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

Intuition How to use this page

Every problem below is stated cleanly, then hidden behind a collapsible Solution callout so you can test yourself first. Problems climb five levels: L1 Recognition (just spot the right term), L2 Application (plug into the SFEE), L3 Analysis (rearrange, chase a sign), L4 Synthesis (combine SFEE with another law), L5 Mastery (a full multi-step aerodynamics scenario). Everything here rests on the parent note SFEE parent note.

The one equation we will keep reaching for (per unit mass): $h_{1} + \frac{1}{2} V_{1}^{2} + g z_{1} + q = h_{2} + \frac{1}{2} V_{2}^{2} + g z_{2} + w_{s}, h = u + p v, h = c_{p} T (perfect gas)$ Symbols: $h$ = enthalpy (J/kg), $V$ = flow speed (m/s), $z$ = height (m), $g = 9.81 m/s^{2}$ , $q$ = heat added per kg, $w_{s}$ = shaft work out per kg, $c_{p}$ = specific heat at constant pressure. Unless stated, use $c_{p} = 1005 J/(kg\cdotK)$ for air.

Level 1 — Recognition

Recall Solution 1.1

WHAT: we want the flow work per unit mass. WHY: to move volume $V_{vol} = v$ (per unit mass) across a face held at pressure $p$ , force $\times$ distance gives pressure $\times$ volume $= p v$ . $flow work per unit mass = p v$ It hides inside enthalpy, because $h = u + p v$ . That is the whole reason $h$ (not $u$ ) rules open-system energy balances — see Closed-system first law of thermodynamics for the $u$ -only version that flow outgrows.

Recall Solution 1.2

WHAT: cross out $q$ , $w_{s}$ , and $g z$ . WHY each dies: adiabatic $\Rightarrow q = 0$ ; no shaft/blade $\Rightarrow w_{s} = 0$ ; a gas is light, so $g z$ over a nozzle length is negligible next to $h \sim 1 0^{5}$ J/kg. Survivors: $h_{1} + \frac{1}{2} V_{1}^{2} = h_{2} + \frac{1}{2} V_{2}^{2} ⟺ h + \frac{1}{2} V^{2} = h_{0} = const .$ This constant $h_{0}$ is stagnation enthalpy, the star of Stagnation properties & isentropic relations.

Level 2 — Application

Recall Solution 2.1

WHAT: apply the reduced SFEE from Ex. 1.2 with $h = c_{p} T$ . $\frac{1}{2} V_{2}^{2} = c_{p} (T_{1} - T_{2}) + \frac{1}{2} V_{1}^{2}$ WHY: the enthalpy drop $c_{p} (T_{1} - T_{2})$ is exactly the energy that reappears as kinetic energy — that is what a nozzle does (converts thermal to kinetic; see Nozzles and diffusers). $\frac{1}{2} V_{2}^{2} = 1005 (150) + \frac{1}{2} (20)^{2} = 150750 + 200 = 150950$ $V_{2} = 2 (150950) = 301900 \approx 549.5 m/s .$ The temperature fell while the speed rose — energy was borrowed from enthalpy.

Recall Solution 2.2

WHAT: SFEE with $q = 0$ , KE and PE $\approx 0$ : $h_{1} = h_{2} + w_{s} \Rightarrow w_{s} = h_{1} - h_{2} .$ WHY the direction: $w_{s}$ is defined as work out; a turbine delivers work, so we expect $w_{s} > 0$ . $w_{s} = 3200 - 2600 = 600 kJ/kg > 0. ✓$ Positive, exactly as a work-producing device should be.

Level 3 — Analysis

Recall Solution 3.1

WHAT: same reduced SFEE, but now solve for temperature: $c_{p} T_{1} + \frac{1}{2} V_{1}^{2} = c_{p} T_{2} + \frac{1}{2} V_{2}^{2} \Rightarrow T_{2} = T_{1} + \frac{V _{1}^{2} - V _{2}^{2}}{2 c _{p}} .$ WHY: a diffuser slows flow, so KE drops and that energy pours back into enthalpy — temperature should rise. $T_{2} = 300 + \frac{25 0 ^{2} - 5 0 ^{2}}{2 ( 1005 )} = 300 + \frac{62500 - 2500}{2010} = 300 + \frac{60000}{2010} = 300 + 29.85$ $T_{2} \approx 329.9 K$ $Δ T = + 29.9 K > 0$ — the gas heats up as it decelerates, the mirror image of the nozzle in Ex. 2.1.

Recall Solution 3.2

WHAT: SFEE $\to h_{1} = h_{2} + w_{s} \Rightarrow w_{s} = c_{p} (T_{1} - T_{2})$ . WHY watch the sign: $w_{s}$ is work out. A compressor consumes work, so the machine should report a negative $w_{s}$ . $w_{s} = 1005 (290 - 460) = 1005 (- 170) = - 170850 J/kg = - 170.85 kJ/kg .$ Negative $\Rightarrow$ we must supply $170.85 kJ/kg$ . The convention caught the direction for us — no need to memorise "compressor takes work," the algebra says it.

Level 4 — Synthesis

Recall Solution 4.1

WHAT: first get density from the ideal-gas law, then feed continuity. WHY two laws: SFEE gave us the speed; to turn speed into a flow rate we need how much mass is packed per cubic metre ( $ρ$ ) and how wide the throat is ( $A$ ) — that is Conservation of mass — continuity equation.

