Review of thermodynamics applied to flow — first law for open systems: Visual walkthrough

Before frame one, three plain-word promises about the symbols you will meet:

Everything below is built only from those seven ideas plus "energy is conserved." Nothing else is assumed.

Step 1 — Corner the honest law: closed-system energy

WHAT. We refuse, for now, to think about pipes. We pick a single blob $δ m$ of gas and follow it. Because it is always the same matter, we may use the one law we trust completely: $change in the blob’s total energy Δ E = heat in Q - all work done b y the blob W_{total}$

WHY. Energy conservation is only simple for a fixed chunk of matter — a closed system. A pipe has mass rushing through it (an open system), which is messy. So we start honest and small, then upgrade.

PICTURE. Look at figure s01: the blob is drawn twice — grey where it starts (just outside inlet 1), magenta where it ends (just outside outlet 2). The machine (violet box) never moves; only our blob does. That is the whole point of "follow the matter."

Figure — Review of thermodynamics applied to flow — first law for open systems

Step 2 — List every kind of energy the blob carries

WHAT. Per kilogram, the blob stores three energies: $e = internal (heat/jiggle) u + kinetic (motion) \frac{1}{2} V^{2} + potential (height) g z$ So $Δ E = δ m (e_{2} - e_{1})$ — the difference between end and start.

WHY. The law in Step 1 needs the change in the blob's energy. We can only compute a change if we first name all the energies that can change. These three are the complete list a moving fluid parcel can store.

PICTURE. Figure s02 stacks the three energies as coloured bars for the blob at state 1 and at state 2. The total height of the stack is what Step 1 tracks; the shape of the stack is free to shift between bars, as long as heat and work account for any change in total height.

Step 3 — The subtle work: pushing the blob through the doors

WHAT. The word $W_{total}$ hides two different jobs. First job: just to enter at 1 and leave at 2, the blob has to be shoved across the boundary. Shoving a volume across a face at pressure $p$ costs work = force × distance = $p \times (volume) = p (v δ m)$ .

At the inlet, the upstream fluid pushes the blob in. That is work done on us → energy gained → $+ p_{1} v_{1}$ on our side.
At the outlet, our blob must push the downstream fluid out of the way. Work done by us → energy spent → $- p_{2} v_{2}$ on our side.

WHY. This "flow work" $p v$ is not optional and not the same as useful work — it is the unavoidable admission fee for crossing a boundary. Miss it and the energy books never balance. This single insight is the reason enthalpy exists.

PICTURE. Figure s03: two doorways. At the inlet an orange piston of the upstream gas pushes the blob in through a slab of length $v_{1}$ (work in). At the outlet the blob pushes a magenta slab of length $v_{2}$ out (work out). The arrows show who pushes whom.

Step 4 — Separate flow work from shaft work

WHAT. The other job inside $W_{total}$ is the shaft work $w_{s}$ : a turbine blade or fan actually connected to the outside world. So the total work per kilogram splits cleanly: $w_{total} = useful, via a blade w_{s} + flow work: out minus in (p_{2} v_{2} - p_{1} v_{1})$

WHY. These are physically different taps. Flow work is the price of entry at the ports; shaft work is real usable energy pulled out through a spinning axle. Confusing them is the classic double-counting error. Keeping them apart is what makes the final formula honest.

PICTURE. Figure s04 labels the two taps on the violet machine: orange doorway arrows (flow work, at the ports) versus a violet spinning shaft on top (shaft work). Same machine, two completely separate energy channels.

Step 5 — Assemble the raw balance

WHAT. Put Steps 1–4 together. Divide the closed-system law by $δ m$ so everything is per kilogram ( $q = Q / δ m$ ): $Δ E / δ m (e_{2} - e_{1}) = q - [w_{s} + (p_{2} v_{2} - p_{1} v_{1})]$ Expand $e = u + \frac{1}{2} V^{2} + g z$ : $(u_{2} + \frac{1}{2} V_{2}^{2} + g z_{2}) - (u_{1} + \frac{1}{2} V_{1}^{2} + g z_{1}) = q - w_{s} - (p_{2} v_{2} - p_{1} v_{1})$

WHY. Nothing new — just substituting Step 2 and Step 4 into Step 1. This is the "before cleanup" equation: everything true is here, but the $p v$ terms are scattered on the wrong side, cluttering the view.

PICTURE. Figure s05 shows the balance as a see-saw: energies of state 1 (plus heat in) on the left pan, energies of state 2 (plus work out) on the right pan, with the two loose $p v$ flow-work blocks sitting awkwardly off to the side — begging to be tidied.

Step 6 — The magic move: glue $p v$ onto $u$ to make enthalpy

WHAT. Slide the flow-work terms across the equals sign so $p_{1} v_{1}$ joins $u_{1}$ , and $p_{2} v_{2}$ joins $u_{2}$ : $\equiv h_{1} (u_{1} + p_{1} v_{1}) + \frac{1}{2} V_{1}^{2} + g z_{1} + q = \equiv h_{2} (u_{2} + p_{2} v_{2}) + \frac{1}{2} V_{2}^{2} + g z_{2} + w_{s}$ The pair $u + p v$ appears so reliably that we give it a name:

WHY. Every time fluid crosses a boundary, $p v$ shows up welded to $u$ . Instead of writing them separately forever, we define their sum once. The equation instantly looks clean — the flow work never has to be mentioned again; it's hiding safely inside $h$ .

