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

3.1.1 · D4 · Physics › Compressible Flow & Aerodynamics › Review of thermodynamics applied to flow — first law for ope

Intuition Is page ko kaise use karein

Neeche har problem pehle clearly diya gaya hai, phir ek collapsible Solution callout ke peeche chhupaaya gaya hai taaki tum pehle khud try kar sako. Problems paanch levels par chadhte hain: L1 Recognition (sirf sahi term dhundho), L2 Application (SFEE mein plug karo), L3 Analysis (rearrange karo, sign chase karo), L4 Synthesis (SFEE ko kisi aur law ke saath combine karo), L5 Mastery (ek poora multi-step aerodynamics scenario). Yahan sab kuch parent note SFEE parent note par tikaa hua hai.

Woh ek equation jo hum baar baar use karenge (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. Jab tak kuch na kaha jaaye, air ke liye $c_{p} = 1005 J/(kg\cdotK)$ use karo.

Level 1 — Recognition

Recall Solution 1.1

WHAT: hum flow work per unit mass dhundh rahe hain. WHY: pressure $p$ par rakhe ek face ke paar volume $V_{vol} = v$ (per unit mass) ko move karne ke liye, force $\times$ distance deta hai pressure $\times$ volume $= p v$ . $flow work per unit mass = p v$ Yeh enthalpy ke andar chhupta hai, kyunki $h = u + p v$ . Yahi poora reason hai ki open-system energy balances mein $u$ ki jagah $h$ raaj karta hai — $u$ -only version ke liye Closed-system first law of thermodynamics dekho jo flow ke liye kam pad jaata hai.

Recall Solution 1.2

WHAT: $q$ , $w_{s}$ , aur $g z$ ko kaato. WHY har ek marta hai: adiabatic $\Rightarrow q = 0$ ; koi shaft/blade nahi $\Rightarrow w_{s} = 0$ ; gas halka hota hai, toh ek nozzle ki length par $g z$ , $h \sim 1 0^{5}$ J/kg ke mukable mein negligible hai. Jo bachte hain: $h_{1} + \frac{1}{2} V_{1}^{2} = h_{2} + \frac{1}{2} V_{2}^{2} ⟺ h + \frac{1}{2} V^{2} = h_{0} = const .$ Yeh constant $h_{0}$ stagnation enthalpy hai, jo Stagnation properties & isentropic relations ka star hai.

Level 2 — Application

Recall Solution 2.1

WHAT: Ex. 1.2 se reduced SFEE apply karo jisme $h = c_{p} T$ hai. $\frac{1}{2} V_{2}^{2} = c_{p} (T_{1} - T_{2}) + \frac{1}{2} V_{1}^{2}$ WHY: enthalpy ka drop $c_{p} (T_{1} - T_{2})$ exactly woh energy hai jo kinetic energy ke roop mein wapas aati hai — yahi ek nozzle karta hai (thermal ko kinetic mein convert karta hai; Nozzles and diffusers dekho). $\frac{1}{2} V_{2}^{2} = 1005 (150) + \frac{1}{2} (20)^{2} = 150750 + 200 = 150950$ $V_{2} = 2 (150950) = 301900 \approx 549.5 m/s .$ Temperature giri jabki speed badhi — energy enthalpy se udhaari li gayi.

Recall Solution 2.2

WHAT: SFEE jisme $q = 0$ , KE aur PE $\approx 0$ hain: $h_{1} = h_{2} + w_{s} \Rightarrow w_{s} = h_{1} - h_{2} .$ WHY direction: $w_{s}$ ko work out define kiya gaya hai; ek turbine work deliver karta hai, toh hum $w_{s} > 0$ expect karte hain. $w_{s} = 3200 - 2600 = 600 kJ/kg > 0. ✓$ Positive, bilkul waisa jaisa ek work-producing device ko hona chahiye.

Level 3 — Analysis

Recall Solution 3.1

WHAT: wohi reduced SFEE, lekin ab temperature ke liye solve karo: $c_{p} T_{1} + \frac{1}{2} V_{1}^{2} = c_{p} T_{2} + \frac{1}{2} V_{2}^{2} \Rightarrow T_{2} = T_{1} + \frac{V _{1}^{2} - V _{2}^{2}}{2 c _{p}} .$ WHY: ek diffuser flow ko slow karta hai, toh KE girta hai aur woh energy wapas enthalpy mein jaati hai — temperature badhni chahiye. $T_{2} = 300 + \frac{25 0 ^{2} - 5 0 ^{2}}{2 ( 1005 )} = 300 + \frac{62500 - 2500}{2010} = 300 + \frac{60000}{2010} = 300 + 29.85$ $T_{2} \approx 329.9 K$ $Δ T = + 29.9 K > 0$ — gas decelerate hone par heat up hoti hai, Ex. 2.1 mein nozzle ka mirror image.

Recall Solution 3.2

WHAT: SFEE $\to h_{1} = h_{2} + w_{s} \Rightarrow w_{s} = c_{p} (T_{1} - T_{2})$ . WHY sign dekho: $w_{s}$ work out hai. Ek compressor work consume karta hai, toh machine ko negative $w_{s}$ report karna chahiye. $w_{s} = 1005 (290 - 460) = 1005 (- 170) = - 170850 J/kg = - 170.85 kJ/kg .$ Negative $\Rightarrow$ hume $170.85 kJ/kg$ supply karna hoga. Convention ne direction apne aap pakad liya — "compressor work leta hai" yaad rakhne ki zaroorat nahi, algebra khud bol deta hai.

