Answer to Question #179883 in Mechanical Engineering for Bryan

Question

A 2.0m long conical diffuser 20cm in diameter at the upstream end has 80cm diameter at the downstream end. At the certain instant the discharge through the diffuser is observed to be 200 L/s of water and is found to increase uniformily at the rate of 50 L/s per second. Estimate the local, convective and total acceleration at a seccion 1.5m from the upstream end.

Answer 1

Distance at section x

$D_x=(\frac{D_2-D_1}{L})x+D_1=(\frac{0.8-0.2}{2})x+0.2=0.2+0.3x$

Given x=1.5x

$D_{1.5}=0.2+0.3(1.5)=0.65m$

Cross section area =$\frac{\pi}{4} \times 0.65^2=0.332 m^2$

$v_{1.5}=\frac{Q}{A}=\frac{200}{0.332} \times 10^{-3}=0.6027 m/s$

Acceleration $a_x=\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial t}$

Local acceleration =$\frac{1}{0.332} \times 0.05=0.1506 m/s^2$

Convective acceleration =$0.6027 \times (\frac{\partial u}{\partial x})_{1.5}$

$V_x=\frac{4}{\frac{\pi}{4}(0.2+0.3x)^2}=\frac{0.2546}{(0.2+0.3x)^2}$

$\frac{\partial u_x}{\partial t}=0.2546 \frac{-2 \times 0.3}{(0.2+0.3 \times 1.5)^2}=-0.55625$

$a_c=-0.6027 \times 0.55625 =-0.3352 m/s^2$

Total acceleration =$0.1506-0.3352=-0.1846 m/s^2$