Energy Considerations, Friction factor

Figure 7.9: Energy balance for a pipe flow.

Consider the pipe flow as in the previous section and perform a energy analysis. With reference to Fig.7.9, we have for a mass balance,

$$\rho_1 V_1 A_1 = \rho_2 V_2 A_2$$ (7.19)

Since the flow is incompressible and it is a constant area pipe, we have,

$$V_1  = V_2 = V$$ (7.20)

Now applying the energy equation for a steady flow,

$$\left( {p \over \gamma} + \alpha {V^2 \over {2g}} + z \right )_1 = \left( {p \over \gamma} + \alpha {V^2 \over {2g}} + z \right )_2 + h_f$$ (7.21)

Note that every term in the above equation has dimension of length. We are considering a fully developed flow, which means that $\alpha_1 = \alpha_2$. We have no external features between (1) and (2) to add in or take away energy. As a result, loss of head is given by,

$$h_f = \left( {p_1 \over \gamma} - {p_2 \over \gamma} \right) + (z_1 -z_2)$$ (7.22)

A force balance in x-direction on the control volume gives,

$$\Delta p (\pi R^2) + \gamma (\pi R^2) L \sin \theta - \tau_w(2\pi R) L = 0$$

$$\noalign{note  that  $L  \sin\theta = \Delta z$}$$

$$\Delta p (\pi R^2) + \gamma (\pi R^2) \Delta z - \tau_w(2\pi R) L = 0$$

$$\noalign{dividing by $\pi R^2 \gamma$, we have}$$

$${{\Delta p} \over \gamma} +\Delta z = {{2 \tau L} \over {\gamma R}}$$ (7.23)

An inspection of Eqns. 7.23 and 7.22 shows that

$$h_f ={{2 \tau_w} \over \gamma} { L \over R}  or  {{4 \tau_w} \over \gamma} { L \over D}$$ (7.24)

Again we remind ourselves that this result is valid for both laminar and turbulent flows.

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