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Giuseppe Mensitieri » 3.1st and 2nd principles of thermodynamics. Integral and local forms. The concept of entropy

1^st law of thermodynamics

THE 1^st LAW OF THERMODYNAMICS

INTEGRAL FORMS OF 1st LAW
We introduce here the following postulate of conservation of energy:

the net rate at which energy flows into a body must be equal to the rate at which energy is accumulated in the body.

This postulate can be put into the following mathematical form:

$\dot U_t=Q_t + W_t\;\;\;\;\;\; \text{eq. (1a)}$

For a rigid body (or for a body undergoing a rigid motion) the work done results in accumulation of kinetic energy only (i.e. W_t = 0).

FOR A RIGID BODY:

$\dot K_t=P_t ; \;\;\; \dot U_t=Q_t\;; \rightarrow$

→ there is no exchange between ‘thermal’and ‘kinetic’ energy!

1^st law of thermodynamics

To deal with the thermodynamics of real materials, one needs to consider the deformation (e.g. change in density) of bodies. There are alternative forms of (eq 1a):

$\dot U_t+\dot K_t=Q_t+P_t\;\;\;\; \text{eq. (1b)}$

In the cases in which the body forces can be expressed as gradient of a scalar quantity which is independent of time:

$\dot U_t+\dot K_t+\dot\Phi_t=Q_t+P_{t, E}\;\;\;\;\; \text{eq.(1c)}$

(if b=-gradΦ, with Φ independent of time).

LOCAL FORMS OF 1^st LAW

Eqs. (1) can be integrated over any finite interval of time to obtain different macroscopic formulations. We can also obtain from the integral forms illustrated by eqs. (1) local forms of 1^st law of thermodynamics. In fact, eq. (1a), which applies to a body of finite size, can be transformed into a field (or local) equation:

$\int_B\dot U\cdot dm =\int_{\partial B}\underline q\cdot d\underline s +\int_BQ\cdot dm +\int_Bw\cdot dm\;\;\;\; \text{eq. (2)}$

1st law of thermodynamics

From the Gauss theorem:

$\int_{\partial B}\underline q \cdot d\underline s =-\int_B(div\; \underline q)\frac {dm}\rho$

where ρ is the local mass density. Equation (2) can be then recast to obtain:

$\dot U=\frac{-(div \underline q)}\rho +Q+w\;\;\;\; \text{eq(3)}$

which is a local form of 1^st law.

Now a question arises:

if $U_t=f(\sigma_t)$

is it also true that $U=f(\sigma)$ ?

The answer is YES, IF THE ‘PRINCIPLE OF LOCAL ACTION’ HOLDS TRUE.

1^st law of thermodynamics

Moreover, if we introduce the following definition:

$q\equiv Q-\frac{div \underline q}\rho$

we consequently derive that:

$\dot U=q+w\;\;\;\;\;\text{eq.(4)}$

Here q is a local rate of heating per unit mass and is related to Q_t by the following expression:

$Q_t=\int_B q \cdot dm$

2^nd law of thermodynamics

THE 2^nd LAW OF THERMODYNAMICS

INTEGRAL FORMS OF 2^nd LAW
The 2^nd law is a formalization of the intuitive concept of irreversibility of natural phenomena. IT IS NOT A CONSERVATION PRINCIPLE! Conservation laws are invariant under a reversal of the direction of time and are expressed in the form of an equality: in fact, 2^nd LAW IS NOT AN EQUALITY!
In the 2^nd law entropy is involved: it is a ‘primitive’ concept. Given a body B, it is postulated that it possesses a total entropy, S_t, which:

is an extensive property;
is a function of state.

The integral form of 2^nd law is expressed as:

$\dot S_t \geq J_t\;\;\;\;\;\text{eq. (5a)}$

where, by definition, J_t is the instantaneous rate of supply of entropy and $S_t=\int_B S\cdot dm$

2^nd law of thermodynamics

1^st and 2^nd laws are coupled if, as postulated below, J_t is related to Q_t.

By definition, we have:

$J_t\equiv \int_{\partial B}\underline j \cdot d\underline s\;\;\;\;\;\;\;\;\; \;\; +\;\; \;\;\;\;\;\;\;\;\; \int_B J\cdot dm$

$\text {eq. (6)}$

$\text{conductive mechanism} + \text{radiative mechanism}$

and we postulate that:

$\underline j\equiv \frac{\underline q}T;\;\;\; J\equiv \frac Q T\;\;\;\text{eq. (7)}$

where j is the entropy flux and J is the radiative entropy density. In these expressions T (temperature) is a ‘primitive’ (undefined) concept.

