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

3.1.1 · D3 · Physics › Compressible Flow & Aerodynamics › Review of thermodynamics applied to flow — first law for ope

Yaad karo woh tool jo hum baar baar use karte hain — Steady-Flow Energy Equation (SFEE), per unit mass:

Scenario matrix

Har flow-energy problem in cells mein se ek (ya combination) hota hai. Har worked example us cell ke saath tagged hai jo woh hit karta hai.

#	Cell class	Kya special hai	Term jo dominate / vanish karta hai	Example
A	Nozzle, adiabatic, no shaft	$q = 0, w_{s} = 0$ ; $h \to K E$	$g z \approx 0$ , $T$ girta hai	Ex 1
B	Compressor, shaft work in	$w_{s} < 0$ (hum supply karte hain)	KE $\approx 0$ , $q = 0$	Ex 2
C	Turbine, shaft work out	$w_{s} > 0$ (hum harvest karte hain)	KE $\approx 0$ , $q = 0$	Ex 3
D	Heat exchanger, no work	$w_{s} = 0$ , $q \neq = 0$	$q \to Δ h$ ( $q$ ka sign)	Ex 4
E	Stagnation / zero-velocity limit	$V_{2} = 0$ (brought to rest)	$h_{0}, T_{0}$ define karta hai	Ex 5
F	Diffuser (nozzle ulta chalaaya)	$V$ girta hai, $T$ badhta hai	Cell A ka sign flip	Ex 6
G	Real-world word problem	words strip karo → terms dhundho	liquid pump, $v Δ p$	Ex 7
H	Exam twist: hidden term	ek "chhota" term actually chhota NAHI hai	$q$ ya KE drop mat karo	Ex 8

Do limiting checks in cells ke andar rehte hain: zero velocity (Cell E, $V \to 0$ ) aur large velocity (Cell A/F, jahan KE, $h$ se compete karta hai).

Cell A — Adiabatic nozzle (enthalpy → kinetic energy)

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

Air ek nozzle mein $T_{1} = 900 K$ , $V_{1} = 50 m/s$ par enter karti hai, aur $T_{2} = 500 K$ par exit karti hai. Adiabatic, no shaft. $V_{2}$ find karo.

Forecast: temperature bahut zyada girti hai ( $400$ K). Guess karo: kya $V_{2}$ zyada $200$ m/s ke paas aayega ya $900$ m/s ke paas?

Step 1 — Dead terms ko khatam karo. $q = 0$ (adiabatic), $w_{s} = 0$ (koi blade nahi), $g z \approx 0$ (gas, chhoti height). SFEE ban jaata hai $h_{1} + \frac{1}{2} V_{1}^{2} = h_{2} + \frac{1}{2} V_{2}^{2}$ . Yeh step kyun? Nozzle mein koi moving part nahi hota aur yeh itna fast hota hai ki heat exchange nahi ho sakta, isliye sirf enthalpy aur KE trade karte hain.

Step 2 — $h = c_{p} T$ substitute karo aur $V_{2}$ solve karo. $\frac{1}{2} V_{2}^{2} = c_{p} (T_{1} - T_{2}) + \frac{1}{2} V_{1}^{2} = 1005 (400) + \frac{1}{2} (50)^{2} = 402000 + 1250 = 403250$ Yeh step kyun? Girta hua $T$ enthalpy release karta hai; woh released energy sirf KE mein ja sakti hai.

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

Verify: Units — $c_{p} Δ T$ hai J/kg = m²/s², toh $m^{2} / s^{2} =$ m/s. ✓ Inlet KE ( $1250$ J/kg) enthalpy release ( $402, 000$ J/kg) ke mukable mein bahut chhota hai, isliye answer essentially $Δ T$ se set hota hai — sensible. $V_{2} \approx 900$ m/s, jo zyada wali guess se match karta hai.

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) — Compressor chalane ke liye work

Air ko adiabatically $T_{1} = 300 K$ se $T_{2} = 560 K$ tak compress kiya jaata hai. Velocities negligible hain. $w_{s}$ per kg find karo. Kya sign hoga, aur iska kya matlab hai?

Forecast: hum air ko squeeze karke garam kar rahe hain. Kya humein work milti hai, ya humein work lagaani padti hai?

