most of the notes on feb 22nd

uni/mmme/1048_thermodynamics_and_fluid_mechanics/fluid_dynamics.md

@@ -384,3 +384,157 @@ Head form:
 \frac{p_s}{\rho g} + z + \frac{v^2}{2g} &= \text{total head} \\
 \frac{p_s}{\rho g} + z &= \text{piezometric head}
 \end{align*}
+
+# Steady Flow Energy Equation (SFEE)
+
+SFEE is a more general equation that can be applied to **any fluid** and also is also takes
+**heat energy** into account.
+This is useful in applications such as a fan heater, jet engines, ICEs, and steam turbines.
+
+The equation deals with 3 types of energy tranfer:
+
+1. Thermal energy transfer (e.g. heat tranfer from central heating to a room)
+2. Work energy transfer (e.g. shaft from car engine that turns wheels)
+3. Energy transfer in fluid flows (e.g. heat energy in a flow, potential energy in a flow, kinetic
+   energy in a flow)
+
+## Derivation of Steady Flow Energy Equation
+
+#### Consider a control volume with steady flows in and out and steady transfers of work and heat.
+
+The properties don't change with time at any any location and are considered uniform over inlet and
+outlet areas $A_1$ and $A_2$.
+
+For steady flow, the mass, $m$, of the fluid **within the control volume** and the total energy, $E$,
+must be constant.
+
+$E$ includes **all forms for energy** but we only consider internal, kinetic, and potential energy.
+
+#### Consider a small time interval $\delta t$.
+
+During $\delta t$, mass $\delta m_1$ enters the control volume and $\delta m_2$ leaves:
+
+![](./images/vimscrot-2022-03-01T22:47:31,932087932+00:00.png)
+
+The specific energy $e_1$ of fluid $\delta m_1$ is the sum of the specific internal energy, specific
+kinetic energy, and specific potential energy:
+
+$$e_1 = u_1 + \frac{v_1^2}{2} gz_1$$
+$$e_2 = u_2 + \frac{v_2^2}{2} gz_2$$
+
+Since the mass is constant in the control volume, $\delta m_1 = \delta m_2$.
+
+#### Applying the First Law of Thermodynamics
+
+The control volume is a system for which $\delta E_1$ is added and $\delta E_2$ is removed::
+
+$$\delta E = \delta E_2 - \delta E_1$$
+
+$E$ is constant so applying the
+[first law of thermodynamics](thermodynamics.html#st-law-of-thermodynamics)
+we know that:
+
+$$\delta Q + \delta W = \delta E$$
+
+We can also say that:
+
+$$\delta E = \delta E_2 - \delta E_1 = \delta m(e_2 - e_1)$$
+
+#### The Work Term
+
+The work term, $\delta W$, is mae up of shaft work **and the work necessary to deform the system**
+(by adding $\delta m_1$ at the inlet and removing $\delta m_2$ at the outlet):
+
+$$\delta W = \delta W_s + \text{net flow work}$$
+
+Work is done **on** the system by the mass entering and **by** the system on the mass leaving.
+
+For example, at the inlet:
+
+![](./images/vimscrot-2022-03-01T22:59:14,129582752+00:00.png)
+
+$$\text{work done on system} = \text{force} \times \text{distance} = p_1A_1\delta x = p_1\delta V_1$$
+
+Knowing this, we can write:
+
+$$\delta W = \delta W_s + (p_1\delta V_1 - p_2\delta V_2)$$
+
+#### Back to the First Law
+
+Substituting these equations:
+
+$$\delta E = \delta E_2 - \delta E_1 = \delta m(e_2 - e_1)$$
+$$\delta W = \delta W_s + (p_1\delta V_1 - p_2\delta V_2)$$
+
+into:
+
+$$\delta Q + \delta W = \delta E$$
+
+gives us:
+
+$$\delta Q + \left[ \delta W_s + (p_1\delta V_1 - p_2\delta V_2)\right] = \delta m (e_2-e_1)$$
+
+Dividing everything by $\delta m$ and with a bit of rearranging we get:
+
+$$q + w_s = e_2-e_1 + \frac{p_2}{\rho_2} - \frac{p_1}{\rho_1}$$
+
+#### Substiute Back for $e$
+
+$$e = u + \frac{v^2}{2} + gz$$
+
+This gives us:
+
+$$q + w_s + \left[ u_2 + \frac{p_2}{\rho_2} + gz_2 + \frac{v_2^2}{2} \right] - \left[ u_1 + \frac{p_1}{\rho_1} + gz_1 + \frac{v_1^2}{2} \right]$$
+
+#### Rearrange and Substitute for Enthalpy
+
+By definition, enthalpy $h = u + pv = u + \frac p \rho$.
+This gives us the equation:
+
+$$q + w_s = (h_2 - h_1) + g(z_2-z_1) + \frac{v_2^2-v_1^2}{2}$$
+
+This equation is in specific energy form.
+
+Multiplying by mass flow rate will give you the power form.
+
+## Application of the Steady Flow Energy Equation
+
+#### Heat Transfer Devices
+
+Like heat exchangers, boilers, condensers, and furnaces.
+
+In this case, $\dot W = 0$, $\delta z ~ 0$, and $\delta v^2 ~ 0$ so the equation can be simplified
+to just
+
+$$\dot Q = \dot m(h_2-h_1) = \dot m c_p(T_2-T_1)$$
+
+#### Throttle Valve
+
+No heat and work transfer.
+Often you can neglect potential and kinetic energy terms, giving you:
+
+$$0 = h_2-h_1)$$
+
+#### Work Transfer Devices
+
+e.g. Turbines, Pumps, Fans, and Compressors
+
+For these there is often no heat transfer ($\dot Q = 0$) and we can neglect potential
+($\delta z ~ 0$) and kinetic ($\delta v^2 ~ 0$) energy terms, giving us the equation
+
+$$\dot W = \dot m (h_2-h_1) = \dot m c_p(T_2-T_1)$$
+
+#### Mixing Devices
+
+e.g. hot and cold water in a shower
+
+In these processes, work and heat transfers are not important and you can often
+neglect potential and kinetic energy terms, giving us the same equation as for the throttle valve
+earlier:
+
+$$0 = h_2-h_1$$
+
+which you may want to write more usefully as:
+
+$$\sum \dot m h_{out} = \sum \dot m h_{in}$$
+