THE PROBLEM
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).
|
T dT/dt r Ts To |
(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
|
T 100 50 0 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.
|
T dT/dt r Ts To 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
|
T 100 50 0 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?