =w_3h_3\\\\\to \frac{V_3A_3}{w}= \frac{V_1 max}{2} \times h_1 -V_2wh_2 \\\\ \to V_3h_3= \frac{V_1 max}{2} \times h_1 -V_2wh_2 \\\\\to 5 \times 0.15 = \frac{V_1}{2} \times 0.5 - 1 \times 0.2 \\\\V_1 = 3.8 \ \frac{m}{s}$