Answer to Question #149123 in Math for Usman

Question

Find the convective acceleration at the middle of a pipe which converges uniformly from 0.6 m diameter to 0.3 m diameter over 3 cm length. The rate of flow is 40 lit/s. If the rate of flow changes uniformly from 40 lit/s to 80 lit/s in 40 seconds, find the total acceleration at the middle of the pipe at 20th second.

Answer 1

$D_1=0.6m, D_2=0.3m,L=3cm=0.03m,$

$Q=40L/s=0.04m^3/s,$

$Q_1=40l/s=0.04m^3/s,Q_2=80l/s=0.08m^3/s$

Case (i)

Flow is one dimensional and hence the velocity components $v=w=0$

Convective acceleration $=u(\partial u/\partial x)$

$A_1=\dfrac{\pi D_1^2}{4}=0.09\pi m^2\approx0.2827m^2$

$A_2=\dfrac{\pi D_2^2}{4}=0.0225\pi m^2\approx0.0707m^2$

$u_1=\dfrac{Q}{A_1}=\dfrac{0.04m^3/s}{0.09\pi m^2}\approx0.1415m/s$

$u_2=\dfrac{Q}{A_2}=\dfrac{0.04m^3/s}{0.0225\pi m^2}\approx0.5657m/s$

As the diameter changes uniformly, the velocity will also change uniformly. The velocity $u$ at any distance $x$ from inlet is given by

$$u = u_1 +\dfrac{u_2 – u_1}{L} x$$

$$\dfrac{\partial u}{\partial x}=\dfrac{u_2-u_1}{L}$$

$$\text{Convective acceleration }=u\cdot \dfrac{u_2-u_1}{L}$$

$x=0.015m$

$$u = 0.1415m/s +\dfrac{0.5657m/s – 0.1415m/s}{0.03m} \cdot0.015m$$

$$=0.3536m/s$$

$$\text{Convective acceleration }$$

$$=0.3536m/s\cdot \dfrac{0.5657m/s – 0.1415m/s}{0.03m}$$

$$=8.3355m/s^2$$

Case (ii)

Total acceleration = (convective + local ) acceleration at t =20 seconds

Rate of flow

$$Q_{t=20}=Q_1+\dfrac{Q_2-Q_1}{40s}\cdot20s$$

$$=0.06m^3/s$$

$$u_1=Q/A_1=0.2122m/s$$

$$u_2=Q/A_2=0.8488m/s$$

The velocity $u$ at any distance $x$ from inlet is given by

$$u = u_1 +\dfrac{u_2 – u_1}{L} x$$

$$u = 0.2122 +21.2207 x$$

$$\dfrac{\partial u}{\partial x}=21.2207s^{-1}$$ $$\text{Convective acceleration }=u\cdot \dfrac{\partial u}{\partial x}$$

$$=(0.2122 +21.2207 x)\cdot21.2207$$

At $x=0.015m$

$$(\text{Convective acceleration })_{x=0.015m}$$

$$=11.2578m/s^2$$

Local acceleration

Diameter at $x=0.015m$ is given by 

$$D = D_1 +\dfrac{D_2 – D_1}{L} x=0.45m$$

$$A=\dfrac{\pi D^2}{4}\approx0.1590m^2$$

$$u_1=Q_1/A=0.2515m/s$$

$$u_2=Q_2/A=0.5030m/s$$

$$\text{Rate of change of velocity}$$

$$=\dfrac{0.5030m/s-0.2515m/s}{40s}=0.0063m/s^2$$

Total acceleration=$=11.2578m/s^2+0.0063m/s^2=11.2641m/s^2$