Subsections

5 Processes

5.1 Control Mass Processes (Closed Systems)

Specific heat and work:

\begin{displaymath}
q_{12} = \frac{Q_{12}}{m}
\quad
w_{12} = \frac{W_{12}}{m}
\qquad
Q_{12} = m\; q_{12}
\quad
W_{12} = m\; w_{12}
\end{displaymath}

Continuity (mass conservation):

\begin{displaymath}
m_1 \left(+ m_{\mbox{\scriptsize added}} \right) = m_2
\end{displaymath}

The first law of thermo (energy conservation):

\begin{displaymath}
E_2 - E_1 = Q_{12} - W_{12}
\qquad (E = U + m {\textstyle \frac12}{\bf V}^2? + m g z?)
\end{displaymath}

Work:

\begin{displaymath}
W_{12} = \int_1^2 p \; {\rm d} V =
\left\{
\begin{arra...
...1}} pV\ln\left(\frac{V_2}{V_1}\right)
\end{array}
\right.
\end{displaymath}

In case of an ideal gas, note that $pV$ can be replaced by $mRT$.

No heat is added or removed if the process is adiabatic.
Exams 3 and final only
More generally:

\begin{displaymath}
Q_{12} =
\left\{
\begin{array}{rl}
\mbox{Adiabatic:}...
...tyle\vphantom{\int_1^2}
T (S_2-S_1)
\end{array}
\right.
\end{displaymath}

The second law for reversible adiabatic (isentropic) processes:

\begin{displaymath}
S_2 = S_1
\end{displaymath}

The second law of thermo for general processes:

\begin{displaymath}
S_{12,\mbox{\scriptsize gen}} + \sum \frac{Q_{k}}{T_{k}}
= S_2 - S_1 \qquad S_{12,\mbox{\scriptsize gen}} \ge 0
\end{displaymath}

End exams 3 and final only

5.2 Rate Equations

The first law as a rate equation:

\begin{displaymath}
\frac{{\rm d}E}{{\rm d}t} = \dot Q - \dot W \qquad
\left...
...bf V}^2}{{\rm d}t}? +
m g \frac{{\rm d}Z}{{\rm d}t}?\right)
\end{displaymath}

Work as a rate equation:

\begin{displaymath}
\dot W = p \dot V
\end{displaymath}

For an ideal gas:

\begin{displaymath}
\frac{{\rm d}u}{{\rm d}t} = C_v \frac{{\rm d}T}{{\rm d}t}
\end{displaymath}

For liquids and solids:

\begin{displaymath}
\dot Q = m C_{(p)} \frac{{\rm d}T}{{\rm d}t}
\end{displaymath}

5.3 Steady State Control Volume Processes

The control volume is assumed to be steady state in all formulae below.

Specific work output and heat added (i.e., per unit mass flowing through):

\begin{displaymath}
w = \frac{\dot W}{\dot m} \quad q = \frac{\dot Q}{\dot m}
\qquad
\dot W = \dot m w \quad \dot Q = \dot m q
\end{displaymath}

In- and outflow velocities and pipe cross-sectional areas:

\begin{displaymath}
\dot m = \dot V/v = A {\bf V}/v \qquad A=\frac\pi 4 D^2
\end{displaymath}

(where ${\bf V}$ is velocity.)

Continuity (mass conservation):

\begin{displaymath}
\sum \dot m_i = \sum \dot m_e
\end{displaymath}

where $\sum$ means sum over all inflow, respectively outflow, points, if there is more than one.

The first law of thermo (energy conservation):

\begin{displaymath}
\sum \dot m_e \left(h_e+ \frac12 {\bf V}_e^2 + g z_e\right...
...t(h_i+ \frac12 {\bf V}_i^2 + g z_i\right) =
\dot Q - \dot W
\end{displaymath}

The kinetic energy and potential energy terms are often ignored. Devices without moving parts do not do work, $\dot W = 0$. Adiabatic devices have no heat transfer, $\dot Q = 0$.

Exams 3 and final only

The second law of thermo for a reversible adiabatic (isentropic) process inside a single entrance and exit CV:

\begin{displaymath}
s_e = s_i
\end{displaymath}

The second law of thermo for a reversible isothermal process inside a single entrance and exit CV:

\begin{displaymath}
q = T \left(s_e - s_i\right)
\end{displaymath}

The second law for a general control volume:

\begin{displaymath}
\sum \dot m_e s_e - \sum \dot m_i s_i
= \sum \frac{\dot ...
...riptsize gen}}
\qquad \dot S_{\mbox{\scriptsize gen}} \ge 0
\end{displaymath}

Specific work done during a reversible process inside a single entrance and exit CV:

\begin{displaymath}
w + {\textstyle \frac12} {\bf V}_e^2? - {\textstyle \frac12} {\bf V}_i^2?
+ g Z_e? - g Z_i? = - \int_i^e v \; {\rm d} p
\end{displaymath}

where

\begin{displaymath}
- \int_i^e v \; {\rm d} p =
\left\{
\begin{array}{rl}
...
...- p v \ln\left(\frac{p_e}{p_i}\right)
\end{array}
\right.
\end{displaymath}

In case of an ideal gas, note that $pv$ can be replaced by $RT$.

Special case of (a) ideal gas, (2) polytropic, $n\ne 1$, (including isentropic constant specific heats, then $n=k$,) and (3) reversible:

\begin{displaymath}
- \int_i^e v \; {\rm d} p =
\frac{nRT_1}{1-n}\left[\left(\frac{P_2}{P_1}\right)^{(n-1)/n}-1\right]
\end{displaymath}

Special case of (a) ideal gas, (2) constant specific heats, (3) polytropic, $n=1$ (isothermal). and (4) reversible:

\begin{displaymath}
- \int_i^e v \; {\rm d} p =
-RT\ln\frac{P_2}{P_1}
\end{displaymath}

Adiabatic turbine and compressor/pump efficiencies:

\begin{displaymath}
\eta_{\mbox{\scriptsize T}} = \frac{w_a}{w_s}
\qquad
\eta_{\mbox{\scriptsize C}} = \frac{w_s}{w_a}
\end{displaymath}

where subscript $s$ indicates an ideal isentropic device with the same entrance conditions and exit pressure and $a$ the actual device.

End exams 3 and final only