The path taken by a fluid particle under a steady flow is a streamline. It is defined as a curve whose tangent at any point is in the direction of the fluid velocity at that point. Consider the path of a particle as shown in Fig.10.7 (a), the curve describes how a fluid particle moves with time. The curve PQ is like a permanent map of fluid flow, indicating how the fluid streams. No two streamlines can cross, for if they do, an oncoming fluid particle can go either one way or the other and the flow would not be steady. Hence, in steady flow, the map of flow is stationary in time.

If we intend to show streamline of every flowing particle, we would end up with a continuum of lines. Consider planes perpendicular to the direction of fluid flow e.g., at three points P, R and Q in Fig.10.7 (b). The plane pieces are so chosen that their boundaries be determined by the same set of streamlines. This means that number of fluid particles crossing the surfaces as indicated at P, R and Q is the same. If area of cross-sections at these points are \( A_p, A_R \text { and } A_Q\) and speeds of fluid particles are

\(v_{\mathrm{p}}, v_{\mathrm{R}}\) and \(v_{\mathrm{Q}}\), then mass of fluid \(\Delta m_{\mathrm{p}}\) crossing at \(A_{\mathrm{p}}\) in a small interval of time \(\Delta t\) is

\(\rho_{\mathrm{p}} A_{\mathrm{p}} v_{\mathrm{p}} \Delta t\).

Similarly mass of fluid \(\Delta m_{\mathrm{R}}\) flowing or crossing at \(A_{\mathrm{R}}\) in a small interval of time \(\Delta t\) is \(\rho_{\mathrm{R}} A_{\mathrm{R}} v_{\mathrm{R}} \Delta t\) and mass of fluid \(\Delta m_{\mathrm{Q}}\) is \(\rho_{\mathrm{Q}} A_{\mathrm{Q}} v_{\mathrm{Q}} \Delta t\) crossing at \(A_Q\).

The mass of liquid flowing out equals the mass flowing in, holds in all cases. Therefore,
\(\rho_{\mathrm{p}} A_{\mathrm{p}} v_{\mathrm{p}} \Delta t=\rho_{\mathrm{R}} A_{\mathrm{R}} v_{\mathrm{R}} \Delta t=\rho_{\mathrm{Q}} A_{\mathrm{Q}} v_{\mathrm{Q}} \Delta t \dots(10.9)\)

For the flow of incompressible fluids

\(\rho_{\mathrm{p}}=\rho_{\mathrm{R}}=\rho_{\mathrm{Q}} \dots(10.10)\)

Equation \((10.9)\) reduces to

\(A_{\mathrm{p}} v_{\mathrm{p}}=A_{\mathrm{R}} v_{\mathrm{R}}=A_{\mathrm{Q}} v_{\mathrm{Q}}\)

which is called the equation of continuity and it is a statement of conservation of mass in flow of incompressible fluids. In general

\(A v\)= constant ……(10.11)

\(A v\) gives the volume flux or flow rate and remains constant throughout the pipe of flow. Thus, at narrower portions where the streamlines are closely spaced, velocity increases and its vice versa. From (Fig 10.7b) it is clear that \(A_{\mathrm{R}}>A_Q\) or \(v_{\mathrm{R}}<v_Q\), the fluid is accelerated while passing from \(\mathrm{R}\) to \(\mathrm{Q}\). This is associated with a change in pressure in fluid flow in horizontal pipes.

Laminar flow and Turbulent flow

Steady flow is achieved at low flow speeds. Beyond a limiting value, called critical speed, this flow loses steadiness and becomes turbulent. One sees this when a fast-flowing stream encounters rocks, small foamy whirlpool-like regions called ‘white water rapids are formed. Figure 10.8 displays streamlines for some typical flows. For example, Fig. 10.8(a) describes a laminar flow where the velocities at different points in the fluid may have different magnitudes but their directions are parallel. Figure 10.8 (b) gives a sketch of turbulent flow.

Example 1: The cylindrical tube of a spray pump has a cross-section of \(8.0 \mathrm{~cm}^2\) one end of which has 40 fine holes each of diameter \(1.0 \mathrm{~mm}\). if the liquid flow inside the tube is \(1.5 \mathrm{~m} \mathrm{~min}^{-1}\), what is the speed of the ejection of the liquid through the holes?

Solution:

Area of the cross-section of the spray pump, \(\mathrm{A}_1=8 \mathrm{~cm}^2=8 \times 10^{-4} \mathrm{~m}^2\) number of holes, \(n=40\)
Diameter of each hole, \(d=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}\)
Radius of each hole, \(r=d / 2=0.5 \times 10^{-3} \mathrm{~m}\)
Area of cross-section of each hole, \(a=\pi r^2=\pi\left(0.5 \times 10^{-3}\right)^2 m^2\)
Total area of 40 holes, \(\mathrm{A}_2=n \times a=40 \times \pi\left(0.5 \times 10^{-3}\right)^2 \mathrm{~m}^2\) \(=31.41 \times 10^{-6} \mathrm{~m}^2\)
speed of flow of liquid inside the tube, \(\mathrm{V}_1=1.5 \mathrm{~m} / \mathrm{min}=0.025 \mathrm{~m} / \mathrm{s}\)
speed of ejection of liquid through the holes \(=V_2\)
According to the law of continuity we have:
\(
\begin{aligned}
& \mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2 \\
& \mathrm{~V}_2=\frac{\mathrm{A}_1 \mathrm{~V}_1}{\mathrm{~A}_2} \\
& =\left(8 \times 10^{-4} \times 0.025\right) 31.61 \times 10^{-6} \\
& =0.633 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)