If the temperature difference between the studied object T, e.g., a cup of coffee and the environment Ts is not very large, the rate of temperature change of the object can be considered proportional to the given temperature difference. This statement can be written in the form of the differential equations as follows:

MATHEMATICAL EQUATIONS OF SYSTEM

where: where r is the cooling coefficient, dt is the time discretization step, the minus sign allows avoiding the unphysical increase in body temperature at T > Ts. This Equation is called Newton's law of thermal conductivity. We consider three models to study it:

• Model 1 - describing the numerical solution of this Equation.

• Model 2, which is a modification of Model 1 that takes into account the case of instantaneous changes of the body temperature, e.g. by 10°C at a given point in time.

• Model 3, in which the cooling coefficient r is “adjusted” in accordance with the experimental data.

THE MODEL 1

The influence diagram (Fig. 1.1) is used for the numerical solution of Eq.1.1, i.e. to obtain the temperature dependence on time T(t).

image2.emf

dT/dt

(01) "dT/dt"= -r*(T-Ts)

(02) FINAL TIME = 60

(03) INITIAL TIME = 0

(04) r=0.1

(05) SAVEPER = TIME STEP

(06) T= INTEG ("dT/dt", To)

(07) TIME STEP = 1

(08) To=100

(09) Ts=25

Fig.1.1. Model 1

Fig.1.2. Equations of Model 1

THE RESULTS OF SIMULATIONS

image3.emf

100

091827364554

Time (Minute)

T : Current.vdfx

The solution plot for Equation 1.1. at r=0.1, Ts=25°C, T(0)=100°C

Task 1

Simulate Model 1 and provide an answer to the following questions:

THE MODEL 2.

The acceleration of the cooling by the addition of the coolant (e.g. milk) is taken into account in Model 2 shown in Fig. 1.3. Let’s assume that in the moment of time time to mix the object cools down by the value of T of mix instantaneously. The graphical solution of the given task is shown in Fig. 1.4.

image5.emf

dT/dt

T of mix

time of mix

<Time>

(01) "dT/dt"= -r*(T-Ts)

(02) FINAL TIME = 60

(03) INITIAL TIME = 0

(04) r= 0.1

(05) SAVEPER = TIME STEP

(06) T= INTEG ("dT/dt"-IF THEN ELSE( time of mix=Time, T of mix, 0), To)

(07) T of mix= 10

(08) time of mix= 5

(09) TIME STEP = 1

(10) To= 100

(11) Ts= 25

Fig.1.3. Model 2

Fig.1.4. Equations of Model 2

THE RESULTS OF SIMULATIONS

image6.emf

100

091827364554

Time (Minute)

T : current

T : Current3.vdfx

Fig. 1.4. The graphical solution of the Model 2 at T of mix=10°C

Task 2

Perform a modelling and answer to the following questions:

Let’s assume that cooling by adding a coolant decreases the temperature by 10°C instantaneously. In this case the temperature sill decrease from 95°C to 75°C faster:

a) if the coolant is added immediately or

b) wait until the temperature decreases to 85°C and add coolant then?

THE MODEL 3

Let’s find the parameters for the Model 1 that are in an agreement with experimental data (Table 1) of the coffee cup cooling at the ambient temperature of Ts=22oC:

Table 1. Experimental data for the coffee cup cooling.

image8.emf

dT/dt

T of mix

time of mix

<Time>

table1

T-experiment

(01) "dT/dt"=-r*(T-Ts)

(02) FINAL TIME = 60

(03) INITIAL TIME = 0

(04) r= 0.1

(05) SAVEPER = TIME STEP

(06) T= INTEG ("dT/dt"-IF THEN ELSE( time of mix=Time, T of mix, 0), To)

(07) T of mix=10

(08) "T-experiment"= table1(Time)

(09) table1( [(0,0)-(15,100)],(0,83),(1,77.7),(2,75.1),(3,73),(4,71.1),(5,69.4),(6,67.8

),(7,66.4),(8,64.7),(9,63.4),(10,62.1),(11,61),(12,59.9),

(13,58.7),(14,57.8),(15,56.6))

(10) time of mix=5

(11) TIME STEP = 1

(12) To= 100

(13) Ts= 25

Fig.1.5. The diagram of the influence of Model 3

Fig.1.6. Equations of Model 3

Figs. 1.7 and 1.8 show the diagrams of the modelled and experimental data at different values of the parameter r (cooling coefficient)

THE RESULTS OF SIMULATIONS

Task 3.

Find r value that fits the corresponding real process the most by performing the simulation of the Model 3.

CONCLUSION

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