Heat engines — efficiency η = 1 − Q_C - Q_H: Exercises

1.7.19 · D4 · Physics › Thermodynamics › Heat engines — efficiency η = 1 − Q_C - Q_H

Kisi bhi symbol se pehle: internal energy $U$ bas working gas mein store hui total thermal energy hai — kitna "garam aur bhara hua" hai. Ek poore cycle mein gas apni exactly starting state pe wapas aa jaati hai, isliye uski stored energy bhi wahan wapas pahunch jaati hai jahan se shuru hui thi; hum is "koi net change nahi" ko $Δ U = 0$ likhte hain, jahan $Δ$ (delta) ka matlab hai "final minus initial".

Poore note mein, saari heats magnitudes (positive numbers) ki tarah likhi gayi hain: $Q_{H}$ woh heat hai jo andar aati hai, $Q_{C}$ woh heat hai jo bahar dump hoti hai, $W$ woh work hai jo bahar jaata hai. Do facts jinpe hum rely karte hain: $W = Q_{H} - Q_{C} η = \frac{W}{Q _{H}} = 1 - \frac{Q _{C}}{Q _{H}}$ aur kisi bhi engine ki ceiling jo hot temperature $T_{H}$ aur cold temperature $T_{C}$ (in kelvin) ke beech chal rahi ho: $η_{Carnot} = 1 - \frac{T _{C}}{T _{H}}$

Figure — Heat engines — efficiency η = 1 − Q_C - Q_H

Upar ki picture har engine ka accountant's view hai: heat red top pipe se andar ( $Q_{H}$ ), useful work ke roop mein ek chhoti si amount green side arrow ke roop mein bahar ( $W$ ), aur bacha hua blue bottom pipe se leak hota hua ( $Q_{C}$ ). Jo bhi andar jaata hai woh bahar aana chahiye — First Law yahan yahi kehta hai. Is teen-arrow picture ko dimaag mein rakho: neeche diye gaye har exercise mein bas unhi teen arrows mein se ek ke baare mein pucha ja raha hai jab baaki do diye gaye hoon.

Level 1 — Recognition

Goal: definition padho, numbers sahi slots mein daalo.

Recall Solution L1-a

Step 1 — Work. Figure ki language mein, hum red pipe ( $Q_{H}$ ) aur blue pipe ( $Q_{C}$ ) jaante hain aur green arrow ( $W$ ) chahiye. Jo bhi heat dump nahi hoti woh work ban jaati hai: $W = Q_{H} - Q_{C} = 800 - 600 = 200 J .$ Yeh step kyun? Ek cycle mein gas apne starting point pe wapas aati hai, isliye uski internal energy unchanged rehti hai, $Δ U = 0$ , aur First Law force karta hai ki net heat ka har joule work ke roop mein nikle. Step 2 — Efficiency. "Jo milta hai usko jo dete hain usse divide karo": $η = \frac{W}{Q _{H}} = \frac{200}{800} = 0.25 = 25%.$ Cross-check. $1 - Q_{C} / Q_{H} = 1 - 600/800 = 0.25$ ✓ — $η$ ke dono forms agree karte hain.

Recall Solution L1-b

Step 1. Temperature ceiling seedha use karo (temperatures already kelvin mein hain): $η_{Carnot} = 1 - \frac{T _{C}}{T _{H}} = 1 - \frac{300}{400} = 0.25 = 25%.$ Kelvin kyun? Yeh formula measure karta hai ki dono reservoirs absolute scale pe ek doosre se kitne door hain, jahan $0 K$ ka matlab hai "bilkul bhi thermal energy nahi". Sirf usi scale pe temperatures ka ek ratio physical meaning rakhta hai.

Level 2 — Application

Goal: unhi dono equations ko rearrange karo taaki woh unknown nikale jo directly nahi diya gaya.

Recall Solution L2-a

Step 1 — Heat in. $η = W / Q_{H}$ ko ulta karo: $Q_{H} = \frac{W}{η} = \frac{500}{0.40} = 1250 MJ .$ Kyun? $η$ batata hai ki $W$ , $Q_{H}$ ka $40%$ hai; ek jaane-maane part se poori cheez recover karne ke liye, fraction se divide karo. Step 2 — Heat dumped. Jo bhi convert nahi hoti woh reject hoti hai: $Q_{C} = Q_{H} - W = 1250 - 500 = 750 MJ .$ Insight. Plant $750 MJ$ waste kar deta hai — us $500 MJ$ se zyada jo woh deliver karta hai. Achhe engines bhi zyada tar heat-dumpers hote hain.