2^nd law of thermodynamics

Based on eq. (6) and (7), the form of eq. (5a) for the 2^nd law becomes:

$\dot S_t \geq \int_{\partial B}\left(\frac{\underline q}{T}\right)\cdot d\underline s +\int_B\left(\frac Q T\right)\cdot dm \;\;\;\;\text{eq. (5b)}$

In writing 1^st and 2^nd law we have introduced four ‘primitive concepts’: heat and energy in the 1st law and entropy and temperature in the 2^nd law.
In the following we will introduce some special forms of the inequality (5).

T not dependent on the point but, possibly, a function of time:

ISOTHERMAL FORM $\dot S_t\geq \frac{Q_t}T$

Adiabatic form (i.e. q=0; Q=0 ), it holds for isolated systems:

ADIABATIC FORM $\dot S_t\geq 0$

2^nd law of thermodynamics

LOCAL FORMS OF 2nd LAW

$\dot S_t\geq \int_{\partial B}\left(\frac{\underline q}T\right)\cdot d \underline s +\int_B\left(\frac Q T\right)dm \rightarrow \int_B\dot S dm \geq \int_B\left[\frac Q T -\frac{div (\underline q / T}{\rho}\right]dm \;\;\;\;\; \text{eq. (5c)}$

From eq. (5c) we obtain the local form of the 2^nd law:

$\dot S\geq \frac Q T -\frac 1 \rho div \left(\frac{\underline q}T\right)\;\;\;\; \text{eq. (8a)}$

As for the case of U, if the principle of local action holds true, $S=f(\sigma)$

Equation (8a) can be recast in the following form:

$\dot S\geq \frac 1 T\left[Q-\frac{div \underline q}{\rho}\right]+q\cdot \frac{\underline \nabla T}{\rho T^2}=\frac q T+\underline q \cdot \frac{\underline \nabla T}{\rho T^2}$

2^nd law of thermodynamics

From the previous expression an alternative expression for the local form of the 2^nd law is obtained:

$\dot S-\frac q T -\underline q\cdot \frac{\underline \nabla T}{\rho T^2}\geq 0\;\;\;\,\text{eq. (8b)}$

Also for the local form of 2^nd law we can introduce expressions for special cases.

Case of locally isothermal body (i.e. $\underline \nabla T=0$ ):

LOCALLY ISOTHERMAL BODY $\dot S \geq \frac q T\;\;\;\; \text{eq. (9)}$

Case of steady state in a stationary body (i.e. $\dot S=0; \dot U=0; w=0$ and hence q = 0):

STATIONARY BODY AT STEADY STATE $\underline q \cdot \underline \nabla T \leq 0\;\;\;\;\text{eq. (10)}$

From eq. (10) one can deduce that in the case of stationary heat conduction in stationary bodies the thermal flux vector must have a positive component in the direction of decreasing T.

2^nd law of thermodynamics

Eq. (8b) can be put in another form by considering that $\dot U =q+w$ :

$\dot S-\frac{(\dot U -w)}T-\underline q\cdot \frac{\underline \nabla T}{\rho T^2}\geq 0\;\;\;\text {eq. (8c)}$

This expression is the basis for the dissipation theory. If we consider a process (σ_t=P(t) or σ = P (t)) at every time the 2^nd law (in each of the proposed forms) can be either satisfied as an INEQUALITY (in such a case the process is INSTANTANEOUSLY IRREVERSIBLE) or as an EQUALITY (in such a case the process is INSTANTANEOUSLY REVERSIBLE).
For instantaneously irreversible processes, we can define a LOCAL VELOCITY OF ENERGY DISSIPATION PER UNIT MASS, Z.
In fact, multiplying eq. (8c) by T, we obtain:

$T\dot S-\dot U+w-\underline q \cdot \frac{\underline \nabla T}{\rho T}\geq 0\rightarrow Z\geq 0\;\;\;\;\text{eq. (8d)}$

where $Z=Z_M+Z_T$ with $Z_M\equiv T\underline S-\dot U+w$ and $Z_T\equiv-\underline q \cdot \frac{\underline \nabla T}{\rho T}$

2^nd law of thermodynamics

We define now the Helmholtz Free Energy, A:

$A\equiv U-TS\Rightarrow \dot A=\dot U-T\dot S-S\dot T$

If we consider the ISOTHERMAL CASE (i.e. $\underline \nabla T=0$ ):

$Z_M\geq 0\Rightarrow -\dot A-S\dot T+w\geq0$

Moreover, if $\dot T=0$ it is also we have $w\geq \dot A$

Consequently:

for isothermal reversible process, with T constant with time

$w=\dot A=\dot U-T\dot S$

for isothermal irreversible process, with T constant with time

$w=Z_M+\dot A=Z_M+\dot U-T\dot S$

2^nd law of thermodynamics

$\dot U-T\dot S$ represents the rate of accumulation of elastic energy.
On the basis of previous expressions, in general, for an isothermal process at constant T, the work is the sum of the non-thermal dissipation rate and of the rate of accumulation of elastic energy.