4. Overview of models and model equations

4.3 Energy and heat

In physics, 'heat' is denoted by the symbol \(Q\) and it is a form of energy exchange between systems or between a system and its environment. There are three ways in which heat transfer can take place: by conduction (or contact), convection (flow), or radiation. Because heat is a form of energy, the energy of an object changes by supplying or extracting heat: $$\Delta E = Q + W$$ The term \(W\) takes into account any work done on the system by its surroundings. This is the 'law of work and energy' including energy transfer by heat; it is also called the 'first law of thermodynamics.

In the models below, heat processes are considered that take place during a time step \(\Delta t\). $$Q = P \cdot \Delta t$$ where \(P\) is the supplied power, i.e. the energy that is absorbed by the object per unit of time or dispersed into the environment.

Below are some examples of models¹ belonging to the subdomains Energy Conversions, and Energy and Interaction in the physics syllabus on Motion and Interaction,^2,3 with corresponding files for the Coach 7 modelling environment.⁴

1. Energy supply-extraction model

The basic dynamic model of supply and extraction of energy (heat, work) is based on the 'law of work and energy' written in terms of the supplied and extracted power \(P_{\rm in}\) and \(P_{\rm out}\). The supply and extraction may depend on time and quantities such as temperature of the system and/or the surroundings.
In the models below, this basic model is used in modelling situations in which energy transformation plays a role.

Formula in Binas:

$$ P = \frac{E}{t}$$

Difference equation:

$$\Delta E = \left( {{P_{{\rm{in}}}} - {P_{{\rm{out}}}}} \right) \cdot \Delta t $$

Text-based model

\(\begin{array}{l} P_{\rm tot}=P_{\rm in} - P_{\rm out}\\ E = E + P_{\rm tot} \cdot dt\\ t = t + dt \end{array}\)

Graphical model



Coach 7

Coach 7

Example

Initial values (SI)

\(\begin{array}{l} {P_{\rm in}} = 100\\ P_{\rm out}= 75\\ E = 0 \\t = 0 \\ dt = 0{.}1 \end{array}\)


2. Heat-temperature model at constant power

In this model, an amount of a substance is heated with a constant power \( P_{\rm in}\); think of an ideal kettle. In this ideal model the absorbed heat is set equal to the supplied energy. The 'specific heat' \(c\) is the amount of heat required to raise the temperature of 1 kg of a substance by 1 K or 1 °C. When the specific heat of water is given, the temperature can be calculated as a function of time.

Formula in Binas:

$$ Q = c \cdot m \cdot \Delta T $$

Difference equation:

$$ \begin{array}{l}Q = {P_{\rm in}} \cdot \Delta t \\ \Delta T = (P_{\rm in}/{c \cdot m}) \cdot \Delta t \end{array}$$

Text-based model

\(\begin{array}{l} T'= P_{\rm in}/(c \cdot m)\\ T = T + T' \cdot dt\\ t = t + dt \end{array}\)

Graphical model



Coach 7

Coach 7

Example

Initial values (SI)

\(\begin{array}{l} {P_{\rm in}} = 2000\\ c = 4186\\ m = 1{.}5\\ T = 20 \\t = 0 \\ dt = 0{.}1 \end{array}\)


3. Greenhouse model

The earth receives radiation from the sun, but also emits radiation itself. The effect on the surface temperature of the earth is calculated with a simple model.^5,6 The sun's incident irradiation amounts to \(S \simeq 1368{\rm W/m}^2.\) The frontal surface of the earth illuminated by the sun is set at \(\pi {R^2}\), where \( R \) the radius of the earth. The total incident irradiated power is expressed in the formula of \(P_{\rm in}\). Herein is \(a \simeq 0{.}37\) the so-called 'albedo factor' which is a measure of the amount of incident radiation that is reflected by the earth.
The earth emits radiation with power \(P_{\rm out}\) that depends on the temperature \(T\) of the surface of the earth according to 'Stefan-Boltzmann law'. The 'emittance factor' \(\varepsilon \simeq 0{.}95\) takes into account the atmosphere blocks part of the emission.⁷

Formulas in Binas:

$$ \begin{array}{l} Q= c \cdot m \cdot \Delta T\\ P = \sigma \cdot A \cdot {T^4} \end{array}$$

Difference equations:

$$ \begin{array}{l}{P_{{\rm{in}}}} = \pi {R^2} \cdot (1 - a) \cdot S\\ {P_{{\rm{out}}}} = 4\pi {R^2} \cdot \sigma \cdot (1 - \varepsilon /2) \cdot {T^4}\\ P_{\rm tot}= P_{\rm in}- P_{\rm out}\\ \Delta E = P_{\rm tot} \cdot \Delta t \\ \Delta T= \Delta E/C_{\rm earth}\end{array}$$

In case \({P_{{\rm{in}}}} = {P_{{\rm{out}}}}\) the amount of radiation received and emitted is equal. The net receipt and emission of radiation causes a temperature change depending on \(C_{\rm earth}\), the average heat capacity per square meter of the surface of the earth. This quantity is set in the example equal to that of the atmosphere.

