Review of thermodynamics applied to flow — first law for open systems: Worked examples

Recall the tool we keep reusing — the Steady-Flow Energy Equation (SFEE), per unit mass:

The scenario matrix

Every flow-energy problem is one (or a combination) of these cells. Each worked example is tagged with the cell it hits.

#	Cell class	What is special	Term that dominates / vanishes	Example
A	Nozzle, adiabatic, no shaft	$q = 0, w_{s} = 0$ ; $h \to K E$	$g z \approx 0$ , $T$ falls	Ex 1
B	Compressor, shaft work in	$w_{s} < 0$ (we supply it)	KE $\approx 0$ , $q = 0$	Ex 2
C	Turbine, shaft work out	$w_{s} > 0$ (we harvest it)	KE $\approx 0$ , $q = 0$	Ex 3
D	Heat exchanger, no work	$w_{s} = 0$ , $q \neq = 0$	$q \to Δ h$ (sign of $q$ )	Ex 4
E	Stagnation / zero-velocity limit	$V_{2} = 0$ (brought to rest)	defines $h_{0}, T_{0}$	Ex 5
F	Diffuser (nozzle run backwards)	$V$ falls, $T$ rises	sign flip of Cell A	Ex 6
G	Real-world word problem	strip words → find terms	liquid pump, $v Δ p$	Ex 7
H	Exam twist: hidden term	one "small" term is NOT small	must NOT drop $q$ or KE	Ex 8

Two limiting checks live inside the cells: zero velocity (Cell E, $V \to 0$ ) and large velocity (Cell A/F, where KE competes with $h$ ).

Cell A — Adiabatic nozzle (enthalpy → kinetic energy)

Worked example Ex 1 (Cell A) — Nozzle exit speed

Air enters a nozzle at $T_{1} = 900 K$ , $V_{1} = 50 m/s$ , and exits at $T_{2} = 500 K$ . Adiabatic, no shaft. Find $V_{2}$ .

Forecast: temperature drops a lot ( $400$ K). Guess: does $V_{2}$ come out closer to $200$ m/s or $900$ m/s?

Step 1 — Kill the dead terms. $q = 0$ (adiabatic), $w_{s} = 0$ (no blade), $g z \approx 0$ (gas, small height). SFEE becomes $h_{1} + \frac{1}{2} V_{1}^{2} = h_{2} + \frac{1}{2} V_{2}^{2}$ . Why this step? A nozzle has no moving part and is too fast to swap heat, so only enthalpy and KE are left to trade.

Step 2 — Substitute $h = c_{p} T$ and solve for $V_{2}$ . $\frac{1}{2} V_{2}^{2} = c_{p} (T_{1} - T_{2}) + \frac{1}{2} V_{1}^{2} = 1005 (400) + \frac{1}{2} (50)^{2} = 402000 + 1250 = 403250$ Why this step? Falling $T$ releases enthalpy; that released energy has nowhere to go but KE.

Step 3 — Take the root. $V_{2} = 2 (403250) = 806500 \approx 898 m/s .$

Verify: Units — $c_{p} Δ T$ is J/kg = m²/s², so $m^{2} / s^{2} =$ m/s. ✓ The inlet KE ( $1250$ J/kg) is tiny next to the enthalpy release ( $402, 000$ J/kg), so the answer is essentially set by $Δ T$ — sensible. $V_{2} \approx 900$ m/s, matching the higher guess.

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

Cell B — Compressor (shaft work in, $w_{s} < 0$ )

Worked example Ex 2 (Cell B) — Work to run a compressor

Air is compressed adiabatically from $T_{1} = 300 K$ to $T_{2} = 560 K$ . Velocities negligible. Find $w_{s}$ per kg. What sign, and what does it mean?

Forecast: we are heating the air by squeezing it. Do we get work out, or must we pay work in?

Step 1 — Simplify SFEE. $q = 0$ , KE $\approx 0$ , $g z \approx 0$ : $h_{1} = h_{2} + w_{s}$ . Why this step? No heat, no motion energy — enthalpy change balances shaft work alone.

Step 2 — Solve. $w_{s} = h_{1} - h_{2} = c_{p} (T_{1} - T_{2}) = 1005 (300 - 560) = - 261300 J/kg = - 261.3 kJ/kg .$ Why this step? Direct substitution; the sign is what carries the physics.

Step 3 — Read the sign. $w_{s} < 0$ : our convention has $w_{s}$ = work out. Negative means work goes in — we must supply $261.3$ kJ per kg. ✓

Verify: Enthalpy went up ( $T$ rose $260$ K). Energy cannot appear from nowhere, so something had to feed it in — the compressor. Sign convention caught it automatically.

