Question 4.SP.4: Figure S4.4 is of a cross section along the centerline of a ......

As a first step we sketch an approximate flow net to provide a general picture of the flow. We note that the flow is symmetric about the pipe axis (axisyrnmetric flow), so the net is not a true two-dimensionalflow net (see Fig. 4.9).

Since D and B are both on the pipe axis, v = 0 and w = 0 due to symmetry, so Eqs. (4.28) for these points reduce to

a_{x}\,=\,u\,\frac{\partial u}{\partial x}+\frac{\partial u}{\partial t},\qquad a_{y}\,=\,0,\qquad a_{z}\,=\,{0}

At section D the streamlines are parallel and hence the area at right angles  to the velocity vectors is a plane circle,

A_{D}\,=\,{\frac{\pi}{4}}\biggl({\frac{8}{12}}\biggr)^{2}\,=\,0.349\,\mathrm{ft}^{2}

So            u\,=\,\frac{Q}{A_{D}}\,=\,\frac{0.1\,+\,0.05t}{0.349}\,=\,\frac{2+t}{6.98}

and          \frac{\partial u}{\partial x}\,=\,0,\;\;\frac{\partial u}{\partial t}\,=\,\frac{1}{6.98}

Thus at t = 5 sec:      V_{D}\,=\,u\,=\,\frac{2+5}{6.98}\,=\,1.003\,\mathrm{fps}

and         a_{D}\,=\,a_{x}\,=\,1.003(0)\,+\,\frac{1}{6.98}\,=\,0.1432\,\mathrm{ftr}/\mathrm{sec}^{2}

At section B, however, the perpendicular flow area is the partial spherical surface through B, with center C and radius r = 2 in (see sketch). By table lookup, or by integration, this area is 2πrh, where h = r – rcos45° = 0.293r. Thus A_B = 2m(0.293r) = 1.840 r²

On the centerline near B,     u\,=\,{\frac{Q}{A_{B}}}\,=\,{\frac{0.1\,+\,0.05t}{1.840r^{2}}}\,=\,{\frac{2+t}{36.8r^{2}}}

and since x = constant – r,

\frac{\partial u}{\partial x}\,=\,-\frac{\partial u}{\partial r}\,=\,-\left[\frac{-2(2+t)}{36.8r^{3}}\right]\,=\,\frac{2\,+\,t}{18.40r^{3}}

and      \frac{\partial u}{\partial t}\,=\,\frac{1}{36.8r^{2}}\,

Thus at r = 2 in and t = 5 sec:

V_{B}\,=\,u\,=\,\frac{2+5}{36.8(2/12)^{2}}\,=\,6.85\,\mathrm{fps}

and            a_{B}\;=\;a_{x}\;=\;6.85\bigg[\frac{2\:+\:5}{18.40(2/12)^{3}}\bigg]\;+\;\frac{1}{36.8(2/12)^{2}}

= 563 (convective) + 0.978 (local)

= 564 ft/sec²

Note: For the flow net shown in the sketch, the velocity at C is infinite  because the flow area at that point is zero. This, of course, cannot occur; in the  real case a jet somewhat similar to that of Fig.11.13 will form downstream of the  nozzle opening.

a_x=\left(u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}+w \frac{\partial u}{\partial z}\right)+\frac{\partial u}{\partial t}           (4.28a)

 a_y=\left(u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}+w \frac{\partial v}{\partial z}\right)+\frac{\partial v}{\partial t}            (4.28b)

 a_z=\left(u \frac{\partial w}{\partial x}+v \frac{\partial w}{\partial y}+w \frac{\partial w}{\partial z}\right)+\frac{\partial w}{\partial t}        (4.28c)