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

WHAT we are doing

HOW: deriving the steady-flow energy equation from scratch

We follow a small mass $\delta m$ as it moves from just outside inlet 1 to just outside outlet 2, and apply the closed-system first law to that travelling lump (it's the same matter throughout, so closed-system law is legal).

Step 1 — Closed-system first law for the lump. $\Delta E = Q - W_{\text{total}}$ Why this step? Energy is conserved for a fixed chunk of matter; that's the only law we're truly sure of, so we start there.

Step 2 — What energy does the lump have? Internal + kinetic + potential, per unit mass: $e = u + \tfrac{1}{2}V^2 + gz$ Why? These are the storable energies the fluid carries; we will need their change between 1 and 2.

Step 3 — Split the work into flow work + shaft work. The total work has two parts:

Flow (displacement) work: the surroundings push the lump in at 1 and it pushes fluid out at 2. To push volume $V_{vol}=v\,\delta m$ across a face at pressure $p$ takes work $p\,V_{vol}=pv\,\delta m$ .
Shaft work $W_s$ : useful work via a shaft/blade (turbine out, compressor in).

So per unit mass: $W_{\text{total}} = w_s + (p_2 v_2 - p_1 v_1)$ Why the sign? At outlet the system does work $p_2v_2$ on the downstream fluid (work out, +). At inlet the upstream fluid does work $p_1v_1$ on our system (so subtract it). This is the heart of the derivation.

Step 4 — Assemble. Per unit mass ( $q=Q/\delta m$ , etc.): $\big(u_2+\tfrac12V_2^2+gz_2\big)-\big(u_1+\tfrac12V_1^2+gz_1\big) = q - w_s - (p_2v_2 - p_1v_1)$

Step 5 — Group $u+pv$ into enthalpy. Move flow-work terms to the left: $(u_2+p_2v_2)+\tfrac12V_2^2+gz_2 = (u_1+p_1v_1)+\tfrac12V_1^2+gz_1 + q - w_s$

The aerodynamics specialisation: adiabatic, no work

In a nozzle/diffuser/free stream: $w_s=0$ (no shaft), $q=0$ (adiabatic, fast flow), $gz$ negligible (gas). SFEE collapses to:

Worked examples

Worked example 1 — Adiabatic nozzle, exit speed

Air enters a nozzle at $T_1=600\text{ K}$ , $V_1=30\text{ m/s}$ ; exits at $T_2=400\text{ K}$ . Find $V_2$ . ( $c_p=1005\,\text{J/kg·K}$ , adiabatic, $w_s=0$ .)

SFEE $\to h_1+\tfrac12V_1^2 = h_2+\tfrac12V_2^2$ . Why drop $q,w_s,gz$ ? Nozzle: no shaft, fast/adiabatic, gas (small $gz$ ). $\tfrac12V_2^2 = c_p(T_1-T_2)+\tfrac12V_1^2 = 1005(200)+\tfrac12(30)^2 = 201000+450$ $V_2=\sqrt{2(201450)} \approx 635\ \text{m/s}.$ Why this step? Falling temperature releases enthalpy, which the flow converts to kinetic energy — exactly what a nozzle is for.

Common mistakes

Active recall

Recall Feynman: explain to a 12-year-old

Imagine a crowded train (the pipe). To get a person into the train you have to push the crowd to make room — that pushing is "flow work." To get a person out the door, the crowd inside pushes them out — that's work too. So each person carries their own backpack of energy ( $u$ ) plus the push needed to squeeze through the door ( $pv$ ). We just glue these together and call the bundle "enthalpy." Now energy in = energy out: heat you add + work a fan adds = the change in everyone's bundles + their running-around energy. When the train speeds through a narrowing tunnel and nobody adds heat, people cool down a little because their bundle energy turned into running-fast energy.

Connections

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

Why does enthalpy replace internal energy in open-system energy balances?

Because mass crossing the boundary requires flow work

pv

, which always pairs with

u

; defining

h=u+pv

absorbs it cleanly.

State the steady-flow energy equation.

h_1+\tfrac12V_1^2+gz_1+q = h_2+\tfrac12V_2^2+gz_2+w_s

What is flow (displacement) work per unit mass at a port of pressure

p

pv

— the work to push unit mass of specific volume

v

across the boundary.

Define stagnation (total) enthalpy.

h_0=h+\tfrac12V^2

; the enthalpy if the flow were brought adiabatically to rest. Constant in adiabatic, no-shaft flow.

For a calorically perfect gas, relate

T_0

and

T

T_0=T+\dfrac{V^2}{2c_p}

, since

h=c_pT

In an adiabatic nozzle, what happens to temperature as velocity increases?

Temperature falls, because

h+\tfrac12V^2=

const so KE is gained at enthalpy's expense.

Sign of shaft work

w_s

for a turbine vs a compressor (work-out convention)?

Turbine:

w_s>0

(work out). Compressor:

w_s<0

(work in).

Difference between flow work and shaft work?

Flow work

pv

pushes fluid through ports (hidden inside

h

); shaft work is extra useful work via a moving boundary/blade — never double-count.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab fluid kisi pipe, nozzle ya turbine ke andar se flow karta hai, toh wo ek open system hota hai — yaani mass boundary cross karti hai. Closed system wala first law $dU=\delta Q-\delta W$ yahan kaafi nahi hai, kyunki fluid ko andar dhakelne aur bahar nikalne mein bhi work lagta hai. Is push ka naam hai flow work $=pv$ . Maza ki baat: yeh $pv$ hamesha internal energy $u$ ke saath chipak jaata hai, isliye hum dono ko jod kar ek naya variable banate hain — enthalpy $h=u+pv$ . Isi wajah se compressible flow ki saari physics enthalpy ke around ghoomti hai.

Final equation, jise Steady Flow Energy Equation (SFEE) kehte hain: $h_1+\tfrac12V_1^2+gz_1+q = h_2+\tfrac12V_2^2+gz_2+w_s$ . Yahan $q$ heat hai jo add hoti hai, aur $w_s$ shaft work hai jo machine bahar nikalti hai (turbine mein positive, compressor mein negative). Yaad rakho: flow work $pv$ already $h$ ke andar chhupa hai, isko dobara mat ginna — yeh sabse common galti hai.

Aerodynamics mein nozzle/free-stream ke liye $q=0$ (adiabatic), $w_s=0$ , aur gas ke liye $gz$ chhota — toh seedha milta hai $h+\tfrac12V^2=h_0=$ constant, jaha $h_0$ stagnation enthalpy hai. Gas perfect ho toh $c_pT+\tfrac12V^2=c_pT_0$ . Iska matlab: jab nozzle mein air tez hoti hai, uska temperature girta hai — kyunki kinetic energy enthalpy se "udhaar" leti hai. Aur jab air kisi body ke nose pe ruk jaati hai, toh garam ho jaati hai ( $T_0$ ) — yahi aerodynamic heating ka root reason hai. Bas yeh ek equation poori compressible flow ki backbone hai.

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

WHAT we are doing

HOW: deriving the steady-flow energy equation from scratch

The aerodynamics specialisation: adiabatic, no work

Worked examples

Common mistakes

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Compressible Flow & Aerodynamics

Connections