Cell C — Turbine (shaft work out, $w_{s} > 0$ )

Worked example Ex 3 (Cell C) — Turbine power output

Hot gas enters a turbine at $T_{1} = 1200 K$ , leaves at $T_{2} = 800 K$ , with mass flow $\overset{m}{˙} = 4 kg/s$ . Adiabatic, KE negligible. Find the shaft work per kg and the total power $\dot{W}_{s}$ .

Forecast: hot gas expands and drives a blade — we harvest energy. Expect $w_{s} > 0$ .

Step 1 — Simplify SFEE (rate form). $q = 0$ , KE $\approx 0$ : $\overset{m}{˙} h_{1} = \overset{m}{˙} h_{2} + \dot{W}_{s}$ . Why this step? A turbine's whole job is shaft work; enthalpy drop supplies it.

Step 2 — Per-unit-mass work. $w_{s} = h_{1} - h_{2} = c_{p} (T_{1} - T_{2}) = 1005 (1200 - 800) = 402000 J/kg = 402 kJ/kg .$ Why this step? Same algebra as the compressor, but $T$ falls, so the sign flips positive.

Step 3 — Multiply by mass flow. $\dot{W}_{s} = \overset{m}{˙} w_{s} = 4 \times 402000 = 1608000 W = 1.608 MW .$ Why this step? Power = energy per kg × kg per second.

Verify: $w_{s} > 0$ — work out, correct for a turbine. Units: (J/kg)(kg/s)=W. ✓ Compare with Ex 2: same $∣ c_{p} Δ T ∣$ arithmetic, opposite sign — the machine's direction lives entirely in the sign of $Δ T$ .

Cell D — Heater with no work (sign of $q$ )

Cell E — Stagnation limit ( $V \to 0$ )

Worked example Ex 5 (Cell E) — Temperature at a nose probe

A fast air stream has $T = 230 K$ and $V = 680 m/s$ . A probe brings it adiabatically to rest. What temperature $T_{0}$ does the probe read?

Forecast: stopping the air converts all its KE into enthalpy → the probe reads hotter than the stream. By how much?

Step 1 — Apply SFEE between "flowing" (state 1) and "at rest" (state 0). Adiabatic, no shaft: $h + \frac{1}{2} V^{2} = h_{0} + \frac{1}{2} (0)^{2}$ . Why this step? "Brought to rest" is exactly the $V_{2} = 0$ degenerate case that defines stagnation enthalpy $h_{0}$ .

Step 2 — Convert to temperature. $T_{0} = T + \frac{V ^{2}}{2 c _{p}} = 230 + \frac{68 0 ^{2}}{2 ( 1005 )} = 230 + \frac{462400}{2010} = 230 + 230.05 = 460.05 K .$ Why this step? All the KE ( $\frac{1}{2} V^{2}$ ) is deposited into enthalpy, raising $T$ .

Verify: The temperature rise $\frac{V ^{2}}{2 c _{p}} \approx 230$ K is huge — this is aerodynamic heating. At $V = 0$ the formula gives $T_{0} = T$ (no rise), the sanity limit. ✓ Links to Stagnation properties & isentropic relations and Speed of sound and Mach number.

Cell F — Diffuser (nozzle run backwards)

Worked example Ex 6 (Cell F) — Diffuser temperature rise

A diffuser slows air from $V_{1} = 800 m/s$ to $V_{2} = 100 m/s$ ; inlet $T_{1} = 250 K$ . Adiabatic, no shaft. Find $T_{2}$ .

Forecast: a diffuser slows the flow, so KE is spent → enthalpy (and $T$ ) must rise. Opposite of the nozzle.

Step 1 — Simplify SFEE. $q = 0, w_{s} = 0, g z \approx 0$ : $h_{1} + \frac{1}{2} V_{1}^{2} = h_{2} + \frac{1}{2} V_{2}^{2}$ . Why this step? Same energy trade as a nozzle, but the velocity term shrinks instead of grows.

Step 2 — Solve for $T_{2}$ . $c_{p} T_{2} = c_{p} T_{1} + \frac{1}{2} (V_{1}^{2} - V_{2}^{2})$ $T_{2} = 250 + \frac{80 0 ^{2} - 10 0 ^{2}}{2 ( 1005 )} = 250 + \frac{640000 - 10000}{2010} = 250 + 313.4 = 563.4 K .$ Why this step? Lost KE ( $V$ drops) is deposited into enthalpy → $T$ climbs.

Verify: $V$ fell, $T$ rose — mirror image of Ex 1's nozzle, confirming Cell A/F symmetry. Compare to Nozzles and diffusers. If $V_{2} = 0$ this reduces to Ex 5's stagnation formula — the two cells connect at the zero-velocity limit. ✓

Cell G — Real-world word problem (a liquid, not a gas)

Worked example Ex 7 (Cell G) — Water pump lifting to a rooftop tank

A pump moves water ( $v = 0.001 m^{3} / kg$ , i.e. density $1000 kg/m^{3}$ ) from a ground reservoir to a tank $z_{2} - z_{1} = 40 m$ higher, raising pressure from $p_{1} = 100 kPa$ to $p_{2} = 500 kPa$ . Water is nearly incompressible and stays at the same temperature; KE change and heat are negligible. Find the shaft work per kg.