Step 1 — SFEE simplify karo. $q = 0$ , KE $\approx 0$ , $g z \approx 0$ : $h_{1} = h_{2} + w_{s}$ . Yeh step kyun? Koi heat nahi, koi motion energy nahi — enthalpy change akele shaft work se balance hota hai.

Step 2 — Solve karo. $w_{s} = h_{1} - h_{2} = c_{p} (T_{1} - T_{2}) = 1005 (300 - 560) = - 261300 J/kg = - 261.3 kJ/kg .$ Yeh step kyun? Direct substitution; sign hi physics carry karta hai.

Step 3 — Sign padho. $w_{s} < 0$ : humara convention hai $w_{s}$ = work out. Negative matlab work in ja raha hai — humein $261.3$ kJ per kg supply karna padega. ✓

Verify: Enthalpy badhI ( $T$ $260$ K badha). Energy kahin se aa nahi sakti, toh kuch toh feed karna padega — compressor ne kiya. Sign convention ne automatically pakad liya.

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

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

Garam gas ek turbine mein $T_{1} = 1200 K$ par enter karti hai, $T_{2} = 800 K$ par nikalta hai, mass flow $\overset{m}{˙} = 4 kg/s$ hai. Adiabatic, KE negligible. Shaft work per kg aur total power $\dot{W}_{s}$ find karo.

Forecast: garam gas expand hoti hai aur blade drive karti hai — hum energy harvest kar rahe hain. Expect karo $w_{s} > 0$ .

Step 1 — SFEE simplify karo (rate form). $q = 0$ , KE $\approx 0$ : $\overset{m}{˙} h_{1} = \overset{m}{˙} h_{2} + \dot{W}_{s}$ . Yeh step kyun? Turbine ka poora kaam shaft work hai; enthalpy drop usse supply karta hai.

Step 2 — Per-unit-mass work. $w_{s} = h_{1} - h_{2} = c_{p} (T_{1} - T_{2}) = 1005 (1200 - 800) = 402000 J/kg = 402 kJ/kg .$ Yeh step kyun? Compressor jaisi hi algebra, lekin $T$ girta hai, isliye sign positive ho jaata hai.

Step 3 — Mass flow se multiply karo. $\dot{W}_{s} = \overset{m}{˙} w_{s} = 4 \times 402000 = 1608000 W = 1.608 MW .$ Yeh step kyun? Power = energy per kg × kg per second.

Verify: $w_{s} > 0$ — work out, turbine ke liye sahi. Units: (J/kg)(kg/s)=W. ✓ Ex 2 se compare karo: same $∣ c_{p} Δ T ∣$ arithmetic, opposite sign — machine ki direction poori tarah $Δ T$ ke sign mein rehti hai.

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

Worked example Ex 4 (Cell D) — Combustor / heat exchanger

Air ek heated duct se flow karti hai: $T_{1} = 400 K$ , $q = + 150 kJ/kg$ add kiya gaya, koi shaft nahi, velocity change negligible. $T_{2}$ find karo.

Forecast: hum heat add kar rahe hain bina koi motion change ke — backpack ( $h$ ) bhaari honi chahiye, toh $T$ badhega.

Step 1 — SFEE simplify karo. $w_{s} = 0$ , KE $\approx 0$ , $g z \approx 0$ : $h_{1} + q = h_{2}$ . Yeh step kyun? Koi machine nahi, koi motion nahi — add ki gayi heat seedha enthalpy mein jaati hai.

Step 2 — $T_{2}$ solve karo. $c_{p} T_{2} = c_{p} T_{1} + q \Rightarrow T_{2} = T_{1} + \frac{q}{c _{p}} = 400 + \frac{150000}{1005} = 400 + 149.3 = 549.3 K .$ Yeh step kyun? Heat ko $c_{p}$ se divide karo taaki energy-per-kg ko temperature rise mein convert kar sako.

Verify: $q > 0 \Rightarrow T_{2} > T_{1}$ . ✓ Agar instead $q = - 150$ kJ/kg (cooling) hota, toh $T_{2} = 250.7$ K milta — same machinery, opposite sign, Cell D ka negative- $q$ half cover karta hai.

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

Worked example Ex 5 (Cell E) — Nose probe par temperature

Ek fast air stream mein $T = 230 K$ aur $V = 680 m/s$ hai. Ek probe use adiabatically rest par le aata hai. Probe kya temperature $T_{0}$ read karta hai?

