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

False — it is perfectly correct for a fixed lump of matter, and the SFEE is derived from it. It is just inconvenient for open systems because we then have to add flow work back in by hand; enthalpy hides that work for us.

False — $pv$ is work done at the boundary to push the packet across a port; it is not stored internal energy. It only becomes bookkeeping-convenient because it always pairs with the stored $u$ to give $h$.

True — $h_0=h+\frac{1}{2}{V}^2$ stays constant, so as the flow accelerates ($V\uparrow$) the static enthalpy $h$ must fall to keep the sum fixed.

True — for a gas, $gz$ over any realistic height is tiny next to $h$ and $\frac{1}{2}{V}^2$, so it is standard to neglect it; for a liquid over large height it may matter.

False — only when $q=0$, $w_s=0$ and $\Delta (\frac{1}{2}{V}^2+gz)=0$. A turbine extracts $w_s>0$, a heater adds $q$, a nozzle trades $h$ for KE — in all of these $h$ changes.

True — since $h=c_{p}T$ with $c_p$ constant, $h$ and $T$ move together; a drop in $h$ is a drop in $T$.

False — flow work $pv$ is already absorbed inside $h$; only shaft work $w_s$ appears explicitly. Listing $pv$ again double-counts it.

True — with $q=0$ the balance gives $w_s=h_1-h_2$; work is done on the gas ($w_s<0$), so $h_2>h_1$ and the gas heats up even without heat addition.

The internal energy $u$ should be enthalpy $h=u+pv$; the writer dropped the flow-work terms $p_1{v}_1$ and $p_2{v}_2$ that turn $u$ into $h$.

With $h_2>h_1$ the value $h_1-h_2$ is negative, so $w_s<0$; the negative sign is exactly what signals work supplied in under the work-out convention.

In an adiabatic nozzle $h+\frac{1}{2}{V}^2$ is constant, so gaining KE costs enthalpy — the air is actually cooler at the fast exit, not hotter.

The nose stops the flow, so the probe reads (nearly) the stagnation temperature $T_0=T+V^2/2c_p$, which is higher than $T$ — this is aerodynamic heating.

The push in at the inlet is work done on the system, so it enters the balance as $-p_1{v}_1$ (equivalently it adds energy to the packet); the writer flipped the sign.

Steady flow with one inlet and one outlet forces ${\dot{m}}_{in}={\dot{m}}_{out}$ by mass conservation (Conservation of mass — continuity equation); a narrower pipe changes velocity, not mass flow rate.

Because moving a packet of specific volume $v$ across a face at pressure $p$ inevitably costs $pv$ of pushing work, and that packet also carries its stored $u$ — the two are inseparable at any port, so nature "pre-bundles" them as $h=u+pv$.

Because defining $h=u+pv$ absorbs the unavoidable flow work, leaving the SFEE clean with no explicit $pv$ terms — the maths mirrors the physics that flow work is always present.

Because energy conservation is only guaranteed for a fixed chunk of matter; we follow one travelling lump (closed) and only afterward reinterpret the result for the fixed control volume.

Because for a perfect gas $h=c_{p}T$, so conserving $h+\frac{1}{2}{V}^2$ becomes a direct trade between $T$ and $V^2$ — every bit of kinetic energy is borrowed from thermal energy.

The convention takes $w_s$ as work out; a turbine delivers work out ($w_s>0$), while a compressor consumes work, i.e. work out is negative ($w_s<0$).

It gives $h_1=h_2$ — enthalpy is conserved even though pressure drops sharply; this is the defining feature of a throttling (isenthalpic) process.

It equals the stagnation enthalpy $h_0$ itself, since $h_0=h+\frac{1}{2}{V}^2$ and the velocity term vanishes — the flow's full energy shows up as static enthalpy.

It reduces to $h\approx \text{const}$, and in the incompressible low-speed limit it recovers Bernoulli's equation as the pressure–velocity trade.

Approximately, yes — "adiabatic" in fast flow means there is no time for meaningful heat exchange per unit mass, so $q\approx 0$ is a good model even without perfect insulation.

Yes along one adiabatic, no-shaft streamline the total enthalpy is conserved; the single-inlet SFEE just needs care to sum each stream's $\dot{m}(h+\frac{1}{2}{V}^2)$ separately.

It stays constant — since $h_0=c_p{T}_0$ and $h_0$ is conserved, the stagnation temperature is unchanged even as static $T$ and $V$ swap (Nozzles and diffusers).