3.29: Untitled Page 55

Chapter 3

maximum depth of the lake to come within 10 cm of its final value. The following information is available:

Year

Surface Area

Maximum Depth

prior to 1941

56,000 acres

181.5 ft.

1970

45,700 acres

164.0 ft.

During these years between 1941 and 1970 the diversion of water from Mono Lake Basin was 56,000 acre‐ft. per year and in 1970 this was increased to 110,000

acre‐ft. per year. The additional water was obtained from wells in the Mono Lake Basin, and as an approximation you can assume that this caused a decrease in

(see Figure 3.9b) by an amount equal to 54,000 acre‐ft. per year. Your 4

m



analysis will lead to an implicit equation for  and the methods described in Appendix B can be used to obtain a solution. Develop a solution based on the following methods:

Part (a). The bisection method

Part (b). The false position method

Part (c). Newton’s method

Part (d). Picard’s method

Part (e). Wegstein’s method

3‐20. During the winter months on many campuses across the country, students can be observed huddled in doorways contemplating an unexpected downpour.

In order not to be accused of idle ways, engineering students will often devote this time to the problem of estimating the speed at which they should run to their next class in order to minimize the unavoidable soaking. This problem has such importance in the general scheme of things that in March of 1973 it became the subject of one of Ann Landers’ syndicated columns entitled “What Way is Wetter?” Following typical Aristotelian logic, Ms. Landers sided with the common sense solution, “the faster you run, the quicker you get there, and the drier you will be.” Clearly a rational analysis is in order and this can be accomplished by means of the macroscopic mass balance for an arbitrary moving control volume.

In order to keep the analysis relatively simple, the running student should be modeled as a cylinder of height h and diameter D as illustrated in Figure 3.20.

The rain should be treated as a continuum with the mass flux of water represented by  v . Here the density  will be equal to the density of water multiplied by the volume fraction of the raindrops and the velocity v will be equal to the velocity of the raindrops. The velocity of the student is given by w,

Single component systems

95

and both v and w should be treated as constants. You should consider the special case in which v and w have no component in the y‐direction, and you should separate your analysis into two parts. In the first part, consider i  v  0 as indicated in Figure 3.20, and in the second part consider the case indicated by i  v  0 . In the first part one needs to consider both i  v  i  w and i  v  i  w .

When i  v  i  w , the runner gets wet on the top and back side. On the other hand, when the

Figure 3.20. Student running in the rain

runner is moving faster than the horizontal component of the rain, the runner gets when on the top and the front side. When i  v  0 the runner only gets wet on the top and the front side and the analysis is somewhat easier. In your search for an extremum, it may be convenient to represent the accumulated mass in a dimensionless form according to

m( t)  m

o

 F parameters



L D h

in which t is equal to the distance run divided by i  w  o u , and

is the initial

o

m

mass of the runner.