Recall Solution L2-b

Step 1 — $Q_{C}$ ko $Q_{H}$ se relate karo. Kyunki $η = 1 - Q_{C} / Q_{H}$ , dumped fraction hai $Q_{C} / Q_{H} = 1 - η = 0.70$ . Toh $Q_{H} = \frac{Q _{C}}{1 - η} = \frac{420}{0.70} = 600 J .$ Yeh step kyun? Humein waste number $Q_{C}$ diya gaya tha, isliye hum waste fraction $1 - η$ pe anchor karte hain, $η$ pe nahi. Step 2 — Work. $W = Q_{H} - Q_{C} = 600 - 420 = 180 J .$ Cross-check. $η = W / Q_{H} = 180/600 = 0.30$ ✓.

Level 3 — Analysis

Goal: ek limit ke against compare karo, ya feasibility ke baare mein reason karo — koi single plug-in nahi.

Recall Solution L3-a

Step 1 — Ceiling nikalo. $η_{Carnot} = 1 - \frac{300}{500} = 1 - 0.60 = 0.40 = 40%.$ Kyun? Koi bhi engine — real ho ya ideal — unhi do temperatures ke beech reversible Carnot value exceed nahi kar sakta. Step 2 — Compare karo. Claimed $0.45 > 0.40$ ceiling → impossible. Yeh maximum se $5$ percentage points zyada hai, jo Second Law violate karega. Physically samjho: Carnot ko beat karne ke liye tumhe ek perfectly reversible engine se bhi kam heat dump karni padegi, jo universe ki entropy ko decrease karega — nature kabhi allow nahi karta.

Recall Solution L3-b

Engine A. $W_{A} = Q_{H} - Q_{C} = 1000 - 700 = 300 J$ (toh $η_{A} = 0.30$ ). Engine B. Pehle uski efficiency: $η_{B} = 1 - 250/500 = 0.50$ . Phir $W_{B} = η_{B} Q_{H} = 0.50 \times 1000 = 500 J .$ Compare karo. $W_{B} = 500 J > W_{A} = 300 J$ : Engine B zyada work karta hai. Kyun matter karta hai: ek hi fuel ( $Q_{H}$ ) se, zyada efficiency ka directly matlab hai zyada work aur kam waste — B sirf $500 J$ waste karta hai jabki A $700 J$ waste karta hai.

Level 4 — Synthesis

Goal: kai relations ko saath silna, kabhi kabhi rate/power ya reversed cycle laana.

Recall Solution L4-a

Idea. Efficiency ek ratio hai, isliye yeh per-cycle energies ki tarah rates pe bhi apply hoti hai: $η = W / Q_{H}$ ke dono sides ko ek cycle ke time se divide karo taaki $η = P / \dot{Q}_{H}$ mile, jahan $\dot{Q}$ ka matlab hai "heat per second". Step 1 — Heat-in rate. $\dot{Q}_{H} = \frac{P}{η} = \frac{600}{0.40} = 1500 MW .$ Step 2 — Dump rate. $\dot{Q}_{C} = \dot{Q}_{H} - P = 1500 - 600 = 900 MW .$ Yeh step kyun? Same energy balance (green arrow plus blue pipe equal red pipe), ab per second padha gaya. Insight. $900 MW$ waste heat continuously environment mein jaati hai — isliye bade plants rivers ya cooling towers ke paas hote hain.

Recall Solution L4-b

Stage 1. $η_{1} = 1 - 400/600 = 1/3$ . Toh $W_{1} = \frac{1}{3} \times 1200 = 400 J, Q_{C, 1} = 1200 - 400 = 800 J .$ Yeh step kyun? $η_{1}$ woh fraction hai jo Stage 1 ki red-pipe heat green-arrow work ke roop mein nikaalti hai, isliye $W_{1}$ pane ke liye $Q_{H}$ ko $η_{1}$ se multiply karo; baaki ( $Q_{H} - W_{1}$ ) Stage 1 ki blue-pipe reject hai. Stage 2. Uska heat input Engine 1 ki reject hai: $Q_{H, 2} = 800 J$ . Uski efficiency $η_{2} = 1 - 300/400 = 0.25$ . Toh $W_{2} = 0.25 \times 800 = 200 J .$ Yeh step kyun? Stage 2 ki red pipe hai Stage 1 ki blue pipe — waste heat ko doosre engine ke fuel ke roop mein recycle kiya ja raha hai — isliye hum phir se us input ko $η_{2}$ se multiply karte hain apna work paane ke liye. Total. $W_{tot} = 400 + 200 = 600 J$ , isliye $η_{overall} = \frac{600}{1200} = 0.50 = 50%.$ Khoobsurat check. $600 K$ se $300 K$ tak seedha ek single Carnot engine $η = 1 - 300/600 = 0.50$ deta hai — exactly same. Ek reversible drop ko reversible stages mein todne se kuch nahi jaata, jo Carnot bound ki ek deep property hai.

