Control Volume Analysis

Figure 7.6: Control Volume analysis of pipe flow.

Let us consider a small element of flow of length l around the centreline of the pipe in the fully developed region as shown in Fig.7.6. The acceleration on the element is zero as the flow is fully developed. The only forces acting upon the fluid element considered are the shear force and the one due to pressure. Let $\tau$ be the shear stress acting and pressure at the left hand end of the element be p. Assuming a pressure drop of $\Delta p$ over the length l the pressure acting upon the right hand end is $p - \Delta p$. A force balance gives,

$$p \pi r^2 - (p - \Delta p) \pi r^2 - \tau 2 \pi r l  =  0$$

$${{\Delta p} \over l} = {{2 \tau} \over r}$$ (7.3)

The above relation establishes that pressure gradient in equilibrium with shear stress is the one that keeps the flow moving. Let us examine Eqn.7.3 in detail. We have $\Delta p$ which is independent of the radial distance r. Of course l is independent of r. Consequently $2 \tau /r$ should also be independent of r. Hence,

$$\tau / r = C;  or   \tau = Cr$$ (7.4)

where C is a constant. Then at the walls of the pipe( i.e., r = D/2, where D is the diameter of the pipe) we have,

$${\tau_w = {{C D} \over 2}},  or  C = {{2 \tau_w}\over D}$$ (7.5)

where $\tau_w$ is the wall shear stress, i.e., shear stress at the wall.

Accordingly, our equation for shear stress Eqn.7.4 becomes

$$\tau = {{2 \tau_w r} \over D}$$ (7.6)

Figure 7.7: Velocity and shear stress distribution in a pipe flow.

Thus we see that the shear stress varies linearly over the cylinder radius (see fig.7.7).

Combining Eqns. 7.3 and 7.6 we get,

$$\Delta p = {{4 l \tau_w} \over D}$$ (7.7)

The Eqn. 7.7 tells us that for a long slender cylinder ($l/D\gg 1$) even a small shear stress can produce a big pressure drop.

Note that we have so far not made any assumption about the flow being laminar or turbulent. The equation 7.7 is therefore very general. In the rest of the present derivation we will assume a laminar, Newtonian flow for which we have,

$$\tau = -\mu{{du}\over {dr}}$$ (7.8)

Let us now combine Eqn. 7.3 with Eqn. 7.8 to obtain an equation for the velocity profile for a laminar flow in a pipe,

$${{du} \over {dr}} = - \left( {{\Delta p} \over {2 \mu l}}\right ) r$$ (7.9)

integrating this equation we have,

$$\int du = -\left( {{\Delta p} \over {2 \mu l}}\right ) \int r dr$$

$$u = -\left( {{\Delta p} \over {2 \mu l}}\right ) {r^2 \over 2} + C'$$

where C' is the constant of integration. This is evaluated from the boundary condition that velocity is zero at the wall. Thus,

$$C'=0,  at   r = {D \over 2}$$

$$C' = \left( {\Delta p} \over {16 \mu l}\right) D^2$$

$$u = \left( {{\Delta p D^2} \over {16 \mu l}}\right ) \left\{ 1 - \left( {{2r} \over D}\right) ^2 \right \}$$

Note that at the centreline we have the maximum velocity. Denoting it by V_c, we have

$$V_c = \left( {\Delta p} \over {16 \mu l}\right) D^2$$

$$u = V_c \left\{ 1 - \left( {{2r} \over D}\right) ^2 \right \}$$

This is the parabolic velocity profile for a laminar flow through a pipe.

(c) Aerospace, Mechanical & Mechatronic Engg. 2005
University of Sydney