Level 4 — Synthesis

Recall Solution 4.1

WHAT: pehle ideal-gas law se density nikalo, phir continuity mein daalo. WHY do laws: SFEE ne speed di; speed ko flow rate mein convert karne ke liye hume chahiye ki har cubic metre mein kitna mass packed hai ( $ρ$ ) aur throat kitna wide hai ( $A$ ) — yeh hai 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) aur continuity (mass) do pillars hain — tum almost hamesha dono saath use karte ho.

Recall Solution 4.2

WHAT: is baar $q$ rakhna hoga — yeh khatam nahi hoga. $c_{p} T_{1} + \frac{1}{2} V_{1}^{2} + q = c_{p} T_{2} + \frac{1}{2} V_{2}^{2} .$ WHY: hume bataya gaya hai ki heat add ho rahi hai, toh $q \neq = 0$ ; reduced adiabatic form yahan galat hoga. $T_{2}$ ke liye solve karo: $T_{2} = T_{1} + \frac{q + \frac{1}{2} ( V _{1}^{2} - V _{2}^{2} )}{c _{p}} = 300 + \frac{50000 + \frac{1}{2} ( 1600 - 3600 )}{1005} .$ $= 300 + \frac{50000 - 1000}{1005} = 300 + \frac{49000}{1005} = 300 + 48.76.$ $T_{2} \approx 348.8 K$ Lagbhag saari heat $T$ badhane mein gayi; tiny KE change ( $- 1000$ J/kg) ne toh kuch khaas fark hi nahi daala.

Level 5 — Mastery

Recall Solution 5.1

WHAT & WHY — part (a): "adiabatically to rest laaya gaya" exactly stagnation state ki definition hai, toh $h + \frac{1}{2} V^{2} = h_{0}$ , matlab $c_{p} T + \frac{1}{2} V^{2} = c_{p} T_{0}$ : $T_{0} = T + \frac{V ^{2}}{2 c _{p}} = 220 + \frac{56 0 ^{2}}{2 ( 1005 )} = 220 + \frac{313600}{2010} = 220 + 156.02.$ $T_{0} \approx 376.0 K$ Static air sirf rokne se 156 K heat up ho jaati hai — yahi aerodynamic heating hai.

Part (b) — Mach number. Speed of sound and Mach number dekho. Pehle 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, jaise numbers ne hint diya tha.

Part (c) — compact form verify karo. $c_{p} = \frac{γ R}{γ - 1}$ use karke: $\frac{V ^{2}}{2 c _{p} T} = \frac{V ^{2} ( γ - 1 )}{2 γ R T} = \frac{γ - 1}{2} \cdot \frac{V ^{2}}{γ R T} = \frac{γ - 1}{2} M^{2} .$ Toh $\frac{T _{0}}{T} = 1 + \frac{V ^{2}}{2 c _{p} T} = 1 + \frac{γ - 1}{2} M^{2}$ . Numerically: $1 + \frac{0.4}{2} (1.884)^{2} = 1 + 0.2 (3.549) = 1.710, \frac{T _{0}}{T} = \frac{376.0}{220} = 1.709. ✓$ Energy equation aur Mach-number picture dono agree karte hain — neeche figure dikhata hai ki kaise wohi $h_{0}$ "static heat" aur "motion" mein split hota hai.

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

Recall Solution 5.2

WHAT — (a) SFEE se speed. Chamber ek reservoir hai, toh $V_{1} \approx 0$ aur $h_{1} = h_{0}$ : $\frac{1}{2} V_{2}^{2} = c_{p} (T_{1} - T_{2}) = 1005 (400) = 402000 \Rightarrow V_{2} = 804000 \approx 896.7 m/s .$ WHY: ek bada enthalpy drop ( $400$ K worth) lagbhag poori tarah kinetic energy mein convert ho jaata hai — yahi banane wala thrust hai (Nozzles and diffusers dekho).

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

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

(d) momentum flux (dominant thrust contribution, pressure-area term ignore karke): $\overset{m}{˙} V_{2} = 12.62 \times 896.7 \approx 11317 N \approx 11.3 kN .$ Har step mein exactly ek conservation law use hua: energy speed ke liye, mass flow rate ke liye, momentum thrust ke liye — compressible flow ka teen-taanga stool.

Wrap-up recall

Recall Har level ke liye ek-line takeaways

L1 — flow work $p v$ $h$ ke andar chhupa hai; $w_{s}$ bahar rehta hai. ::: Kabhi $p v$ double-count mat karo. L2 — reduced SFEE $h + \frac{1}{2} V^{2} =$ const temperature drop ko speed mein badalta hai. ::: Nozzle cool karta hai, speed up karta hai. L3 — ek fixed equation $w_{s} = h_{1} - h_{2}$ sign ko device ka naam lene deti hai. ::: Negative $w_{s}$ = work in. L4 — SFEE (energy) ko continuity (mass) $\overset{m}{˙} = ρ A V$ ke saath pair karo. ::: Speed akela flow rate nahi hai. L5 — reservoir ⇒ $h_{1} = h_{0}$ ; energy→speed, mass→ $\overset{m}{˙}$ , momentum→thrust. ::: Teen laws, ek 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