Level 5 — Mastery

Goal: poore multi-step problems jahan ek galti cascade karti hai. Dhyan se padho.

Recall Solution L5-a

Step 1 — Work per cycle. $W = η Q_{H} = 0.22 \times 2500 = 550 J .$ Step 2 — Cycles per second. $2000 cycles/min \div 60 = 33. \overline{3} cycles/s$ . Step 3 — Power. Power work per second hai: $P = W \times (cycles/s) = 550 \times \frac{2000}{60} = 18333. \overline{3} W \approx 18.33 kW .$ Pehle 60 se kyun divide karein? Power per second honi chahiye; cycle count per minute diya gaya tha, isliye multiply karne se pehle rate convert karo. Step 4 — Heat dumped per second. Per cycle $Q_{C} = Q_{H} - W = 2500 - 550 = 1950 J$ . Per second: $\dot{Q}_{C} = 1950 \times \frac{2000}{60} = 65000 W = 65 kW .$ Sanity: $\dot{Q}_{H} = 2500 \times \frac{2000}{60} \approx 83.33 kW$ , aur sach mein $P + \dot{Q}_{C} = 18.33 + 65 = 83.33 kW$ ✓.

Recall Solution L5-b

Step 1 — Carnot ceiling. $η_{Carnot} = 1 - \frac{350}{700} = 0.50 = 50%.$ Step 2 — Second-law efficiency = actual ÷ ideal: $\frac{η}{η _{Carnot}} = \frac{0.35}{0.50} = 0.70 = 70%.$ Matlab: engine in temperatures ke beech ek perfect reversible engine jo kar sakta tha uska $70%$ capture karta hai. Step 3 — Highest allowed $T_{C}$ . $0.45$ ki real efficiency tabhi legal hai jab ceiling kam se kam $0.45$ ho. Ceiling ko target ke barabar set karne se borderline (highest allowed) $T_{C}$ milta hai: $1 - \frac{T _{C}}{700} = 0.45 ⟹ \frac{T _{C}}{700} = 0.55 ⟹ T_{C} = 700 \times 0.55 = 385 K .$ Yeh step kyun? $T_{C}$ kam karna ceiling raise karta hai (bada temperature gap = efficiency ke liye zyada room). Isliye koi bhi $T_{C} \leq 385 K$ ceiling ko $\geq 0.45$ rakhta hai aur $η = 0.45$ permissible banata hai; $385 K$ se upar yeh impossible ho jaata hai. Final answer. Highest allowed cold temperature $T_{C} = 385 K$ hai. Abhi ka $350 K$ ( $0.50$ ceiling deta hua) already $0.45$ target ke liye room rakhta hai.

Answer key (numbers only)

Recall Saare final answers reveal karo

L1-a: $W = 200 J$ , $η = 0.25$
L1-b: $η = 0.25$
L2-a: $Q_{H} = 1250 MJ$ , $Q_{C} = 750 MJ$
L2-b: $Q_{H} = 600 J$ , $W = 180 J$
L3-a: ceiling $0.40$ ; claim $0.45$ impossible hai (by $0.05$ exceed karta hai)
L3-b: $W_{A} = 300 J$ , $W_{B} = 500 J$ → B jeet jaata hai
L4-a: $\dot{Q}_{H} = 1500 MW$ , $\dot{Q}_{C} = 900 MW$
L4-b: $W_{tot} = 600 J$ , $η_{overall} = 0.50$
L5-a: $W = 550 J$ , $P \approx 18.33 kW$ , $\dot{Q}_{C} = 65 kW$
L5-b: ceiling $0.50$ ; second-law efficiency $0.70$ ; highest allowed $T_{C} = 385 K$

Connections

First Law of Thermodynamics — upar har $W = Q_{H} - Q_{C}$ balance.
Second Law of Thermodynamics — woh ceiling jo L3-a ke claim ko kill karti hai.
Carnot Cycle — $η = 1 - T_{C} / T_{H}$ ka source jo L1, L3, L4, L5 mein use hua.
Entropy — kyun staged reversible engines (L4-b) kuch nahi khoते.
Refrigerators and Heat Pumps — same balance ulta chalaya gaya.
PV diagrams and Work — woh loop area jiska value woh $W$ hai jo hum baar baar compute karte hain.