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Giuseppe Mensitieri » 2.Energy , Work and Heating

Fundamental concepts: energy and heating

ENERGY AND HEATING

$\mathfrak{B}$ is endowed with an amount of energy, U_t, which is a function of state (no claim is made that U_t is the total energy of the body!). Conversely, kinetic energy is not a function of state. U_t is an absolutely additive function of mass (i.e. it is an extensive quantity).

U≡ INTERNAL ENERGY DENSITY

$U_{t}=\int _{ \mathfrak{B}}{U}{dm}$

Energy can flow from one body to another by two mechanisms:

HEATING
WORK

Fundamental concepts: energy and heating

HEATING

$\left(Q \right)_t\equiv$ INSTANTANEOUS RATE OF ENERGY INFLUX INTO THE BODY DUE TO HEATING

There are two mechanism of heating:

CONDUCTION MECHANISM:
$\left(Q_t \right)_{COND}=\int _{\partial \mathfrak{B}}\underline{q}\cdot \underline{ds}$

where $\underline{q}$ is the conductive heat flux vector and $\underline{ds}$ is the inward pointing area vector.

RADIATION MECHANISM:
$\left(Q_t \right)_{COND}=\int _{\partial \mathfrak{B}}{Q}{dm}$

where Q is the local rate of radiation heat inflow per unit mass.

Fundamental concepts: energy and heating

$\underline{q}$ and Q are ‘local‘ quantities. $Q_{t}$ is not a function of state. It is defined as:

$Q_{t}=\left( Q_t\right)_{COND}+ \left( Q_t\right)_{RAD}$

In fact, as we will see, $\underline{q}$ is a function of local state, σ, while Q is not

Fundamental concepts: work and kinetic energy

FORCES

Power, P, is the rate at which a force $\underline{F}$ is doing work acting on a material point, v which moves at velocity:

$P = \underline{F}\cdot\underline{v}$

There are different types of forces:

CONTACT FORCES:
acting on the surface of the body, whose power is indicated as P_t,E

BODY FORCES:
acting directly on elementary masses within the body, whose power is indicated as P_t,B

$P_{t,\mathfrak{B}}= \int_{\mathfrak{B}}\underline{b}\cdot\underline{v}dm$

Fundamental concepts: work and kinetic energy

Here $\underline{b}$ is the field acceleration (e.g. acceleration of gravity). If $\underline{b}=-grad\,\phi$ , with φ independent of time (e.g. φ = g.h) , then:

$\underline{b}\cdot\underline{v}=-\underline{v}\,\cdot grad\,\phi=-\,\dot{\phi}$

Here we have used the notation:

$\dot{a}\equiv\frac{\partial a}{\partial t}+\underline v\cdot grad\,a$

to indicate the substantial derivative. We finally obtain:

$P_{t,\mathfrak{B}}=-\dot{\phi}_{t}$

since, from Reynolds’ Transport Theorem (see note 1 at the end of lecture) we have:

$\dot{\phi}_{t}=\int _{\mathfrak{B}}\dot{\phi}\,dm$

with
$\phi_{t}=\int _{\mathfrak{B}}\phi\,dm\,\equiv\,potential \:energy$

Fundamental concepts: work and kinetic energy

Should $\underline{b}$ not admit a time-independent potential, one could not even define a potential energy but the power of body forces would still be defined by:

$\int _{\mathfrak{B}}\left( \underline{b}\cdot\underline{v}\right)\,dm$

Fundamental concepts: work and kinetic energy

KINETIC ENERGY

We pass now to define the Kinetic Energy, K_t:

$K_{t}\equiv \frac{1}{2}\int _{\mathfrak{B}}\left( \underline{v}\cdot\underline{v}\right)\,dm\: \: \: \Rightarrow \: \: \:\dot{K}_{t}= \int _{\mathfrak{B}}\left( \underline{v}\cdot\underline{\dot{v}}\right)\,dm ~~~~~~~~~~~~~~ \text{eq. (1)}$

A fundamental theorem of classical mechanics states that for a rigid body, or for any body which is instantaneously undergoing a rigid motion, we have:

$P_{t}=\dot{K}_{t}$

For more details on that see note 2 at the end of the lecture. The previous equation is a purely mechanical result: it does not derive from any principle of energy conservation! This equation applies only to rigid body motions. In fact, rigid body motions can never result in the transformation of energy absorbed as heat into works, and viceversa. Rigid body motions are thermodynamically degenerate.

