Laminar flow in circular ducts is not one-dimensional, but we may still use the equivalent mass-average velocity \(V=U_{m} / 2\) from equations (2.10) and (2.11) in our onedimensional formulations. This velocity came from solving \(\int u d A \equiv\left(\pi r_{0}^{2}\right) V\). Now, in the energy equation (2.36) we encounter the integral \(\int ho e(\mathbf{V} \cdot \hat{n}) d A\) over the surface in Figure 2.1, where \(e\) has the kinetic energy content given in equation (1.20); this is equivalent to calculating \(\int u^{3} d A=2.0\left(\pi r_{0}^{2}\right) V^{3}\) where \(V\) relates to \(U_{m}\) as indicated above. Verify that for laminar flow the resulting kinetic energy terms in equation (2.49) need to be multiplied by the factor 2.0 as mentioned above and in the text. Assume that the density remains constant along each cross section (Figure P2.4).

Transcribed Image Text:

n Surface Figure P2.4 u = Um [1-(r/ro)]