Table 9.12

The results of solving for h(t) are shown in Table 9.12.

Figure 9.14 shows the graph obtained for h against t and, as we would expect, the response of the gutter follows the rain profile, rising to a maximum at around t = 28 s and then subsiding to zero after about 51 s. Again a slight mathematical problem occurs near the end because, as h becomes very small, we are almost dividing by zero in our differential equation for h(t).

9.7 Turf

Context

Grass court tennis matches are always susceptible to interruption due to rain. A water-proof cover is not often available and so play can only be resumed when

Figure 9.14

the top layer of turf has dried out sufficiently. This means that either the rainwater has soaked right through to the subsoil, or it has evaporated back into the atmosphere after it has stopped raining. A number of mechanical devices can be tried to speed up the drying process but, to avoid damaging the turf, it is often best to let the turf dry out naturally. Can a mathematical model be set up to represent the drying process?

Problem statement

Given a local rain shower, is it possible to predict when play can be resumed? In particular, suppose that the turf is dry to begin with when it suddenly starts to rain and continues at a constant rate of half an hour. Suppose also that the amount of rain collected in this time is 1.8 cm. (Note that this is a depth and not a volume until you multiply by the area of catchment.)

Formulate a mathematical model

The problem is somewhat like that in section 9.6 in that we have a flow principle again:

We list the factors in Table 9.13.

Note that Q is measured in metres and becomes an actual volume of water when multiplied by the area of the turf. Similarly e and s are measured in metres per second and, when multiplied by A, then

become volume flow rates measured in cubic metres per second.

Now consider the model in Figure 9.15. Since the turf is dry to begin with, we have an initial condition Q ( t = 0) = 0. With reference to the ‘flow principle’ above, it is necessary to model the inflow, outflow and turf holding capacity. The inflow is easily done as it is the product of the rainfall rate and the area of turf considered. Hence

(9.9)

For the outflow rate, we have to decide how the water will disappear from the turf. This will be through drainage into the subsoil, and through evaporation back into the atmosphere. While it is still raining, it is unlikely that there will be any evaporation; so only the soak-away rate has to be modelled. We shall take it to be proportional to the amount of water currently present in the turf, i.e.

(9.10)

Once it has stopped raining, water continues to soak away through the subsoil and can now also evaporate. The rate of evaporation will depend on the local air temperature and humidity. Also the water that evaporates will come from the top surface of the wet turf. To keep the model simple, the

Table 9.13

Figure 9.15

evaporation rate will be taken as proportional to the amount of water in the turf, i.e.

Now a differential equation can be formed for Q ( t ) and is expressed in two stages according to whether it is still raining or not:

(9.11)

This model could be made more realistic but also more complicated by considering alternative expressions for the soak-away and evaporation rates.

To see whether the solution of equation (9.11) gives an adequate result for the water held in the turf, we obtain Q ( t ) by integrating the equation, first supplying data for r ( t ), a and b. Note in passing that the turf thickness D has not come into the problem. The rainfall rate r ( t ) was stated earlier to be constant over a period of half an hour, i.e.

To proceed with the development of the model, we must put in suitable values for the proportionality constants a and b, but what values are ‘suitable’? At this point, which very often occurs in modelling, we have to choose from a number of alternative courses of action. One is to look deeper into the theoretical principles involved and to hope to find appropriate values of a and b from scientific considerations. Another option is to continue with a and b as parameters with unspecified values, to obtain the mathematical solution and then to compare the model's predictions with real data. It will then be a question of deriving the values of a and b for the best fit to the data (see chapter 8). For the purpose of obtaining a specific answer for this example, we shall avoid these issues here and insert particular values of a = 0.001 s −1 and b = 0.0005 s −1 .

Obtain the mathematical solution

From equation (9.11), if we take 1 m 2 of turf then A = 1 and the differential equations to be solved are

(9-12) (9-13)

Both these are simple linear first-order equations and the solution can be obtained without difficulty. Integrating equation (9.12) gives

(9.14)

This holds up to t = 1800, at which time Q ( t ) 0.00835.

Now solving equation (9.13) in the same way gives

where B is an integration constant to be calculated from the condition Q (1800) = 0.00835. Substituting, we get B = 0.124. Hence,

(9.15)

Interpret the solution

Equation (9.15) predicts how the quantity of water in the turf decreases after the end of the shower. Our problem was to find when play can be resumed, which presumably means when the turf is dry again; so Q = 0. However, the expression in equation (9.15) is a negative exponential function and Q ( t ) will never actually be zero according to this model. We can make an assumption that, when the water