Fundamental concepts: work and kinetic energy

Equation (1) refers only to the nonpolar case. An analogous rotational contribution arises for the polar case, when also power related to work of distributed body couples have to be considered. In fact, with reference to rotational motions, it can be written:

$P_{t,rot}=\int _{\mathfrak{B}}\left(\underline{c}\cdot\underline{\omega}\right)dm$

Where $\underline{c}\,dm=\underline{\dot{\omega}}dI$ consequently:

$P_{t,rot}=\int_{\mathfrak{B}}\left(\underline{\omega}\cdot\dot{\underline{\omega}}\right)dI=\dot{K}_{t,rot}$

For a body which is NOT undergoing a rigid body motion, we have:

$W_{t}=P_{t}-\dot{K}_{t}$

Fundamental concepts: work and kinetic energy

In the previous equation W_t is the ‘net’ rate of work done on the body. It is worth noting that body forces contribute nothing to W_t(see note 2 at the end of the lecture): the work of body forces goes entirely into increasing K_t. The following expression can be adopted for W_t:

$W_t=\int_B w\cdot dm$

where w is the ‘net’ rate of work per unit mass done on a differential body element of mass dm.
It is important to underline here that both W_tand Q_t are NOT functions of state. In fact dW_t and dQ_t, defined respectively as

$dW_t=W_t\cdot dt; \;\;\;\;dQ_t=Q_t\cdot dt$

are not exact differential forms. We recall here that in order the differential form $M(x,y)dx+N(x,y)dy$ to be ‘exact’, it should happen that:

$\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}$

i.e.

$dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy$

Note 1: Reynold’s transport theorem

THE REYNOLD’S TRANSPORT THEOREM

On the material derivative of a volume integral
We illustrate here how to evaluate the material derivative of a quantity which consists in an integral extended to a volume, i.e.:

$\phi_t=\int_B\phi \;dm \rightarrow \frac{d_{(m)}\phi_t}{dt}=\frac{d_{(m)}}{dt}\int_B\phi\; dm$

Here the integral is evaluated on the volume occupied by the body $\mathfrak{B}$ . If we the integral has to be evaluated on a volume V which is fixed in the space with time, we have:

$\frac d {dt}\int_V\rho\;dV=\int_V\frac{\partial\rho}{\partial t}dV\;\;\;\;\;\text{eq. }(\alpha)$

However, if both the integrand and the integration volume, containing a fixed mass, are changing with time, expression (α) is no more valid. Now the problem can be stated as follows: we need to evaluate the material derivative of a volume integral which measures the rate of change of the total amount of a quantity transported by a certain amount of mass. In the following $\psi=A\cdot \rho$ , where A is a quantity per unit mass while ψ is a quantity per unit volume.

Note 1: Reynold’s transport theorem

$\begin{Bmatrix}\begin{bmatrix}Rate\:of\:increase\\of\:the\:instantaneous\\amount\:of\:A\\in\:the\:fixed\:volume\:V\end{bmatrix}=\begin{bmatrix}Rate\:of\:increase\:of\:A\\ associated\:t\:the\\ material\:located\\instantaneoulsy\:in\:the\\control\:volume\end{bmatrix}-\begin{bmatrix}Net\:flux\:rate\:of\:A\\ toward\:external\:regions\\ transported\:by\:mass\\ crossing\:the\:control\\ surface\\ \end{bmatrix}\\\int_{V}\frac{\partial \left (\rho A \right )}{\partial t}dV \: \: \: \: \: \: \: \: \: \: \: \: \: \:\:\:\:\: =\:\:\:\:\:\:\:\:\:\:\frac{d_{\left (m \right )}}{dt}\int_{V}\rho \, A\, dV\:\:\:\:\:\:\:\:\:\:\:\:\:\:-\:\:\:\:\:\:\:\:\:\:\:\:\:\int _{S}\rho \, A\, \left (\underline{v}\cdot\underline{n} \right )dS \end{Bmatrix}$

Which can be recast in the following form:

$\frac{d_{\left (m \right )}}{dt}\int_{V}\rho \, A\, dV=\int_{V}\frac{\partial \left (\rho A \right )}{\partial t}dV+\int _{S}\rho \, A\,v_{j}\,n_{j}dS \;\;\;\;\;\;\text{(eq. n1.1)}$

which is one of the forms of the Reynolds’ Transport Theorem.

