Question

A non-viscous liquid of constant density $1000{\text{kg/m}}^3$ flows in a streamline motion along a tube of variable cross-section. The tube is kept inclined in a vertical plane, as shown in the figure and the velocity of the liquid at point $A$ is $1\text{m/s}$. The area of cross-section of the tube at two points $A$ and $B$ at heights $2\text{m}$ and $5\text{m}$ are $4\times {10}^{-3}{\text{m}}^2$ and $8\times {10}^{-3}{\text{m}}^2$ respectively. Then:

• A. outlet velocity of liquid at point B is $0.5\text{m/s}$.
• B. work done by pressure per unit volume is $4.956\times {10}^4{\text{J/m}}^3$
• C. work done by gravity per unit volume is $-3\times {10}^4{\text{J/m}}^3$
• D. work done by gravity per unit volume is $-6\times {10}^4{\text{J/m}}^3$

Answer 1

Answer: (A), (C)

work done by gravity per unit volume is $-3\times {10}^4{\text{J/m}}^3$

Given:
Density of water $\rho$ = $1000{\text{kg/m}}^3$
Area of cross section at point $A\left(a_1\right)=4\times {10}^{-3}{\text{m}}^2$
Area of cross section at point $B\left(a_2\right)=8\times {10}^{-3}{\text{m}}^2$

Let us assume that the velocities of the liquid at points A and B are $V_1$ and $V_2$.

(a). Applying the continuity equation at point A and B, we get
$a_1{V}_1=a_2{V}_2$
$\Rightarrow V_2=\frac{a_1}{a_2}{V}_1=\frac{4\times {10}^{-3}}{8\times {10}^{-3}}\times 1$
$\Rightarrow V_2=0.5\text{m/s}$

(b). Applying Bernoulli's theorem at point A and B
$P_1+\frac{1}{2}\rho V_1^2+\rho g{h}_1=P_2+\frac{1}{2}\rho V_2^2+\rho g{h}_2$

where each term represents a contribution to the energy per unit volume.
Hence, work done by pressure per unit volume is
$P_1-P_2=\frac{1}{2}\rho (V_2^2-V_1^2)+\rho g(h_2-h_1)$
$=\frac{1}{2}\times 1000({0.5}^2-1^2)+1000\times 10\times (5-2)$
$=2.9625\times {10}^{4}J/m^3$

(c). Work done per unit volume by gravity
$W_g=\rho g{h}_1-\rho g{h}_2=\rho g(h_1-h_2)=1000\times 10\times (2-5)=-3\times {10}^4{\text{J/m}}^3$
So, options (a) and (c) are correct.