Step 1 — density. $ρ_{2} = \frac{p _{2}}{R T _{2}} = \frac{80000}{287 \times 350} = \frac{80000}{100450} \approx 0.7964 kg/m^{3} .$ Step 2 — continuity. $\overset{m}{˙} = ρ_{2} A_{2} V_{2} = 0.7964 \times 0.01 \times 549.5 \approx 4.376 kg/s .$ $\overset{m}{˙} \approx 4.38 kg/s$ SFEE (energy) and continuity (mass) are the two pillars — you almost always use them together.

Recall Solution 4.2

WHAT: keep $q$ this time — it does not vanish. $c_{p} T_{1} + \frac{1}{2} V_{1}^{2} + q = c_{p} T_{2} + \frac{1}{2} V_{2}^{2} .$ WHY: we are told heat is added, so $q \neq = 0$ ; the reduced adiabatic form would be wrong here. Solve for $T_{2}$ : $T_{2} = T_{1} + \frac{q + \frac{1}{2} ( V _{1}^{2} - V _{2}^{2} )}{c _{p}} = 300 + \frac{50000 + \frac{1}{2} ( 1600 - 3600 )}{1005} .$ $= 300 + \frac{50000 - 1000}{1005} = 300 + \frac{49000}{1005} = 300 + 48.76.$ $T_{2} \approx 348.8 K$ Almost all the heat went into raising $T$ ; the tiny KE change ( $- 1000$ J/kg) barely dented it.

Level 5 — Mastery

Recall Solution 5.1

WHAT & WHY — part (a): "brought adiabatically to rest" is exactly the definition of the stagnation state, so $h + \frac{1}{2} V^{2} = h_{0}$ , i.e. $c_{p} T + \frac{1}{2} V^{2} = c_{p} T_{0}$ : $T_{0} = T + \frac{V ^{2}}{2 c _{p}} = 220 + \frac{56 0 ^{2}}{2 ( 1005 )} = 220 + \frac{313600}{2010} = 220 + 156.02.$ $T_{0} \approx 376.0 K$ The static air is heated 156 K just by being stopped — that is aerodynamic heating.

Part (b) — Mach number. See Speed of sound and Mach number. First the local sound speed: $a = γ R T = 1.4 \times 287 \times 220 = 88396 \approx 297.3 m/s .$ $M = \frac{V}{a} = \frac{560}{297.3} \approx 1.884.$ Supersonic, as the numbers hinted.

Part (c) — verify the compact form. Using $c_{p} = \frac{γ R}{γ - 1}$ : $\frac{V ^{2}}{2 c _{p} T} = \frac{V ^{2} ( γ - 1 )}{2 γ R T} = \frac{γ - 1}{2} \cdot \frac{V ^{2}}{γ R T} = \frac{γ - 1}{2} M^{2} .$ So $\frac{T _{0}}{T} = 1 + \frac{V ^{2}}{2 c _{p} T} = 1 + \frac{γ - 1}{2} M^{2}$ . Numerically: $1 + \frac{0.4}{2} (1.884)^{2} = 1 + 0.2 (3.549) = 1.710, \frac{T _{0}}{T} = \frac{376.0}{220} = 1.709. ✓$ The energy equation and the Mach-number picture agree — the figure below shows how the same $h_{0}$ splits between "static heat" and "motion."

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

Recall Solution 5.2

WHAT — (a) speed from SFEE. Chamber is a reservoir, so $V_{1} \approx 0$ and $h_{1} = h_{0}$ : $\frac{1}{2} V_{2}^{2} = c_{p} (T_{1} - T_{2}) = 1005 (400) = 402000 \Rightarrow V_{2} = 804000 \approx 896.7 m/s .$ WHY: a huge enthalpy drop ( $400$ K worth) converts almost entirely into kinetic energy — that is thrust in the making (see Nozzles and diffusers).

(b) density from the ideal-gas law: $ρ_{2} = \frac{p _{2}}{R T _{2}} = \frac{101000}{287 \times 500} = \frac{101000}{143500} \approx 0.7038 kg/m^{3} .$

(c) mass flow from continuity: $\overset{m}{˙} = ρ_{2} A_{2} V_{2} = 0.7038 \times 0.02 \times 896.7 \approx 12.62 kg/s .$

(d) momentum flux (dominant thrust contribution, ignoring pressure-area term): $\overset{m}{˙} V_{2} = 12.62 \times 896.7 \approx 11317 N \approx 11.3 kN .$ Every step used exactly one conservation law: energy for speed, mass for flow rate, momentum for thrust — the three-legged stool of compressible flow.

Wrap-up recall

Recall One-line takeaways per level

L1 — flow work $p v$ hides inside $h$ ; $w_{s}$ stays outside. ::: Never double-count $p v$ . L2 — reduced SFEE $h + \frac{1}{2} V^{2} =$ const turns temperature drop into speed. ::: Nozzle cools, speeds up. L3 — one fixed equation $w_{s} = h_{1} - h_{2}$ lets the sign name the device. ::: Negative $w_{s}$ = work in. L4 — pair SFEE (energy) with continuity (mass) $\overset{m}{˙} = ρ A V$ . ::: Speed alone isn't flow rate. L5 — reservoir ⇒ $h_{1} = h_{0}$ ; energy→speed, mass→ $\overset{m}{˙}$ , momentum→thrust. ::: Three laws, one nozzle.

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