PICTURE. Figure s06: the loose $p v$ blocks from Step 5 snap onto the $u$ bars, forming taller "enthalpy" columns $h_{1}$ and $h_{2}$ . Same energy, tidier accounting.

Step 7 — Edge case: the aerodynamic collapse (no shaft, no heat, gas)

WHAT. For a nozzle, diffuser, or free stream we switch off three terms:

$w_{s} = 0$ — no blades, just a duct.
$q = 0$ — the flow is so fast there is no time for heat to leak (adiabatic).
$g z$ negligible — a light gas barely notices height.

The SFEE shrinks to $h + \frac{1}{2} V^{2} = h_{0} = const, h_{0} = ==stagnation enthalpy==$ and with $h = c_{p} T$ for a calorically perfect gas: $c_{p} T + \frac{1}{2} V^{2} = c_{p} T_{0} \Rightarrow \frac{T _{0}}{T} = 1 + \frac{V ^{2}}{2 c _{p} T} .$

WHY. These aren't approximations pulled from air — each dropped term corresponds to a physical feature the device lacks. What survives is the master trade: enthalpy $\leftrightarrow$ kinetic energy. Speeding the gas up must cool it down. (Slowing it down — as at a probe nose — heats it: that is the $T_{0}$ of a Mach-2 stream.)

PICTURE. Figure s07: as the duct narrows and $V$ rises left-to-right, the orange kinetic bar grows while the violet enthalpy bar shrinks by exactly the same amount — their sum (the dashed line $h_{0}$ ) stays flat. Temperature, riding on $h$ , falls.

Step 8 — Degenerate checks: does the formula survive extreme inputs?

WHAT. Push the SFEE to its corners and confirm it stays sensible.

No flow at all ( $V_{1} = V_{2} = 0$ , no heat, one port sealed): SFEE gives $h_{1} = h_{2} + w_{s}$ . This is just a stationary closed device balance — nothing crossed, so it reduces to the closed-system law in enthalpy form. ✓
Incompressible, slow, no work, no heat ( $v =$ const, small $V$ ): with $h = u + p v$ and $u$ near-constant, the balance becomes $p_{1} v + \frac{1}{2} V_{1}^{2} + g z_{1} = p_{2} v + \frac{1}{2} V_{2}^{2} + g z_{2}$ — the Bernoulli equation. ✓
Turbine vs compressor (signs): a turbine gives work → $w_{s} > 0$ ; a compressor absorbs work → $w_{s} < 0$ . In Example 2 of the parent, $w_{s} = c_{p} (T_{1} - T_{2}) = 1005 (300 - 500) = - 201 kJ/kg$ , correctly negative. ✓

WHY. A law you trust must degrade gracefully into the simpler laws you already know at its boundaries. SFEE passing all three checks is your evidence the bundling in Step 6 was legitimate, not a trick.

PICTURE. Figure s08: three mini-panels — a sealed box (→ closed law), a smooth incompressible tube (→ Bernoulli), and a shaft with $\pm$ arrows (→ turbine/compressor sign) — showing the one equation wearing three familiar disguises.

The one-picture summary

The whole film in a single frame: a blob enters at 1, gets pushed in by flow work $p_{1} v_{1}$ , flows through a machine that may add heat $q$ and remove shaft work $w_{s}$ , then pushes its way out at 2 paying flow work $p_{2} v_{2}$ . Bundle each $u + p v$ into $h$ and the balance reads clean: what goes in must come out.

$h_{1} + \frac{1}{2} V_{1}^{2} + g z_{1} + q = h_{2} + \frac{1}{2} V_{2}^{2} + g z_{2} + w_{s}$

Recall Feynman retelling — the whole walkthrough in plain words

We didn't want to think about a whole windy pipe, so we cheated: we glued our eyes to one tiny puff of air and rode along with it. Riding a fixed puff, energy is simple — heat in, minus work it does, equals how much its energy changed (Step 1). The puff carries three kinds of energy: heat-jiggle inside it, running-fast energy, and how-high-up energy (Step 2). Then we noticed the sneaky part: just to squeeze in the front door, the crowd behind shoves the puff (they do work on it); and to get out the back door, the puff shoves the crowd ahead (it does work). That shove-through-the-door cost is "flow work," $p v$ , and it's completely separate from any useful work a fan or turbine blade might add or take (Steps 3–4). We wrote everything down (Step 5) and saw the door-cost $p v$ was always stuck next to the inside energy $u$ — so we just glued them together forever and named the bundle enthalpy, $h = u + p v$ (Step 6). Now the books are tidy: bundle-in + motion + height + heat = bundle-out + motion + height + shaft work. For a plain duct with no blades and no time to lose heat, all that survives is a trade: the puff speeds up only by spending its bundle, so it cools down as it races (Step 7). And when we shove the equation into weird corners — nothing flowing, or slow water, or a blade one way vs the other — it politely turns into the old closed-system law, or Bernoulli, or the right turbine/compressor sign (Step 8). One honest law, many disguises.

Connections

Parent: First Law for Open Systems
Stagnation properties & isentropic relations
Speed of sound and Mach number
Conservation of mass — continuity equation
Closed-system first law of thermodynamics
Bernoulli equation as low-speed limit of SFEE
Nozzles and diffusers