Note 1: Reynold’s transport theorem

Since
$\frac {\partial\left( \rho A\right)} {\partial t}=A\frac {\partial\rho} {\partial t}+\rho\frac {\partial A} {\partial t}$

and

$\int _{S}\rho A\left ( \underline{v}\cdot\underline{n} \right )}ds=\int _{V}div \left({\rho A\underline{v}}\right)\,dV$

it follows that:

$\frac{d_{\left(m \right)}}{dt}\int \rho A\,dV=\int _{V}\left [ \rho \frac{\partial A}{\partial t}+A\frac{\partial \rho }{\partial t}+\rho \underline{v}\cdot \underline {\nabla} A +A\underline{v}\,\cdot\underline {\nabla} \rho +\rho A\left( \underline {\nabla}\cdot\underline{v}\right)\right ]\,dV$

Now, it can be demonstrated that, in view of the local form of the differential mass balance, we have:

$A\frac {\partial\rho} {\partial t}+A\left(\underline{\nabla}\cdot\rho \underline{v}}\right)=0$

Note 1: Reynold’s transport theorem

Consequently, one can put eq. n1.1 it the form

$\frac{d_{\left(m\right)}}{dt}\left(\int_{R\left(m\right)}\rho\,A\,dV\right)=\int_{R\left(m\right)}\rho\,\frac{d_{\left(m\right)}A}{dt}\,dV\,\,\,\Rightarrow \,\,\,\frac{d_{\left(m\right)}}{dt}\left(\int_{R\left(m\right)}\,A\,dm \right )=\int_{R\left( m\right)}\,\frac{d_{\left(m\right)}A}{dt}\,dm$

which is another form of the Reynold’s transport theorem. It is worth noting that, instead, we have:

$\frac{d_{\left(m\right)}}{dt}\left(\int_{R\left(m\right)}\rho\,A\,dV\right)\neq \int_{R\left( m\right)}\,\frac{d_{\left(m\right)}\left (\rho A \right )}{dt}\,dm$

Note 2: Kinetic energy and deformable bodies

KINETIC ENERGY AND THE POWER OF BODY AND CONTACT FORCES FOR DEFORMABLE BODIES

In the case of a deformable body, the following relationships hold between the power of body and contact forces and kinetic energy:

$P_{t,B}=\int_B(\underline b \cdot \underline v)dm; \;\;\; \underbar b=\underline{\dot v}\Rightarrow P_{t,B}=\int_B(\underline v \cdot \underline {\dot v})dm=\dot K_{t,B}\;\;\; \text{eq. n2.1}$

$P_{t,E}=\int_{\partial B}(\underline f \cdot \underline v)ds;\;\;\; \Rightarrow \;\; P_{t,E}=\int_{\partial B}\underline v \cdot (\underline{\underline T}\cdot \underline n)ds=\int_Bdiv (\underline{\underline T}\cdot \underline v)dV\;\;\; \text{(eq. n2.2)}$

where $\underline{\underline T}$ is the stress tensor.

Since

$div\left(\underline{\underline{T}}\cdot\underline{v}\right) = \underline{v}\cdot div \underline{\underline{T}}+\underline{\underline{T}}:\underline{\nabla}\,\underline{v}$

we obtain from eq. n2.2:

$P_{t,E}=\int_B(\underline v \cdot div\;\underline{\underline T})dV+\int_B(\underline{\underline T} :\underline \nabla\; \underline v)dV=\dot K_{t,E}+\dot U_t$

$P_{t,E}=\;\;\;\;\;\;\;\;\;\dot K_{t,E} \;\;\;\;\;\;\;\;\; + \;\;\;\;\;\;\;\;\;\;\;\ \dot U_t$

Note 2: Kinetic energy and deformable bodies

Hence we finally obtain:

$P_t=P_{t,B}+P_{t,E}=\dot K_{t,B}+\dot K_{t,E}+\dot U_t=\dot K_t+\dot U_t$

We note here that, in the case of fluids, the term $\underline{\underline{T}}:\underline{\nabla}\,\underline{v}$ contains two distinct contributions. In fact, the total stress tensor can be put in the form:

$\underline{\underline T}=P^*\underline{\underline I}+\underline{\underline S}$

where P^* is the thermodynamic pressure while $\underline{\underline{S}}$ is the dynamical stress tensor, i.e. that part of the stress tensor which arises when the fluid is in an out-of-equilibrium state. Consequently we have that:

$\underline{\underline{T}}:\underline{\nabla}\,\underline{v}=P^{*}\left (\underline{\nabla}\cdot\underline{v} \right )+\underline{\underline{S}}:\underline{\nabla}\,\underline{v}$

In the previous expression:

$P^{*}\left (\underline{\nabla}\cdot\underline{v} \right )$ represents the rate of reversibile increase of internal energy per unit volume due to isotropic compression of the fluid.

$\underline{\underline{S}}:\underline{\nabla}\,\underline{v}$ represents the rate of increase of internal energy due to the dynamical part of the stress tensor. As an example, in the case of a viscoelastic fluid, this term include i) a rate of irreversible increase of internal energy due to viscous dissipation (related to changes of both shape and volume of the fluid element) and ii) a rate of reversible increase of internal energy due to elastic deformation of the fluid element.