Forecast: for a liquid, "enthalpy change" is really just the pressure-push $v Δ p$ (since $u$ barely changes at constant $T$ ), plus the lift $g Δ z$ . Expect $w_{s} < 0$ (we pay).

Step 1 — Start from full SFEE. $q = 0$ , $Δ K E \approx 0$ : $h_{1} + g z_{1} = h_{2} + g z_{2} + w_{s}$ . Why this step? Real pumps do change height, so $g z$ is NOT droppable here (unlike gases).

Step 2 — For an incompressible liquid at constant $T$ , $Δ u \approx 0$ , so $h_{2} - h_{1} = (u_{2} + p_{2} v) - (u_{1} + p_{1} v) \approx v (p_{2} - p_{1}) .$ Why this step? With $u$ frozen, the only enthalpy change is the flow-work term $v Δ p$ . This is the seed of Bernoulli equation as low-speed limit of SFEE.

Step 3 — Assemble and solve. $w_{s} = (h_{1} - h_{2}) + g (z_{1} - z_{2}) = - v (p_{2} - p_{1}) - g (z_{2} - z_{1})$ $= - 0.001 (500000 - 100000) - 9.81 (40) = - 400 - 392.4 = - 792.4 J/kg .$ Why this step? Both raising pressure and raising height cost work; both appear with a minus.

Verify: $w_{s} < 0$ — work in, correct for a pump. ✓ Split check: pressure term $= 400$ J/kg, lift term $= 392.4$ J/kg — comparable, so dropping either would be wrong. Units: $v Δ p = m^{3} / kg \cdot Pa = J/kg$ . ✓

Cell H — Exam twist (a "small" term is not small)

Worked example Ex 8 (Cell H) — Cooled nozzle: don't drop

$q$ ! A high-temperature nozzle is deliberately cooled (film cooling): $T_{1} = 1000 K$ , $V_{1} = 100 m/s$ , exit $V_{2} = 700 m/s$ , and heat $q = - 80 kJ/kg$ is removed. No shaft. Find $T_{2}$ . The trap: a careless student assumes "nozzle ⇒ adiabatic" and drops $q$ .

Forecast: removing heat and speeding up both drain enthalpy, so $T_{2}$ should fall more than an adiabatic nozzle would predict.

Step 1 — Keep every non-zero term. $w_{s} = 0$ , $g z \approx 0$ , but $q = - 80$ kJ/kg stays: $h_{1} + \frac{1}{2} V_{1}^{2} + q = h_{2} + \frac{1}{2} V_{2}^{2} .$ Why this step? The problem explicitly removes heat — "nozzle" does NOT license $q = 0$ here. This is the whole twist.

Step 2 — Solve for $T_{2}$ . $c_{p} T_{2} = c_{p} T_{1} + \frac{1}{2} (V_{1}^{2} - V_{2}^{2}) + q$ $c_{p} T_{2} = 1005 (1000) + \frac{1}{2} (10 0^{2} - 70 0^{2}) + (- 80000)$ $= 1005000 + \frac{1}{2} (10000 - 490000) - 80000 = 1005000 - 240000 - 80000 = 685000$ $T_{2} = \frac{685000}{1005} = 681.6 K .$ Why this step? All three drains (heat out, KE up) subtract from enthalpy.

Step 3 — Show the wrong answer. If you (wrongly) drop $q$ : $c_{p} T_{2} = 1005000 - 240000 = 765000 \Rightarrow T_{2} = 761.2$ K. That's $79.6$ K too high — exactly $q / c_{p}$ . Why show it? To see precisely how much the dropped term was worth.

Verify: Correct $T_{2} = 681.6$ K; the difference from the naive answer is $761.2 - 681.6 = 79.6 K \approx 80000/1005$ . ✓ Lesson: never drop a term the problem hands you a number for.

Active recall

Recall Which cell is which?

Nozzle speeds up, $T$ falls ::: Cell A Compressor, $w_{s} < 0$ (work in) ::: Cell B Turbine, $w_{s} > 0$ (work out) ::: Cell C Duct with heat added, no work ::: Cell D Flow brought to rest ( $V \to 0$ ) defines $T_{0}$ ::: Cell E Diffuser slows flow, $T$ rises ::: Cell F Liquid pump: $Δ h \approx v Δ p$ plus lift $g Δ z$ ::: Cell G Cooled nozzle where you must keep $q$ ::: Cell H

Connections

Review of thermodynamics applied to flow — first law for open systems (index 3.1.1)
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