Forecast: air ko rokne se uski saari KE enthalpy mein convert hoti hai → probe stream se zyada garam read karta hai. Kitna zyada?

Step 1 — SFEE "flowing" (state 1) aur "at rest" (state 0) ke beech apply karo. Adiabatic, no shaft: $h + \frac{1}{2} V^{2} = h_{0} + \frac{1}{2} (0)^{2}$ . Yeh step kyun? "Brought to rest" exactly woh $V_{2} = 0$ degenerate case hai jo stagnation enthalpy $h_{0}$ ko define karta hai.

Step 2 — Temperature mein convert karo. $T_{0} = T + \frac{V ^{2}}{2 c _{p}} = 230 + \frac{68 0 ^{2}}{2 ( 1005 )} = 230 + \frac{462400}{2010} = 230 + 230.05 = 460.05 K .$ Yeh step kyun? Saari KE ( $\frac{1}{2} V^{2}$ ) enthalpy mein deposit hoti hai, $T$ badhata hai.

Verify: Temperature rise $\frac{V ^{2}}{2 c _{p}} \approx 230$ K bahut bada hai — yahi aerodynamic heating hai. $V = 0$ par formula $T_{0} = T$ deta hai (koi rise nahi), sanity limit. ✓ Links to Stagnation properties & isentropic relations aur Speed of sound and Mach number.

Cell F — Diffuser (nozzle ulta chalaaya)

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

Ek diffuser air ko $V_{1} = 800 m/s$ se $V_{2} = 100 m/s$ par slow karta hai; inlet $T_{1} = 250 K$ hai. Adiabatic, no shaft. $T_{2}$ find karo.

Forecast: ek diffuser flow ko slow karta hai, isliye KE kharach hoti hai → enthalpy (aur $T$ ) badhni chahiye. Nozzle ka ulta.

Step 1 — SFEE simplify karo. $q = 0, w_{s} = 0, g z \approx 0$ : $h_{1} + \frac{1}{2} V_{1}^{2} = h_{2} + \frac{1}{2} V_{2}^{2}$ . Yeh step kyun? Same energy trade nozzle jaisi, lekin velocity term badhne ki jagah ghatti hai.

Step 2 — $T_{2}$ solve karo. $c_{p} T_{2} = c_{p} T_{1} + \frac{1}{2} (V_{1}^{2} - V_{2}^{2})$ $T_{2} = 250 + \frac{80 0 ^{2} - 10 0 ^{2}}{2 ( 1005 )} = 250 + \frac{640000 - 10000}{2010} = 250 + 313.4 = 563.4 K .$ Yeh step kyun? Khoyi KE ( $V$ girta hai) enthalpy mein deposit hoti hai → $T$ chadhta hai.

Verify: $V$ gira, $T$ badha — Ex 1 ke nozzle ka mirror image, Cell A/F symmetry confirm karta hai. Nozzles and diffusers se compare karo. Agar $V_{2} = 0$ ho toh yeh Ex 5 ke stagnation formula mein reduce ho jaata hai — donon cells zero-velocity limit par connect hote hain. ✓

Cell G — Real-world word problem (ek liquid, gas nahi)

Worked example Ex 7 (Cell G) — Water pump rooftop tank tak lift karta hai

Ek pump water ( $v = 0.001 m^{3} / kg$ , yaani density $1000 kg/m^{3}$ ) ko ek ground reservoir se ek tank tak le jaata hai jo $z_{2} - z_{1} = 40 m$ oopar hai, pressure $p_{1} = 100 kPa$ se $p_{2} = 500 kPa$ tak raise hota hai. Water nearly incompressible hai aur same temperature par rehti hai; KE change aur heat negligible hain. Shaft work per kg find karo.

Forecast: liquid ke liye, "enthalpy change" actually sirf pressure-push $v Δ p$ hai (kyunki $u$ constant $T$ par barely change karta hai), plus lift $g Δ z$ . Expect karo $w_{s} < 0$ (hum pay karte hain).

Step 1 — Full SFEE se start karo. $q = 0$ , $Δ K E \approx 0$ : $h_{1} + g z_{1} = h_{2} + g z_{2} + w_{s}$ . Yeh step kyun? Real pumps height change karte hain, isliye $g z$ drop nahi kar sakte (gases ke unlike).

