Balance of Energy

◦︎ First Law of Thermodynamics

◦︎ Conservation of Energy

One of the most important laws in physics and engineering

✨ 19.9.2024, 22:51

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✏️ 28.8.2025, 10:31

The balance of energy (power), the first law of thermodynamcics can be concisely written as

\begin{align} \label{eq:j1soax99c1} \dot{U} + \dot{K} = Q + L_a \end{align}

where \(U\) is the internal energy of the system, \(K\) the kinetic energy, \(Q\) the heat exchange with the environment and \(L_a\) the mechanical power exchange.

with Balance of Mechanical Energy

\begin{align} \dot{U} = Q + L_i \end{align}

Gathering all the contributions, Internal Energy, Heat and power of internal forces

\begin{align} \int_{ \mathcal{R}_t} \dot{u} \; \rho \;\textrm{d}v = - \int_{ \mathcal{R}_t} \textrm{div}\,\textbf{q} \; \textrm{d}v + \int_{ \mathcal{R}_t} r \; \textrm{d}v + \int_{ \mathcal{R}_t}\textbf{T} \cdot \textbf{D}\;\textrm{d}v \end{align}

or in terms of a localized balance

\begin{align} \rho \;\dot{u} = -\textrm{div}\,\textbf{q} + \textbf{T} \cdot \textbf{D} + r \end{align}

1. Control Volume

\begin{align} \int_{\mathcal{R}_{c}} \frac{\partial \left( u \; \rho \right)} {\partial t}\,\textrm{d}v &+ \int_{\partial \mathcal{R}_t} u \; \rho \;\textbf{v}\cdot \textbf{n}\;\textrm{d}a + \int_{\mathcal{R}_{c}} \frac{ 1 }{ 2 } \frac{\partial}{\partial t} \left( \rho \; \textbf{v} \cdot \textbf{v} \right) \;\textrm{d}v + \int_{\partial \mathcal{R}_c} \frac{ 1 }{ 2 } \left( \rho \; \textbf{v} \cdot \textbf{v} \right) \;\textbf{v}\cdot \textbf{n} \;\textrm{d}a \\ &= \int_{ \partial\mathcal{R}_c} q \; \textrm{d}a + \int_{ \mathcal{R}_c} r \; \textrm{d}v + \int_{ \partial\mathcal{R}_c} \textbf{v} \cdot\textbf{t} \;\textrm{d}a + \int_{ \mathcal{R}_c} \textbf{b}\cdot \textbf{v} \;\textrm{d}v \end{align}

or by sorting of volume and surface integrals

\begin{align} \int_{\mathcal{R}_{c}} \frac{\partial } {\partial t} \left( u \; \rho +\frac{ 1 }{ 2 }\;\rho \; \textbf{v} \cdot \textbf{v} \right) \;\textrm{d}v &+ \int_{\partial \mathcal{R}_c} \left( u \; \rho + \frac{ 1 }{ 2 }\;\rho \; \textbf{v} \cdot \textbf{v} \right) \;\textbf{v}\cdot \textbf{n}\;\textrm{d}a \\ &= \int_{ \mathcal{R}_c} \left( r + \textbf{b}\cdot \textbf{v} \right) \; \textrm{d}v + \int_{ \partial\mathcal{R}_c} \left( q + \textbf{v} \cdot\textbf{t} \right) \; \textrm{d}a \end{align}

1.1. Stationary Processes without Body Forces

\begin{align} \int_{\partial \mathcal{R}_c} \left( u \; \rho + \frac{ 1 }{ 2 }\;\rho \; \textbf{v} \cdot \textbf{v} \right) \;\textbf{v}\cdot \textbf{n}\;\textrm{d}a = \int_{ \partial\mathcal{R}_c} \left( q + \textbf{v} \cdot\textbf{t} \right) \; \textrm{d}a \end{align}

According to the split of the stress vector this can be further developed by

1.1.1. Adiabatic and no change of mass flow internal energy

\begin{align} \int_{\partial \mathcal{R}_c} \left( \frac{ 1 }{ 2 }\;\rho \; \textbf{v} \cdot \textbf{v} \right) \;\textbf{v}\cdot \textbf{n}\;\textrm{d}a = - \int_{ \partial\mathcal{R}_c} p\;\textbf{v} \cdot\textbf{n} \; \textrm{d}a + \int_{ \partial\mathcal{R}_c} \textbf{v} \cdot\textbf{t}_T \; \textrm{d}a \end{align}

Defining the last term as the extracted or applied power of the system, e.g. power extracted by a rotating shaft, \eqref{eq:j1soax99c1}

\begin{align} P := \int_{ \partial\mathcal{R}_c} \textbf{v} \cdot\textbf{t}_T \; \textrm{d}a \end{align}

then

\begin{align} \int_{\partial \mathcal{R}_c} \left( p + \frac{ 1 }{ 2 }\;\rho \; \textbf{v} \cdot \textbf{v} \right) \;\textbf{v}\cdot \textbf{n}\;\textrm{d}a = P \end{align}

Links from this note

Balance of Mechanical Energy
✨ 19.9.2024, 21:27
✏️ 20.8.2025, 17:07