Answer to Question #139254 in Mechanical Engineering for Yashwanth Kumar Reddy

Question

A conical diffusing section diverges uniformly from 0.1 m diameter to 0.2 m diameter
over a length of 1 metre. Find the local and convective acceleration at the middle of
the diffuser. Consider the following two cases:
(i) rate of flow is 100 litres/sec at it remains constant,
(ii) rate of flow varies uniformly from 100 lit/sec to 200 litre/sec in 5 sec. and time of
interest is when t = 2 sec.
Velocity at any cross section, perpendicular to the flow direction, may be assumed to be
uniform.

Answer 1

Here inital diameter of section is 0.1 m and as it diverges so diameter at other end is 0.2 m, length of section =1 m, rate of flow = 100 ltr/s

(i) In case first flow rate is constant

Q= $100\times 10^{-3} m^3/s$

Let at some distance x the diameter of pipe will be

$D_x=D_1+\frac{D_2-D_1}{L}x=0.1+\frac{0.2-0.1}{1}x=0.1+0.1x=0.1(1+x)$

Area=A=$\frac{3.14}{4}D_x^2= 0.785(0.1(1+x))^2$

velocity of flow=u= $\frac{Q}{A}= \frac{0.01}{0.00785(1+x)^2}$

In steady state condition we know that local acceleration will be zero

$\frac{\delta u}{\delta t}=0$

and the value of convective acceleration will be

$a=u\frac{\delta u}{\delta x}=$$\frac{0.01}{0.00785(1+x)^2} \times (\frac{\delta (\frac{0.01}{0.00785(1+x)^2 )})}{\delta x})$

at mid point , x=0.5

$a=$ $\frac{-2\times 0.5 \times 0.01}{0.00785^2(1+0.5)^2}= 42.7 \frac{m}{s^2}$

(b) when rate of flow varies then local acceleration will not be zero it will have some value too, at t=2 sec

$a= \frac{0.01}{0.00785(1+x)^2} \frac{\delta Q}{\delta t}$ =1.132 m/s^2

and here as discharge also changing from 100 l/s to 200 l/s

so convective acceleration

$a=\frac{-2\times 0.14 \times 0.01}{0.00785^2(1+0.14)^2}= 83.77 \frac{m}{s^2}$