Text-based model

\(\begin{array}{l} {P_{{\rm{in}}}} = \pi {R^2} \cdot (1 - a) \cdot S \\ {P_{{\rm{out}}}} = 4\pi {R^2} \cdot \sigma \cdot (1 - \varepsilon /2) \cdot {T^4}\\ P_{\rm tot}= P_{\rm in}- P_{\rm out}\\ E = E + P_{\rm tot}\cdot dt\\ T = E/C_{\rm earth}\\{\rm Temperature (Celsius)}= T-273{,}15\\ {\rm Time(year)}=t/3{,}15\cdot 10^7 \\t =\ t + dt \end{array}\)

Graphical model



Coach 7

Coach 7

Example

Initial values (SI)

\(\begin{array}{l} S = 1368\\ \sigma=5{.}67 \cdot 10^{-8}\\ a= 0{.}37\\ \varepsilon=0{.}95\\R=6{.}37 \cdot 10^{6}\\ C_{\rm earth}=2 \cdot 10^{21}\\ E=0\\ t = 0 \\ dt = 5000 \end{array}\)


4. Model of cooling

Heat loss is an everyday phenomenon, for example the cooling of hot coffee/tea in a closed travel mug. For this purpose, a simple model can be constructed on the basis that the most important contribution to the cooling is provided by the heat conduction through the insulating wall of the mug. This is described by the given formula in Binas for heat conduction. Herein denotes \(\lambda\) the heat conductance, \(A \) the area of the cooling object, and \(d\) the thickness of the insulating wall. The energy loss is proportional to the difference in temperatures between the body \(T\) and its surroundings \(T_{\rm env}\) From this energy loss follows the decrease in temperature as $$\Delta T = - k \cdot (T - {T_{{\rm{env}}}}) \cdot \Delta t$$ This relation is known as 'Newton's law of cooling'.


Formulas in Binas:

$$ \begin{array}{l} Q = c \cdot m \cdot \Delta T \\ P = \lambda \cdot A \cdot \Delta T/d \end{array}$$

Difference equations:

$$ \begin{array}{l} a=\lambda \cdot A/d \\ k= a/(c \cdot m) \\ {P_{\rm out}} = a \cdot (T - T_{\rm env}) \\ \Delta E= - P_{\rm out}\cdot \Delta t \\ \Delta T= T' \cdot \Delta t\\ T'= - k \cdot (T - T_{\rm env}) \end{array} $$

In the example, the temperature curve is computed with parameter values that describe the cooling of coffee/tea well.⁸ The specific heat is that of water.

Text-based model

\(\begin{array}{l} {P_{\rm out}} = a \cdot (T - T_{\rm env}) \\ T'= -{P_{\rm out}}/(c \cdot m) \\ T = T + T'\cdot dt\\ t = t + dt \end{array}\)

Graphical model



Coach 7

Coach 7

Example

Initial values (SI)

\(\begin{array}{l} a=1{.}42\\ c =4180\\ m = 0{.}33\\ T_{\rm env} = 21 \\ T=95 \\ t = 0 \\ dt = 0{.}1 \end{array}\)


5. Power-velocity model

In this model, the 'law of work and energy ' \( \Delta {E_{\rm k}} = {W_{\rm tot}} \) is used to compute the velocity and displacement of an object, for example the Buckeye Bullet, an electric car that is especially designed to break speed records.⁹ It is assumed that this car is propelled by a constant power \( P_{\rm car}\) but also is subject to friction, mainly in the form of air resistance with force \(F_{\rm d} \). The rolling resistance of the car is neglected.
After 90 seconds, at time \(t_{\rm brake}\), the parachute opens for braking and the car stops as a result of the braking force \(F_{\rm brake}\) and lack of propulsion. In the example, the parameter values of the sample exam Buckeye Bullet⁹ are used.