Step 2 — Incompressible liquid ke liye constant $T$ par, $Δ u \approx 0$ , isliye $h_{2} - h_{1} = (u_{2} + p_{2} v) - (u_{1} + p_{1} v) \approx v (p_{2} - p_{1}) .$ Yeh step kyun? $u$ frozen hone par, sirf flow-work term $v Δ p$ se enthalpy change hota hai. Yahi Bernoulli equation as low-speed limit of SFEE ka beej hai.

Step 3 — Assemble karo aur solve karo. $w_{s} = (h_{1} - h_{2}) + g (z_{1} - z_{2}) = - v (p_{2} - p_{1}) - g (z_{2} - z_{1})$ $= - 0.001 (500000 - 100000) - 9.81 (40) = - 400 - 392.4 = - 792.4 J/kg .$ Yeh step kyun? Pressure raise karna aur height raise karna dono work cost karte hain; dono minus ke saath appear hote hain.

Verify: $w_{s} < 0$ — work in, pump ke liye sahi. ✓ Split check: pressure term $= 400$ J/kg, lift term $= 392.4$ J/kg — comparable hain, isliye kisi ek ko drop karna galat hoga. Units: $v Δ p = m^{3} / kg \cdot Pa = J/kg$ . ✓

Cell H — Exam twist (ek "chhota" term chhota nahi hai)

Worked example Ex 8 (Cell H) — Cooled nozzle:

$q$ mat drop karo! Ek high-temperature nozzle ko deliberately cool kiya jaata hai (film cooling): $T_{1} = 1000 K$ , $V_{1} = 100 m/s$ , exit $V_{2} = 700 m/s$ , aur heat $q = - 80 kJ/kg$ remove ki jaati hai. No shaft. $T_{2}$ find karo. Trap yeh hai: ek careless student assume karta hai "nozzle ⇒ adiabatic" aur $q$ drop kar deta hai.

Forecast: heat remove karna aur speed up karna dono enthalpy drain karte hain, isliye $T_{2}$ ek adiabatic nozzle se zyada girna chahiye.

Step 1 — Har non-zero term rakhein. $w_{s} = 0$ , $g z \approx 0$ , lekin $q = - 80$ kJ/kg rehta hai: $h_{1} + \frac{1}{2} V_{1}^{2} + q = h_{2} + \frac{1}{2} V_{2}^{2} .$ Yeh step kyun? Problem explicitly heat remove karta hai — "nozzle" yahan $q = 0$ ka license nahi deta. Yahi poora twist hai.

Step 2 — $T_{2}$ solve karo. $c_{p} T_{2} = c_{p} T_{1} + \frac{1}{2} (V_{1}^{2} - V_{2}^{2}) + q$ $c_{p} T_{2} = 1005 (1000) + \frac{1}{2} (10 0^{2} - 70 0^{2}) + (- 80000)$ $= 1005000 + \frac{1}{2} (10000 - 490000) - 80000 = 1005000 - 240000 - 80000 = 685000$ $T_{2} = \frac{685000}{1005} = 681.6 K .$ Yeh step kyun? Teeno drains (heat out, KE up) enthalpy se subtract hote hain.

Step 3 — Galat answer dikhao. Agar tum (galat tarike se) $q$ drop karo: $c_{p} T_{2} = 1005000 - 240000 = 765000 \Rightarrow T_{2} = 761.2$ K. Yeh $79.6$ K zyada hai — exactly $q / c_{p}$ . Yeh kyun dikhaya? Taaki dekh sako ki dropped term kitne ka tha.

Verify: Correct $T_{2} = 681.6$ K; naive answer se difference $761.2 - 681.6 = 79.6 K \approx 80000/1005$ hai. ✓ Lesson: kisi bhi term ko mat drop karo jiske liye problem ne number diya ho.

Active recall

Recall Kaun sa cell kaun sa hai?

Nozzle speed up karta hai, $T$ girta hai ::: Cell A Compressor, $w_{s} < 0$ (work in) ::: Cell B Turbine, $w_{s} > 0$ (work out) ::: Cell C Duct jisme heat add ho, koi work nahi ::: Cell D Flow brought to rest ( $V \to 0$ ) $T_{0}$ define karta hai ::: Cell E Diffuser flow slow karta hai, $T$ badhta hai ::: Cell F Liquid pump: $Δ h \approx v Δ p$ plus lift $g Δ z$ ::: Cell G Cooled nozzle jahan $q$ rakhna zaroori hai ::: 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