Formulas in Binas:

$$\begin{array}{l}{E_{\rm k}} = \frac{1}{2}m \cdot {v^2} \\ {F_{\rm d}} = k \cdot {v^2} \\ P=F \cdot v \end{array}$$

Difference equation:

$$\begin{array}{l} P_{\rm d}={F_{\rm d}} \cdot v\\P_{\rm brake}=F_{\rm brake} \cdot v\\ {P_{\rm tot}} = P_{\rm car} - P_{\rm d} - P_{\rm brake} \\ \Delta {E_{\rm k}} = {P_{\rm tot}} \cdot \Delta t \end{array} $$

Text-based model

\(\begin{array}{l} {\rm If} \; t < t_{\rm brake} \\ {\rm Then}\\ \quad k= 0{.}1 \\ \quad P_{\rm car}=P_{\rm max}\\ \quad P_{\rm brake}=0 \\ {\rm Else}\\ \quad k=1{.}0 \\ \quad P_{\rm car}=0 \\ \quad P_{\rm brake}=F_{\rm brake} \cdot v\\ {\rm EndIf}\\ {P_{\rm d}} = k \cdot {v^3}\\ {P_{\rm tot}} = P_{\rm car} - {P_{\rm d}} -P_{\rm brake}\\ {E_{\rm k}} = {E_{\rm k}} + {P_{\rm tot}} \cdot dt\\ v = \sqrt {2{E_{\rm k}}/m} \\ x = x + v \cdot dt\\ t = t + dt \end{array}\)

Graphical model



Coach 7

Coach 7

Example

Initial values (SI)

\(\begin{array}{l} m=1740\\ F_{\rm brake}=2500\\ P_{\rm max}= 3{.}75 \cdot 10^5\\ t_{\rm brake}=90\\ {E_{\rm k}} = 0\\x=0\\ t =0 \\ dt = 0{.}1 \end{array}\)


6. Power balance model

The speed of a sprinter can be computed via the power balance equation relating the change in kinetic energy \(E_{\rm k}\) in a time step \(\Delta t\), the propulsive power \(P_{\rm s}\) of the sprinter's muscles and the internal frictional power of the sprinter \(P_{\rm f}\) leading to heat transfer. Here we neglect the air resistance during sprinting.
It is assumed that from the start the maximum propulsive power \( P_{\rm max}\) is reached via a model of exponentially restricted growth, according to the formula given below, and that the frictional power depends quadratic on the speed \(v\) (in other words, the internal friction is proportional to the speed). The assumption for the propulsive power can be consider as the assumption that only the aerobic power of muscles is of interest at the beginning of the sprint.¹⁰ In more realistic models, the anaerobic power and also aerodynamic effects are taken into account.¹¹

Formulas in Binas:

$$\begin{array}{l} \Delta E_{\rm k} = W_{\rm tot}\\ P = W/t\end{array}$$

Difference equation:

$$\begin{array}{l} P_{\rm p} = P_{\rm max } \cdot \left( {1 - \exp \left( { - k_{\rm s} \cdot t} \right)} \right)\\ P_{\rm f} = k_{\rm f} \cdot {v^2}\\ \Delta {E_{\rm k}} = (P_{\rm s} - P_{\rm f}) \cdot \Delta t \end{array}$$

As is common in biomechanics, computations are done per unit of mass; so \({E_k} = {\textstyle{1 \over 2}}{v^2}\).

In the example \(k_{\rm s}\) are \(k_{\rm f}\) are equal, and the initial values are chosen such that there is a good agreement between model results and experimaental data of the 100 m sprint of Carl Lewis at the athletics world championship in 1987.

Text-based model

\( \begin{array}{l} P_{\rm f} = k_{\rm f} \cdot {v^2}\\ P_{\rm s} = P_{\rm max } \cdot \left( 1 - \exp \left( - {k_{\rm s}} \cdot t \right) \right)\\ P_{\rm tot} = P_{\rm s} - P_{\rm f} \\ E_{\rm k} = E_{\rm k} + P_{\rm tot} \cdot dt \\ v = \sqrt{2{E_{\rm k}}}\\ x= x + v \cdot dt\\ t = t + dt \end{array}\)

Graphical model



Coach 7

Coach 7

Example

Initial values (SI)

\(\begin{array}{l} P_{\rm max }=106\\ k_{\rm f} = 0{.}75\\k_{\rm s} = 0{.}75\\ E_{\rm k} = 0\\ x = 0\\ t = 0\\ dt = 0{.}01 \end{array}\)